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Title: Solution manual of Engineering Economics 7th Ed
Description: Solution manual of Engineering Economics 7th Ed by Leland Blank and Anthony

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Se v e nth Ed ition
n th

ENGINEERING
ECONOMY

Se v e n th Ed ition

ENGINEERING
ECONOMY

Leland Blank, P
...

Texas A & M University
American University of Sharjah, United Arab Emirates

Anthony Tarquin, P
...

University of Texas at El Paso

TM

TM

ENGINEERING ECONOMY: SEVENTH EDITION
Published by McGraw-Hill, a business unit of The McGraw-Hill Companies, Inc
...
Copyright © 2012 by The McGraw-Hill Companies, Inc
...
Previous editions
© 2005, 2002, and 1998
...
,
including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance
learning
...

This book is printed on recycled, acid-free paper containing 10% postconsumer waste
...
Massar
Senior Marketing Manager: Curt Reynolds
Development Editor: Lorraine K
...
Rolwes
Cover Designer: Studio Montage, St
...

Library of Congress Cataloging-in-Publication Data
Blank, Leland T
...
— 7th ed
...
cm
...

ISBN-13: 978-0-07-337630-1 (alk
...
Engineering economy
...
Tarquin, Anthony J
...
Title
...
4
...
15—dc22
2010052297

www
...
com

This book is dedicated to Dr
...
Sheppard, Jr
...

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CONTENTS
Preface to Seventh Edition

LEARNING
STAGE 1
Chapter 1

xiii

THE FUNDAMENTALS
Foundations of Engineering Economy
1
...
2
1
...
4
1
...
6
1
...
8
1
...
10

Chapter 2

Factors: How Time and Interest Affect Money
PE

2
...
2
2
...
4
2
...
6
2
...
1
3
...
3

Chapter 4

Engineering Economics: Description and
Role in Decision Making
Performing an Engineering Economy Study
Professional Ethics and Economic Decisions
Interest Rate and Rate of Return
Terminology and Symbols
Cash Flows: Estimation and Diagramming
Economic Equivalence
Simple and Compound Interest
Minimum Attractive Rate of Return
Introduction to Spreadsheet Use
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—Renewable Energy Sources for Electricity Generation
Case Study—Refrigerator Shells

Calculations for Uniform Series That Are Shifted
Calculations Involving Uniform Series and Randomly Placed Single Amounts
Calculations for Shifted Gradients
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—Preserving Land for Public Use

Nominal and Effective Interest Rates
PE

4
...
2
4
...
4
4
...
6
4
...
8
4
...
1
5
...
3
5
...
5

Advantages and Uses of Annual Worth Analysis
Calculation of Capital Recovery and AW Values
Evaluating Alternatives by Annual Worth Analysis
AW of a Permanent Investment
Life-Cycle Cost Analysis
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—The Changing Scene of an Annual Worth Analysis

Rate of Return Analysis: One Project
7
...
2
7
...
4
7
...
6

Chapter 8

Progressive Example—Water for Semiconductor Manufacturing Case
Formulating Alternatives
Present Worth Analysis of Equal-Life Alternatives
Present Worth Analysis of Different-Life Alternatives
Future Worth Analysis
Capitalized Cost Analysis
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—Comparing Social Security Beneﬁts

Annual Worth Analysis
6
...
2
6
...
4
6
...
1
8
...
3
8
...
5
8
...
7

Chapter 9

All-in-One Spreadsheet Analysis (Optional)
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—ROR Analysis with Estimated Lives That Vary
Case Study—How a New Engineering Graduate Can Help His Father

Beneﬁt/Cost Analysis and Public Sector Economics
PE

9
...
2
9
...
4
9
...
6

Progressive Example—Water Treatment Facility #3 Case
Public Sector Projects
Beneﬁt/Cost Analysis of a Single Project
Alternative Selection Using Incremental B/C Analysis
Incremental B/C Analysis of Multiple, Mutually Exclusive Alternatives
Service Sector Projects and Cost-Effectiveness Analysis
Ethical Considerations in the Public Sector
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—Comparing B/C Analysis and CEA of Trafﬁc Accident Reduction

LEARNING
STAGE 2

218
219
220
225
226
227

228
229
230
235
238
242
246
250
251
252
258
259

EPILOGUE: SELECTING THE BASIC ANALYSIS TOOL

LEARNING
STAGE 3

ix

MAKING BETTER DECISIONS

Chapter 10

Project Financing and Noneconomic Attributes
10
...
2
10
...
4
10
...
6
10
...
1
11
...
3
11
...
5
11
...
1
12
...
3
12
...
5

Chapter 13

Breakeven and Payback Analysis
13
...
2
13
...
4

LEARNING
STAGE 4
Chapter 14

Effects of Inﬂation
Understanding the Impact of Inﬂation
Present Worth Calculations Adjusted for Inﬂation
Future Worth Calculations Adjusted for Inﬂation
Capital Recovery Calculations Adjusted for Inﬂation
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—Inﬂation versus Stock and Bond Investments

Cost Estimation and Indirect Cost Allocation
15
...
2
15
...
4
15
...
6
15
...
8

Chapter 16

Breakeven Analysis for a Single Project
Breakeven Analysis Between Two Alternatives
Payback Analysis
More Breakeven and Payback Analysis on Spreadsheets
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—Water Treatment Plant Process Costs

322
323
325
327
329
332
334
334
338

340
341
345
348
352
355
355
361
363

ROUNDING OUT THE STUDY

14
...
2
14
...
4

Chapter 15

An Overview of Capital Rationing among Projects
Capital Rationing Using PW Analysis of Equal-Life Projects
Capital Rationing Using PW Analysis of Unequal-Life Projects
Capital Budgeting Problem Formulation Using Linear Programming
Additional Project Ranking Measures
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions

Understanding How Cost Estimation Is Accomplished
Unit Method
Cost Indexes
Cost-Estimating Relationships: Cost-Capacity Equations
Cost-Estimating Relationships: Factor Method
Traditional Indirect Cost Rates and Allocation
Activity-Based Costing (ABC) for Indirect Costs
Making Estimates and Maintaining Ethical Practices
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—Indirect Cost Analysis of Medical Equipment Manufacturing Costs
Case Study—Deceptive Acts Can Get You in Trouble

Depreciation Methods
16
...
2
16
...
4
16
...
6

Depletion Methods
Chapter Summary
Appendix
16A
...
2 Switching between Depreciation Methods
16A
...
1
17
...
3
17
...
5
17
...
7
17
...
9

Chapter 18

Sensitivity Analysis and Staged Decisions
18
...
2
18
...
4
18
...
6

Chapter 19

Determining Sensitivity to Parameter Variation
Sensitivity Analysis Using Three Estimates
Estimate Variability and the Expected Value
Expected Value Computations for Alternatives
Staged Evaluation of Alternatives Using a Decision Tree
Real Options in Engineering Economics
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—Sensitivity to the Economic Environment
Case Study—Sensitivity Analysis of Public Sector Projects—Water Supply Plans

More on Variation and Decision Making under Risk
19
...
2
19
...
4
19
...
1
A
...
3
A
...
5
A
...
1
B
...
3

The Balance Sheet
Income Statement and Cost of Goods Sold Statement
Business Ratios

561
561
562
563

Appendix C

Code of Ethics for Engineers

566

Appendix D

Alternate Methods for Equivalence Calculations

569

D
...
2

Appendix E

Glossary of Concepts and Terms
E
...
2

Reference Materials
Factor Tables
Photo Credits
Index

Using Programmable Calculators
Using the Summation of a Geometric Series

579
581
610
611

Important Concepts and Guidelines
Symbols and Terms

569
570

573
573
576

PREFACE TO SEVENTH EDITION
This edition includes the time-tested approach and topics of previous editions and introduces signiﬁcantly new print and electronic features useful to learning about and successfully applying the exciting ﬁeld of engineering economics
...
Learning to understand, analyze, and manage the money side of any
project is vital to its success
...
This book is a great help to the learner and the instructor in accomplishing these goals by using easy-to-understand language, simple graphics, and online features
...
Plus the supporting online materials
are new and updated to enhance the teaching and learning experience
...
Students
should be at the sophomore level or above with a basic understanding of engineering concepts
and terminology
...

Practitioners and professional engineers who need a refresher in economic analysis and cost
estimation will ﬁnd this book very useful as a reference document as well as a learning medium
...
Each chapter starts with a statement of purpose and a speciﬁc learning outcome (ABET style) for each section
...

The appendices are important elements of learning for this text:
Appendix A
Appendix B
Appendix C
Appendix D
Appendix E

Spreadsheet layout and functions (Excel is featured)
Accounting reports and business ratios
Code of Ethics for Engineers (from NSPE)
Equivalence computations using calculators and geometric series; no tables
Concepts, guidelines, terms, and symbols for engineering economics

There is considerable ﬂexibility in the sequencing of topics and chapters once the ﬁrst six
chapters are covered, as shown in the progression graphic on the next page
...
Foundations
2
...
More Factors
4
...
Present Worth
6
...
Rate of Return
8
...
Benefit/Cost

10
...
Replacement
12
...
Breakeven and
Payback
14
...
Estimation
16
...
After-Tax
18
...
Risk and Simulation

after Learning Stage 2 (Chapter 9) is completed
...
The progression graphic can help in the design of the course content and
topic ordering
...
This behavioral-based
approach sensitizes the reader to what
is ahead, leading to improved
understanding and learning
...

SECTION

TOPIC

LEARNING OUTCOME

3
...

3
...

3
...

CONCEPTS AND GUIDELINES
Time value of money

It is a well-known fact that money makes money
...

This is the most important concept in engineering economy
...
Appendix E includes a brief
description of each fundamental concept
...
6

Numerous in-chapter examples
throughout the book reinforce the
basic concepts and make
understanding easier
...

A dot-com company plans to place money in a new venture capital fund that currently returns
18% per year, compounded daily
...
7], with r ϭ 0
...

0
...
716%
Effective i% per year ϭ 1 ϩ ——
365
(b) Here r ϭ 0
...

0
...
415%
Effective i% per 6 months ϭ 1 ϩ ——
182

(

)

(

)

PROGRESSIVE EXAMPLES
PE
Water for Semiconductor Manufacturing Case: The worldwide contribution of
semiconductor sales is about $250 billion
per year, or about 10% of the world’s
GDP (gross domestic product)
...
Depending upon the
type and size of fabrication plant (fab),
the need for ultrapure water (UPW) to
manufacture these tiny integrated circuits
is high, ranging from 500 to 2000 gpm
(gallons per minute)
...
Potable water
obtained from purifying seawater or
brackish groundwater may cost from
$2 to $3 per 1000 gallons, but to obtain
UPW on-site for semiconductor manufacturing may cost an additional $1 to $3 per
1000 gallons
...
5 billion to
construct, with approximately 1% of this
total, or $25 million, required to provide
the ultrapure water needed, including
the necessary wastewater and recycling
equipment
...
It is fortunate to

Several chapters include a progressive
example—a more detailed problem statement
introduced at the beginning of the chapter and
expanded upon throughout the chapter in
specially marked examples
...

have the option of desalinated seawater
or puriﬁed groundwater sources in the
location chosen for its new fab
...

Source
Equipment ﬁrst
cost, $M

Seawater
(S)
Ϫ20

Groundwater
(G)
Ϫ22

Ϫ0
...
3

Salvage value, % of
ﬁrst cost

5

10

Cost of UPW, $ per
1000 gallons

4

5

AOC, $M per year

Angular has made some initial estimates
for the UPW system
...
2)
PW analysis of different-life alternatives (Section 5
...
5)
Problems 5
...
34

bla76302_ch04_094-126
...
indd 212

An icon in the margin indicates the
availability of an animated voice-over slide
presentation that summarizes the material in
the section and provides a brief example for
learners who need a review or prefer videobased materials
...

3
...
In this case several methods can be used to ﬁnd the equivalent present worth P
...

• Use the F͞P factor to ﬁnd the future worth of each disbursement in year 13, add them, and
then ﬁnd the present worth of the total, using P ϭ F(P͞F,i,13)
...

• Use the P͞A factor to compute the “present worth” P3 ϭ A(P͞A,i,10) (which will be located
in year 3, not year 0), and then ﬁnd the present worth in year 0 by using the (P͞F,i,3) factor
...
8
Breakeven

Chris and her father just purchased a ROR Ϸofﬁce building for $160,000 that is in need of a lot
Incremental small 17%
of repairs, but is located in a prime commercial area of the city
...
are $18,000 the ﬁrst year, increasing by $1000 per year thereafter
...
4
...

SPREADSHEETS

Solution
12/11/10 6:52 PM

Figure 13–11 shows the annual costs (column B) and the sales prices if the building is kept 2
or 3 years (columns C and E, respectively)
...
These results bracket the payback
period for each retention period 7–12 price
...

Figure and sales

Spreadsheet between 3 and ROIC (column D)
...
6
...

(b) At a sales price of $370,000, the 8% return payback period is between 5 and 6 years (column F)
...

bla76302_ch07_172-201
...
Cell tags or full cells detail
built-in functions and relations developed
to solve a speciﬁc problem
...
8

bla76302_ch13_340-364
...
indd 73

FE EXAM AND COURSE
REVIEWS
Each chapter concludes with several
multiple-choice, FE Exam–style
problems that provide a simpliﬁed
review of chapter material
...

12/7/10 7:26 AM

ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
8
...
39 In comparing mutually exclusive alternatives by
the ROR method, you should:
(a) Find the ROR of each alternative and pick
the one with the highest ROR
(b) S l
h
l
i
h
i
l

8
...
PEC has a capacity of approximately 1300 MW (megawatts) of power, of
which 277 MW, or about 21%, is from renewable sources
...
A constant
question is how much of PEC’s generation capacity should be
from renewable sources, especially given the environmental
issues with coal-generated electricity and the rising costs of
hydrocarbon fuels
...

Consider yourself a member of the board of directors of
PEC
...

As such, you do not represent a speciﬁc district within the
entire service area; all other directors do represent a speciﬁc
district
...

Information
Here are some data that you have obtained
...
Electricity generation cost estimates are national
in scope, not PEC-speciﬁc, and are provided in cents per
kilowatt-hour (¢/kWh)
...
5
4
...
1
4
...
5

Reasonable Average
7
...
6
8
...
8

National average cost of electricity to residential customers: 11¢/kWh

PEC average cost to residential customers: 10
...
92 ¢/kWh (renewable sources)
Expected life of a generation facility: 20 to 40 years (it is
likely closer to 20 than 40)
Time to construct a facility: 2 to 5 years
Capital cost to build a generation facility: $900 to $1500
per kW
You have also learned that the PEC staff uses the wellrecognized levelized energy cost (LEC) method to determine
the price of electricity that must be charged to customers to
break even
...
The LEC formula, expressed
in dollars per kWh for (t ϭ 1, 2,
...
If you wanted to know more about the new arrangement with the wind farm in south Texas for the additional 60 MW per year, what types of questions would
you ask of a staff member in your ﬁrst meeting with
him or her?
2
...

What about the ethical aspects of the government’s allowance for these plants to continue polluting the atmosphere
with the emissions that may cause health problems for
citizens and further the effects of global warming? What
types of regulations, if any, should be developed for PEC
(and other generators) to follow in the future?

New and updated case studies at the
end of most chapters present realworld, in-depth treatments and
exercises in the engineering
profession
...

ACKNOWLEDGMENT OF CONTRIBUTORS
It takes the input and efforts of many individuals to make signiﬁcant improvements in a textbook
...

Paul Askenasy, Texas Commission on Environmental Quality
Jack Beltran, Bristol-Myers Squibb
Robert Lundquist, Ohio State University
William Peet, Infrastructure Coordination, Government of Niue
Sallie Sheppard, Texas A&M University
We thank the following individuals for their comments, feedback, and review of material to assist
in making this edition a real success
...
Dacquisto, Gonzaga University
Houshang Darabi, University of Illinois at Chicago
Freddie Davis, West Texas A&M University
Edward Lester Dollar, Southern Polytechnic State University
Ted Eschenbach, University of Alaska
Clara Fang, University of Hartford
Abel Fernandez, University of the Paciﬁc
Daniel A
...
Fries, University of Central Florida
Nathan Gartner, University of Massachusetts–Lowell
Johnny R
...
Johnson, Valparaiso University
Justin W
...
Madrid, New Mexico State University
Saeed Manafzadeh, University of Illinois at Chicago
Quamrul Mazumder, University of Michigan–Flint
Deb McAvoy, Ohio University
Gene McGinnis, Christian Brothers University
Bruce V
...
Rider, Ohio Northern University
John Ristroph, University of Louisiana at Lafayette
Saeid L
...
Song, Norfolk State University
James Stevens, University of Colorado at Colorado Springs
John A
...
Sutton, Purdue University
Pete Weiss, Valparaiso University

xx

Acknowledgment of Contributors

Greg Wiles, Southern Polytechnic State University
Richard Youchak, University of Pittsburgh at Johnstown
William A
...
We hope you ﬁnd the contents of this edition helpful in
your academic and professional activities
...
com
atarquin@utep
...
When you have completed stage 1, you will be
able to understand and work problems that account for the
time value of money, cash ﬂows occurring at different times with
different amounts, and equivalence at different interest rates
...

The factors commonly used in all engineering economy computations are introduced and applied here
...
Also, after these chapters, you
should be comfortable using many of the spreadsheet functions
...
A checkmark icon in the margin indicates that a new concept or guideline
is introduced at this point
...

SECTION

TOPIC

LEARNING OUTCOME

1
...

1
...

1
...

1
...

1
...

1
...

1
...

1
...

1
...

1
...

T

he need for engineering economy is primarily motivated by the work that engineers
do in performing analyses, synthesizing, and coming to a conclusion as they work on
projects of all sizes
...
These decisions involve the fundamental elements of cash ﬂows of money, time,
and interest rates
...

1
...

Most decisions involve money, called capital or capital funds, which is usually limited in
amount
...

Engineers play a vital role in capital investment decisions based upon their ability and experience
to design, analyze, and synthesize
...
Engineering economy deals with the
economic factors
...
Mathematical techniques
simplify the economic evaluation of alternatives
...
Therefore, besides applications to projects in your future jobs, what you learn
from this book and in this course may well offer you an economic analysis tool for making
personal decisions such as car purchases, house purchases, major purchases on credit, e
...
,
furniture, appliances, and electronics
...

People make decisions; computers, mathematics, concepts, and guidelines assist people in
their decision-making process
...
Therefore, the numbers used in engineering economy are best estimates of what is expected to occur
...
In short, the variation between an amount or time estimated now and that observed
in the future is caused by the stochastic (random) nature of all economic events
...
Example 1
...

EXAMPLE 1
...
He found the average cost (to
the nearest dollar) to be $570 per repair from data taken over a 5-year period
...
However, the last 3 years of costs are higher and
more consistent with an average of $631
...

If the analysis is to use the most recent data and trends, a range of, say, Ϯ5% of $630 is recommended
...

The criterion used to select an alternative in engineering economy for a speciﬁc set of estimates
is called a measure of worth
...
This is the
concept of the time value of money
...
The time value of money explains the change
in the amount of money over time for funds that are owned (invested) or owed (borrowed)
...

The time value of money is very obvious in the world of economics
...
If we borrow money today, in one form or another, we expect to return the original
amount plus some additional amount of money
...
For example,
assume you invested $4975 exactly 3 years ago in 53 shares of IBM stock as traded on the New
York Stock Exchange (NYSE) at $93
...
You expect to make 8% per year appreciation,
not considering any dividends that IBM may declare
...
25 per share for a total of $6744
...
This increase in value represents a
rate of return of 10
...
(These type of calculations are explained later
...

1
...
Implementing a structured procedure is the best approach to select the best solution to the problem
...

2
...

4
...

Collect relevant, available data and deﬁne viable solution alternatives
...

Identify an economic measure of worth criterion for decision making
...
2

Performing an Engineering Economy Study

5
...

6
...

7
...

Technically, the last step is not part of the economy study, but it is, of course, a step needed to
meet the project objective
...
Accordingly, steps 5 and 6 may result in selection
of an alternative different from the economically best one
...
This occurs when projects are independent of one another
...
Figure 1–1 illustrates the steps above for
one alternative
...

Problem Description and Objective Statement A succinct statement of the problem and
primary objective(s) is very important to the formation of an alternative solution
...
The objectives may be to generate the forecasted electricity

Step in
study

1

Problem
description

Objective
statement

Available data
2

Alternatives for
solution

3

Cash ﬂows and
other estimates

4

Measure of worth
criterion

5

Engineering
economic analysis

6

Best alternative
selection

7

One or more approaches
to meet objective

Expected life
Revenues
Costs
Taxes
Project ﬁnancing

Implementation
and monitoring

PW, ROR, B/C, etc
...

New engineering
economy study
begins

5

6

Chapter 1

Foundations of Engineering Economy

needed for 2015 and beyond, plus to not exceed all the projected emission allowances in these
future years
...
Words, pictures, graphs, equipment and service descriptions, simulations, etc
...
The best estimates for parameters are also part of the alternative
...
If changes in income (revenue)
may occur, this parameter must be estimated
...
For example, if two alternatives are
described and analyzed, one will likely be selected and implementation initiated
...

Cash Flows All cash ﬂows are estimated for each alternative
...
When cash ﬂow estimates for speciﬁc parameters are expected to
vary signiﬁcantly from a point estimate made now, risk and sensitivity analyses (step 5) are
needed to improve the chances of selecting the best alternative
...
Estimation of
costs is discussed in Chapter 15, and the elements of variation (risk) and sensitivity analysis are
included throughout the text
...
The result of the analysis will be one or more numerical values; this can be
in one of several terms, such as money, an interest rate, number of years, or a probability
...

Before an economic analysis technique is applied to the cash ﬂows, some decisions about
what to include in the analysis must be made
...
Federal, state or provincial, county, and city taxes will impact the costs of every
alternative
...
If taxes and inﬂation are expected to impact all alternatives equally, they
may be disregarded in the analysis
...
Also, if the impact of inﬂation over time is important
to the decision, an additional set of computations must be added to the analysis; Chapter 14
covers the details
...
For example, if alternative A has a rate of return (ROR) of
15
...
9% per year, B is better economically
...
There are many possible noneconomic factors;
some typical ones are
•
•
•
•
•

Market pressures, such as need for an increased international presence
Availability of certain resources, e
...
, skilled labor force, water, power, tax incentives
Government laws that dictate safety, environmental, legal, or other aspects
Corporate management’s or the board of director’s interest in a particular alternative
Goodwill offered by an alternative toward a group: employees, union, county, etc
...

At times, only one viable alternative is identiﬁed
...
The do-nothing alternative maintains the status quo
...
3

Professional Ethics and Economic Decisions

Whether we are aware of it or not, we use criteria every day to choose between alternatives
...
But how did you
deﬁne best? Was the best route the safest, shortest, fastest, cheapest, most scenic, or what? Obviously, depending upon which criterion or combination of criteria is used to identify the best, a
different route might be selected each time
...
Thus, when there are
several ways of accomplishing a stated objective, the alternative with the lowest overall cost or
highest overall net income is selected
...
3 Professional Ethics and Economic Decisions
Many of the fundamentals of engineering ethics are intertwined with the roles of money and
economics-based decisions in the making of professionally ethical judgments
...
For example, Chapter 9, Beneﬁt/Cost Analysis and Public Sector Economics, includes material on the ethics of public project contracts and public policy
...

The terms morals and ethics are commonly used interchangeably, yet they have slightly
different interpretations
...
Ethical practices can be evaluated by
using a code of morals or code of ethics that forms the standards to guide decisions and
actions of individuals and organizations in a profession, for example, electrical, chemical,
mechanical, industrial, or civil engineering
...

Universal or common morals These are fundamental moral beliefs held by virtually all people
...

It is possible for actions and intentions to come into conﬂict concerning a common moral
...
After their collapse on September 11,
2001, it was apparent that the design was not sufﬁcient to withstand the heat generated by the
ﬁrestorm caused by the impact of an aircraft
...
However, their design
actions did not foresee this outcome as a measurable possibility
...
These usually parallel the common morals in that stealing, lying, murdering, etc
...

It is quite possible that an individual strongly supports the common morals and has excellent
personal morals, but these may conﬂict from time to time when decisions must be made
...
If he or she does not
know how to work some test problems, but must make a certain minimum grade on the ﬁnal
exam to graduate, the decision to cheat or not on the ﬁnal exam is an exercise in following or
violating a personal moral
...
The code
states the commonly accepted standards of honesty and integrity that each individual is expected
to demonstrate in her or his practice
...

Although each engineering profession has its own code of ethics, the Code of Ethics for
Engineers published by the National Society of Professional Engineers (NSPE) is very commonly used and quoted
...
Here are three examples from
the Code:
“Engineers, in the fulﬁllment of their duties, shall hold paramount the safety, health, and welfare of the public
...
1)
“Engineers shall not accept ﬁnancial or other considerations, including free engineering designs, from material or equipment suppliers for specifying their product
...
5
...
”
(section III
...
b)
As with common and personal morals, conﬂicts can easily rise in the mind of an engineer
between his or her own ethics and that of the employing corporation
...
Suppose the engineer has worked for years in a military defense contractor’s
facility and does the detailed cost estimations and economic evaluations of producing ﬁghter
jets for the Air Force
...
Although the employer and the engineer are not violating any ethics code,
the engineer, as an individual, is stressed in this position
...
Conﬂicts such as this can place individuals in real dilemmas with no or mostly unsatisfactory
alternatives
...
Many money-related situations, such as those that follow, can have ethical dimensions
...

• Family or personal connections with individuals in a company offer unfair or insider information that allows costs to be cut in strategic areas of a project
...

While the system is operating:

• Delayed or below-standard maintenance can be performed to save money when cost overruns
exist in other segments of a project
...

• Safety margins are compromised because of cost, personal inconvenience to workers, tight
time schedules, etc
...
245–8)
...
Some 500,000 persons were exposed to inhalation of this deadly gas that burns moist parts of the body
...
Although Union Carbide owned the facility, the Indian government
had only Indian workers in the plant
...
However, one of
the surprising practices that caused unnecessary harm to workers was the fact that masks, gloves,
and other protective gear were not worn by workers in close proximity to the tanks containing
MIC
...

Many ethical questions arise when corporations operate in international settings where the
corporate rules, worker incentives, cultural practices, and costs in the home country differ from
those in the host country
...
3

Professional Ethics and Economic Decisions

other cost-reducing factors
...

It is important to understand that the translation from universal morals to personal morals and
professional ethics does vary from one culture and country to another
...
In some societies and cultures, corruption in the process of contract making is common and
often “overlooked” by the local authorities, who may also be involved in the affairs
...
Find and punish the
individuals involved
...

EXAMPLE 1
...
He has been a registered professional engineer (PE) for the last 15 years
...
Carol offered to pay for
his time and talent, but Jamie saw no reason to take money for helping with data commonly
used by him in performing his job at Burris
...

Yesterday, Jamie was called into his supervisor’s ofﬁce and told that Burris had not received
the contract award in Sharpstown, where a metro system is to be installed
...
This job was
greatly needed by Burris, as the country and most municipalities were in a real economic
slump, so much so that Burris was considering furloughing several engineers if the Sharpstown
bid was not accepted
...
Jamie was astounded and angry
...
The numbers used by the competitor to win the Sharpstown award were the same numbers that Jamie had prepared for Burris
on this bid, and they closely matched the values that he gave Carol when he helped her
...
As a result,
Jamie was escorted out of his ofﬁce and the building within one hour and told to not ask anyone
at Burris for a reference letter if he attempted to get another engineering job
...

Refer to the NSPE Code of Ethics for Engineers (Appendix C) for speciﬁc points of concern
...
Some of these mistakes, oversights, and possible code
violations are summarized here
...
1
...
”

9

10

Chapter 1

Foundations of Engineering Economy

Carol

• Did not share the intended use of Jamie’s work
• Did not seek information from Jamie concerning his employer’s intention to bid on the
same project as her client
• Misled Jamie in that she did not seek approval from Jamie to use and quote his information
and assistance
• Did not inform her client that portions of her work originated from a source employed by a
possible bid competitor
• Likely violated, at least, Code of Ethics for Engineers section III
...
a, which reads, “Engineers shall, whenever possible, name the person or persons who may be individually responsible for designs, inventions, writings, or other accomplishments
...

1
...
Computationally, interest is the difference
between an ending amount of money and the beginning amount
...
There are always two perspectives to an amount of interest—interest paid
and interest earned
...
Interest is paid when a person or organization borrowed money (obtained a loan) and repays a larger amount over time
...
The numerical values and formulas used are the same for both perspectives, but
the interpretations are different
...
1]

When interest paid over a speciﬁc time unit is expressed as a percentage of the principal, the result is called the interest rate
...
2]

The time unit of the rate is called the interest period
...
Shorter time periods can be used, such as 1% per month
...
If only the rate is stated,
for example, 8
...

Loan

Bank

Repayment
ϩ interest

Loan

Borrower

Investor

Repayment
ϩ interest

(a)

Figure 1–2
(a) Interest paid over time to lender
...

(b)

Corporation

1
...
3
An employee at LaserKinetics
...
Determine the interest amount and the interest rate paid
...
Apply Equation [1
...

Interest paid ϭ $10,700 Ϫ 10,000 ϭ $700
Equation [1
...

$700
Percent interest rate ϭ ———— ϫ 100% ϭ 7% per year
$10,000

EXAMPLE 1
...
, plans to borrow $20,000 from a bank for 1 year at 9% interest for new
recording equipment
...
(b) Construct a column graph that shows the original loan amount and total amount due after 1 year
used to compute the loan interest rate of 9% per year
...
2] for interest accrued
...
09) ϭ $1800
The total amount due is the sum of principal and interest
...
2]: $1800 interest, $20,000 original loan
principal, 1-year interest period
...
Example 1
...

Comment
Note that in part (a), the total amount due may also be computed as
Total due ϭ principal(1 ϩ interest rate) ϭ $20,000(1
...

11

12

Chapter 1

Foundations of Engineering Economy

From the perspective of a saver, a lender, or an investor, interest earned (Figure 1–2b) is the
ﬁnal amount minus the initial amount, or principal
...
3]

Interest earned over a speciﬁc period of time is expressed as a percentage of the original amount
and is called rate of return (ROR)
...
4]

The time unit for rate of return is called the interest period, just as for the borrower’s perspective
...

The term return on investment (ROI) is used equivalently with ROR in different industries and
settings, especially where large capital funds are committed to engineering-oriented programs
...
2] and [1
...

EXAMPLE 1
...

(b) Calculate the amount of interest earned during this time period
...

If X is the original deposit,
Total accrued ϭ deposit ϩ deposit(interest rate)
$1000 ϭ X ϩ X(0
...
05) ϭ 1
...
38
1
...
3] to determine the interest earned
...
38 ϭ $47
...
3 to 1
...
When more than one interest period is involved, e
...
, the amount of interest after 3 years, it is necessary to state whether the interest is accrued on a simple or compound
basis from one period to the next
...

Since inﬂation can signiﬁcantly increase an interest rate, some comments about the fundamentals of inﬂation are warranted at this early stage
...
That is, $10 now will not purchase the same amount of gasoline
for your car (or most other things) as $10 did 10 years ago
...

Inﬂation

In simple terms, interest rates reﬂect two things: a so-called real rate of return plus the expected
inﬂation rate
...

The safest investments (such as government bonds) typically have a 3% to 4% real rate of
return built into their overall interest rates
...

Clearly, inﬂation causes interest rates to rise
...
And from the vantage point of the saver or investor in a ﬁxed-interest account,

1
...
Inﬂation means that cost and revenue
cash ﬂow estimates increase over time
...
We see the effect of inﬂation in that money
purchases less now than it did at a previous time
...

Commonly, engineering economy studies assume that inﬂation affects all estimated values
equally
...
However, if inﬂation were explicitly taken into account, and it was reducing the value of money at, say, an average of 4% per year,
then it would be necessary to perform the economic analysis using an inﬂated interest rate
...
32% per year using the relations derived in Chapter 14
...
5 Terminology and Symbols
The equations and procedures of engineering economy utilize the following terms and symbols
...

P ϭ value or amount of money at a time designated as the present or time 0
...
Also F is called future worth (FW)
and future value (FV); dollars
A ϭ series of consecutive, equal, end-of-period amounts of money
...
It should be clear that a present value P represents a
single sum of money at some time prior to a future value F or prior to the ﬁrst occurrence of an
equivalent series amount A
...
e
...
Both conditions must
exist before the series can be represented by A
...
Unless stated otherwise, assume that the rate applies throughout the entire n years or interest periods
...

All engineering economy problems involve the element of time expressed as n and interest
rate i
...

Additional symbols used in engineering economy are deﬁned in Appendix E
...
6
Today, Julie borrowed $5000 to purchase furniture for her new house
...
Determine the engineering economy symbols and
their value for each option
...

(b) One payment 3 years from now with interest based on 7% per year
...

P ϭ $5000

i ϭ 5% per year

n ϭ 5 years

Aϭ?

(b) Repayment requires a single future amount F, which is unknown
...
7
You plan to make a lump-sum deposit of $5000 now into an investment account that pays 6%
per year, and you plan to withdraw an equal end-of-year amount of $1000 for 5 years, starting
next year
...
Deﬁne the engineering economy symbols involved
...

P ϭ $5000
A ϭ $1000 per year for 5 years
F ϭ ? at end of year 6
i ϭ 6% per year
n ϭ 5 years for the A series and 6 for the F value

EXAMPLE 1
...
(a) Identify the symbols, and
(b) calculate the amount that had to be deposited exactly 1 year ago to earn $5000 in interest
now, if the rate of return is 6% per year
...

Pϭ?
i ϭ 6% per year
n ϭ 1 year
F ϭ P ϩ interest ϭ ? ϩ $5000
(b) Let F ϭ total amount now and P ϭ original amount
...
Now we can determine P
...
1] through [1
...

F ϭ P ϩ Pi
The $5000 interest can be expressed as
Interest ϭ F – P ϭ (P ϩ Pi) – P
ϭ Pi
$5000 ϭ P(0
...
33
0
...
6

15

Cash Flows: Estimation and Diagramming

1
...
All cash ﬂows occur during speciﬁc time periods, such as 1 month, every 6 months, or 1 year
...
For
example, a payment of $10,000 once every year in December for 5 years is a series of 5 outgoing
cash ﬂows
...
Engineering economy bases its computations on the timing, size, and direction of cash ﬂows
...
A plus sign indicates a cash inﬂow
...
A negative or minus sign indicates a cash outﬂow
...

Cash ﬂow

Of all the steps in Figure 1–1 that outline the engineering economy study, estimating cash ﬂows
(step 3) is the most difﬁcult, primarily because it is an attempt to predict the future
...
As you scan these, consider how the cash inﬂow
or outﬂow may be estimated most accurately
...
2 million in capital expenditures for a water recycling unit
All of these are point estimates, that is, single-value estimates for cash ﬂow elements of an
alternative, except for the last revenue and cost estimates listed above
...
For the initial chapters, we will utilize point estimates
...

Once all cash inﬂows and outﬂows are estimated (or determined for a completed project), the
net cash ﬂow for each time period is calculated
...
5]
[1
...

At the beginning of this section, the timing, size, and direction of cash ﬂows were mentioned
as important
...

The end-of-period convention means that all cash inﬂows and all cash outﬂows are assumed to
take place at the end of the interest period in which they actually occur
...

End-of-period convention

16

Chapter 1

Figure 1–4

Foundations of Engineering Economy

Year 1

Year 5

A typical cash ﬂow time
scale for 5 years
...

Cash flow

i = 4% per year

1

2

3

4

5

Year

–

In assuming end-of-period cash ﬂows, it is important to understand that future (F) and uniform
annual (A) amounts are located at the end of the interest period, which is not necessarily
December 31
...
7 the lump-sum deposit took place on July 1, 2011, the withdrawals will take place on July 1 of each succeeding year for 6 years
...

The cash ﬂow diagram is a very important tool in an economic analysis, especially when the
cash ﬂow series is complex
...
The diagram includes what is known, what is estimated, and what is
needed
...

Cash ﬂow diagram time t ϭ 0 is the present, and t ϭ 1 is the end of time period 1
...
The time scale of Figure 1–4 is set up for 5 years
...

While it is not necessary to use an exact scale on the cash ﬂow diagram, you will probably
avoid errors if you make a neat diagram to approximate scale for both time and relative cash ﬂow
magnitudes
...
A
vertical arrow pointing up indicates a positive cash ﬂow
...
We will use a bold, colored arrow to indicate what is unknown
and to be determined
...
The interest rate is also indicated on the diagram
...
The arrow
for the unknown value is generally drawn in the opposite direction from the other cash ﬂows;
however, the engineering economy computations will determine the actual sign on the F value
...
Assume you
borrow $8500 from a bank today to purchase an $8000 used car for cash next week, and you plan
to spend the remaining $500 on a new paint job for the car two weeks from now
...
The cash ﬂow signs and amounts for
these perspectives are as follows
...
6

Cash Flows: Estimation and Diagramming
$8500

1

2
Week

0
$500

$8000

Figure 1–6
Cash ﬂows from perspective of borrower for loan and purchases
...
For your perspective,
all three cash ﬂows are involved and the diagram appears as shown in Figure 1–6 with a time scale
of weeks
...

EXAMPLE 1
...
Carla Ramos, a lead engineer for Mexico and Central
American operations, plans expenditures of $1 million now and each of the next 4 years just
for the improvement of ﬁeld-based pressure-release valves
...

Solution
Figure 1–7 indicates the uniform and negative cash ﬂow series (expenditures) for ﬁve periods,
and the unknown F value (positive cash ﬂow equivalent) at exactly the same time as the ﬁfth
expenditure
...
Therefore, the last negative cash ﬂow occurs at the end of the fourth year, when F
also occurs
...
This addition demonstrates that year 0 is the end-of-period
point for the year Ϫ1
...
9
...
10
An electrical engineer wants to deposit an amount P now such that she can withdraw an equal
annual amount of A1 ϭ $2000 per year for the ﬁrst 5 years, starting 1 year after the deposit, and
a different annual withdrawal of A2 ϭ $3000 per year for the following 3 years
...
5% per year?

17

18

Chapter 1

Foundations of Engineering Economy

Solution
The cash ﬂows are shown in Figure 1–8
...
The withdrawals (positive cash inﬂow) for the A1 series occur at the end of years 1 through 5, and A2
occurs in years 6 through 8
...
5%
P=?

Figure 1–8
Cash ﬂow diagram with two different A series, Example 1
...

EXAMPLE 1
...
The annual rental income
from the compressor has been $750
...
The company plans to sell the compressor at the end of next year for
$150
...

Solution

Let now be time t ϭ 0
...
5]
...
Present worth P is located at year 0
...
11
...
7

Economic Equivalence

19

1
...
Before we delve into the economic aspects, think of the many types of equivalency we
may utilize daily by transferring from one scale to another
...
370 inches ϭ 1 meter
100 centimeters ϭ 1 meter
1000 meters ϭ 1 kilometer
1 kilometer ϭ 0
...
Consider the equivalency of a speed of 110 kilometers per hour (kph) into miles per minute using conversions between distance and time scales
with three-decimal accuracy
...
609 kilometers
1 hour ϭ 60 minutes
110 kph ϭ 68
...
365 mph ϭ 1
...
Note that throughout these statements, the fundamental relations of 1 mile ϭ 1
...
If a fundamental relation changes, the entire equivalency is in error
...

Economic equivalence is a combination of interest rate and time value of money to determine the different amounts of money at different points in time that are equal in economic
value
...

Amount accrued ϭ 100 ϩ 100(0
...
06) ϭ $106
If someone offered you a gift of $100 today or $106 one year from today, it would make no difference which offer you accepted from an economic perspective
...
However, the two sums of money are equivalent to each other only when the
interest rate is 6% per year
...

In addition to future equivalence, we can apply the same logic to determine equivalence for
previous years
...
06 ϭ $94
...
From these illustrations, we can state the following: $94
...

The fact that these sums are equivalent can be veriﬁed by computing the two interest rates for
1-year interest periods
...
66
$94
...

Economic equivalence

20

Chapter 1

Foundations of Engineering Economy
i = 6% per year

Amount, $

100
94
...
00 interest
$5
...

EXAMPLE 1
...
In general, batteries are stored throughout the year,
and a 5% cost increase is added each year to cover the inventory carrying charge for the distributorship owner
...
Make the calculations
necessary to show which of the following statements are true and which are false about battery
costs
...
60 one year from now
...

(c) A $38 cost now is equivalent to $39
...

(d) A $3000 cost now is equivalent to $2887
...

(e) The carrying charge accumulated in 1 year on an investment of $20,000 worth of
batteries is $1000
...
05) ϭ $102
...
60; therefore, it is false
...
60͞1
...
57 $98
...
00͞1
...
24 $200; therefore, it is false
...
05) ϭ $39
...

(d) Cost now is 2887
...
05) ϭ $3031
...

(e) The charge is 5% per year interest, or $20,000(0
...

Comparison of alternative cash ﬂow series requires the use of equivalence to determine when
the series are economically equal or if one is economically preferable to another
...
Example 1
...

EXAMPLE 1
...
He wants to borrow $10,000 now and repay it
over the next 1 or 2 years
...
Howard received
2-year repayment options from banks A and B
...
8

Simple and Compound Interest
Amount to pay, $ per year
Year

Bank A

Bank B

1

−5,378
...
00

2

−5,378
...
00

Total paid

−10,756
...
00

After reviewing these plans, Howard decided that he wants to repay the $10,000 after only
1 year based on the expected increased revenue
...

Now Howard has three options and wonders which one to take
...
(This is determined by using computations that you will learn in Chapter 2
...

The brother-in-law repayment plan requires a total of $600 in interest 1 year later plus the
principal of $10,000, which makes the interest rate 6% per year
...
Even though the sum of money repaid is smaller, the timing of the cash ﬂows
and the interest rate make it less desirable
...
The interest
rate, timing, and economic equivalence must be considered
...
8 Simple and Compound Interest
The terms interest, interest period, and interest rate (introduced in Section 1
...

However, for more than one interest period, the terms simple interest and compound interest become important
...
The total simple interest over several periods is computed as
Simple interest ϭ (principal)(number of periods)(interest rate)
I ϭ Pni

[1
...

EXAMPLE 1
...
The loan is for 3 years at 10% per year simple interest
...
10) ϭ $10,000
Total interest for 3 years from Equation [1
...
10) ϭ $30,000

21

22

Chapter 1

Foundations of Engineering Economy

The amount due after 3 years is
Total due ϭ $100,000 ϩ 30,000 ϭ $130,000
The interest accrued in the ﬁrst year and in the second year does not earn interest
...

In most ﬁnancial and economic analyses, we use compound interest calculations
...
Thus,
compound interest means interest on top of interest
...
Now the
interest for one period is calculated as
Compound interest ϭ (principal ϩ all accrued interest)(interest rate)

[1
...

(

jϭtϪ1

It ϭ P ϩ

⌺
jϭ1

)

IJ (i)

[1
...
15
Assume an engineering company borrows $100,000 at 10% per year compound interest and
will pay the principal and all the interest after 3 years
...
Graph the interest and total owed for each year, and compare with
the previous example that involved simple interest
...
8]
...
10) ϭ $10,000
100,000 ϩ 10,000 ϭ $110,000
110,000(0
...
10) ϭ $12,100
121,000 ϩ 12,100 ϭ $133,100

The repayment plan requires no payment until year 3 when all interest and the principal, a total
of $133,100, are due
...
The differences due to compounding are clear
...

Note that while simple interest due each year is constant, the compounded interest due
grows geometrically
...
For example, if the loan is for 10 years, not 3, the extra paid for compounding interest
may be calculated to be $59,374
...
8

23

Simple and Compound Interest
0

Ϫ10

1

I

2

I

3

I

0

Year

Ϫ10

I is constant

Ϫ11

1

2

3

I increases
geometrically

I
I

Ϫ12

Amount owed (ϫ $1000)

Amount owed (ϫ $1000)

I
Ϫ100

Ϫ110

Ϫ120

Ϫ100

Ϫ110

Ϫ120
Geometric
increase

Arithmetic
increase
Ϫ130

Ϫ130

Ϫ140

Ϫ140
(a)

(b)

Figure 1–11
Interest I owed and total amount owed for (a) simple interest (Example 1
...
15)
...
15
is to utilize the fact that compound interest increases geometrically
...
In this case, the total amount due at the end of each
year is
Year 1: $100,000(1
...
10)2 ϭ $121,000
Year 3: $100,000(1
...
The general
form of the equation is
Total due after n years ‫ ؍‬principal(1 ؉ interest rate)n years
‫ ؍‬P(1 ؉ i)n

Year

[1
...
Equation [1
...
This fundamental relation will be used many times in the upcoming chapters
...
This also shows that there are
many ways to take into account the time value of money
...
16
Table 1–1 details four different loan repayment plans described below
...

• Plan 1: Pay all at end
...
Interest accumulates each year on the total of principal and all accrued interest
...
The accrued interest is paid each
year, and the entire principal is repaid at the end of year 5
...
The accrued interest and one-ﬁfth
of the principal (or $1000) are repaid each year
...

• Plan 4: Pay equal amount of interest and principal
...
Since the loan balance decreases at a rate slower than that in plan 3 due to the equal
end-of-year payments, the interest decreases, but at a slower rate
...
00
432
...
56
503
...
20

$5400
...
00
6298
...
44
7346
...
64

$5000
...
00
5832
...
56
6802
...
64

Plan 2: Pay Interest Annually; Principal Repaid at End
0
1
2
3
4
5

$400
...
00
400
...
00
400
...
00
5400
...
00
5400
...
00

Total

$Ϫ400
...
00
Ϫ400
...
00
– 5400
...
00
5000
...
00
5000
...
00

$Ϫ7000
...
00
320
...
00
160
...
00

$5400
...
00
3240
...
00
1080
...
00
Ϫ1320
...
00
Ϫ1160
...
00

$5000
...
00
3000
...
00
1000
...
00

Plan 4: Pay Equal Annual Amount of Interest and Principal
0
1
2
3
4
5
Total

$400
...
82
258
...
65
92
...
00
4479
...
43
2411
...
28

$−1252
...
28
−1252
...
28
– 1252
...
40

$5000
...
72
3227
...
15
1159
...
9

Minimum Attractive Rate of Return

(a) Make a statement about the equivalence of each plan at 8% compound interest
...
Comment on the total amounts repaid for the two plans
...
The difference in the total amounts repaid can be explained by the time value of money and by the partial repayment of principal prior to
year 5
...
64 at the end of year 5
Plan 2 $400 per year for 4 years and $5400 at the end of year 5
Plan 3 Decreasing payments of interest and partial principal in years 1 ($1400)
through 5 ($1080)
Plan 4 $1252
...
This amount covers accrued interest and a partial
amount of principal repayment
...
Since
the annual accrued interest of $400 is paid each year and the principal of $5000 is repaid
in year 5, the schedule is exactly the same as that for 8% per year compound interest, and
the total amount repaid is the same at $7000
...
Any deviation from this schedule will
cause the two plans and amounts to differ
...
9 Minimum Attractive Rate of Return
For any investment to be proﬁtable, the investor (corporate or individual) expects to receive more
money than the amount of capital invested
...
The deﬁnition of ROR in Equation [1
...

Engineering alternatives are evaluated upon the prognosis that a reasonable ROR can be
expected
...

25

26

Chapter 1

Minimum Attractive Rate
of Return (MARR)

Cost of capital

Foundations of Engineering Economy

The Minimum Attractive Rate of Return (MARR) is a reasonable rate of return established
for the evaluation and selection of alternatives
...
MARR is also referred to as the hurdle rate,
cutoff rate, benchmark rate, and minimum acceptable rate of return
...
In the United
States, the current U
...
Treasury Bill return is sometimes used as the benchmark safe rate
...
The MARR is not a rate that is
calculated as a ROR
...

To develop a foundation-level understanding of how a MARR value is established and used
to make investment decisions, we return to the term capital introduced in Section 1
...
Although
the MARR is used as a criterion to decide on investing in a project, the size of MARR is fundamentally connected to how much it costs to obtain the needed capital funds
...
The interest, expressed as a percentage rate per
year, is called the cost of capital
...
Alternatively, you might
choose to use your credit card and pay off the balance on a monthly basis
...
Or, you could use funds from your savings account that
earns 5% per year and pay cash
...
The 9%, 15%, and 5% rates are your cost of capital estimates to raise the
capital for the system by different methods of capital ﬁnancing
...

Rate of return,
percent

Expected rate of return on
a new proposal

Range for the rate of return on
accepted proposals, if other
proposals were rejected
for some reason

All proposals must offer
at least MARR to
be considered

MARR

Rate of return on
“safe investment”

Figure 1–12
Size of MAAR relative to other rate of return values
...
10

27

Introduction to Spreadsheet Use

In general, capital is developed in two ways—equity ﬁnancing and debt ﬁnancing
...
Chapter 10 covers these in greater detail, but
a snapshot description follows
...
Individuals can use their own cash, savings, or investments
...

Debt ﬁnancing The corporation borrows from outside sources and repays the principal and interest according to some schedule, much like the plans in Table 1–1
...
Individuals, too, can utilize debt
sources, such as the credit card (15% rate) and bank options (9% rate) described above
...
If the HDTV is purchased with 40% credit card money at 15% per year and 60% savings
account funds earning 5% per year, the weighted average cost of capital is 0
...
6(5) ϭ
9% per year
...
So the inequality
ROR Ն MARR Ͼ WACC

[1
...
Exceptions may be government-regulated requirements
(safety, security, environmental, legal, etc
...

Often there are many alternatives that are expected to yield a ROR that exceeds the MARR as
indicated in Figure 1–12, but there may not be sufﬁcient capital available for all, or the project’s
risk may be estimated as too high to take the investment chance
...
The expected rate of return on the unfunded project is called the opportunity
cost
...
Numerically, it is the largest rate of return of all the projects not accepted
(forgone) due to the lack of capital funds or other resources
...
e
...

As an illustration of opportunity cost, refer to Figure 1–12 and assume a MARR of 12% per
year
...
Meanwhile, proposal B has a ROR ϭ 14
...
Since proposal A is not undertaken due to the lack of capital, its estimated ROR of 13%
is the opportunity cost; that is, the opportunity to make an additional 13% return is forgone
...
10 Introduction to Spreadsheet Use
The functions on a computer spreadsheet can greatly reduce the amount of hand work for equivalency computations involving compound interest and the terms P, F, A, i, and n
...
However, as cash ﬂow series become more complex, the spreadsheet
offers a good alternative
...
Appendix A is a primer on using spreadsheets and Excel
...
Appendix A also includes a section on spreadsheet layout that is useful when the
economic analysis is presented to someone else—a coworker, a boss, or a professor
...
The functions are great supplemental tools, but they do not replace the understanding of engineering economy relations, assumptions, and techniques
...

To ﬁnd the present value P: ‫ ؍‬PV(i%, n, A, F)
To ﬁnd the future value F: ‫ ؍‬FV(i%, n, A, P)
To ﬁnd the equal, periodic value A: ‫ ؍‬PMT(i%, n, P, F)
To ﬁnd the number of periods n: ‫ ؍‬NPER(i%, A, P, F)
To ﬁnd the compound interest rate i: ‫ ؍‬RATE( n, A, P, F)
To ﬁnd the compound interest rate i: ‫ ؍‬IRR(ﬁrst_cell:last_cell)
To ﬁnd the present value P of any series: ‫ ؍‬NPV(i%, second_cell:last_cell) ؉ ﬁrst_cell
If some of the parameters don’t apply to a particular problem, they can be omitted and zero is
assumed
...
If the
parameter omitted is an interior one, the comma must be entered
...
In all cases, the function must be preceded by an equals sign (ϭ) in the
cell where the answer is to be displayed
...
6a, where the
equivalent annual amount A is unknown, as indicated by A ϭ ?
...
) To ﬁnd A using a spreadsheet
function, simply enter the PMT function ϭ PMT(5%,5,5000)
...
The answer ($1154
...
The answer may appear in red and in parentheses, or with a minus sign on your screen
to indicate a negative amount from the perspective of a reduction in the account balance
...
6b
...
The FV function appears in the formula bar; and many examples throughout this text will include cell tags, as shown here, to indicate the format of
important entries
...
Once set up, the spreadsheet can be
used to perform sensitivity analysis for estimates that are subject to change
...
(Note: The spreadsheet examples may
be omitted, if spreadsheets are not used in the course
...
)

ϭ PMT(5%,5,5000)

Figure 1–13
Use of spreadsheet functions PMT and FV, Example 1
...

ϭ FV(7%,3,,5000)

1
...
17
A Japan-based architectural ﬁrm has asked a United States–based software engineering group
to infuse GPS sensing capability via satellite into monitoring software for high-rise structures
in order to detect greater than expected horizontal movements
...
The inclusion of accurate GPS data is estimated to increase annual revenue over
that for the current software system by $200,000 for each of the next 2 years, and by $300,000
for each of years 3 and 4
...
Develop spreadsheets to answer the questions
below
...

(b) Repeat part (a) if estimated revenue increases from $300,000 to $600,000 in years 3 and 4
...
This will decrease the real rate
of return from 8% to 3
...

Solution by Spreadsheet
Refer to Figure 1–14a to d for the solutions
...
(Actually, all the questions can be
answered on one spreadsheet by changing the numbers
...
)
The Excel functions are constructed with reference to the cells, not the values themselves, so that sensitivity analysis can be performed without function changes
...
For example, the
8% rate in cell B2 will be referenced in all functions as B2, not 8%
...

See Appendix A for additional information about using cell referencing and building
spreadsheet relations
...
As an illustration, for year 3 the interest I3 and revenue plus interest R3 are
I3 ϭ (cumulative revenue through year 2)(rate of return)
ϭ $416,000(0
...

Cell C8 relation for I3: ϭ F7*B2
Cell E8 relation for CF3: ϭ B8 ϩ C8
The equivalent amount after 4 years is $1,109,022, which is comprised of $1,000,000 in
total revenue and $109,022 in interest compounded at 8% per year
...

(b) To determine the effect of increasing estimated revenue for years 3 and 4 to $600,000,
use the same spreadsheet and change the entries in cells B8 and B9 as shown in
Figure 1–14c
...

(c) Figure 1–14d shows the effect of changing the original i value from 8% to an inﬂationadjusted rate of 3
...
[Remember to return to the
$300,000 revenue estimates for years 3 and 4 after working part (b)
...

29

30

Chapter 1

Foundations of Engineering Economy

(a) Total interest and revenue for base case, year 4

(b) Spreadsheet relations for base case

Revenue changed
(c) Totals with increased revenue in years 3 and 4

Rate of
return
changed

(d) Totals with inflation of 4% per year considered

Figure 1–14
Spreadsheet solutions with sensitivity analysis, Example 1
...

Comment
Later we will learn how to utilize the NPV and FV Excel ﬁnancial functions to obtain the same
answers determined in Figure 1–14, where we developed each basic relation
...

Problems

31

CHAPTER SUMMARY
Engineering economy is the application of economic factors and criteria to evaluate alternatives,
considering the time value of money
...

The concept of equivalence helps in understanding how different sums of money at different
times are equal in economic terms
...
This power of compounding is very noticeable, especially over
extended periods of time, and for larger sums of money
...
The MARR is always higher than the return from a safe investment
and the cost to acquire needed capital
...
1 List the four essential elements involved in decision making in engineering economic analysis
...
2 What is meant by (a) limited capital funds and
(b) sensitivity analysis?
1
...

1
...

Ethics
1
...
Management wants to return some of the
engineering design work to the United States
rather than export all of it to India, where the primary design work has been accomplished for the
last decade
...

Stefanie and her design team were selected as a
test case to determine the quality and speed of the
design work they could demonstrate on a more
fuel-efﬁcient diesel locomotive
...
One of her team members had a
great design idea on a key element that will improve fuel efﬁciency by approximately 15%
...
S
...

Although reluctant at ﬁrst, Stefanie did go forward
with a design that included the efﬁciency improvement, and no mention of the origin of the idea was
made at the time of the oral presentation or documentation delivery
...

Consult the NSPE Code of Ethics for Engineers
(Appendix C) and identify sections that are points
of concern about Stefanie’s decisions and actions
...
6 Consider the common moral precept that stealing
is wrong
...
One of Hector’s buddies takes a
high-energy drink from a six-pack on the shelf,
opens it, drinks it, and returns the empty can to the
package, with no intention of paying for it
...
Others do it all the time
...
Personally, Hector believes this is a form
of stealing
...

1
...
He has a strong belief in the universal moral that it is wrong to do serious harm to
another person
...

1
...
Although he had a
passing score prior to the ﬁnal, his ﬁnal grade was
so low that he has now ﬂunked the entire year and
will likely have to extend his graduation another
semester or two
...
He did realize during the
semester that he was doing something that even
he considered wrong morally and ethically
...
The classroom was reconﬁgured for the ﬁnal exam in a way
that he could not get any answers from classmates,
and cell phones were collected prior to the exam,
thus removing texting possibilities to friends outside the classroom who might help him on the
ﬁnal exam
...
The question to Claude is,
“What have you been doing throughout this year
to make passing scores repeatedly, but demonstrate such a poor command of Spanish on the
ﬁnal exam?”
From an ethical viewpoint, what options does
Claude have in his answer to this question? Also,
discuss some of the possible effects that this experience may have upon Claude’s future actions and
moral dilemmas
...
9 RKI Instruments borrowed $3,500,000 from a private equity ﬁrm for expansion of its manufacturing
facility for making carbon monoxide monitors/
controllers
...
What
was the interest rate on the loan?

Foundations of Engineering Economy

1
...
The
terms of the loan were such that the company could
pay interest only at the end of each year for up to 5
years, after which the company would have to pay
the entire amount due
...
11 Which of the following 1-year investments has the
highest rate of return?
(a) $12,500 that yields $1125 in interest,
(b) $56,000 that yields $6160 in interest, or
(c) $95,000 that yields $7600 in interest
...
12 A new engineering graduate who started a consulting business borrowed money for 1 year to furnish
the ofﬁce
...
However, because the new graduate had not built up a
credit history, the bank made him buy loan-default
insurance that cost 5% of the loan amount
...

What was the effective interest rate the engineer
paid for the loan?
1
...
14 The symbol P represents an amount of money at a
time designated as present
...
Explain what each symbol stands for: PW, PV, NPV, DCF, and CC
...
15 Identify the four engineering economy symbols
and their values from the following problem statement
...

Thompson Mechanical Products is planning to
set aside $150,000 now for possibly replacing its
large synchronous reﬁner motors whenever it becomes necessary
...
16 Identify the four engineering economy symbols and
their values from the following problem statement
...
17 Identify the four engineering economy symbols
and their values from the following problem statement
...

A green algae, Chlamydomonas reinhardtii,
can produce hydrogen when temporarily deprived
of sulfur for up to 2 days at a time
...
4 million to commercialize the process
...
18 Identify the four engineering economy symbols
and their values from the following problem statement
...

Vision Technologies, Inc
...

The company expects to spend $100,000 per year
for labor and $125,000 per year for supplies before
a product can be marketed
...
21 Many credit unions use semiannual interest periods
to pay interest on customer savings accounts
...

50
70
—
120
20
—
150
90
—
—
40
110

1
...

Month

Receipts,
$1000

Disbursements,
$1000

Jan
Feb
Mar
Apr
May
June
July
Aug
Sept
Oct
Nov
Dec

500
800
200
120
600
900
800
700
900
500
400
1800

300
500
400
400
500
600
300
300
500
400
400
700

1
...

1
...

1
...
20 Identify the following as cash inﬂows or outﬂows
to commercial air carriers: fuel cost, pension plan
contributions, fares, maintenance, freight revenue, cargo revenue, extra-bag charges, water and
sodas, advertising, landing fees, seat preference
fees
...

Atlas Long-Haul Transportation is considering installing Valutemp temperature loggers in all
of its refrigerated trucks for monitoring temperatures during transit
...
25 Construct a cash ﬂow diagram that represents the
amount of money that will be accumulated in
15 years from an investment of $40,000 now at an
interest rate of 8% per year
...
26 At an interest rate of 15% per year, an investment
of $100,000 one year ago is equivalent to how
much now?

34

Chapter 1

1
...
A company
that makes Ethernet adapters is planning to expand
its production facility at a cost of $1,000,000 one
year from now
...
If the interest rate is 15% per year,
how much of a discount is the company getting?
1
...
One year ago, each
share of stock was worth $40
...
At
what interest rate would the ﬁrm’s offer be equivalent to the worth of the stock last year?
1
...
The company said that although it could not
give bonuses this year, it would give each engineer two bonuses next year, the regular one of
$8000 plus an amount equivalent to the $8000
that each engineer should have gotten this year
...
30 University tuition and fees can be paid by using
one of two plans
...

On-time: Pay total amount due when classes start
...

(a) How much is paid in the early-bird plan?
(b) What is the equivalent amount of the savings
compared to the on-time payment at the time
that the on-time payment is made?
Simple and Compound Interest
1
...
32 Iselt Welding has extra funds to invest for future
capital expansion
...
33 To ﬁnance a new product line, a company that
makes high-temperature ball bearings borrowed
$1
...
If the com-

Foundations of Engineering Economy

pany repaid the loan in a lump sum amount after
2 years, what was (a) the amount of the payment
and (b) the amount of interest?
1
...
e
...
If the interest rate on the bonds
was 9% per year, how much does the company
have to pay the bond holders? The face value
(principal) of the bonds is $6,000,000
...
35 A solid waste disposal company borrowed money
at 10% per year interest to purchase new haulers
and other equipment needed at the companyowned landﬁll site
...
36 If interest is compounded at 20% per year, how
long will it take for $50,000 to accumulate to
$86,400?
1
...
If a
person deposits $10,000 at 10% per year simple
interest, what compound interest rate would yield
the same amount of money in 3 years?
MARR and Opportunity Cost
1
...

1
...

1
...

1
...
A staff member for the chief ﬁnancial ofﬁcer
used key words to identify the projects and then
listed them in order of projected rate of return as
shown below
...
4
19
13
...
6
8
...
42 State the purpose for each of the following built-in
spreadsheet functions
...
43 What are the values of the engineering economy
symbols P, F, A, i, and n in the following functions?
Use a question mark for the symbol that is to be

35

determined
...
44 Write the engineering economy symbol that corresponds to each of the following spreadsheet
functions
...
45 In a built-in spreadsheet function, if a certain parameter is not present, (a) under what circumstances can it be left blank and (b) when must a
comma be entered in its place?
1
...
Sheryl receives simple interest and
Marcelly gets compound interest
...
Assume no withdrawals or further deposits are made during the 4 years
...
47 The concept that different sums of money at different points in time can be said to be equal to each
other is known as:
(a) Evaluation criterion
(b) Equivalence
(c) Cash ﬂow
(d ) Intangible factors
1
...
49 All of the following are examples of cash outﬂows,
except:
(a) Asset salvage value
(b) Income taxes
(c) Operating cost of asset
(d ) First cost of asset
1
...
51 The following annual maintenance and operation
(M&O) costs for a piece of equipment were collected over a 5-year period: $12,300, $8900,
$9200, $11,000, and $12,100
...
In conducting a sensitivity analysis, the
most reasonable range of costs to use (i
...
, percent
from the average) is:
(a) Ϯ5% (b) Ϯ11% (c) Ϯ17% (d) Ϯ25%
1
...
53 Assume that you and your best friend each have
$1000 to invest
...
Your
friend invests her money at a bank that pays 10%
per year simple interest
...
54 The time it would take for a given sum of money to
double at 4% per year simple interest is closest to:
(a) 30 years (b) 25 years
(c) 20 years (d) 10 years

36

Chapter 1

Foundations of Engineering Economy

1
...
56 To ﬁnance a new project costing $30 million, a
company borrowed $21 million at 16% per year
interest and used retained earnings valued at 12%
per year for the remainder of the investment
...
5% (b) 13
...
8% (d) 15
...
PEC has a capacity of approximately 1300 MW (megawatts) of power, of
which 277 MW, or about 21%, is from renewable sources
...
A constant
question is how much of PEC’s generation capacity should be
from renewable sources, especially given the environmental
issues with coal-generated electricity and the rising costs of
hydrocarbon fuels
...

Consider yourself a member of the board of directors of
PEC
...

As such, you do not represent a speciﬁc district within the
entire service area; all other directors do represent a speciﬁc
district
...

Information
Here are some data that you have obtained
...
Electricity generation cost estimates are national
in scope, not PEC-speciﬁc, and are provided in cents per
kilowatt-hour (¢/kWh)
...
5
4
...
1
4
...
5

Reasonable Average
7
...
6
8
...
8

National average cost of electricity to residential customers: 11¢/kWh

PEC average cost to residential customers: 10
...
92 ¢/kWh (renewable sources)
Expected life of a generation facility: 20 to 40 years (it is
likely closer to 20 than 40)
Time to construct a facility: 2 to 5 years
Capital cost to build a generation facility: $900 to $1500
per kW
You have also learned that the PEC staff uses the wellrecognized levelized energy cost (LEC) method to determine
the price of electricity that must be charged to customers to
break even
...
The LEC formula, expressed
in dollars per kWh for (t ϭ 1, 2,
...
If you wanted to know more about the new arrangement with the wind farm in south Texas for the additional 60 MW per year, what types of questions would
you ask of a staff member in your ﬁrst meeting with
him or her?
2
...

What about the ethical aspects of the government’s allowance for these plants to continue polluting the atmosphere
with the emissions that may cause health problems for
citizens and further the effects of global warming? What
types of regulations, if any, should be developed for PEC
(and other generators) to follow in the future?

Case Study
3
...
27¢/kWh for
this year
...
You did learn the
following:
This is year t ϭ 11 for LEC computation purposes
n ϭ 25 years
i ϭ 5% per year

E11 ϭ 5
...
22 ¢/kWh (last year’s breakeven
cost to customers)
From these sketchy data, can you determine the value of unknowns in the LEC relation for this year? Is it possible to
determine if the wind farm addition of 60 MW makes any
difference in the electricity rate charged to customers? If not,
what additional information is necessary to determine the
LEC with the wind source included?

CASE STUDY
REFRIGERATOR SHELLS
Background
Large refrigerator manufacturers such as Whirlpool, General
Electric, Frigidaire, and others may subcontract the molding of
their plastic liners and door panels
...
Because of improvements in mechanical properties, the molded plastic can sustain increased vertical and horizontal loading, thus signiﬁcantly reducing the need
for attached metal anchors for some shelving
...

The company president wants a recommendation on whether
Innovations should offer the new technology to the major manufacturers and an estimate of the necessary capital investment
to enter this market
...
At this stage,
you are not expected to perform a complete engineering economic analysis, for not enough information is available
...

Information
Some information useful at this time is as follows:
• The technology and equipment are expected to last about
10 years before new methods are developed
...

• The expected returns on capital investment used for the
last three new technology projects were compound rates of
15%, 5%, and 18%
...

• Equity capital ﬁnancing beyond $5 million is not possible
...

• Annual operating costs have been averaging 8% of ﬁrst
cost for major equipment
...
2 million
...
You label these options as alternatives A and B
...
Use the ﬁrst four steps of the decision-making process
to generally describe the alternatives and identify what
economic-related estimates you will need to complete
an engineering economy analysis for the president
...
Identify any noneconomic factors and criteria to be considered in making the alternative selection
...
During your inquiries about alternative B from its manufacturer, you learn that this company has already produced
a prototype molding machine and has sold it to a company
in Germany for $3 million (U
...
dollars)
...
The company is willing to sell time on the
equipment to Innovations immediately to produce its own
shells for U
...
delivery
...
Consider this as alternative C, and develop the estimates necessary to evaluate C
at the same time as alternatives A and B
...

SECTION

TOPIC

LEARNING OUTCOME

2
...

2
...

2
...

2
...

2
...

2
...

2
...

T

he cash ﬂow is fundamental to every economic study
...

This chapter develops derivations for all the commonly used engineering economy factors
that take the time value of money into account
...
Spreadsheet functions are used in order to rapidly work with cash ﬂow series
and to perform sensitivity analysis
...

PE
The Cement Factory Case: Votorantim
Cimentos North America, Inc
...
The plant will
be called Houston American Cement, or
HAC
...

The plant investment, expected
to amount to $200 million, has been
planned for 2012; however, it is currently
delayed due to the economic downturn
in construction
...
All
analysis will use a planning horizon of
5 years commencing when the plant
begins operation
...
1)
Uniform series factors (2
...
3)
Arithmetic gradient factors (2
...
6)
Determining unknown n values (2
...
1 Single-Amount Factors (F͞P and P͞F )
The most fundamental factor in engineering economy is the one that determines the amount
of money F accumulated after n years (or periods) from a single present worth P, with interest
compounded one time per year (or period)
...
Therefore, if an amount P is invested at time t ϭ 0, the amount F1 accumulated
1 year hence at an interest rate of i percent per year will be
F1 ϭ P ϩ Pi
ϭ P(1 ϩ i)
where the interest rate is expressed in decimal form
...

F2 ϭ F1 ϩ F1i
ϭ P(1 ϩ i) ϩ P(1 ϩ i)i

[2
...
1], will be
F3 ϭ F2 ϩ F2i

40

Chapter 2

Factors: How Time and Interest Affect Money
F = given

F=?

i = given

0

1

2

n–2

i = given

n–1

n

0

P = given

1

2

n–2

n–1

n

P=?
(a)

(b)

Figure 2–1
Cash ﬂow diagrams for single-payment factors: (a) ﬁnd F, given P, and (b) ﬁnd P, given F
...
To ﬁnd F, given P,
F ‫ ؍‬P(1 ؉ i)n

[2
...
This is the conversion factor that, when multiplied by P, yields
the future amount F of an initial amount P after n years at interest rate i
...

Reverse the situation to determine the P value for a stated amount F that occurs n periods
in the future
...
2] for P
...
3]

The expression (1 ؉ i)Ϫn is known as the single-payment present worth factor (SPPWF), or the
P͞F factor
...
The cash ﬂow diagram is shown in Figure 2–1b
...

A standard notation has been adopted for all factors
...
It is always in the general form (X͞Y,i,n)
...
For example, F͞P
means ﬁnd F when given P
...

Using this notation, (F͞P,6%,20) represents the factor that is used to calculate the future
amount F accumulated in 20 periods if the interest rate is 6% per period
...
The
standard notation, simpler to use than formulas and factor names, will be used hereafter
...
This
information is also included inside the front cover
...
1

Single-Amount Factors (F͞P and P͞F )

To simplify routine engineering economy calculations, tables of factor values have been prepared for interest rates from 0
...
These tables, found at the rear of the book, have a colored edge for easy identiﬁcation
...

The word discrete in the title of each table emphasizes that these tables utilize the end-of-period
convention and that interest is compounded once each interest period
...
For
example, the value of the factor (P͞F,5%,10) is found in the P͞F column of Table 10 at period 10
as 0
...
This value is determined by using Equation [2
...

1
(P͞F,5%,10) ϭ ————
(1 ϩ i)n
1
ϭ ————
(1
...
6139
1
...
4]

A present amount P is determined using the PV function with the format
‫ ؍‬PV(i%,n,,F)

[2
...
Refer to Appendix A or Excel online help for more
information on the use of FV and PV functions
...
1
Sandy, a manufacturing engineer, just received a year-end bonus of $10,000 that will be invested
immediately
...
Find the amount of funds that will be available in 20 years by using
(a) hand solution by applying the factor formula and tabulated value and (b) a spreadsheet function
...
The symbols and values are
P ϭ $10,000

Fϭ?

i ϭ 8% per year

n ϭ 20 years

(a) Factor formula: Apply Equation [2
...
Rounding to four decimals, we have
F ϭ P(1 ϩ i)n ϭ 10,000(1
...
6610)
ϭ $46,610
Standard notation and tabulated value: Notation for the F͞P factor is (F͞P,i%,n)
...
6610)
ϭ $46,610
Table 13 provides the tabulated value
...

(b) Spreadsheet: Use the FV function to ﬁnd the amount 20 years in the future
...
4]; the numerical entry is ϭ FV(8%,20,,10000)
...
57) displayed
...
) The FV function
has performed the computation in part (a) and displayed the result
...

41

42

Chapter 2

Factors: How Time and Interest Affect Money

PE

EXAMPLE 2
...
Delays beyond the anticipated implementation year of 2012 will require additional money to construct the factory
...

(a) The equivalent investment needed if the plant is built in 2015
...

Solution
Figure 2–2 is a cash ﬂow diagram showing the expected investment of $200 million ($200 M)
in 2012, which we will identify as time t ϭ 0
...

Figure 2–2

F3 = ?

Cash ﬂow diagram for
Example 2
...

P−4 = ?

−4 −3
2008

2009

−2

−1

0

1

2

2010

2011

2012

2013

2014

3
2015

t
Year

$200 M

(a) To ﬁnd the equivalent investment required in 3 years, apply the F͞P factor
...

F3 ϭ P(F͞P,i,n) ϭ 200(F͞P,10%,3) ϭ 200(1
...
2 ($266,200,000)
Now, use the FV function on a spreadsheet to ﬁnd the same answer, F3 ϭ $266
...

(Refer to Figure 2–3, left side
...
2
...
To determine the
equivalent cost 4 years earlier, consider the $200 M in 2012 (t ϭ 0) as the future value F
and apply the P͞F factor for n ϭ 4 to ﬁnd PϪ4
...
) Table 15 supplies
the tabulated value
...
6830)
ϭ $136
...

This equivalence analysis indicates that at $136
...

2
...
2 Uniform Series Present Worth Factor and
Capital Recovery Factor (P͞A and A͞P)
The equivalent present worth P of a uniform series A of end-of-period cash ﬂows (investments)
is shown in Figure 2–4a
...
3],
and summing the results
...

(1 ϩ i)1
(1 ϩ i)2
(1 ϩ i)3
1
1
ϩ A ———— ϩ A ————
(1 ϩ i)n
(1 ϩ i)nϪ1

[

The terms in brackets are the P͞F factors for years 1 through n, respectively
...

[

1
1
1
1
1
P ϭ A ———— ϩ ———— ϩ ———— ϩ
...
6]

To simplify Equation [2
...
This results in Equation [2
...
Now
subtract the two equations, [2
...
7], and simplify to obtain the expression for P when
i 0 (Equation [2
...

[ (1 ϩ i)

1
1
1
1
1
P
——— ϭ A ———— ϩ ———— ϩ ———— ϩ
...
ϩ (1 ϩ1i)
Ϫi P ϭ A
1
[ (1 ϩ1i) Ϫ (1 ϩ i) ]
1ϩi

(1 ϩ i)

1
1
1
1
1
——— P ϭ A ———— ϩ ———— ϩ
...
7]

]

————
1

[

1
A
P ϭ —— ———— Ϫ 1
Ϫi (1 ϩ i)n

]

[

(1 ؉ i)n ؊ 1
P ‫ ؍‬A ——————
i(1 ؉ i)n

]

i

0

[2
...
8] is the conversion factor referred to as the uniform series
present worth factor (USPWF)
...
The cash ﬂow diagram is Figure 2–4a
...

2

n–2

A=?
(b)

n–1

n

44

Chapter 2

Factors: How Time and Interest Affect Money

TABLE 2–2
Notation
(P͞A,i,n)
(A͞P,i,n)

P͞A and A͞P Factors: Notation and Equations

Factor
Name

Factor
Formula

Uniform series
present worth
Capital recovery

P͞A

Standard
Notation Equation

Excel
Function

(1 ϩ i)n Ϫ 1
i(1 ϩ i)
i(1 ϩ i)n
—————
(1 ϩ i)n − 1

Find/Given

P ϭ A(P͞A,i,n)

ϭ PV(i%,n,A)

A ϭ P(A͞P,i,n)

ϭ PMT(i%,n,P)

—————
n

A͞P

To reverse the situation, the present worth P is known and the equivalent uniform series
amount A is sought (Figure 2–4b)
...
Solve Equation [2
...
9]

The term in brackets is called the capital recovery factor (CRF), or A͞P factor
...

Placement of P

The P͞A and A͞P factors are derived with the present worth P and the ﬁrst uniform annual
amount A one year (period) apart
...

The factors and their use to ﬁnd P and A are summarized in Table 2–2 and inside the front cover
...
Tables at the end of
the text include the factor values
...
4641
...

Spreadsheet functions can determine both P and A values in lieu of applying the P͞A and A͞P
factors
...
The format, is
‫ ؍‬PV(i%,n,A,F)

[2
...
The format is
‫ ؍‬PMT(i%, n,P,F)

[2
...

EXAMPLE 2
...
The
present worth is
P ϭ 600(P͞A,16%,9) ϭ 600(4
...
90
The PV function ϭ PV(16%,9,600) entered into a single spreadsheet cell will display the
answer P ϭ ($2763
...

2
...
4 The Cement Factory Case
As mentioned in the chapter introduction of this case, the Houston American Cement plant
may generate a revenue base of $50 million per year
...
With money worth
10% per year, address the following question from the president: Will the initial investment
be recovered over the 5-year horizon with the time value of money considered? If so, by how
much extra in present worth funds? If not, what is the equivalent annual revenue base required
for the recovery plus the 10% return on money? Use both tabulated factor values and spreadsheet functions
...
The cash ﬂow diagram is similar to Figure 2–4a, where the
ﬁrst A value occurs 1 year after P
...
7908)
ϭ $189
...

To determine the minimum required to realize a 10% per year return, use the A͞P factor
...

A ϭ 200(A͞P,10%,5) ϭ 200(0
...
76 per year
The plant needs to generate $52,760,000 per year to realize a 10% per year return over
5 years
...
Figure 2–5 shows
the use of ϭ PV(i%,n,A,F) on the left side to ﬁnd the present worth and the use of
ϭ PMT(i%,n,P,F) on the right side to determine the minimum A of $52,760,000 per year
...
The minus sign placed before
each function name forces the answer to be positive, since these two functions always display
the answer with the opposite sign entered on the estimated cash ﬂows
...
4
...
3 Sinking Fund Factor and Uniform Series Compound
Amount Factor (A͞F and F͞A)
The simplest way to derive the A͞F factor is to substitute into factors already developed
...
3] is substituted into Equation [2
...

][

i(1 ϩ i)n
1
A ϭ F ———— —————
n
(1 ϩ i) (1 ϩ i)n − 1

[

[

i
A ‫ ؍‬F —————
(1 ؉ i)n ؊ 1

]

]

[2
...
12] is the A͞F or sinking fund factor
...
This is shown graphically in Figure 2–6a, where A is a uniform annual investment
...
The last A value and F occur at the same time
...
12] can be rearranged to ﬁnd F for a stated A series in periods 1 through n (Figure 2–6b)
...
13]

The term in brackets is called the uniform series compound amount factor (USCAF), or F͞A factor
...
It is important to remember that the future amount F occurs in the same period as the last A
...
They are (F͞A,i,n) and
(A͞F,i,n)
...

As a matter of interest, the uniform series factors can be symbolically determined by using an
abbreviated factor form
...

Using the factor formulas, we have
(1 ϩ i)n Ϫ 1
(1 ϩ i)n Ϫ 1
(F͞A,i,n) ϭ [(1 ϩ i)n] —————— ϭ ——————
n
i
i(1 ϩ i)

[

]

For solution by spreadsheet, the FV function calculates F for a stated A series over n years
...
14]

The P may be omitted when no separate present worth value is given
...
The
format is
‫ ؍‬PMT(i%,n,P,F)

[2
...

F = given

F=?
i = given

i = given
0

1

2

n–2

n–1

n

A=?
(a)

Figure 2–6
Cash ﬂow diagrams to (a) ﬁnd A, given F, and (b) ﬁnd F, given A
...
3

Sinking Fund Factor and Uniform Series Compound Amount Factor (A͞F and F͞A)

TABLE 2–3
Notation
(F͞A,i,n)
(A͞F,i,n)

47

F͞A and A͞F Factors: Notation and Equations

Factor
Name

Find/Given

Uniform series
compound amount
Sinking fund

Factor
Formula

Standard Notation
Equation

(1 ϩ i)n Ϫ 1 F ϭ A(F͞A,i,n)
i
i
————— A ϭ F(A͞F,i,n)
(1 ϩ i)n − 1

F͞A

—————

A͞F

Excel
Functions
ϭ FV(i%,n,A)
ϭ PMT(i%,n,F)

EXAMPLE 2
...
Ford capital earns at a
rate of 14% per year
...
In $1000 units, the F value in year 8 is found
by using the F͞A factor
...
2328) ϭ $13,232
...
5
...
6 The Cement Factory Case
Once again, consider the HAC case presented at the outset of this chapter, in which a projected
$200 million investment can generate $50 million per year in revenue for 5 years starting
1 year after start-up
...
Now the president would like the answers to a couple of new questions
about the estimated annual revenues
...

(a) What is the equivalent future worth of the estimated revenues after 5 years at 10% per year?
(b) Assume that, due to the economic downturn, the president predicts that the corporation
will earn only 4
...

What is the required amount of the annual revenue series over the 5-year period to be economically equivalent to the amount calculated in (a)?

Solution

(a) Figure 2–6b is the cash ﬂow diagram with A ϭ $50 million
...
We use tabulated values and the spreadsheet
function to ﬁnd F in year 5
...
In $1 million units, the
future worth of the revenue series is
F ϭ 50(F͞A,10%,5) ϭ 50(6
...
255 ($305,255,000)

PE

48

Chapter 2

Factors: How Time and Interest Affect Money

ϭ ϪPMT(4
...
5% for the cement factory case,
Example 2
...

If the rate of return on the annual revenues were 0%, the total amount after 5 years would
be $250,000,000
...

Spreadsheet: Apply the FV factor in the format ϭ ϪFV(10%,5,50) to determine F ϭ
$305
...
Because there is no present amount in this computation, P is omitted
from the factor
...
(As before, the minus sign forces the FV function
to result in a positive value
...
He wants the revenue stream to generate the equivalent that it would at a 10% per year return, that is, $305
...
5% per year return is achievable
...
Since the factor
tables do not include 4
...
In $1 million units,
0
...
255(A͞F,4
...
255 —————— ϭ 305
...
18279)
(1
...
798

[

]

The annual revenue requirement grows from $50 million to nearly $55,800,000
...
6% each year
...
5% and F ϭ $305
...
We can use the cell reference method
(described in Appendix A) for the future amount F
...
798 per year (in $1 million units)
...
4 Factor Values for Untabulated i or n Values
Often it is necessary to know the correct numerical value of a factor with an i or n value that is
not listed in the compound interest tables in the rear of the book
...

• Use the formula listed in this chapter or the front cover of the book,
• Use an Excel function with the corresponding P, F, or A value set to 1
...

When the formula is applied, the factor value is accurate since the speciﬁc i and n values are
input
...
Additionally, the formulas become more complex
when gradients are introduced, as you will see in the following sections
...
For example, the P͞F factor is determined using the PV function with A omitted (or set to 0) and
F ϭ 1, that is, PV(i%,n,,1) or PV(i%,n,0,1)
...
Functions to determine the six common factors are
as follows
...
4

Factor Values for Untabulated i or n Values
Factor

To Do This
Find P, given F
...

Find P, given A
...

Find F, given A
...

P͞F
F͞P
P͞A
A͞P
F͞A
A͞F

Excel Function
ϭ ϪPV(i%,n,,1)
ϭ ϪFV(i%,n,,1)
ϭ ϪPV(i%,n,1)
ϭ ϪPMT(i%,n,1)
ϭ ϪFV(i%,n,1)
ϭ ϪPMT(i%,n,,1)

Figure 2–9 shows a spreadsheet developed explicitly to determine these factor values
...
The
values for i ϭ 3
...
As we already know, these same functions will
determine a ﬁnal P, A, or F value when actual or estimated cash ﬂow amounts are entered
...
Also interpolation introduces some
level of inaccuracy, depending upon the distance between the two boundary values selected for
i or n, as the formulas themselves are nonlinear functions
...
Refer to Figure 2–10 for a graphical description of the following explanation
...
Second, ﬁnd the corresponding tabulated factor values
(f1 and f2)
...

Figure 2–9
Enter requested i and n

Figure 2–10

Factor value
axis
f2

c

f

Linear interpolation in factor
value tables
...

Table

a

Known
x1

Required
x
b

Known i or n
x2
axis

49

50

Chapter 2

Factors: How Time and Interest Affect Money

(x – x1)
f ϭ f1 ϩ ———— (f2 – f1)
(x2 – x1)
a
f ϭ f1 ϩ — c ϭ f1 ϩ d
b

[2
...
17]

The value of d will be positive or negative if the factor is increasing or decreasing, respectively,
in value between x1 and x2
...
7
Determine the P͞A factor value for i ϭ 7
...

Solution
Factor formula: Apply the formula from inside the front cover of the book for the P͞A factor
...
0775)10 Ϫ 1
1
...
75%,10) ϭ ————— ϭ ——————— ϭ ————
n
i(1 ϩ i)
0
...
0775)10 0
...
78641
Spreadsheet: Utilize the spreadsheet function in Figure 2–9, that is, ϭ ϪPV(7
...
78641
...
Apply the Equation [2
...
17] sequence, where x is the interest rate i, the bounding interest rates are
i1 ϭ 7% and i2 ϭ 8%, and the corresponding P͞A factor values are f1 ϭ (P͞A,7%,10) ϭ 7
...
7101
...
75 Ϫ 7)
f ϭ f1 ϩ ——— (f2 – f1) ϭ 7
...
7101 Ϫ 7
...
0236 ϩ (0
...
3135) ϭ 7
...
2351
ϭ 6
...
2351
...
75% and 10 years, plus it takes more calculations than using the formula or spreadsheet function
...

2
...
The cash ﬂow series of maintenance costs involves a
constant gradient, which is $5000 per year
...
The amount of change is called the gradient
...
In the
case of a gradient, each year-end cash ﬂow is different, so new formulas must be derived
...
This is convenient because in actual applications, the
base amount is usually signiﬁcantly different in size compared to the gradient
...
Assume these cost $2500; that is, $2500 is the base
amount
...
5

51

Arithmetic Gradient Factors (P͞G and A͞G)
0

1

2

3

4

n

n–1

Time

Figure 2–11
Cash ﬂow diagram of an
arithmetic gradient series
...
If you estimate that total costs will increase by $200 each year, the amount the second
year is $2700, the third $2900, and so on to year n, when the total cost is 2500 ϩ (n Ϫ 1)200
...
Note that the gradient ($200) is ﬁrst observed between year 1 and year 2, and the base amount ($2500 in year 1) is not equal to the gradient
...

G ϭ constant arithmetic change in cash ﬂows from one time period to the next; G may be positive
or negative
...
18]

It is important to realize that the base amount deﬁnes a uniform cash ﬂow series of the size A that
occurs eash time period
...
If the base amount is ignored, a generalized arithmetic (increasing) gradient cash ﬂow diagram is as shown in Figure 2–12
...
This is called a conventional gradient
...
8
A local university has initiated a logo-licensing program with the clothier Holister, Inc
...
Determine the gradient and construct a cash ﬂow diagram that identiﬁes
the base amount and the gradient series
...
18], solved for G, determines the arithmetic gradient
...

52

Chapter 2

Factors: How Time and Interest Affect Money
CF9 =
$200,000
$185,000
$170,000
$155,000
$140,000
$125,000
$110,000
G = $15,000

CF1 = $95,000
$80,000

0

1

2

3

4

5

6

7

8

9

Year

Figure 2–13
Diagram for gradient series, Example 2
...

The cash ﬂow diagram (Figure 2–13) shows the base amount of $80,000 in years 1 through 9
and the $15,000 gradient starting in year 2 and continuing through year 9
...
The addition of the two results in PT
...
19]

where PA is the present worth of the uniform series only, PG is the present worth of the gradient
series only, and the ϩ or Ϫ sign is used for an increasing (ϩG) or decreasing (ϪG) gradient,
respectively
...
20]

Three factors are derived for arithmetic gradients: the P͞G factor for present worth, the A͞G
factor for annual series, and the F͞G factor for future worth
...
We use the single-payment present worth factor (P͞F,i,n), but the same result can be obtained by using the F͞P, F͞A, or P͞A factor
...

P ϭ G(P͞F,i,2) ϩ 2G(P͞F,i,3) ϩ 3G(P͞F,i,4) ϩ
...

[

3
nϪ2
nϪ1
1
2
P ϭ G ———— ϩ ———— ϩ ———— ϩ
...
21]

Multiplying both sides of Equation [2
...
ϩ ———— ϩ ———— [2
...
21] from Equation [2
...

[

] [

n
1
1
1
1
iP ϭ G ———— ϩ ———— ϩ
...
23]

The left bracketed expression is the same as that contained in Equation [2
...
Substitute the closed-end form of the P͞A factor from Equation [2
...
5

53

Arithmetic Gradient Factors (P͞G and A͞G)
PG = ?
i = given
0

1

2

3

4

n

n–1

0

1

2

3

4

n–1

n

G
2G

3G
(n – 2)G
(n – 1)G
(b)

(a)

Figure 2–14
Conversion diagram from an arithmetic gradient to a present worth
...
23] and simplify to solve for P G , the present worth of the gradient series
only
...
24]

Equation [2
...
Figure 2–14a is converted into the
equivalent cash ﬂow in Figure 2–14b
...
25]

Remember: The conventional arithmetic gradient starts in year 2, and P is located in year 0
...
24] expressed as an engineering economy relation is
PG ϭ G(P͞G,i,n)

[2
...
19] to calculate total present worth
...

The equivalent uniform annual series AG for an arithmetic gradient G is found by multiplying
the present worth in Equation [2
...
In standard notation form, the
equivalent of algebraic cancellation of P can be used
...
27]

which is the rightmost term in Equation [2
...
The expression in brackets in Equation [2
...
This factor
converts Figure 2–15a into Figure 2–15b
...
Factor values
are tabulated in the two rightmost columns of factor values at the rear of this text
...

There is no direct, single-cell spreadsheet function to calculate PG or AG for an arithmetic
gradient
...
General formats for these functions are
‫ ؍‬NPV(i%, second_cell:last_cell) ؉ ﬁrst_cell
‫ ؍‬PMT(i%, n, cell_with_PG)

[2
...
29]

The word entries in italic are cell references, not the actual numerical values
...
2, for a description of cell reference formatting
...
10
...
The resulting factor,
(F͞G,i,n), in brackets, and engineering economy relation is

[( ) (

) ]

n
1 (1 ϩ i) – 1
FG ϭ G — ————— Ϫ n
i
i

EXAMPLE 2
...
At a recent meeting, the engineers estimated that a total of
$500,000 will be deposited at the end of next year into an account for the repair of old and
safety-questionable bridges throughout the area
...
Determine the
equivalent (a) present worth and (b) annual series amounts, if public funds earn at a rate
of 5% per year
...
According to Equation [2
...
The total present worth PT
occurs in year 0
...

In $1000 units, the total present worth is
PT ϭ 500(P͞A,5%,10) ϩ 100(P͞G,5%,10)
ϭ 500(7
...
6520)
ϭ $7026
...
5

Arithmetic Gradient Factors (P͞G and A͞G)
0

1

2

$500

3

$600

4

$700

5

$800

6

$900

7

$1000

8

$1100

9

$1200

10

$1300

$1400

Figure 2–16
Cash ﬂow series with a conventional arithmetic gradient (in $1000 units),
Example 2
...

PG = ?

PA = ?
A = $500
1 2
9 10

G = $100
1 2
9 10

+
$100
Base

Gradient $900

PT = ?

PT = PA + PG

1

2

$500

$600

3

$700

4

$800

5

$900

6

$1000

7

$1100

8

$1200

9

$1300

10

$1400

Figure 2–17
Partitioned cash ﬂow diagram (in $1000 units), Example 2
...

(b) Here, too, it is necessary to consider the gradient and the base amount separately
...
20] and occurs in years 1 through 10
...
0991)
ϭ $909
...
Any other cash ﬂows must be considered separately
...
In this case, considering round-off error,
AT ϭ PT (A͞P,5%,10) ϭ 7026
...
12950)
ϭ $909
...
10 The Cement Factory Case

The announcement of the HAC cement factory states that the $200 million (M) investment is
planned for 2012
...
Further investigation may determine, for example, that the $200 M is a present worth in the year 2012 of anticipated investments during the next 4 years (2013 through 2016)
...
As before, assume the time
value of money for investment capital is 10% per year to answer the following questions using
tabulated factors and spreadsheet functions, as requested below
...

(b) Given the planned investment series, what is the equivalent annual amount that will be
invested from 2013 to 2016? Use both tabulated factors and spreadsheet functions
...
) What must be the amount of
yearly constant decrease through 2016 to have a present worth of exactly $200 M in
2012, provided $100 M is expended in 2013? Use a spreadsheet
...
Figure 2–18 diagrams the cash
ﬂows with the shaded area showing the constantly declining investment each year
...

Tabulated factors: Equation [2
...
Money is expressed in $1 million units
...
30]

ϭ 100(3
...
3781)
ϭ $207
...
5 M
...
Figure 2–19 shows the entries and function
NPV(i%,second_cell:last_cell)
...
The result displayed in cell C9, $207
...
(Note that the NPV function does not consider two separate series of
cash ﬂows as is necessary when using tabulated factors
...
5 M
...
First, apply Equation [2
...
Both relations are illustrated here, in $1 million units,
PT ϭ ?

Figure 2–18
Cash ﬂow diagram for decreasing gradient in $1 million units, Example 2
...

i = 10% per year
2013

0

2014

2015

2016

Year

1

2

3

4

Time

Base
A ϭ $100
$25

$50

$100

Gradient
G ϭ $Ϫ25

2
...
10a and b
...
20]:
AT ϭ 100 – 25(A͞G,10%,4) ϭ 100 Ϫ 25(1
...
471

($65,471,000 per year)

Use PT :
AT ϭ 207
...
537(0
...
471 per year
Spreadsheet: Apply the PMT function in Equation [2
...
471 per year (Figure 2–19)
...
It is an excellent tool to apply
when one cell entry must equal a speciﬁc value and only one other cell can change
...
This is the same as stating PT ϭ 200 in Equation [2
...
All other parameters retain their current value
...
When OK is clicked, the solution is displayed; G ϭ $Ϫ26
...

Refer to Figure 2–20 again
...
721 M, the equivalent total present worth invested over the 4 years
will be exactly $200 M
...
10c
...
721 to make
present worth exactly $200

58

Chapter 2

Factors: How Time and Interest Affect Money

2
...
This change occurs
every year on top of a starting amount in the ﬁrst year of the project
...

A geometric gradient series is a cash ﬂow series that either increases or decreases by a constant
percentage each period
...

g ϭ constant rate of change, in decimal form, by which cash ﬂow values increase or decrease
from one period to the next
...

A1 ϭ initial cash ﬂow in year 1 of the geometric series
P g ϭ present worth of the entire geometric gradient series, including the initial amount
A1
Note that the initial cash ﬂow A1 is not considered separately when working with geometric
gradients
...
The relation to determine the total present
worth Pg for the entire cash ﬂow series may be derived by multiplying each cash ﬂow in Figure 2–21a by the P͞F factor 1͞(1 ϩ i)n
...
ϩ ——————
(1 ϩ i)n
(1 ϩ i)1
(1 ϩ i)2
(1 ϩ i)3

[

(1 ϩ g)2
(1 ϩ g)nϪ1
1ϩ g
1
ϭ A1 ——— ϩ ———— ϩ ———— ϩ
...
31]

Multiply both sides by (1 ϩ g)͞(1 ϩ i), subtract Equation [2
...

[

(

)

1ϩg n
1 Ϫ ———
1ϩi
Pg ϭ A1 ———————
iϪg

]

g

i

[2
...
32] is the (P͞A,g,i,n) or geometric gradient series present
worth factor for values of g not equal to the interest rate i
...
31] and observe that the term 1/(1 + i) appears n times
...

n

A1(1 – g)n – 1

2
...
ϩ ———
(1 ϩ i) (1 ϩ i) (1 ϩ i)
(1 ϩ i)
nA1
Pg ϭ ———
(1 ϩ i)

)
[2
...

The equation for Pg and the (P͞A,g,i,n) factor formula are
Pg ϭ A1(P͞A,g,i,n)

(

)

1؉g n
1 ؊ ———
1؉i
(P͞A,g,i,n) ‫—————— ؍‬
i−g
n
———
1؉i

[2
...
35]

g‫؍‬i

It is possible to derive factors for the equivalent A and F values; however, it is easier to determine
the Pg amount and then multiply by the A͞P or F͞P factor
...
Once the cash ﬂows are entered, P and A are determined using the NPV and PMT
functions, respectively
...
11
A coal-ﬁred power plant has upgraded an emission control valve
...
The maintenance cost is expected to be high at $1700 the ﬁrst year, increasing by 11% per year thereafter
...

Solution by Hand
The cash ﬂow diagram (Figure 2–22) shows the salvage value as a positive cash ﬂow and all
costs as negative
...
35] for g i to calculate Pg
...

PT ϭ Ϫ8000 Ϫ Pg ϩ 200(P͞F,8%,6)
1 Ϫ (1
...
08)6
ϭ Ϫ8000 Ϫ 1700 ——————— ϩ 200(P͞F,8%,6)
0
...
11
ϭ Ϫ8000 Ϫ 1700(5
...
11
...
11)
$1700(1
...
11)3
$1700(1
...
11)5

Placement of
Gradient Pg

60

Chapter 2

Factors: How Time and Interest Affect Money

Solution by Spreadsheet
Figure 2–23 details the spreadsheet operations to ﬁnd the geometric gradient present worth Pg
and total present worth PT
...
Cell tags detail the relations for the
second and third components; the ﬁrst cost occurs at time 0
...
If this factor is used repeatedly, it is worthwhile using cell reference
formatting so that A1, i, g, and n values can be changed and the correct value is always obtained
...

Present worth of salvage
ϭ ϪPV(8%,6,,200)

Present worth of maintenance costs, Eq
...
35]
ϭ Ϫ1700* ((1-((1
...
08))^6)/(0
...
11))

Figure 2–23
Geometric gradient and total present worth calculated via spreadsheet, Example 2
...

PE

EXAMPLE 2
...
The revenue
series estimate of $50 million annually is quite optimistic, especially since there are many
other cement product plants operating in Florida and Georgia on the same limestone deposit
...
) Therefore, it is important to be sensitive in our analysis to possibly declining and increasing revenue series, depending upon the longer-term success of the
plant’s marketing, quality, and reputation
...
Determine the present worth and future worth equivalents of all revenues during this 5-year time
frame at the same rate used previously, that is, 10% per year
...
In year 1, A1 ϭ $50 M and revenues decrease in year 5 to
A1(1 Ϫ g)nϪ1 ϭ 50 M(1 Ϫ 0
...
88)4 ϭ $29
...
[2
...
10 and g ϭ Ϫ0
...
In $1 million units,

[

(

)

]

0
...
10
Pg ϭ 50 ——————— ϭ 50[3
...
10 Ϫ (Ϫ0
...
80
F ϭ 152
...
80(1
...
08
This means that the decreasing revenue stream has a 5-year future equivalent worth of
$246
...
If you look back to Example 2
...
7

Determining i or n for Known Cash Flow Values

uniform revenue series of $50 M annually is $305
...
In conclusion, the 12% declining
geometric gradient has lowered the future worth of revenue by $59
...

2
...
An example for which i is sought
may be stated as follows: A company invested money to develop a new product
...
There are several ways to ﬁnd an unknown i or n value, depending upon the
nature of the cash ﬂow series and the method chosen to ﬁnd the unknown
...
The most
difﬁcult and complex involves ﬁnding i or n for irregular cash ﬂows mixed with uniform and
gradient series utilizing solution by hand and calculator
...

Single Amounts—P and F Only
Hand or Calculator Solution Set up the equivalence relation and (1) solve for the variable
using the factor formula, or (2) ﬁnd the factor value and interpolate in the tables
...
(See below and Appendix A for details
...

Spreadsheet Solution Use the IRR or RATE function to ﬁnd i or the NPER function to ﬁnd n
...

Spreadsheet Solution Use the IRR or RATE function to ﬁnd i or the NPER function to ﬁnd
n
...
)
Besides the PV, FV, and NPV functions, other spreadsheet functions useful in determining i
are IRR (internal rate of return) and RATE, and NPER (number of periods) to ﬁnd n
...
In all three
of these functions, at least one cash ﬂow entry must have a sign opposite that of others in order
to ﬁnd a solution
...
36]

To use IRR to ﬁnd i, enter all cash ﬂows into contiguous cells, including zero values
...
37]

The single-cell RATE function ﬁnds i when an A series and single P and/or F values are involved
...

[2
...
13
If Laurel made a $30,000 investment in a friend’s business and received $50,000 5 years later,
determine the rate of return
...

1
P ϭ F(P͞F,i,n) ϭ F ————
(1 ϩ i)n
1
30,000 ϭ 50,000 ————
(1 ϩ i)5
1
0
...
2
i ϭ —— Ϫ 1 ϭ 0
...
76%)
0
...

( )

P ϭ F(P͞F,i,n)
30,000 ϭ 50,000(P͞F,i,5)
(P͞F,i,5) ϭ 0
...
6000 for n ϭ 5 lies between 10% and 11%
...
76%
...
14
Pyramid Energy requires that for each of its offshore wind power generators $5000 per year
be placed into a capital reserve fund to cover unexpected major rework on ﬁeld equipment
...
What rate of return did this practice provide to the company? Solve by hand and
spreadsheet
...
Either the A͞F or F͞A factor can be used
...
0500
From the A͞F interest tables for 15 years, the value 0
...
By interpolation, i ϭ 3
...

F = $100,000 Figure 2–24
Diagram to determine the rate
of return, Example 2
...

i= ?
0

1

2

3

4

5

6

7

8

A = $5000

9

10

11

12

13

14 15

2
...

A single-cell solution using the RATE function can be applied since A ϭ $Ϫ5000 occurs
each year and F ϭ $100,000 takes place in the last year of the series
...
98%
...

The IRR function is much better for answering “what if ” questions
...
In any cell enter the IRR function
...
98% is displayed
...
The IRR function does not need these numbers, but it makes the cash ﬂow entry activity easier and more
accurate
...

Figure 2–25
Use of RATE and IRR
functions to determine
i value for a uniform
series, Example 2
...

i using RATE function
ϭ RATE(15,-5000,,100000)

i using IRR function
ϭ IRR(E2:E17))

EXAMPLE 2
...
All analysis thus far has taken place at 10% per year; however, the parent company has made it clear that its other international plants are able to show a 20% per year return
on the initial investment
...

Solution
If hand solution is utilized, the present worth relation can be established and the n values
interpolated in the tables for each of the three rate of return values
...
00
This is a good opportunity to utilize a spreadsheet and repeated NPER functions from
Equation [2
...
Figure 2–26 shows the single-cell
ϭ NPER(i%,50,Ϫ200) function for each rate of return
...
15
...
Capability in using these formulas and their
standard notation manually and with spreadsheets is critical to complete an engineering economy
study
...
Additionally, you can solve
for rate of return i or time n
...
1 Look up the numerical value for the following factors from the interest tables
...
(P͞F,6%,8)
2
...
(A͞G,15%,20)
4
...
(P͞G,35%,15)
Determination of F, P, and A
2
...
, afford
to spend now on an energy management system if
the software will save the company $21,300 per
year for the next 5 years? Use an interest rate of
10% per year
...
The
cost is $985,000, and a $100,000 deposit will hold
one of the ﬁrst 100 “cars
...
At an interest rate of 10% per
year, what is the effective total cost of the PAV in
year 3?
2
...
If the fund earned
interest at 6% per year, how much was in the account 14 years after it was started?

2
...
The distributor of the
inclinometers is temporarily overstocked and is offering them at a 40% discount from the regular cost
of $142
...
Assume the interest rate is
10% per year
...
6 One of the biggest vulnerabilities in a control system is network devices, such as Ethernet-based
network switches that are located in unsecured
locations and accessible to everyone
...
The company is considering expanding its manufacturing
lines now or doing it in 3 years
...
9 million, what equivalent amount
could the company afford to spend in 3 years? The
interest rate is 15% per year
...
4 The Moller Skycar M400 is a ﬂying car known
as a personal air vehicle (PAV) that is expected

2
...
If the new equipment will cost $220,000 to purchase and install,
how much must the company save each year for
3 years in order to justify the investment, if the
interest rate is 10% per year?
2
...
of Carnegie, Pennsylvania, makes a
control pinch valve that provides accurate, repeatable control of abrasive and corrosive slurries, outlasting gate, plug, ball, and even satellite coated
valves
...

2
...
3 million for its corporate headquarters, what
must the building be worth in 10 years? The company expects all expenditures to earn a rate of return of at least 18% per year
...
10 CGK Rheosystems makes high-performance rotational viscometers capable of steady shear and
yield stress testing in a rugged, compact footprint
...
11 Five years ago a consulting engineer purchased a
building for company ofﬁces constructed of bricks
that were not properly ﬁred
...
Because of the problem with the
bricks, the selling price of the building was 25%
below the price of comparable, structurally sound
buildings
...
This resulted in restoring the building to its fair market value
...
12 Metso Automation, which manufactures addressable quarter-turn electric actuators, is planning to
set aside $100,000 now and $150,000 one year
from now for possible replacement of the heating
and cooling systems in three of its larger manufacturing plants
...
13 Syringe pumps often fail because reagents adhere
to the ceramic piston and deteriorate the seal
...

One of Trident’s customers expects to reduce
downtime by 30% as a result of the new seal design
...
14 China spends an estimated $100,000 per year on
cloud seeding efforts, which includes using antiaircraft guns and rocket launchers to ﬁll the sky
with silver iodide
...
If the yields of cash crops
will increase by 4% each year for the next 3 years
because of extra irrigation water captured behind
dams during cloud seeding, what is the maximum
amount the farmers should spend now on the cloud
seeding activity? The value of the cash crops without the extra irrigation water would be $600,000
per year
...

2
...
07 million to improve
(i
...
, deepen) a retention basin and reconstruct the
spillway that was severely damaged in a ﬂood
2 years ago
...
If the projects are assumed to
have a 20-year life, what is the annual worth of the
savings at an interest rate of 6% per year?
2
...
5 miles per gallon for cars and light trucks by
the year 2016
...
If a person purchases a new car in
2012 and keeps it for 5 years, how much must be
saved in fuel costs each year to justify the extra
cost? Use an interest rate of 8% per year
...
17 In an effort to reduce childhood obesity by reducing the consumption of sugared beverages,
some states have imposed taxes on soda and
other soft drinks
...

However, if taxes were increased to 18 cents on
the dollar, Sturm calculated they would make a

66

Chapter 2

Factors: How Time and Interest Affect Money

signiﬁcant difference
...
Use an
interest rate of 6% per year
...
18 The Texas Tomorrow Fund (TTF) is a program
started in 1996 in Texas wherein parents could prepay their child's college tuition when the child was
young
...
Later, the Texas
legislature allowed universities to set their own tuition rates; tuition costs jumped dramatically
...

If the TTF fund grew at a rate of 4% per year,
while tuition costs increased at 7% per year, determine the state’s shortfall when a newborn enters
college 18 years later
...
19 Henry Mueller Supply Co
...
e
...
Annual cash ﬂows are
shown in the table below
...

Year
Income, $1000
Cost, $1000

1

2

3

4

5

6

7

8

200 200 200 200 200 200 200 200
90 90 90 90 90 90 90 90

2
...
If the new units will cost $350,000, how
much should the company set aside each year, if
the account earns 10% per year?

calculated values, assuming the formula-calculated
value is the correct one
...
25 Proﬁts from recycling paper, cardboard, aluminum, and glass at a liberal arts college have increased at a constant rate of $1100 in each of the
last 3 years
...
26 A report by the Government Accountability Ofﬁce (GAO) shows that the GAO expects the
U
...
Postal Service to lose a record $7 billion at
the end of this year, and if the business model is
not changed, the losses will total $241 billion by
the end of year 10
...
27 Rolled ball screws are suitable for high-precision
applications such as water jet cutting
...
Determine the equivalent annual cost at an
interest rate of 8% per year
...
21 Find the numerical value of the following factors
using (a) interpolation and (b) the formula
...
(A͞P,13%,15)
2
...
22 Find the numerical value of the following factors
using (a) interpolation, (b) the formula, and (c) a
spreadsheet function
...
(F͞P,14%,62)
2
...
23 For the factor (F͞P,10%,43), ﬁnd the percent difference between the interpolated and formulacalculated values, assuming the formula-calculated
value is the correct one
...
24 For the factor (F͞A,15%,52), ﬁnd the percent difference between the interpolated and formula-

2
...
7-m-diameter milling head that
emits low vibration and processes stress-relieved
aluminum panels measuring up to 6000 mm long
...
If the company offers to repay the loan with $60,000 in year 1 and
amounts increasing by $10,000 each year through
year 5, how much can the company borrow at an
interest rate of 10% per year?
2
...

(a) What is the amount of the cash ﬂow in year 3?
(b) What is the future worth of the entire cash
ﬂow series in year 10? Let i ϭ10% per year
...
30 For the cash ﬂows below, determine the amount in year 1, if the annual worth in years 1 through 9 is $601
...

Year

1

2

3

4

5

Cost, $1000

A

A ϩ 30

A ϩ 60

A ϩ 90

A ϩ 120

2
...
1 billion
available 5 years from now to ﬁnance production of a handheld “electronic brain” that, based
on your behavior, will learn how to control
nearly all the electronic devices in your home,
such as the thermostat, coffee pot, TV, and
sprinkler system
...
If the amount set
aside at the end of year 1 is $50 million, how
much will the constant increase G have to be
each year? Assume the investment account
grows at a rate of 18% per year
...
32 Tacozza Electric, which manufactures brush dc
servomotors, budgeted $75,000 per year to pay for
certain components over the next 5 years
...

Geometric Gradient
2
...
Calculate the ﬁrst
two annual worth factor values, that is, A values for
n ϭ 1 and 2, that would be in a 10% interest table
for a growth rate of 4% per year
...
34 Determine the present worth of a geometric gradient series with a cash ﬂow of $50,000 in year 1 and
increases of 6% each year through year 8
...

2
...

Contract 1 has a cost of $10,000 in year 1; costs
will escalate at a rate of 4% per year for 10 years
...

2
...
A new contract
between the two entities resulted in a reduction in
future price increases in the cost of the water from

6

7

A ϩ 150 A ϩ 180

8

9

A ϩ 210

A ϩ 240

8% per year to 4% per year for the next 20 years
...

2
...
Use
an interest rate of 6% per year
...
38 Gesky Industrial Products manufactures brushless
blowers for boilers, food service equipment, kilns,
and fuel cells
...
What was the interest rate on the loan? Use
hand and spreadsheet solutions
...
39 If the value of Jane’s retirement portfolio increased from $170,000 to $813,000 over a 15-year
period, with no deposits made to the account over
that period, what annual rate of return did she
make?
2
...
According to
Consumer Credit Counseling Service, a homeowner with a $100,000 mortgage and a 520 credit
score will pay $110,325 more in interest charges
over the life of a 30-year loan than a homeowner
with the same mortgage and a credit score of 720
...
41 During a period when the real estate market in
Phoenix, Arizona, was undergoing a signiﬁcant
downturn, CSM Consulting Engineers made an
agreement with a distressed seller to purchase an
ofﬁce building under the following terms: total
price of $1
...
CSM was able to make this deal because
of poor market conditions at the time of purchase,
and, at the same time, planning to sell the building
in 4 years (when market conditions would probably be better) and move to a larger ofﬁce building
in Scottsdale, Arizona
...
9 million, what
rate of return per year did the company make on
the investment?
2
...
The contract required the company to repay the investors through an innovative mechanism called faux dividends, a series of
uniform annual payments over a ﬁxed period of
time
...
43 Bessimer Electronics manufactures addressable
actuators in one of its Maquiladora plants in
Mexico
...
If the company does
make the annual investments, what rate of return
will it realize?
2
...
Your boss saw a report submitted by the chief ﬁnancial ofﬁcer (CFO)
that said the equivalent annual worth of maintaining the equipment used in producing the resins was
$48,436 over the last 5 years
...
Your boss
thought $48,436 was too high, so she asked you to
determine what interest rate the CFO used in making the calculations
...
45 Acme Bricks, a masonry products company, wants
to have $600,000 on hand before it invests in new
conveyors, trucks, and other equipment
...
46 An engineer who was contemplating retirement
had $1
...
However, a severe recession caused his portfolio to decrease to only 55% of the original amount, so he
kept working
...
6 million value?

Factors: How Time and Interest Affect Money

2
...

If you invest $200,000 of the company’s money in
a natural gas well that is expected to provide income of $29,000 per year, how long must the well
produce at that rate in order to get the money back
plus a rate of return of 10% per year?
2
...
She invested in
a stock fund that averaged a 12% rate of return over
that period
...
49 A mechanical engineering graduate who wanted
to have his own business borrowed $350,000
from his father as start-up money
...
If the engineer was able to pay his father
$15,000 in year 1, $36,700 in year 2, and amounts
increasing by $21,700 each year, how many
years did it take for the engineer to repay the
loan?
2
...
The cost at the
end of the next year (year 1) is expected to be
$13,000
...
51 In cleaning out some ﬁles that were left behind by
the engineer who preceded you in your current job,
you found an old report that had a calculation for
the present worth of certain maintenance costs for
state highways
...
03)͞(1 ϩ 0
...
06 Ϫ 0
...
What is its value?
2
...
If the cash ﬂow in
year 1 is $35,000 and the gradient amount is
$19,000, what is the value of n at an interest rate of
10% per year?
2
...
If the cash ﬂow in
year 1 is $25,000 and the gradient increase is
18% per year, what is the value of n? The interest
rate is 10% per year
...
54 The amount of money that Diamond Systems can
spend now for improving productivity in lieu of
spending $30,000 three years from now at an interest rate of 12% per year is closest to:
(a) $15,700
(b) $17,800
(c) $19,300
(d ) $21,350
2
...
If the conveyor belt resulted in cost
savings of $4200 per year, the length of time it
would take for the company to recover its investment at 8% per year is closest to:
(a) Less than 9 years
(b) 9 to 10 years
(c) 11 to 12 years
(d ) Over 12 years
2
...
S
...
If U
...
Garment’s operating cost per machine is $22,000 for year 1 and increases by a
constant $1000 per year through year 5, what is
the equivalent uniform annual cost per machine
for the 5 years at an interest rate of 8% per year?
(a) $23,850
(b) $24,650
(c) $25,930
(d ) Over $26,000
2
...
58 At i ϭ 4% per year, A for years 1 through 6 of the
cash ﬂows shown below is closest to:
(a) $300
(b) $560
(c) $800
(d ) $1040
0

1

2

3

4

5

6

$300
$400

Years

2
...
60 A small construction company is considering the
purchase of a used bulldozer for $61,000
...
61 The cost of lighting and maintaining the tallest
smokestack in the United States (at a shuttered
ASARCO reﬁnery) is $90,000 per year
...
62 An enthusiastic new engineering graduate plans to
start a consulting ﬁrm by borrowing $100,000 at
10% per year interest
...
63 An engineer who believed in “save now and play
later” wanted to retire in 20 years with $1
...

At 10% per year interest, to reach the $1
...
64 The cost of a border fence is $3 million per
mile
...
6 million
(b) $4
...
9 million
(d ) Over $5
...
65 An investment of $75,000 in equipment that
will reduce the time for machining self-locking
fasteners will save $20,000 per year
...
66 The number of years required for an account to accumulate $650,000 if Ralph deposits $50,000 each
year and the account earns interest at a rate of 6%
per year is closest to:
(a) 13 years
(b) 12 years
(c) 11 years
(d) 10 years
2
...
, manufactures cleaning nozzles
for reverse-pulse jet dust collectors
...
The rate of return per year on the investment is closest to:
(a) 20%
(b) 18%
(c) 16%
(d) Less than 15%

Factors: How Time and Interest Affect Money

2
...
If the cost in
year 1 was $26,000 and it increased by $2000 per
year through year 5, the present worth of the costs
at an interest rate of 10% per year is closest to:
(a) $102,900
(b) $112,300
(c) $122,100
(d) $195,800
2
...
If
your investments earn 10% per year, the amount
you will have at the end of year 20 is closest to:
(a) $242,568
(b) $355,407
(c) $597,975
(d) $659,125
2
...
If income in year 1 was $300,000 and it decreased by
$30,000 per year through year 4, the annual worth
of the income at 10% per year is closest to:
(a) $310,500
(b) $258,600
(c) $203,900
(d) $164,800
2
...
She has gotten interested in the major effects that
time and interest rates have on the amount of money necessary to do things and the signiﬁcant growth in the amount of
money when a large number of years are considered
...

The four situations are described here
...
Manhattan Island was purchased in 1626 for $24
...

B
...

Case Study

C
...

Sundara estimated the annual rate of return must be
quite good, especially given that she is lucky to earn 4%
per year on her own investments these days
...
A friend who is not good with money, went to a pawn
shop and borrowed $200 for one week and paid $30 in
interest
...
However, she did
not know whether the interest was simple or compounded monthly, and how much may be owed were
this loan not paid off for 1 year
...
What is the annual interest rate for each situation? Include both the annual simple and the compound rates
for situation D
...
Calculate and observe the total amount of money involved in each situation at the end of the time periods
compared to the starting amount
...
Think of a situation for yourself that may be similar to
any of those above
...

71

CHAPTER 3

Combining
Factors and
Spreadsheet
Functions

L E A R N I N G

O U T C O M E S

Purpose: Use multiple factors and spreadsheet functions to ﬁnd equivalent amounts for cash ﬂows that have nonstandard placement
...
1

Shifted series

• Determine the P, F or A values of a series
starting at a time other than period 1
...
2

Shifted series and single cash
ﬂows

• Determine the P, F, or A values of a shifted series
and randomly placed single cash ﬂows
...
3

Shifted gradients

• Make equivalence calculations for shifted
arithmetic or geometric gradient series that
increase or decrease in size of cash ﬂows
...
For
a given sequence of cash ﬂows, there are usually several correct ways to determine the equivalent present worth P, future worth F, or annual worth A
...

3
...
In this case several methods can be used to ﬁnd the equivalent present worth P
...

• Use the F͞P factor to ﬁnd the future worth of each disbursement in year 13, add them, and
then ﬁnd the present worth of the total, using P ϭ F(P͞F,i,13)
...

• Use the P͞A factor to compute the “present worth” P3 ϭ A(P͞A,i,10) (which will be located
in year 3, not year 0), and then ﬁnd the present worth in year 0 by using the (P͞F,i,3) factor
...
For Figure 3–1, the “present worth” obtained using the P͞A factor is located in year 3
...
Note that a P value is always located
1 year or period prior to the beginning of the ﬁrst series amount
...
The most common
mistake made in working problems of this type is improper placement of P
...

Placement of P

To determine a future worth or F value, recall that the F͞A factor derived in Section 2
...
Figure 3–3 shows the location
of the future worth when F͞A is used for Figure 3–1 cash ﬂows
...

Placement of F

It is also important to remember that the number of periods n in the P͞A or F͞A factor is equal
to the number of uniform series values
...
Figures 3–2 and 3–3 show Figure 3–1 renumbered to determine n ϭ 10
...

A = $50

Figure 3–2

P3 = ?

0

1

2

3

4
1

5
2

6
3

7
4

8
5

A = $50

9
6

10

11

12

7

8

9

13
10

Year
n

Location of present worth
and renumbering for n for
the shifted uniform series
in Figure 3–1
...

F=?

0

1

2

3

4
1

5
2

6

7

3

4

8
5

9

10

12 13

Year

7

6

11
8

9

n

10

A = $50

As stated above, several methods can be used to solve problems containing a uniform series
that is shifted
...
Speciﬁc steps should be followed to avoid errors:
1
...

3
...

5
...

Locate the present worth or future worth of each series on the cash ﬂow diagram
...

Draw another cash ﬂow diagram representing the desired equivalent cash ﬂow
...

These steps are illustrated below
...
1
The offshore design group at Bechtel just purchased upgraded CAD software for $5000 now
and annual payments of $500 per year for 6 years starting 3 years from now for annual upgrades
...
The symbol PA is used throughout this chapter
to represent the present worth of a uniform annual series A, and PA represents the present worth
'
at a time other than period 0
...
The
correct placement of PA and the diagram renumbering to obtain n are also indicated
...
Also, n ϭ 6, not 8, for the P͞A factor
...

PA ϭ $500(P͞A,8%,6)
'
Since PA is located in year 2, now ﬁnd PA in year 0
...
1
...

PT ϭ P0 ϩ PA
ϭ 5000 ϩ 500(P͞A,8%,6)(P͞F,8%,2)
ϭ 5000 ϩ 500(4
...
8573)
ϭ $6981
...
1

Calculations for Uniform Series That Are Shifted

The more complex that cash ﬂow series become, the more useful are the spreadsheet functions
...
The NPV function, like the PV function, determines
the P values, but NPV can handle any combination of cash ﬂows directly from the cells
...
Use the format
NPV(i%, second_cell:last_cell) ؉ ﬁrst_cell
First_cell contains the cash ﬂow for year 0 and must be listed separately for NPV to correctly
account for the time value of money
...

The easiest way to ﬁnd an equivalent A over n years for a shifted series is with the PMT function, where the P value is from the NPV function above
...

PMT(i%, n, cell_with_P,F)
Alternatively, the same technique can be used when an F value was obtained using the FV function
...
”
It is very fortunate that any parameter in a spreadsheet function can itself be a function
...
The format is
PMT(i%, n, NPV(i%,second_cell:last_cell) ؉ ﬁrst_cell,F)

[3
...
All three of these functions are illustrated in Example 3
...

EXAMPLE 3
...
If the machine will be recalibrated for each of 6 years starting 3 years after purchase, calculate the 8-year equivalent
uniform series at 16% per year
...

Solution by Hand
Figure 3–5a and b shows the original cash ﬂows and the desired equivalent diagram
...
Then either the A͞P factor or the
A͞F factor can be used
...

Present worth method
...
) Calculate PA for the shifted series in year 2,
'
followed by PT in year 0
...

P'A ϭ 8000(P͞A,16%,6)
PT ϭ PA(P͞F,16%,2) ϭ 8000(P͞A,16%,6)(P͞F,16%,2)
'
ϭ 8000(3
...
7432) ϭ $21,907
...

A' ϭ PT (A͞P,16%,8) ϭ $5043
...
(Refer to Figure 3–5a
...

F ϭ 8000(F͞A,16%,6) ϭ $71,820
The A͞F factor is now used to obtain A' over all 8 years
...
20

75

76

Chapter 3
PT = ?

0

Combining Factors and Spreadsheet Functions
Ј
PA = ?

1 2

F=?

3

4

5

6

7 8

i = 16% per year

0

1

2

3

4

A = $8000

6

7

8

AЈ = ?

(a)

5

(b)

ϭ ϪPMT(16%,8,B12)

ϭ ϪPMT(16%,8,NPV(16%,B4:B11) ϩ B3
ϭ NPV(16%,B4:B11) ϩ B3
(c)

Figure 3–5
(a) Original and (b) equivalent cash ﬂow diagrams; and (c) spreadsheet functions to determine
P and A, Example 3
...

Solution by Spreadsheet
(Refer to Figure 3–5c
...
Use the NPV function to display P ϭ $21,906
...

There are two ways to obtain the equivalent A over 8 years
...
(1) Enter the PMT function making direct reference to
the P value (see cell tag for D͞E5), or (2) use Equation [3
...

3
...
1 are applied to the uniform series and the single-amount formulas are applied to the one-time cash ﬂows
...
3 and 3
...
For spreadsheet solutions, it is necessary to enter the net cash
ﬂows before using the NPV and other functions
...
3
An engineering company in Wyoming that owns 50 hectares of valuable land has decided to
lease the mineral rights to a mining company
...
The engineering
company makes a proposal to the mining company that it pay $20,000 per year for 20 years
beginning 1 year from now, plus $10,000 six years from now and $15,000 sixteen years from
now
...
2

Calculations Involving Uniform Series and Randomly Placed Single Amounts

Solution
The cash ﬂow diagram is shown in Figure 3–6 from the owner’s perspective
...

PT ϭ 20,000(P͞A,16%,20) ϩ 10,000(P͞F,16%,6) ϩ 15,000(P͞F,16%,16)
ϭ $124,075
Note that the $20,000 uniform series starts at the end of year 1, so the P͞A factor determines the
present worth at year 0
...
3
...

Then you obtain the A value by multiplying P or F by the appropriate A͞P or A͞F factor
...
4 illustrates this procedure
...
4
A design-build-operate engineering company in Texas that owns a sizable amount of land
plans to lease the drilling rights (oil and gas only) to a mining and exploration company
...
e
...
Utilize engineering economy relations by
hand and by spreadsheet to determine the ﬁve equivalent values listed below at 16% per year
...

2
...

4
...

Total present worth PT in year 0
Future worth F in year 22
Annual series over all 22 years
Annual series over the ﬁrst 10 years
Annual series over the last 12 years

Solution by Hand
Figure 3–7 presents the cash ﬂows with equivalent P and F values indicated in the correct years
for the P͞A, P͞F, and F͞A factors
...
PT in year 0: First determine P' of the series in year 2
...

77

78

Chapter 3

Combining Factors and Spreadsheet Functions

P'A ϭ 20,000(P͞A,16%,20)
PT ϭ P'A(P͞F,16%,2) ϩ 10,000(P͞F,16%,6) ϩ 15,000(P͞F,16%,16)
ϭ 20,000(P͞A,16%,20)(P͞F,16%,2) ϩ 10,000(P͞F,16%,6)
ϩ 15,000(P͞F,16%,16)
ϭ $93,625

[3
...
F in year 22: To determine F in year 22 from the original cash ﬂows (Figure 3–7), ﬁnd
F for the 20-year series and add the two F values for the two single amounts
...

F ϭ 20,000(F͞A,16%,20) ϩ 10,000(F͞P,16%,16) ϩ 15,000(F͞P,16%,6)

[3
...
A over 22 years: Multiply PT ϭ $93,625 from (1) above by the A͞P factor to determine
an equivalent 22-year A series, referred to as A1–22 here
...
16635) ϭ $15,575

[3
...
In this
case, the computation is A1–22 ϭ F(A͞F,16%,22) ϭ $15,575
...
4
...
A over years 1 to 10: This and (5), which follows, are special cases that often occur in
engineering economy studies
...
This occurs when a deﬁned study
period or planning horizon is preset for the analysis
...
) To determine the equivalent A series for years 1 through 10 only (call it A1–10),
the PT value must be used with the A͞P factor for n ϭ 10
...

A1–10 ϭ PT (A͞P,16%,10) ϭ 93,625(0
...
5]

5
...
This transforms Figure 3–7 into the 12-year series A11–22 in Figure 3–8b
...
03241) ϭ $79,457

[3
...
This is another demonstration of the time value of money
...
The $20,000 series and
the two single amounts have been entered into separate columns, B and C
...
2

79

Calculations Involving Uniform Series and Randomly Placed Single Amounts
Figure 3–8

A1–10 = ?

1

2

3

4

5

6

7

8

9

10

11

21

22 Year

20

21 22

Cash ﬂows of Figure 3–7
converted to equivalent
uniform series for
(a) years 1 to 10 and
(b) years 11 to 22
...

Example 3
...

values are all entered so that the functions will work correctly
...
To prepare for sensitivity
analysis, the functions are developed using cell reference format or global variables, as indicated in the column E function
...

1
...
The sum of these in F14 is PT ϭ $93,622, which
corresponds to the value in Equation [3
...
Alternatively, PT can be determined directly
via the sum of two NPV functions, shown in row 15
...
The FV function in row 18 uses the P value in F14 (preceded by a minus sign) to determine F twenty-two years later
...
3]
...
To ﬁnd the 22-year A series starting in year 1, the PMT function in row 21 references
the P value in cell F14
...
4] to
obtain A1–22
...
The cell reference format is
ϭ PMT(D1,22,−(NPV(D1,B6:B27)ϩB5 ϩ NPV(D1,C6:C27)ϩC5))
...
and 5
...
These are both true for the
series requested here
...
5] and [3
...

Comment
Remember that round-off error will always be present when comparing hand and spreadsheet
results
...
Also, be very careful when constructing spreadsheet functions
...
Always check
your function entries carefully before touching
...
3 Calculations for Shifted Gradients
In Section 2
...
The P͞G factor, Equation [2
...

The present worth of an arithmetic gradient will always be located two periods before the
gradient starts
...

The relation A ϭ G(A͞G,i,n) was also derived in Section 2
...
The A͞G factor in Equation [2
...
Recall that the base amount must be treated separately
...

A conventional gradient series starts between periods 1 and 2 of the cash ﬂow sequence
...
The n value in the P͞G and A͞G
factors for a shifted gradient is determined by renumbering the time scale
...
The n value for the gradient factor is determined by
the renumbered period where the last gradient increase occurs
...
Example 3
...

EXAMPLE 3
...
has tracked the average inspection cost on a robotics manufacturing line for
8 years
...
Analyze the gradient increase, using the P͞G factor
...
Figure 3–10b and c partitions these two
series
...
It is
clear that n ϭ 5 for the P͞G factor
...

3
...
5
...
19]
...

PT ϭ PA ϩ PG ϭ 100(P͞A,i,8) ϩ 50(P͞G,i,5)(P͞F,i,3)

It is important to note that the A͞G factor cannot be used to ﬁnd an equivalent A value in periods 1 through n for cash ﬂows involving a shifted gradient
...
To ﬁnd the equivalent annual series in years 1 through 10 for the gradient series
only, ﬁrst ﬁnd the present worth PG of the gradient in actual year 5, take this present worth back
to year 0, and annualize the present worth for 10 years with the A͞P factor
...

To ﬁnd the equivalent A series of a shifted gradient through all the n periods, ﬁrst ﬁnd the
present worth of the gradient at actual time 0, then apply the (A͞P, i, n) factor
...

81

82

Chapter 3

Combining Factors and Spreadsheet Functions

EXAMPLE 3
...

0

1

2

3

$50

$50

4

5

6

7

$50
$70
$90
$110
$130

Figure 3–12
Diagram of a shifted gradient, Example 3
...

Solution

The base amount annual series is AB ϭ $50 for all 7 years (Figure 3–13)
...
The gradient n is 5
...

P0 ϭ PG(P͞F,i,2) ϭ 20(P͞G,i,5)(P͞F,i,2)

Annualize the gradient present worth from year 1 through year 7 to obtain AG
...

AT ϭ 20(P͞G,i,5)(P͞F,i,2)(A͞P,i,7) ϩ 50
For a spreadsheet solution, enter the original cash ﬂows into adjacent cells, say, B3 through
B10, and use an embedded NPV function in the PMT function
...

P0 = ?

0

PG = ?

1

2

AT = ?

3

0
$50

1

$50

4

5
3

6

7

4

Year

$50

AB = $50

2

5

Gradient n

$70
$90
G = $20

$110
$130

Figure 3–13
Diagram used to determine A for a shifted gradient, Example 3
...

In Section 2
...
The factor was derived to ﬁnd the
present worth in year 0, with A1 in year 1 and the ﬁrst gradient appearing in year 2
...
Refer
to Figure 2–21 as a refresher for the cash ﬂows
...
35] is the formula used for the factor
...

3
...
7
Chemical engineers at a Coleman Industries plant in the Midwest have determined that a
small amount of a newly available chemical additive will increase the water repellency of
Coleman’s tent fabric by 20%
...
He expects
the annual price to increase by 12% per year thereafter for the next 8 years
...
Use i ϭ 15% per year to determine the equivalent total present
worth for all these cash ﬂows
...
The total present worth PT is found using g ϭ 0
...
15
...
34] and [2
...

PT ϭ 35,000 ϩ A(P͞A,15%,4) ϩ A1(P͞A,12%,15%,9)(P͞F,15%,4)
1 Ϫ (1
...
15)9
ϭ 35,000 ϩ 7000(2
...
5718)
0
...
12
ϭ 35,000 ϩ 19,985 ϩ 28,247

[

]

ϭ $83,232
Note that n ϭ 4 in the (P͞A,15%,4) factor because the $7000 in year 5 is the cash ﬂow of the
initial amount A1
...
If cells B1 through B14
are used, the function to ﬁnd P ϭ $83,230 is
NPV(15%,B2:B14)ϩB1
The fastest way to enter the geometric series is to enter $7000 for year 5 (into cell B6) and set
up each succeeding cell multiplied by 1
...

PT = ?

0

Pg = ?

1

2

3 4

5
0

$7000

i = 15% per year

6

7

8

9

10 11 12 13

1

2

3

4

5

6

7

8

9

Year
Geometric
gradient n

$7840

$35,000

$17,331
12% increase
per year

Figure 3–14
Cash ﬂow diagram including a geometric gradient with g ϭ 12%, Example 3
...

Decreasing arithmetic and geometric gradients are common, and they are often shifted gradient series
...
Equivalence computations for present worth P and annual worth A are basically the
same as discussed in Chapter 2, except for the following
...

For shifted, decreasing gradients:
• The base amount A (arithmetic) or initial amount A1 (geometric) is the largest amount in the
ﬁrst year of the series
...

• The amount used in the factors is –G for arithmetic and –g for geometric gradient series
...

Figure 3–15 partitions a decreasing arithmetic gradient series with G ϭ $−100 that is shifted
1 year forward in time
...

PT ϭ $800(P͞F,i,1) ϩ 800(P͞A,i,5)(P͞F,i,1) Ϫ 100(P͞G,i,5)(P͞F,i,1)

EXAMPLE 3
...
This type of new-technology glass uses electrochrome coating to allow rapid adjustment to sun and dark in building glass, as well as assisting
with internal heating and cooling cost reduction
...
All cash ﬂow estimates are
in $1000 units, and the interest rate expectation is 8% per year
...

Years 6 through 10: No new investment and no withdrawals
...

3
...
If the withdrawal series is over- or underfunded, what is the exact amount available in
year 11, provided all other estimates remain the same?

Solution by Hand
Figure 3–16 presents the cash ﬂow diagram and the placement of the equivalent P values used
in the solution
...

Pg,10 = ?

Pg,0 = ?

$20,000
A1 = $20,000
g = Ϫ0
...
8
...
The present worth in year 0 is
PG ϭ Ϫ[7000(P͞A,8%,5) Ϫ 1000(P͞G,8%,5)]
ϭ $Ϫ20,577
Withdrawal series: Decreasing, shifted geometric series starting in year 12 with A1 ϭ $20,000,
g ϭ Ϫ0
...
If the present worth in year 10 is identiﬁed as Pg,10, the
present worth in year 0 is Pg,0
...
35] for the (P͞A,Ϫ20%,8%,5) factor
...
7]

}

1 ϩ (Ϫ0
...
08
ϭ 20,000 ———————— (0
...
08 Ϫ (Ϫ0
...
7750)(0
...
Either additional funds
must be invested or less must be withdrawn to make the series equivalent at 8% per year
...
7] and set Pg,0 ϭ ϪPG ϭ 20,577
...
7750)(0
...

85

86

Chapter 3

Combining Factors and Spreadsheet Functions

ϭ NPV(8%,B4:B18) + B3
(a)

(b)

Figure 3–17
Spreadsheet solution, Example 3
...
(a) Cash ﬂows and NPV function and (b) Goal Seek to determine initial withdrawal amount in year 11
...
To determine if the investment series will cover the withdrawal series, enter
the cash ﬂows (in column B using the functions shown in column C) and apply the NPV function shown in the cell tag to display PT ϭ $ϩ5130 directly
...

The Goal Seek tool is very handy in determining the initial withdrawal amount that results
in PT ϭ 0 (cell B19)
...

Each succeeding withdrawal is 80% of the previous one
...
However, now
establish the entry in B4 as the changing cell
...

CHAPTER SUMMARY
In Chapter 2, we derived the equations to calculate the present, future, or annual worth of speciﬁc
cash ﬂow series
...
For example, when a
uniform series does not begin in period 1, we still use the P͞A factor to ﬁnd the “present worth”
of the series, except the P value is located one period ahead of the ﬁrst A value, not at time 0
...

With this information, it is possible to solve for P, A, or F for any conceivable cash ﬂow series
...
Though
spreadsheet solutions are fast, they do remove some of the understanding of how the time value
of money and the factors change the equivalent value of money
...
1 Industrial Electric Services has a contract with the
U
...
Embassy in Mexico to provide maintenance
for scanners and other devices in the building
...
e
...
2 Civil engineering consulting ﬁrms that provide services to outlying communities are vulnerable to a
number of factors that affect the ﬁnancial condition
of the communities, such as bond issues and real
estate developments
...
At the end of that time, a
mild recession slowed the development, so the parties signed another contract for $190,000 per year
for 2 more years
...

3
...
Determine the
present worth at an interest rate of 12% per year
...
Show (a) hand and
(b) spreadsheet solutions
...
4 Standby power for water utility pumps and other
electrical devices is provided by diesel-powered
generators
...
The utility estimates that by switching to gas,
it will save $22,000 per year, starting 3 years from
now
...

3
...

i = 10% per year
0

1

2

$200

4

5

6

7

8

$90
$200

3

$90

Year

$90

$200

3
...
56 per 1000 gallons
...
28 per 1000 gallons
...
) If Fort
Bliss uses 2 billion gallons of water each year,
what is the present worth of the discount for a 20year period at an interest rate of 6% per year?
3
...
One of the latest schemes
for big-time athletics is the “sports mortgage
...
In return, the seats themselves
will stay locked in at current-year prices
...

A fan plans to purchase the sports mortgage along
with a current-season ticket and pay for both now,
then buy a ticket each year for the next 30 years
...
8 The cash ﬂow associated with making self-locking
fasteners is shown below
...

8

9

Income, $1000 20 20 20 20 30 30 30 30 30
Cost, $1000
8 8 8 8 12 12 12 12 12

Year

0

1

2

3

4

5

6

7

30
25

3
...
In an effort to pay off the loan quickly, the
company made four payments in years 1 through
4, with each payment being twice as large as the
preceding one
...
10 Revenue from the sale of ergonomic hand tools
was $300,000 in years 1 through 4 and $465,000 in
years 5 through 9
...

3
...
They believe that they will need
$2,000,000 in year 20
...
They already
have $25,000 in their investment account
...
12 Costs associated with the manufacture of miniature high-sensitivity piezoresistive pressure transducers are $73,000 per year
...

Using an interest rate of 10% per year, determine
(a) the equivalent annual cost of the manufacturing
operations and (b) the equivalent annual savings in
years 1 through 5
...
13 Calculate the equivalent annual cost in years 1
through 9 of the following series of disbursements
...
Show (a)
hand and (b) spreadsheet solutions
...
14 For the cash ﬂows below, ﬁnd the value of x that
makes the equivalent annual worth in years 1
through 7 equal to $300 per year
...
Show solutions (a) by hand
and (b) using the Goal Seek tool
...
15 Precision Instruments, Inc
...
The company borrowed
$10,000,000 with the understanding that it would
make a $2,000,000 payment at the end of year 1
and then make equal annual payments in years 2
through 5 to pay off the loan
...
16 A construction management company is examining
its cash ﬂow requirements for the next 7 years
...
Speciﬁcally, the company
expects to spend $6000 one year from now, $9000
three years from now, and $10,000 six years from
now
...
17 Find the equivalent annual worth for the cash
ﬂows shown, using an interest rate of 12% per
year
...

Year
Cash Flow, $

1

2

3

4

5

6

7

8

9

20

20

20

20

60

60

60

60

60

3
...

Year

Income, $͞Year

Expense, $͞Year

0
1–4
5–10

0
700
2000

Ϫ2500
Ϫ200
Ϫ300

3
...
The club
will pay $1000 per year plus make $350,000 in
improvements at the park
...
M
...
If the club
makes $100,000 worth of improvements now
and then $50,000 worth each year for the next
5 years, what is the equivalent annual cost of
the lease to the club at an interest rate of 10%
per year?
3
...
In return,
season ticket prices stay locked in at currentyear prices, and the package can be sold in the
secondary market, while taking a tax write-off
for donating to a school
...
The first
payment is made now (i
...
, beginning-of-year
payment), and an additional nine payments are
to be made at the end of each year for the next
9 years
...
What is the total amount of the payment
each year in years 0 through 9? Use an interest
rate of 10% per year
...
21 The expansion plans of Acme Granite, Stone &
Brick call for the company to add capacity for a
new product in 5 years
...
If the company sets aside $55,000 now
and $90,000 in year 2, what uniform annual
amount will it have to put in an account in years 3
through 5 to have the $360,000? Assume the account earns interest at 8% per year
...
22 For the cash ﬂows shown, calculate the future
worth in year 8 using i ϭ 10% per year
...
24 New actuator element technology enables engineers to simulate complex computer-controlled
movements in any direction
...
25 Austin Utilities is planning to install solar panels
to provide some of the electricity for its groundwater desalting plant
...
The ﬁrst phase will cost $4 million in
year 1 and $5 million in year 2
...
Let i ϭ
10% per year
...

3
...

i = 10% per year
1

2

3

4

5

x

x

x

6

7

8

2x

2x

Year

x

2x

300

3
...
If you make annual
deposits of a uniform amount A into the account
that earns interest at a rate of 7% per year, how
many years from now will it be until the value of
the account is equal to 10 times the value of a
single deposit?

0

2

6

100

Cash Flow, $

1

1

x

Year

0

3

4

5

6

7

8

Random Single Amounts and Uniform Series
3
...
The replacement process will cost the
company $50,000 three years from now
...

3
...
If the
company spent only $42,000 in year 1, what uniform annual amount should the company expect to
spend in each of the next 4 years to expend the
entire budget? Assume the company uses an interest rate of 10% per year
...
30 A recently hired chief executive ofﬁcer wants to
reduce future production costs to improve the company’s earnings, thereby increasing the value of the
company’s stock
...
By how much must annual
costs decrease in years 3 through 7 to recover the
investment plus a return of 12% per year?
3
...
Use i ϭ 10% per year
...
27 For the cash ﬂows shown in the diagram, determine the value of x that will make the future worth
in year 8 equal to $Ϫ70,000
...
32 For the following series of income and expenses,
ﬁnd the equivalent value in year 9 at an interest rate
of 12% per year
...

3
...

Year

1

9

10

Years

Income, $

Expense, $

Cash Flow, $

13 13 13 13 16 19 22 25 28

31

0
1–6
7–9
10–16

0
9,000
28,000
38,000

Ϫ70,000
Ϫ13,000
Ϫ14,000
Ϫ19,000

3
...
The contract
price is $1
...
The payment plan is Z dollars
now, 2Z dollars in year 2, and 3Z dollars in years 3
through 5
...
34 A foursome of entrepreneurial electrical engineering graduates has a plan to start a new solar power
equipment company based on STE (solar thermal
electric) technology
...
Within the agreement, the loan is
to be repaid by allocating 80% of the company’s
proﬁts each year for the ﬁrst 4 years to the investors
...
The company’s business plan indicates that they expect to make no
proﬁt for the ﬁrst year, but in years 2 through 4, they
anticipate proﬁts to be $1
...
If the
investors accept the deal at an interest rate of 15%
per year, and the business plan works to perfection,
what is the expected amount of the last loan payment (in year 5)?
Shifted Gradients
3
...
Let i ϭ 10% per year
...
37

2

3

4

5

6

7

8

A low-cost noncontact temperature measuring tool
may be able to identify railroad car wheels that are in
need of repair long before a costly structural failure
occurs
...

3
...

Year

1

2

3

4

5

6

7

8

9

10

Cash Flow, $

90 90 90 85 80 75 70 65 60 55

3
...

There will be no income in years 1 and 2, but in
year 3 income will be $250,000, and thereafter it
will increase according to an arithmetic gradient
through year 15
...

3
...
Use an interest rate of 10% per year
...
41 The cost associated with manufacturing highperformance lubricants closely follows the cost of
crude oil
...
4 million in
years 1 through 3, after which the cost increased by
3% per year through this year
...
e
...

Show both (a) hand and (b) spreadsheet solutions
...
42 Find the future worth in year 10 of $50,000 in
year 0 and amounts increasing by 15% per year
through year 10 at an interest rate of 10% per year
...
43 The cost of tuition at public universities has been
steadily increasing for many years
...
8 million for the ﬁrst 3 years
...
77 million in year 4,
$1
...
What is the
equivalent annual worth of the pumping costs over
the 9 years at an interest rate of 12% per year?

4 years for all students who ﬁnished in the top 3%
of their class
...
If the tuition for the ﬁrst 4 years will be
$7200 per year and it increases by 5% per year for
the next 5 years, what is the present worth of the
tuition cost at an interest rate of 8% per year?
3
...
The deal was structured such that the
private equity ﬁrm received $3 million immediately after the deal was closed (in year 0) through
the sale of some assets
...
36 million, and it is projected to increase
by 12% each year through expansion of the customer base
...

3
...
Assume
i ϭ 10% per year
...
49 Income from the mining of mineral deposits usually decreases as the resource becomes more difﬁcult to extract
...

Use an interest rate of 18% per year
...

3
...

i = 10% per year
0

1

2

3

4

5

6

7

8

Year

$220

Shifted Decreasing Gradients

$270
$320

3
...

Year

0

Cash
Flow, $

1

2

3

0 8000 8000 8000ϪG

4

5

8000Ϫ2G

8000Ϫ3G

3
...

Year

Cash Flow, $

Year

Cash Flow, $

0
1
2
3

850
800
750
700

4
5
6
7

650
600
550
500

3
...
51 Income from the sale of application software (apps)
is usually constant for several years and then decreases quite rapidly as the market gets close to
saturation
...
Determine the equivalent annual income in years 1
through 7, using an interest rate of 10% per year
...
52 Determine the future worth in year 10 of a cash
ﬂow series that starts in year 0 at $100,000 and
decreases by 12% per year
...

92

Chapter 3

Combining Factors and Spreadsheet Functions

ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
3
...
If the company
plans to deposit money each year, starting now, the
equation that represents the deposit each year at
8% per year interest is:
(a) 1,900,000(A͞F,8%,3)
(b) 1,900,000(A͞F,8%,4)
(c) 1,900,000 ϩ 1,900,000(A͞F,8%,3)
(d) 1,900,000 ϩ 1,900,000(A͞F,8%,2)
3
...
55 For the diagram shown, the respective values of n
for the following equation are:
P0 ϭ 100(P͞A,10%,n)(P͞F,10%,n)
(a) 6 and 1
(b) 6 and 2
(c) 7 and 1
(d) 7 and 2

3
...
Assume the interest
rate is 8% per year
...
59 The net present worth in year 0 of the following
series of incomes and expenses at 8% per year is
closest to:
Years

Income, $

(a)
(c)

2

3

4

5

6

7

3
...
If the company plans to deposit money into an account each year for 4 years
beginning 2 years from now (ﬁrst deposit is in year 2)
to pay for the expansion, the equation that represents
the amount of the deposit at 9% per year interest is:
(a) A ϭ 10,000,000(A͞F,9%,5)
(b) A ϭ 10,000,000(A͞F,9%,4)
(c) A ϭ 10,000,000(A͞P,9%,4)
(d) A ϭ 10,000,000(A͞F,9%,4)(P͞F,9%,1)

3
...
60 For the cash ﬂows shown, the equivalent annual
worth in periods 1 through 5 at an interest rate of
10% per year is closest to:

(a)
(c)
$100

Ϫ1000
Ϫ100
Ϫ200

$14,300
$16,100

i = 10% per year

1

12,000
700
900

0
1–6
7–11

P0 = ?

0

Expenses, $

$1120
$1350

(b)
(d)

$1240
$1490

3
...
62 In order to have cash available for unforeseen
emergencies, Baring Systems, a military contractor, wants to have $2,000,000 in a contingency
fund 4 years from now
...
Its mission is to
ensure the preservation of the natural resources, while providing necessary, but minimal, development for recreational
use by the public
...

A southern U
...
state, which has long-term groundwater
problems, has asked the TPL to manage the purchase of
10,000 acres of aquifer recharge land and the development of
three parks of different use types on the land
...
Total
annual purchase amounts are expected to decrease 25% each
year through year 5 and then cease for this particular project
...
5 million citizens immediately to the southeast of this acreage relies heavily on the aquifer’s water
...

The bond interest rate is an effective 7% per year
...
Increases in construction costs are expected to be $100,000 each year through
year 6
...
Use the bond
issue funds to assist with this purchase
...

• Raise the remaining project funds over the next 2 years in
equal annual amounts
...

Case Study Exercises
1
...
If the TPL did agree to fund all costs except the $3 million bond proceeds now available, determine the equivalent annual amount that must be raised in years 4
through 6 to supply all remaining project funds
...

3
...
tpl
...
Identify some
economic and noneconomic factors that you believe
must be considered when the TPL is deciding to purchase land to protect it from real estate development
...

SECTION

TOPIC

LEARNING OUTCOME

4
...

4
...

4
...

4
...

4
...

4
...

4
...

4
...

4
...

I

n all engineering economy relations developed thus far, the interest rate has been
a constant, annual value
...
In fact,
weekly, daily, and even continuous compounding may be experienced in some project evaluations
...
—have interest rates compounded for a time period shorter than
1 year
...

This chapter explains how to understand and use nominal and effective interest rates
in engineering practice and in daily life situations
...

PE
The Credit Card Offer Case: Today, Dave
received a special offer of a new credit
card from Chase Bank linked with the
major airline that he ﬂies frequently
...
The bonus package
includes extra airline points (once the
ﬁrst purchase is made), priority airport
check-in services (for 1 year), several
free checked-bag allowances (for up
to 10 check-ins), extra frequent-ﬂyer
points on the airline, access to airline
lounges (provided he uses the card
on a set time basis), plus several other
rewards (rental car discounts, cruise trip
amenities, and ﬂoral order discounts)
...

Dave has a credit card currently with
a bank that he is planning to leave due
to its poor customer service and high
monthly fees
...

In the page that accompanies the
offer letter, “pricing information” is
included
...
A summary of
several of these rates and fees follows
...
24% per year (sum of the current U
...

Government prime rate of 3
...
99%, which is the APR added to determine
the balance transfer APR for Chase Bank)
19
...
99% per year (maximum penalty APR)

$85; free the ﬁrst year
$5 or 3% of each transfer, whichever is greater
$10 or 3% of each advance, whichever is greater
$39 each occurrence, if balance exceeds $250
$39 each occurrence
$39 each occurrence

*All APR rates are variable, based on a current 3
...
99% added to
determine purchase͞balance transfer APR; with 15
...
99% added to determine penalty APR
...
If no minimum payment is received
within 60 days, the penalty APR applies to all outstanding balances and all future transactions
on the account
...
1)

Effective annual interest rates (4
...
6)

4
...
Here we discuss nominal and effective interest rates, which
have the same basic relationship
...
For example, if an
interest rate is expressed as 1% per month, the terms nominal and effective interest rates must
be considered
...
The interest amounts for loans, mortgages, bonds, and stocks
are commonly based upon interest rates compounded more frequently than annually
...
In our own personal ﬁnances, we manage
most cash disbursements and receipts on a nonannual time basis
...
First, consider a nominal interest rate
...
By deﬁnition,
r ‫ ؍‬interest rate per time period ؋ number of periods
[4
...
1]
...
5% per month is the same as each of the following nominal rates
...
1]

What This Is

24 months
12 months
6 months
3 months

1
...
5 ϫ 12 ϭ 18%
1
...
5 ϫ 3 ϭ 4
...
” These nominal rates are calculated in the same way that simple rates
are calculated using Equation [1
...

After the nominal rate has been calculated, the compounding period (CP) must be included in the interest rate statement
...
5%
per month
...
5% per quarter, compounded monthly
...

Effective interest rate i

An effective interest rate i is a rate wherein the compounding of interest is taken into
account
...

The most common form of interest rate statement when compounding occurs over time periods
shorter than 1 year is “% per time period, compounded CP-ly,” for example, 10% per year, compounded monthly, or 12% per year, compounded weekly
...
If the CP is not mentioned, it is understood to

4
...
For example, an interest rate of
“1
...
An
equivalent effective rate statement, therefore, is 1
...

All of the following are effective interest rate statements because either they state they are
effective or the compounding period is not mentioned
...

Statement

CP

What This Is

i ϭ 10% per year
i ϭ effective 10% per year,
compounded monthly
1
i ϭ 1_% per month
2

CP not stated; CP ϭ year
CP stated; CP ϭ month

Effective rate per year
Effective rate per year

1
i ϭ effective 1_% per month,
2
compounded monthly
i ϭ effective 3% per quarter,
compounded daily

CP not stated; CP ϭ month

Effective rate per month

CP stated; CP ϭ month

Effective rate per month; terms effective
and compounded monthly are redundant
Effective rate per quarter

CP stated; CP ϭ day

All nominal interest rates can be converted to effective rates
...

All interest formulas, factors, tabulated values, and spreadsheet functions must use an effective
interest rate to properly account for the time value of money
...
This is the same as the nominal rate
...
25% on a monthly basis
...
This is the same as an effective rate
...
As we will learn in the following sections, the effective
rate is always greater than or equal to the nominal rate, and similarly APY Ն APR
...

Interest period (t)—The period of time over which the interest is expressed
...
The time unit of 1 year is
by far the most common
...

Compounding period (CP)—The shortest time unit over which interest is charged or earned
...
If CP is not stated, it is assumed to be the same as the interest
period
...
If the compounding period CP and the time period t are the same, the compounding frequency is 1, for example, 1% per month, compounded monthly
...
It has an interest period t of
1 year, a compounding period CP of 1 month, and a compounding frequency m of 12 times per
year
...

In previous chapters, all interest rates had t and CP values of 1 year, so the compounding frequency was always m ϭ 1
...
Now, it will be necessary to express a nominal rate as an
effective rate on the same time base as the compounding period
...
2]

As an illustration, assume r ϭ 9% per year, compounded monthly; then m ϭ 12
...
2] is used to obtain the effective rate of 9%͞12 ϭ 0
...

97

98

Chapter 4

Nominal and Effective Interest Rates

Note that changing the interest period t does not alter the compounding period, which is 1 month
in this illustration
...
5% per 6 months,
compounded monthly, are two expression of the same interest rate
...
1
Three different bank loan rates for electric generation equipment are listed below
...

(a) 9% per year, compounded quarterly
...

(c) 4
...

Solution
Apply Equation [4
...

The graphic in Figure 4–1 indicates the effective rate per CP and how the interest rate is distributed over time
...
25%

Quarter

(b) 9% per
year

Compounding
Frequency (m)

Month

( )

Distribution over Time Period t

4

...
75%

0
...
75%

1

12

...
75%

3

...
75%

1

...
25%

...
25%

...
25%

...
25%

...
173%

(c) 4
...
173%
1

12 14 16

26 Week

Figure 4–1
Relations between interest period t, compounding period CP, and effective interest rate per CP
...
Basically
there are three ways to express interest rates, as detailed in Table 4–1
...
For the ﬁrst format, a nominal interest rate is given and the
compounding period is stated
...
In the second format, the stated rate is identiﬁed as effective (or APY could also be used),
so the rate is used directly in computations
...
This rate
is effective over a compounding period equal to the stated interest period of 1 year in this case
...

TABLE 4–1

Various Ways to Express Nominal and Effective Interest Rates

Format of Rate Statement

Example of Statement

What about the Effective Rate?

Nominal rate stated,
compounding period stated

8% per year, compounded
quarterly

Find effective rate for any time
period (next two sections)

Effective rate stated

Effective 8
...
243% per year
directly for annual cash ﬂows

Interest rate stated, no
compounding period stated

8% per year

Rate is effective for CP equal to stated
interest period of 1 year; ﬁnd effective rate for all other time periods

4
...
2 The Credit Card Offer Case
As described in the introduction to this case, Dave has been offered what is described as a credit
card deal that should not be refused—at least that is what the Chase Bank offer letter implies
...
24% is an annual rate, with no compounding period mentioned
...
Therefore, we should conclude that the CP is 1 year, the same as the annual interest
period of the APR
...

(a) First, determine the effective interest rates for compounding periods of 1 year and
1 month so Dave knows some effective rates he might be paying when he transfers the
$1000 balance from his current card
...
What is the amount of the total balance he owes?
Now, Dave looks a little closer at the ﬁne print of the “pricing information” sheet and discovers
a small-print statement that Chase Bank uses the daily balance method (including new transactions) to determine the balance used to calculate the interest due at payment time
...

Solution

(a) The interest period is 1 year
...
2] for both CP values of 1 year (m ϭ 1
compounding period per year) and 1 month (m ϭ 12 compounding periods per year)
...
24͞1 ϭ 14
...
24͞12 ϭ 1
...

Amount owed after 1 month ϭ 1000 ϩ 1000(0
...
03(1000)
ϭ 1000 ϩ 11
...
87
Including the $30 fee, this represents an interest rate of (41
...
187%
for only the 1-month period
...
2], now with m ϭ 365 compounding periods per year
...
24͞365 ϭ 0
...
2 Effective Annual Interest Rates
In this section, effective annual interest rates are calculated
...
For example,
we will learn that a nominal 18% per year, compounded quarterly is the same as an effective rate of
19
...

The symbols used for nominal and effective interest rates are
r ϭ nominal interest rate per year
CP ϭ time period for each compounding
m ϭ number of compounding periods per year
i ϭ effective interest rate per compounding period ϭ r͞m
ia ϭ effective interest rate per year
The relation i ϭ r͞m is exactly the same as Equation [4
...

99

PE

100

Chapter 4

Nominal and Effective Interest Rates

Figure 4–2

P(1 + i)m = P(1 + ia)

Future worth calculation
at a rate i, compounded m
times in a year
...
Like compound interest, an effective interest rate at any point
during the year includes (compounds) the interest rate for all previous compounding periods
during the year
...
We set P ϭ $1 for
simpliﬁcation
...
Since interest may be compounded several times during the year, use the effective annual
rate symbol ia to write the relation for F with P ϭ $1
...
The effective rate i per CP must be compounded through all m periods
to obtain the total effect of compounding by the end of the year
...
The effective annual interest rate formula
for ia is
ia ‫ ؉ 1( ؍‬i)m ؊ 1

[4
...
3] calculates the effective annual interest rate ia for any number of compounding
periods per year when i is the rate for one compounding period
...
3] can be
solved for i to determine the effective interest rate per compounding period
...
4]

As an illustration, Table 4–2 utilizes the nominal rate of 18% per year for different compounding periods (year to week) to determine the effective annual interest rate
...
Table 4–3 summarizes the effective annual
rate for frequently quoted nominal rates using Equation [4
...
A standard of 52 weeks and
365 days per year is used throughout
...
8
...
34615

1
...
5

9

18

Rate per
Compound
Period, i%

9%

2

4
...
34615% in each

5

1
...
5%

9

4

4
...
5%

2

9%

1

18%

Distribution of i over the Year of
Compounding Periods

r ϭ 18% per year, compounded CP-ly

Effective Annual Interest Rates Using Equation [4
...
0034615)52 Ϫ 1 ϭ 19
...
015)12 Ϫ 1 ϭ 19
...
045)4 Ϫ 1 ϭ 19
...
09)2 Ϫ 1 ϭ 18
...
18)1 Ϫ 1 ϭ 18%

Effective Annual
Rate, ia ‫ ؉ 1( ؍‬i)m ؊ 1

102

Chapter 4

Nominal and Effective Interest Rates

TABLE 4–3 Effective Annual Interest Rates for Selected Nominal Rates
Nominal
Rate r %

Semiannually
(m ‫)2 ؍‬

Quarterly
(m ‫)4 ؍‬

Monthly
(m ‫)21 ؍‬

Weekly
(m ‫)25 ؍‬

Daily
(m ‫)563 ؍‬

0
...
50
1
...
50
2
3
4
5
6
7
8
9
10
12
15
18
20
25
30
40
50

0
...
501
1
...
506
2
...
023
4
...
063
6
...
123
8
...
203
10
...
360
15
...
810
21
...
563
32
...
000
56
...
250
0
...
004
1
...
015
3
...
060
5
...
136
7
...
243
9
...
381
12
...
865
19
...
551
27
...
547
46
...
181

0
...
501
1
...
510
2
...
042
4
...
116
6
...
229
8
...
381
10
...
683
16
...
562
21
...
073
34
...
213
63
...
250
0
...
005
1
...
020
3
...
079
5
...
180
7
...
322
9
...
506
12
...
158
19
...
093
28
...
869
48
...
479

0
...
501
1
...
511
2
...
045
4
...
126
6
...
247
8
...
417
10
...
745
16
...
714
22
...
390
34
...
150
64
...
250
0
...
005
1
...
020
3
...
081
5
...
184
7
...
329
9
...
517
12
...
183
19
...
140
28
...
986
49
...
872

EXAMPLE 4
...
She purchased Southwest stock for $6
...
14 per share
...
Help Janice understand exactly what she earned in terms of (a) effective annual rate and
(b) effective rate for quarterly compounding, and for monthly compounding
...

Solution
(a) The effective annual rate of return ia has a compounding period of 1 year, since the
stock purchase and sales dates are exactly 1 year apart
...
90 per share and using the deﬁnition of interest rate in Equation [1
...
24
ia ϭ ———————————— ϫ 100% ϭ —— ϫ 100% ϭ 90
...
90
(b) For the effective annual rates of 90
...
43%,
compounded monthly, apply Equation [4
...

Quarter:
Month:

m ϭ 4 times per year

1͞4

i ϭ (1
...
17472 Ϫ 1 ϭ 0
...
472% per quarter, compounded quarterly
...
9043) Ϫ 1 ϭ 1
...
05514
This is 5
...

Comment
Note that these quarterly and monthly rates are less than the effective annual rate divided by
the number of quarters or months per year
...
43%͞12 ϭ
7
...
This computation is incorrect because it neglects the fact that compounding
takes place 12 times during the year to result in the effective annual rate of 90
...

The spreadsheet function that displays the result of Equation [4
...
The format is

4
...
5]

Note that the rate entered in the EFFECT function is the nominal annual rate r% per year, not
the effective rate i% per compounding period
...
3]
...
25% per year, compounded
quarterly, and you want to ﬁnd the effective annual rate ia
...
25%,4)
to display ia ϭ 5
...
This is the spreadsheet equivalent of Equation [4
...
25͞4 ϭ 1
...

ia ϭ (1 ϩ 0
...
05354

(5
...

The NOMINAL spreadsheet function ﬁnds the nominal annual rate r
...
6]

This function is designed to display only nominal annual rates
...
For example, if the effective annual rate
is 10
...
381%,4) to display r ϭ 10% per year, compounded quarterly
...
1]
...
5%
...

EXAMPLE 4
...
24% per year, or 1
...
He will transfer
a balance of $1000 and plans to pay it and the transfer fee of $30, due at the end of the ﬁrst
month
...
Dave accepts the employment offer, and in his hurried, excited departure, he forgets to send the credit
card service company a change of address
...
2 to be $1041
...

(a) If this situation continues for a total of 12 months, determine the total due after 12 months
and the effective annual rate of interest Dave has accumulated
...
99% per year after one late
payment of the minimum payment amount, plus a late payment fee of $39 per occurrence
...

Solution
(a) Because Dave did not pay the ﬁrst month’s amount, the new balance of $1041
...
99%͞12 ϭ 2
...
The ﬁrst
3 months and last 2 months are detailed below
...

PE

Chapter 4

Nominal and Effective Interest Rates

ϭ SUM(B6:D6)
ϭ J4/12

ϭ K4/12

Month 1: transfer fee
Months 2-12: late-payment fee

Interest rate per month
Month 1: 1
...
499%

Figure 4–3
Monthly amounts due for a credit card, Example 4
...

Month 1:
Month 2:
Month 3:

1000 ϩ 1000(0
...
87
1041
...
87(0
...
91
1106
...
91(0
...
57

...
25 ϩ 1689
...
02499) ϩ 39 ϭ $1770
...
46 ϩ 1770
...
02499) ϩ 39 ϭ $1853
...
71 after 12 periods
...
71 ϭ 1000(F͞P, i,12) ϭ 1000(1 ϩ i)
1͞12

1 ϩ i ϭ (1
...
05278

i ϭ 5
...
3] to determine the effective
annual rate of 85
...

12

ia ϭ (1 ϩ i )m Ϫ 1 ϭ (1
...
85375

(85
...
24% (or
1
...
207% per year, compounded monthly
...
3], with a small
rounding error included,
ia ϭ (1 ϩ i)m Ϫ 1ϭ (1
...
15207
First, Dave will not pay at the stated rate of 14
...
207%
...
99% and (2) the monthly
fees of $39 for not making a payment
...
99% per year
...
207% to 85
...

Comment
This is but one illustration of why the best advice to an individual or company in debt is to
spend down the debt
...

4
...
3] is applied to ﬁnd ia the result is usually not an integer
...
There are
alternative ways to ﬁnd the factor value
...

• Use the spreadsheet function with ia (as discussed in Section 2
...

• Linearly interpolate between two tabulated rates (as discussed in Section 2
...

4
...
3] in Section 4
...
We can generalize this equation to determine the effective interest
rate for any time period (shorter or longer than 1 year)
...
7]

where i ‫ ؍‬effective rate for speciﬁed time period (say, semiannual)
r ‫ ؍‬nominal interest rate for same time period (semiannual)
m ‫ ؍‬number of times interest is compounded per stated time period (times per
6 months)
The term r͞m is always the effective interest rate over a compounding period CP, and m is
always the number of times that interest is compounded per the time period on the left of the
equals sign in Equation [4
...
Instead of ia, this general expression uses i as the symbol for the
effective interest rate, which conforms to the use of i throughout the remainder of this text
...
5 and 4
...

EXAMPLE 4
...
An engineer is on a
Tesla committee to evaluate bids for new-generation coordinate-measuring machinery to be
directly linked to the automated manufacturing of high-precision vehicle components
...
To get a clear understanding of ﬁnance costs, Tesla management asked the engineer to determine the effective
semiannual and annual interest rates for each bid
...
8% per year, compounded monthly

(a) Determine the effective rate for each bid on the basis of semiannual periods
...

(c) Which bid has the lowest effective annual rate?

Solution
(a) Convert the nominal rates to a semiannual basis, determine m, then use Equation [4
...
For bid 1,
r ϭ 9% per year ϭ 4
...
045 2
Effective i% per 6 months ϭ 1 ϩ ——— Ϫ 1 ϭ 1
...
55%
2
Table 4–4 (left section) summarizes the effective semiannual rates for all three bids
...
5
Semiannual Rates

Annual Rates

Bid

Nominal r per
6 Months, %

CP per
6 Months, m

Equation [4
...
7],
Effective i, %

1
2
3

4
...
0
4
...
55
6
...
48

9
12
8
...
31
12
...
16

(b) For the effective annual rate, the time basis in Equation [4
...
For bid 1,
r ϭ 9% per year

m ϭ 4 quarters per year

(

)

0
...
0931 Ϫ 1 ϭ 9
...

(c) Bid 3 includes the lowest effective annual rate of 9
...
48% when interest is compounded monthly
...
6
A dot-com company plans to place money in a new venture capital fund that currently returns
18% per year, compounded daily
...
7], with r ϭ 0
...

0
...
716%
Effective i% per year ϭ 1 ϩ ——
365
(b) Here r ϭ 0
...

0
...
415%
Effective i% per 6 months ϭ 1 ϩ ——
182

(

)

(

)

4
...

The payment period (PP) is the length of time between cash ﬂows (inﬂows or outﬂows)
...
It is important to determine if PP ϭ CP, PP Ͼ CP, or PP Ͻ CP
...
These time periods are
shown in Figure 4–4
...

r = nominal 8% per year, compounded semiannually
CP
6 months

Figure 4–4
One-year cash ﬂow diagram for a monthly payment period (PP) and
semiannual compounding
period (CP)
...
5

TA BLE 4–5 Section References for Equivalence Calculations Based on
Payment Period and Compounding Period Comparison
Length
of Time

Involves
Single Amounts
(P and F Only)

Involves Uniform Series
or Gradient Series
(A, G, or g)

PP ϭ CP
PP > CP
PP < CP

Section 4
...
5
Section 4
...
6
Section 4
...
7

As we learned earlier, to correctly perform equivalence calculations, an effective interest rate
is needed in the factors and spreadsheet functions
...
The next three sections (4
...
7) describe procedures to determine correct i and n values for engineering economy factors
and spreadsheet functions
...
Table 4–5 provides the section reference
...
5 and 4
...

A general principle to remember throughout these equivalence computations is that when
cash actually ﬂows, it is necessary to account for the time value of money
...
After 3
months there is no cash ﬂow and no need to determine the effect of quarterly compounding
...

4
...
In virtually
all situations, PP will be equal to or greater than CP
...
If the rate is 8% per year, for example, PP ϭ CP ϭ 1 year
...
The procedures to perform equivalence computations are the same for
both situations, as explained here
...
Method 1 is easier to apply, because the interest tables
in the back of the text can usually provide the factor value
...
For spreadsheets, either method is acceptable; however, method 1 is usually easier
...
The relations to calculate P and
F are
P ϭ F(P͞F, effective i% per CP, total number of periods n)

[4
...
9]

For example, assume that the stated credit card rate is nominal 15% per year, compounded
monthly
...
To ﬁnd P or F over a 2-year span, calculate the effective monthly
rate of 15%͞12 ϭ 1
...
Then 1
...

Any time period can be used to determine the effective interest rate; however, the interest rate
that is associated with the CP is typically the best because it is usually a whole number
...

Method 2: Determine the effective interest rate for the time period t of the nominal rate, and
set n equal to the total number of periods, using this same time period
...
8] and [4
...
For a credit card rate of 15% per year, compounded monthly, the
time period t is 1 year
...
15
Effective i% per year ϭ 1 ϩ ——
12

)

12

Ϫ 1 ϭ 16
...
25%,24) ϭ 0
...
076%,2) ϭ 0
...

EXAMPLE 4
...
The company sells compost produced by garbage-to-compost plants in
the United States and Vietnam
...
Find the
amount in the account now (after 10 years) at an interest rate of 12% per year, compounded
semiannually
...
Both methods are illustrated to ﬁnd F in year 10
...
There are n ϭ (2)(number of years) semiannual periods for each cash ﬂow
...
9] is
F ϭ 1000(F͞P,6%,20) ϩ 3000(F͞P,6%,12) ϩ 1500(F͞P,6%,8)
ϭ 1000(3
...
0122) ϩ 1500(1
...
634 million)

Express the effective annual rate, based on semiannual compounding
...
12 2
Effective i% per year ϭ 1 ϩ —— Ϫ 1 ϭ 12
...
Use the factor formula (F͞P,i,n) ϭ (1
...
9] to obtain the same answer as above
...
36%,10) ϩ 3000(F͞P,12
...
36%,4)
ϭ 1000(3
...
0122) ϩ 1500(1
...
634 million)
F=?

0

1

2

3

4

$1000

5

6

$1500
$3000

Figure 4–5
Cash ﬂow diagram, Example 4
...

7

8

9

10

Year

Equivalence Relations: Series with PP Ն CP

4
...
03%,24)

Cash Flow Series

Interest Rate

$500 semiannually
for 5 years
$75 monthly for
3 years
$180 quarterly for
15 years
$25 per month
increase for
4 years
$5000 per quarter
for 6 years

4
...
This also establishes the time unit of the effective interest rate
...
The
n value is the total number of quarters
...
This is a direct application of the following general guideline:
When cash ﬂows involve a series (i
...
, A, G, g) and the payment period equals or exceeds the
compounding period in length:
•
•

Find the effective i per payment period
...

In performing equivalence computations for series, only these values of i and n can be used in
interest tables, factor formulas, and spreadsheet functions
...

Table 4–6 shows the correct formulation for several cash ﬂow series and interest rates
...

EXAMPLE 4
...
What is the equivalent total amount after the last payment, if these funds are taken
from a pool that has been returning 8% per year, compounded quarterly?

Solution
The cash ﬂow diagram is shown in Figure 4–6
...
Applying the guideline, we need to determine an effective semiannual interest rate
...
7] with r ϭ 4% per 6-month
period and m ϭ 2 quarters per semiannual period
...
04 2
Effective i% per 6 months ϭ 1 ϩ —— Ϫ 1 ϭ 4
...
04%
...
04% seems reasonable, since we expect the effective rate to be slightly
higher than the nominal rate of 4% per 6-month period
...
The relation for F is
F ϭ A(F͞A,4
...
3422)
ϭ $9171
...
3422 using a spreadsheet, enter the FV function from
Figure 2–9, that is, ϭ ϪFV(4
...
Alternatively, the ﬁnal answer of $9171
...
04%,14,500)
...
8
...
9 Credit Card Offer Case

In our continuing credit card saga of Dave and his job transfer to Africa, let’s assume he did
remember that the total balance is $1030, including the $30 balance transfer fee, and he wants
to set up a monthly automatic checking account transfer to pay off the entire amount in 2 years
...
99% per year
...

Solution

The monthly A series is needed for a total of n ϭ 2(12) ϭ 24 payments
...
24%͞12 ϭ 1
...

Solution by hand:

Use a calculator or hand computation to determine the A͞P factor value
...
187%,24) ϭ 1030(0
...
57 per month for 24 months
Solution by Spreadsheet: Use the function ϭ ϪPMT(1
...
04813 to determine A for n ϭ 24 payments
...
187%,24,1030) to directly display the required monthly payment of A ϭ $Ϫ49
...

The effective annual interest rate or APY is computed using Equation [4
...
24%
per year, compounded monthly, and m ϭ 12 times per year
...
1424
Effective i per year ϭ 1 ϩ ———
12

)

12

– 1 ϭ 1
...
207% per year
This is the same effective annual rate ia determined in Example 4
...

Equivalence Relations: Series with PP Ն CP

4
...
10
The Scott and White Health Plan (SWHP) has purchased a robotized prescription fulﬁllment
system for faster and more accurate delivery to patients with stable, pill-form medication for
chronic health problems, such as diabetes, thyroid, and high blood pressure
...
The expected life is 10 years
...
Find this semiannual A value
both by hand and by spreadsheet, if capital funds are evaluated at 8% per year, using two different compounding periods:
Rate 1
...

Rate 2
...

Solution
Figure 4–7 shows the cash ﬂow diagram
...
This pattern makes the solution by hand quite involved if the P͞F factor, not the P͞A
factor, is used to ﬁnd P for the 10 annual $200,000 costs
...

Solution by hand—rate 1: Steps to ﬁnd the semiannual A value are summarized below:
PP ϭ CP at 6 months; ﬁnd the effective rate per semiannual period
...

Number of semiannual periods n ϭ 2(10) ϭ 20
...
, 20 periods because the costs are annual,
not semiannual
...

[⌺
20

P ϭ 3,000,000 ϩ 200,000

(P͞F,4%,k)

kϭ2,4

]

ϭ 3,000,000 ϩ 200,000(6
...

2

4

6

8

1

2

3

4

A per 6 months = ?
10
12
14

16

18

20

8

9

10

0

6 months
5

6

7

$200,000 per year
i1 = 8%, compounded semiannually
i2 = 8%, compounded monthly

P = $3 million

Figure 4–7
Cash ﬂow diagram with two different compounding periods, Example 4
...

Years

111

112

Chapter 4

Nominal and Effective Interest Rates

Solution by hand—rate 2: The PP is 6 months, but the CP is now monthly; therefore, PP Ͼ CP
...
7] is applied with
r ϭ 4% and m ϭ 6 months per semiannual period
...
04
Effective semiannual i ϭ 1 ϩ ———
6
P ϭ 3,000,000 ϩ 200,000

[

)

6

Ϫ 1 ϭ 4
...
067%,k)

kϭ2,4

]

ϭ 3,000,000 ϩ 200,000(6
...
067%,20) ϭ $320,064
Now, $320,064, or $1286 more semiannually, is required to cover the more frequent compounding of the 8% per year interest
...
067%
...

Solution by spreadsheet—rates 1 and 2: Figure 4–8 presents a general solution for the problem
at both rates
...
They continue the
cash ﬂow pattern of $200,000 every other 6 months through cell B32
...
This allows some
sensitivity analysis to be performed for different PP and CP values
...
This technique works well for spreadsheets once PP and CP are entered in the time unit of the CP
...
The ﬁnal A values in D14 ($318,784) and F14 ($320,069) are
the same (except for rounding) as those above
...
10
...
7 Equivalence Relations: Single Amounts
and Series with PP Ͻ CP
If a person deposits money each month into a savings account where interest is compounded
quarterly, do all the monthly deposits earn interest before the next quarterly compounding time?
If a person's credit card payment is due with interest on the 15th of the month, and if the full payment is made on the 1st, does the ﬁnancial institution reduce the interest owed, based on early
payment? The usual answers are no
...
7

compounded, bank loan were made early by a large corporation, the corporate ﬁnancial ofﬁcer
would likely insist that the bank reduce the amount of interest due, based on early payment
...
The timing of cash ﬂow transactions between compounding
points introduces the question of how interperiod compounding is handled
...

For a no-interperiod-interest policy, negative cash ﬂows (deposits or payments, depending on the
perspective used for cash ﬂows) are all regarded as made at the end of the compounding period,
and positive cash ﬂows (receipts or withdrawals) are all regarded as made at the beginning
...
This procedure can signiﬁcantly alter the distribution of cash
ﬂows before the effective quarterly rate is applied to ﬁnd P, F, or A
...
5 and 4
...
Example 4
...
Of course, noneconomic factors may be present
...
11
Last year AllStar Venture Capital agreed to invest funds in Clean Air Now (CAN), a start-up
company in Las Vegas that is an outgrowth of research conducted in mechanical engineering at
the University of Nevada–Las Vegas
...
The venture fund manager
generated the cash ﬂow diagram in Figure 4–9a in $1000 units from AllStar’s perspective
...
The receipts were unexpected this ﬁrst year; however, the product has great promise,
and advance orders have come from eastern U
...
plants anxious to become zero-emission
coal-fueled plants
...
How much is AllStar in the “red” at the end of the year?
Receipts from CAN
$90

$90

$120
$45
1

0
0

1

2

3

4

5

6

$75

8

9

10
$50

11

12

Month

$100

$150

7

Year

Payments to CAN
$200
(a)
$180

0

1
0

1

2

3

$165

2
4

5

6

3
7

8

9

4
10

11

12
$50

Quarter
Month

$150
$175
$200
F=?
(b)

Figure 4–9
(a) Actual and (b) moved cash ﬂows (in $1000) for quarterly compounding periods using no interperiod
interest, Example 4
...

113

114

Chapter 4

Nominal and Effective Interest Rates

Solution
With no interperiod interest considered, Figure 4–9b reﬂects the moved cash ﬂows
...
Calculate
the F value at 12%͞4 ϭ 3% per quarter
...

If PP Ͻ CP and interperiod compounding is earned, then the cash ﬂows are not moved, and the
equivalent P, F, or A values are determined using the effective interest rate per payment period
...
The effective interest rate formula will have an m value less than 1, because there is
only a fractional part of the CP within one PP
...
When the nominal rate is 12% per year, compounded
quarterly (the same as 3% per quarter, compounded quarterly), the effective rate per PP is
Effective weekly i% ϭ (1
...
228% per week

4
...

Continuous compounding is present when the duration of CP, the compounding period, becomes
inﬁnitely small and m, the number of times interest is compounded per period, becomes inﬁnite
...

As m approaches inﬁnity, the effective interest rate Equation [4
...
First, recall the deﬁnition of the natural logarithm base
...
71828ϩ
h
ϱ

h

[4
...
7] as m approaches inﬁnity is found by using r͞m ϭ 1͞h, which makes
m ϭ hr
...
11]

Equation [4
...
As an illustration, if the nominal annual r ϭ 15% per year, the effective
continuous rate per year is
i% ϭ e0
...
183%
For convenience, Table 4–3 includes effective continuous rates for the nominal rates listed
...
5] or [4
...
A value of 10,000 or higher provides sufﬁcient
accuracy
...
12
...
8

Effective Interest Rate for Continuous Compounding

EXAMPLE 4
...

(b) An investor requires an effective return of at least 15%
...
5%, or 0
...
By Equation [4
...
015 − 1 ϭ 1
...
18 per year is
i% per year ϭ er − 1 ϭ e0
...
722%
(b) Solve Equation [4
...

er − 1 ϭ 0
...
15
ln er ϭ ln 1
...
13976
Therefore, a rate of 13
...
The general formula to ﬁnd the nominal rate, given the effective
continuous rate i, is r ϭ ln(1 ϩ i)
...
5% and annual rate
r ϭ 18% with a large m to display effective i values
...
5%,10000)
ϭ EFFECT(18%,10000)

effective i ϭ 1
...
722% per year

(b) Use the function in Equation [4
...
976% per year, compounded continuously
...
13
Engineers Marci and Suzanne both invest $5000 for 10 years at 10% per year
...

Solution
Marci: For annual compounding the future worth is
F ϭ P(F͞P,10%,10) ϭ 5000(2
...
11], ﬁrst ﬁnd the effective i per year for use in the F͞P factor
...
10 Ϫ 1 ϭ 10
...
517%,10) ϭ 5000(2
...
For comparison, daily compounding yields an effective rate of 10
...
517% for continuous compounding
...
Examples of costs are energy and water costs, inventory costs, and labor costs
...
In these cases, the economic
analysis can be performed for continuous cash ﬂow (also called continuous funds ﬂow) and the
continuous compounding of interest as discussed above
...
In fact, the monetary differences for continuous cash ﬂows relative
to the discrete cash ﬂow and discrete compounding assumptions are usually not large
...

4
...
Loan rates may increase from one year to another
...
The mortgage rate is slightly adjusted annually to reﬂect the age of the loan, the current cost of mortgage
money, etc
...
If the variation in i is large, the equivalent values
will vary considerably from those calculated using the constant rate
...

To determine the P value for future cash ﬂow values (Ft) at different i values (it) for each year t,
we will assume annual compounding
...
Using standard notation and the P͞F factor,
P ‫ ؍‬F1(P͞F,i1,1) ؉ F2(P͞F,i1,1)(P͞F,i2,1) ؉
...
(P͞F,in,1)

[4
...
12] is the expression for the present worth of the future cash ﬂow
...
(P͞F,in,1)

[4
...
Since the equivalent P has been
determined numerically using the varying rates, this new equation will have only one unknown,
namely, A
...
14 illustrates this procedure
...
14
CE, Inc
...
The net proﬁt from the equipment for each of
the last 4 years has been decreasing, as shown below
...
The return has been increasing
...
Take the annual variation of rates of return into
account
...
Equation [4
...
Since for both years 1 and 2 the net proﬁt is $70,000 and the
annual rate is 7%, the P͞A factor can be used for these 2 years only
...
8080) ϩ 35(0
...
7284)](1000)
ϭ $172,816

[4
...
14
...
14], set it equal to P ϭ $172,816, and solve for A
...
See Figure 4–10 for the cash ﬂow diagram
transformation
...
8080) ϩ (0
...
7284)] ϭ A[3
...
25% is used, the result is A ϭ $52,467
...

When there is a cash ﬂow in year 0 and interest rates vary annually, this cash ﬂow must be
included to determine P
...
This is accomplished by
inserting the factor value for (P͞F,i0,0) into the relation for A
...
00
...
In this case, the
A value is determined using the F͞P factor, and the cash ﬂow in year n is accounted for by including the factor (F͞P,in,0) ϭ 1
...

CHAPTER SUMMARY
Since many real-world situations involve cash ﬂow frequencies and compounding periods other
than 1 year, it is necessary to use nominal and effective interest rates
...

r m
Effective i ‫1 ؊ ) — ؉ 1 ( ؍‬
m

Year

118

Chapter 4

Nominal and Effective Interest Rates

The m is the number of compounding periods (CP) per interest period
...

All engineering economy factors require the use of an effective interest rate
...
If only single amounts (P and F)
are present, there are several ways to perform equivalence calculations using the factors
...
This requires that the relative lengths of PP
and CP be considered as i and n are determined
...

From one year (or interest period) to the next, interest rates will vary
...

PROBLEMS
Nominal and Effective Rates
4
...

4
...

4
...

4
...

4
...
2% per year, compounded monthly; (d) effective
3
...

Given Interest Rate

Desired Interest Rate

1% per month
3% per quarter
2% per quarter
0
...
1% per 6 months

Nominal rate per year
Nominal rate per 6 months
Nominal rate per year
Nominal rate per quarter
Nominal rate per 2 years

4
...
5% per year, compounded monthly?
4
...

4
...
Determine the APY
...

4
...
87% compounded quarterly, determine (a) the effective
quarterly rate and (b) the nominal annual rate
...
6 Identify the following interest rate statements as
either nominal or effective: (a) 1
...
6% per month, compounded weekly; (e) 0
...

4
...
A less well-known company in
the chip business has been growing fast enough
that the company uses a minimum attractive rate of
return of 60% per year
...

4
...
(Assume 4 weeks͞month
...
13 An interest rate of 21% per year, compounded
every 4 months, is equivalent to what effective rate
per year? Show hand and spreadsheet solutions
...
14 An interest rate of 8% per 6 months, compounded
monthly, is equivalent to what effective rate per
quarter?
4
...
The owner was confused by the terminology and asked you to help
...
16 In ‘N Out Payday Loans advertises that for a fee of
only $10, you can immediately borrow up to $200
for one month
...
17 A government-required truth-in-lending document
showed that the APR was 21% and the APY was
22
...
Determine the compounding frequency at
which the two rates are equivalent
...
18 Julie has a low credit rating, plus she was furloughed from her job 2 months ago
...
Since she is a little
short on money to pay her rent, she decided to borrow $100 from a loan company, which will charge
her only $10 interest if the $110 is paid no more
than 1 week after the loan is made
...
19 Assume you deposit 25% of your monthly check
of $5500 into a savings account at a credit union
that compounds interest semiannually
...
20 Interest is compounded quarterly, and singlepayment cash ﬂows (that is, F and P) are separated
by 5 years
...
21 When interest is compounded quarterly and a
uniform series cash ﬂow of $4000 occurs every
6 months, what time periods on i and n must be
used?
4
...
If the replacement is expected to take

119

place in 3 years, how much will the company have
in its investment set-aside account? Assume the
company can achieve a rate of return of 12% per
year, compounded quarterly
...
23 Wheeling-Pittsburgh Steel is investigating whether
it should replace some of its basic oxygen furnace
equipment now or wait to do it later
...
e
...
How much can the company afford to
spend now, if its minimum attractive rate of return
is 1
...
24 How much can Wells Fargo lend to a developer
who will repay the loan by selling 6 view lots at
$190,000 each 2 years from now? Assume the
bank will lend at a nominal 14% per year, compounded semiannually
...
25 How much will be in a high-yield account at the
National Bank of Arizona 12 years from now if
you deposit $5000 now and $7000 ﬁve years from
now? The account earns interest at a rate of 8% per
year, compounded quarterly
...
26 Loadstar Sensors is a company that makes load͞
force sensors based on capacitive sensing technology
...
If the company
has already set aside $12 million in an investment
account for the expansion, how much more must
the company add to the account next year (i
...
,
1 year from now) so that it will have the $28 million 4 years from now? The account earns interest
at 12% per year, compounded quarterly
...
27 A structural engineering consulting company is
examining its cash ﬂow requirements for the next
6 years
...
Speciﬁcally, the company expects to spend $21,000 two
years from now, $24,000 three years from now,
and $10,000 ﬁve years from now
...
28 Irvin Aerospace of Santa Ana, California, was
awarded a 5-year contract to develop an advanced
space capsule airbag landing attenuation system
for NASA’s Langley Research Center
...
What is the present worth of the
contract at 16% per year, compounded quarterly, if
the quarterly cost in years 1 through 5 is $2 million
per quarter?

120

Chapter 4

4
...
(A drill
collar is the heavy tubular connection between a
drill pipe and a drill bit
...
30 In 2010, the National Highway Trafﬁc Safety Administration raised the average fuel efﬁciency standard to 35
...
The rules will cost
consumers an average of $926 extra per vehicle in
the 2016 model year
...
How much will the monthly savings in the
cost of gasoline have to be to recover Yolanda’s
extra cost? Use an interest rate of 0
...

4
...
S
...
3 million of the $87 million capital cost for a
desalting plant constructed and operated by El Paso
Water Utilities (EPWU)
...
85 per 1000 gallons
for 20 years
...
3 million capital cost
is amortized at an interest rate of 6% per year,
compounded monthly
...
32 Beginning in 2011, city hall, administrative ofﬁces, and municipal courts in the city of El Paso,
Texas, will go on a 10 hour͞day, 4-day workweek
from the beginning of May through the end of September
...
If this work
schedule continues for the next 10 years, what is
the future worth of the savings at the end of that
time (i
...
, end of year 2020)? Use an interest rate
of 0
...

4
...

Prices ranged from $999 for steel models such as
Dad Remembered to $3199 for the Sienna Bronze
casket
...
An individual purchased a Sienna
Bronze casket and made 12 equal monthly payments (in months 1 through 12) at no interest
...
34 NRG Energy plans to construct a giant solar plant
in Santa Teresa, New Mexico, to supply electricity
to West Texas Electric
...
It will provide enough
power for 30,000 homes
...
35 Many college students have Visa credit cards that
carry an interest rate of “simple 24% per year”
(that is, 2% per month)
...
25
...
25 per month and
adds no other charges to the card?
4
...
S
...
After working for a year or so,
he found himself in ﬁnancial trouble, and he borrowed $500 from a friend in the ﬁnance department at his ofﬁce
...
The two
got separated doing different jobs, and 1 year went
by
...

(a) What does Bart now owe his friend? (b) What
effective annual interest rate did Bart pay on this
$500 loan?
4
...
The
fee will be reduced by $10 each month of the
2-year contract
...
Assume AT&T pays $499 for an
iPhone that it sells for $199
...
e
...
5% per month on its $300 investment in a customer who terminates the contract
after 12 months?
4
...
5 years from now at an
interest rate of 2% per month?

Problems

4
...
The
company plans to borrow $3
...
How
much will the company have to get in a lump-sum
payment when the project is over in order to earn
24% per year, compounded quarterly, on its investment?
4
...

Year

1

2

3

4

5

Cash Flow, $

200,000

0

350,000

0

400,000

4
...

F=?
i = 1% per month

0

1

2

3

4

5

$30

$30

$30

$30

$30

6

$50

$50

8 Years

7

$30

$50

121

Lee antiscalant) for use at its nanoﬁltration water
conditioning plant
...
If the chemical cost is $11 per day,
determine the equivalent cost per month at an interest rate of 12% per year, compounded monthly
...

4
...
S
...
What is
the future worth of the income (after the 2½ years)
at an interest rate of 6% per year, compounded
quarterly? Assume there is no interperiod compounding
...
47 The Autocar E3 refuse truck has an energy recovery system developed by Parker Hanniﬁn
LLC that is expected to reduce fuel consumption
by 50%
...

(The truck recharges the accumulators when it
brakes
...
How much can a private wastehauling company afford to spend now on the
recovery system, if it wants to recover its investment in 3 years plus a return of 14% per year,
compounded semiannually? Assume no interperiod compounding
...
42 According to the Government Accountability Ofﬁce (GAO), if the U
...
Postal Service does not
change its business model, it will lose $480 million
next month and $500 million the month after that,
and the losses will increase by $20 million per
month for the next 10 years
...
25% per month, what is the equivalent uniform
amount per month of the losses through year 10?

Continuous Compounding

4
...
What is the present worth of
the maintenance costs at an interest rate of 10%
per year, compounded quarterly?

4
...
3% per month, compounded continuously?

Equivalence When PP Ͻ CP
4
...

4
...
48 What effective interest rate per year is equal to
1
...

4
...
6% per month, compounded continuously?

4
...
If the company wants
to make an effective 25% per year, compounded
continuously, what nominal daily rate of return has
to be realized? Assume 365 days per year
...
52 U
...
Steel is planning a plant expansion that is expected to cost $13 million
...

122

Chapter 4

Nominal and Effective Interest Rates

4
...
What is the present worth of the costs at
an interest rate of 10% per year, compounded
continuously?

nanowires can be extruded by blasting the carbon
nanotubes with an electron beam
...
7 million in year 1,
$2
...
4 million in year 3 to
develop the technology, determine the present
worth of the investments in year 0, if the interest
rate in year 1 is 10% and in years 2 and 3 it is 12%
per year
...
54 Many small companies use accounts receivable as
collateral to borrow money for continuing operations and meeting payrolls
...
25% per month after
4 months, how much will the company owe at the
end of 1 year?

4
...

P=?

i = 10%

4
...
If the interest rate was 8%
per year for the ﬁrst 3 years and then it increased to
10% in years 4 and 5, what is the equivalent future
worth (in year 5) of the maintenance cost? Show
hand and spreadsheet solutions
...
56 By ﬁlling carbon nanotubes with miniscule wires
made of iron and iron carbide, incredibly thin

6

$160

ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
4
...
59 An interest rate is an effective rate under all of the
following conditions, except when:
(a) The compounding period is not stated
(b) The interest period and compounding period
are the same
(c) The interest statement says that the interest
rate is effective
(d) The interest period is shorter than the compounding period
4
...
61 An interest rate of 1
...
5% per month
(b) 4
...
0% per quarter, compounded continuously
9% per 6 months

4
...
68% per year, compounded monthly,
is the closest to:
(a) 12% per year
(b) 12% per year, compounded annually
(c) 1% per month
(d) 1% per month, compounded annually
4
...
02%
(b) 12
...
31%
(d) 12
...
64 If you make quarterly deposits for 3 years into an
account that compounds interest at 1% per month,
the value of n in the F͞A factor that will determine
F at the end of the 3-year period is:
(a) 3
(b) 12
(c) 36
(d) None of these

123

Additional Problems and FE Exam Review Questions

4
...

1
...
5% per quarter, compounded
monthly
(a) Both are nominal rates
...

(c) Rate 1 is effective and rate 2 is nominal
...

4
...
e
...

(b) The compounding period is equal to the payment period
...

(d) The compounding period is longer than the
payment period
...
67 An engineer who is saving for her retirement plans
to deposit $500 every quarter, starting one quarter
from now, into an investment account
...
68 A company that makes ﬂange-mount, motorized rotary potentiometers expects to spend $50,000 for a
certain machine 4 years from now
...
69 For the cash ﬂow diagram shown, the unit of the
payment period (PP) is:
(a) Months
(b) Quarters
(c) Semiannual (d) Years

F=?
i = 1% per month

0

1

2

3

4

5

6

7

8

1

9

10

11

2

12 Quarter
3

Year

$400 $400
$420
$440
$460
$480
$500

4
...
At an interest rate of effective
10% per year, compounded semiannually, the equation that represents the equivalent annual worth A in
years 1 through 5 is:
(a) A ϭ 10,000(P͞F,10%,2)(A͞P,10%,5)
ϩ 10,000(A͞F,10%,5)
(b) A ϭ 10,000(A͞P,10%,4) ϩ 10,000(A͞F,10%,5)
(c) A ϭ 10,000(P͞F,5%,2)(A͞P,5%,10)
ϩ 10,000(A͞F,5%,10)
(d) A ϭ [10,000(F͞P,10%,5)
ϩ 10,000](A͞F,10%,5)
4
...
If you
want to know the total after 4 years, the value of n
you should use in the F͞A factor is:
(a) 2
(b) 4
(c) 8
(d) 12
4
...
At an interest rate of 14% per year,
compounded semiannually, the uniform amount
that must be deposited into a sinking fund every
6 months is closest to:
(a) $21,335
(b) $24,825
(c) $27,950
(d) $97,995

124

Chapter 4

Nominal and Effective Interest Rates

CASE STUDY
IS OWNING A HOME A NET GAIN OR NET LOSS OVER TIME?
Background
The Carroltons are deliberating whether to purchase a house
or continue to rent for the next 10 years
...
Plus, the high school
that their children attend is very good for their college prep
education, and they all like the neighborhood where they
live now
...

If the Carroltons do not buy a house, they will continue
to rent the house they currently occupy for $2700 per
month
...
Additionally, they will add to this investment at the
end of each year the same amount as the monthly 15-year
mortgage payments
...

Information
Two ﬁnancing plans using ﬁxed-rate mortgages are currently
available
...

Plan

Description

A

30-year ﬁxed rate of 5
...
0% per year interest;
10% down payment

B

Other information:
• Price of the house is $330,000
...

• Up-front fees (origination fee, survey fee, attorney’s fee,
etc
...

Any money not spent on the down payment or monthly payment will be invested and return at a rate of 6% per year
(0
...

The Carroltons anticipate selling the house after 10 years
and plan for a 10% increase in price, that is, $363,000 (after
all selling expenses are paid)

Case Study Exercises
1
...
No taxes are considered on proceeds from the
savings or investments
...
The Carroltons
decided to use the largest future worth after 10 years to
select the best of the plans
...

Plan A analysis: 30-year ﬁxed-rate loan
Amount of money required for closing costs:
Down payment (10% of $330,000)
Up-front fees (origination fee, attorney’s
fee, survey, ﬁling fee, etc
...
25%͞12 ϭ 0
...

A ϭ 297,000(A͞P,0
...
005522)
ϭ $1640
Add the T&I of $500 for a total monthly payment of
PaymentA ϭ $2140 per month
The future worth of plan A is the sum of three future
worth components: remainder of the $40,000 available
for the closing costs (F1A); left-over money from that
available for monthly payments (F2A); and increase in
the house value when it is sold after 10 years (F3A)
...

F1A ϭ (40,000 Ϫ 36,000)(F͞P,0
...
5%,120)
ϭ $116,354

Case Study

Net money from the sale in 10 years (F3A) is the difference between the net selling price ($363,000) and the
remaining balance on the loan
...
4375%,120)
Ϫ 1640(F͞A,0
...
6885) Ϫ 1640(157
...
Perform this analysis if all estimates remain the same,
except that when the house sells 10 years after purchase,
the bottom has fallen out of the housing market and the
net selling price is only 70% of the purchase price, that
is, $231,000
...
An engineering economic
analysis evaluates cash ﬂow estimates for parameters such as initial
cost, annual costs and revenues, nonrecurring costs, and possible
salvage value over an estimated useful life of the product; process,
or service
...

After completing these chapters, you will be able to evaluate
most engineering project proposals using a well-accepted economic
analysis technique, such as present worth, future worth, capitalized
cost, life-cycle costing, annual worth, rate of return, or beneﬁt /cost
analysis
...

Important note: If depreciation and͞or after-tax analysis is to be
considered along with the evaluation methods in Chapters 5 through
9, Chapter 16 and/or Chapter 17 should be covered, preferably after
Chapter 6
...

SECTION

TOPIC

LEARNING OUTCOME

5
...

5
...

5
...

5
...

5
...

A

future amount of money converted to its equivalent value now has a present worth
(PW) that is always less than that of the future cash ﬂow, because all P͞F factors have
a value less than 1
...
For this reason, present
worth values are often referred to as discounted cash ﬂows (DCF), and the interest rate is referred to as the discount rate
...
Up to this point, present worth computations have been
made for one project or alternative
...
Two additional applications are covered here—future worth and capitalized cost
...

To understand how to organize an economic analysis, this chapter begins with a description of independent and mutually exclusive projects as well as revenue and cost
alternatives
...
This industry produces the microchips used in many
of the communication, entertainment,
transportation, and computing devices
we use every day
...
Ultrapure water is
obtained by special processes that commonly include reverse osmosis͞deionizing
resin bed technologies
...

A fab costs upward of $2
...

A newcomer to the industry, Angular
Enterprises, has estimated the cost proﬁles for two options to supply its anticipated fab with water
...
The initial cost estimates for the UPW system are
given below
...
5

Ϫ0
...

Life of UPW equipment 10 years
UPW needs

1500 gpm

Operating time

16 hours per
day for 250 days
per year

This case is used in the following topics
(Sections) and problems of this chapter:
PW analysis of equal-life alternatives
(Section 5
...
3)
Capitalized cost analysis (Section 5
...
20 and 5
...
1 Formulating Alternatives
The evaluation and selection of economic proposals require cash ﬂow estimates over a stated
period of time, mathematical techniques to calculate the measure of worth (review Example 1
...
From all the

Chapter 5

Present Worth Analysis

Mandates
n
tio

ea

s

ma

or
Inf

Id

130

Experience

Estimates
Plans

Proposals
Not viable

1

B

Not viable
2

D

m

C

E

Either
of these

Mutually
exclusive
alternatives
1
2
m
+
DN

Nature
of proposals

Independent
projects

Select
all
justified

DN
1
2
…

…

Select
only
one

…

A

Viable

DN = do nothing

m

Types of cash flow estimates
* Costs only

* Revenues and costs
Revenue alternative

Cost alternative

Perform evaluation and make selection

Figure 5–1
Progression from proposals to economic evaluation to selection
...
This progression is detailed in Figure 5–1
...
Once the obviously nonviable ideas
are eliminated, the remaining viable proposals are ﬂeshed out to form the alternatives to be
evaluated
...

The nature of the economic proposals is always one of two types:
Mutually exclusive alternatives: Only one of the proposals can be selected
...

Independent projects: More than one proposal can be selected
...

The do-nothing (DN) proposal is usually understood to be an option when the evaluation is
performed
...
2

131

Present Worth Analysis of Equal-Life Alternatives

The DN alternative or project means that the current approach is maintained; nothing new
is initiated
...

Do nothing

If it is absolutely required that one or more of the deﬁned alternatives be selected, do nothing is
not considered
...

Mutually exclusive alternatives and independent projects are selected in completely different ways
...
Only one is chosen, and the
rest are rejected
...
For independent projects one, two or more, in
fact, all of the projects that are economically justiﬁed can be accepted, provided capital funds
are available
...

Independent projects are evaluated one at a time and compete only with the DN project
...
When performed correctly as described in each chapter, any
of the techniques will reach the same conclusion of which alternative or alternatives to select
...

A parallel can be developed between independent and mutually exclusive evaluation
...
Zero, one, two, or more may be selected
...
This number includes the DN alternative, as shown in Figure 5–1
...
Commonly, in real-world applications, there
are restrictions, such as an upper budgetary limit, that eliminate many of the 2m alternatives
...

Chapter 12 treats independent projects with a budget limitation; this is called capital budgeting
...
Cash ﬂow estimates determine whether the alternatives are revenue- or cost-based
...
Deﬁnitions for these types follow:
Revenue: Each alternative generates cost (cash outﬂow) and revenue (cash inﬂow) estimates,
and possibly savings, also considered cash inﬂows
...

Cost: Each alternative has only cost cash ﬂow estimates
...
These are
also referred to as service alternatives
...
Differences in evaluation methodology are
detailed in each chapter
...
2 Present Worth Analysis of Equal-Life Alternatives
The PW comparison of alternatives with equal lives is straightforward
...
The present worth method is quite popular in industry because all
future costs and revenues are transformed to equivalent monetary units NOW; that is, all future
cash ﬂows are converted (discounted) to present amounts (e
...
, dollars) at a speciﬁc rate of return,
which is the MARR
...
The required conditions and evaluation procedure are as follows:
If the alternatives have the same capacities for the same time period (life), the equal-service
requirement is met
...

Equal-service
requirement

132

Chapter 5

Present Worth Analysis

For mutually exclusive (ME) alternatives, whether they are revenue or cost alternatives, the following guidelines are applied to justify a single project or to select one from several alternatives
...

Two or more alternatives: Select the alternative with the PW that is numerically largest,
that is, less negative or more positive
...

Note that the guideline to select one alternative with the lowest cost or highest revenue uses
the criterion of numerically largest
...
The selections below correctly apply the guideline for two alternatives A and B
...
The selection guideline is as follows:
Independent project
selection

One or more independent projects: Select all projects with PW Ն 0 at the MARR
...

All PW analyses require a MARR for use as the i value in the PW relations
...

EXAMPLE 5
...
During lab research, three equal-service machines need to be evaluated
economically
...
The MARR is
10% per year
...
The salvage values are considered a “negative” cost, so a ϩ sign
precedes them
...
)
The PW of each machine is calculated at i ϭ 10% for n ϭ 8 years
...

PWE ϭ Ϫ4500 Ϫ 900(P͞A,10%,8) ϩ 200(P͞F,10%,8) ϭ $Ϫ9208
PWG ϭ Ϫ3500 Ϫ 700(P͞A,10%,8) ϩ 350(P͞F,10%,8) ϭ $Ϫ7071
PWS ϭ Ϫ6000 Ϫ 50(P͞A,10%,8) ϩ 100(P͞F,10%,8) ϭ $Ϫ6220
The solar-powered machine is selected since the PW of its costs is the lowest; it has the
numerically largest PW value
...
3

133

Present Worth Analysis of Different-Life Alternatives

EXAMPLE 5
...
With the options of seawater or groundwater sources,
it is a good idea to determine if one system is more economical than the other
...

Solution
An important ﬁrst calculation is the cost of UPW per year
...
44 M per year
Groundwater: (5͞1000)(1500)(60)(16)(250) ϭ $1
...
In $1 million units:
PW relation: PW ϭ ﬁrst cost Ϫ PW of AOC Ϫ PW of UPW ϩ PW of salvage value
PWS ϭ Ϫ20 Ϫ 0
...
44(P͞A,12%,10) ϩ 0
...
5(5
...
44(5
...
3220)
ϭ $Ϫ30
...
3(P͞A,12%,10) Ϫ 1
...
10(22)(P͞F,12%,10)
ϭ Ϫ22 Ϫ 0
...
6502) Ϫ 1
...
6502) ϩ 2
...
3220)
ϭ $Ϫ33
...
52 M
...
3 Present Worth Analysis of Different-Life
Alternatives
When the present worth method is used to compare mutually exclusive alternatives that have
different lives, the equal-service requirement must be met
...
2 is followed, with one exception:
The PW of the alternatives must be compared over the same number of years and must end
at the same time to satisfy the equal-service requirement
...
A fair comparison requires that PW values represent
cash ﬂows associated with equal service
...
The equal-service requirement is
satisﬁed by using either of two approaches:
LCM: Compare the PW of alternatives over a period of time equal to the least common
multiple (LCM) of their estimated lives
...

This approach does not necessarily consider the useful life of an alternative
...

For either approach, calculate the PW at the MARR and use the same selection guideline as
that for equal-life alternatives
...
For example, lives of 3 and 4 years are compared over a 12-year period
...
Additionally, the LCM approach requires that some assumptions be made about
subsequent life cycles
...
The service provided will be needed over the entire LCM years or more
...
The selected alternative can be repeated over each life cycle of the LCM in exactly the same
manner
...
Cash ﬂow estimates are the same for each life cycle
...
If the cash ﬂows are expected to change by any other rate, then the PW analysis must
be conducted using constant-value dollars, which considers inﬂation (Chapter 14)
...
For the study period approach, a time horizon is chosen over
which the economic analysis is conducted, and only those cash ﬂows which occur during that time
period are considered relevant to the analysis
...
An estimated market value at the end of the study period must be made
...
The
study period approach is often used in replacement analysis (Chapter 11)
...

EXAMPLE 5
...
, plans to purchase new cut-and-ﬁnish equipment
...

Vendor A
First cost, $
Annual M&O cost, $ per year
Salvage value, $
Life, years

Vendor B

Ϫ15,000
Ϫ3,500
1,000
6

Ϫ18,000
Ϫ3,100
2,000
9

(a) Determine which vendor should be selected on the basis of a present worth comparison,
if the MARR is 15% per year
...
If a study period of 5 years is used and the salvage values are not expected to
change, which vendor should be selected?

Solution
(a) Since the equipment has different lives, compare them over the LCM of 18 years
...
These are years 6 and 12 for vendor A and year 9 for B
...
Calculate PW at 15% over 18 years
...
3

Present Worth Analysis of Different-Life Alternatives

135

PWA = ?
$1000
1

2

$1000

6

12

$1000
16

17 18

Year

$3500
$15,000

$15,000

$15,000
Vendor A

PWB = ?
$2000
1

2

$2000
16

9

17 18

Year

$3100
$18,000

$18,000
Vendor B

Figure 5–2
Cash ﬂow diagram for different-life alternatives, Example 5
...

Vendor B is selected, since it costs less in PW terms; that is, the PWB value is numerically larger than PWA
...
The PW analysis is
PWA ϭ Ϫ15,000 Ϫ 3500(P͞A,15%,5) ϩ 1000(P͞F,15%,5)
ϭ $Ϫ26,236
PWB ϭ Ϫ18,000 Ϫ 3100(P͞A,15%,5) ϩ 2000(P͞F,15%,5)
ϭ $Ϫ27,397
Vendor A is now selected based on its smaller PW value
...
In situations such as
this, the standard practice of using a ﬁxed study period should be carefully examined to
ensure that the appropriate approach, that is, LCM or ﬁxed study period, is used to satisfy
the equal-service requirement
...
4 Water for Semiconductor Manufacturing Case
When we discussed this case in the introduction, we learned that the initial estimates of equipment life were 10 years for both options of UPW (ultrapure water)—seawater and groundwater
...
However, it is expected that, instead of complete replacement, a total refurbishment of the equipment for $10 M after 5 years will extend the life through
the anticipated 10th year of service
...
2
...

PE

136

Chapter 5

Present Worth Analysis

Seawater option
ϭ 〈OC ϩ UPW cost/year
ϭ Ϫ0
...
44
Groundwater option
ϭ 〈OC ϩ UPW cost/year
ϭ Ϫ0
...
80
Both options
Salvage values
included here

Salvage values are the
same after 5 and 10 years

Figure 5–3
PW analyses using LCM and study period approaches for water for semiconductor manufacturing case,
Example 5
...

Solution
A spreadsheet and the NPV function are a quick and easy way to perform this dual analysis
...

LCM of 10 years: In the top part of the spreadsheet, the LCM of 10 years is necessary to satisfy the equal-service requirement; however, the ﬁrst cost in year 5 is the refurbishment cost of
$Ϫ10 M, not the $Ϫ20 M expended in year 0
...
94 M in year 5 accounts for the continuing AOC and annual UPW cost of
$Ϫ1
...
The NPV functions shown on the spreadsheet
determine the 12% per year PW values in $1 million units
...
31

PWG ϭ $Ϫ33
...

Study period of 5 years: The lower portion of Figure 5–3 details a PW analysis using the
second approach to evaluating different-life alternatives, that is, a speciﬁc study period, which
is 5 years in this case study
...

Again the economic decision is reversed as the 12% per year PW values favor the seawater
option
...
43

PWG ϭ $Ϫ28
...
Both are correct answers given the decision of how the equal-service requirement is met
...

5
...
The PW values are
Option S:
Option G:

n ϭ 5 years, PWS ϭ $Ϫ26
...

n ϭ 10 years, PWG ϭ $Ϫ33
...

For independent projects, use of the LCM approach is unnecessary since each project is compared to the do-nothing alternative, not to each other, and satisfying the equal-service requirement is not a problem
...

5
...
The n value in the F͞P
factor is either the LCM value or a speciﬁed study period
...

Future worth analysis over a speciﬁed study period is often utilized if the asset (equipment, a
building, etc
...
Suppose
an entrepreneur is planning to buy a company and expects to trade it within 3 years
...
Example 5
...
Another excellent application of FW analysis is for projects that will come
online at the end of a multiyear investment period, such as electric generation facilities, toll roads,
airports, and the like
...

The selection guidelines for FW analysis are the same as for PW analysis; FW Ն 0 means the
MARR is met or exceeded
...

EXAMPLE 5
...
There was a net loss of £10 million at the end of year 1 of ownership
...
This means that breakeven
net cash ﬂow was achieved this year
...

(a) The British conglomerate has just been offered £159
...
Use FW analysis to determine if the MARR will be
realized at this selling price
...
5 million
...

FW3 ϭ Ϫ75(F͞P,25%,3) Ϫ 10(F͞P,25%,2) Ϫ 5(F͞P,25%,1) ϩ 159
...
36 ϩ 159
...
86 million
No, the MARR of 25% will not be realized if the £159
...

ME alternative
selection

138

Chapter 5

Present Worth Analysis
FW = ?

£159
...
5
...

(b) Determine the future worth 5 years from now at 25% per year
...
The A͞G and F͞A factors are applied to the arithmetic gradient
...
81 million
The offer must be for at least £246
...
This is approximately
3
...

5
...
A perpetual
or inﬁnite life is the effective planning horizon
...
The economic worth of these types of projects or
endowments is evaluated using the present worth of the cash ﬂows
...

The formula to calculate CC is derived from the PW relation P ϭ A(P͞A,i%,n), where n ϭ ϱ
time periods
...
We replace the symbols P and PW with
CC as a reminder that this is a capitalized cost equivalence
...
1]

Solving for A or AW, the amount of new money that is generated each year by a capitalization
of an amount CC is
AW ‫ ؍‬CC (i)

[5
...
Equation [5
...
If $20,000 is invested now (this is the

5
...
This leaves the
original $20,000 to earn interest so that another $2000 will be accumulated the next year
...
An annual operating cost of $50,000
and a rework cost estimated at $40,000 every 12 years are examples of recurring cash ﬂows
...

The procedure to determine the CC for an inﬁnite sequence of cash ﬂows is as follows:
1
...

2
...
This is their CC value
...
Find the A value through one life cycle of all recurring amounts
...
) Add this to all other uniform
amounts (A) occurring in years 1 through inﬁnity
...

4
...
This is an
application of Equation [5
...

5
...

Drawing the cash ﬂow diagram (step 1) is more important in CC calculations than elsewhere,
because it helps separate nonrecurring and recurring amounts
...

EXAMPLE 5
...
The director wants to know the total equivalent cost of all future costs incurred to purchase the software system
...

The system has an installed cost of $150,000 and an additional cost of $50,000 after
10 years
...
In addition, there is expected to be a recurring major upgrade cost of $15,000
every 13 years
...

Solution
(a) The ﬁve-step procedure to ﬁnd CC now is applied
...
Draw a cash ﬂow diagram for two cycles (Figure 5–5)
...
Find the present worth of the nonrecurring costs of $150,000 now and $50,000 in year
10 at i ϭ 5%
...

CC1 ϭ Ϫ150,000 Ϫ 50,000(P͞F,5%,10) ϭ $Ϫ180,695
0

2

4

6

8

10

12

14

20

26

$5000
$8000
$15,000

i = 5% per year
$50,000

$150,000

Figure 5–5
Cash ﬂows for two cycles of recurring costs and all nonrecurring amounts, Example 5
...

$15,000

Year

139

140

Chapter 5

Present Worth Analysis

3 and 4
...
1]
...
05 ϭ $Ϫ16,940
There are several ways to convert the annual software maintenance cost series to A
and CC values
...
05 ϭ $Ϫ100,000
Second, convert the step-up maintenance cost series of $−3000 to a capitalized cost
CC4 in year 4, and ﬁnd the present worth in year 0
...
)
Ϫ3,000
CC4 ϭ ———— (P͞F,5%,4) ϭ $Ϫ49,362
0
...
The total capitalized cost CCT for Haverty County Transportation Authority is the sum
of the four component CC values
...
2] determines the AW value forever
...
05) ϭ $17,350
Correctly interpreted, this means Haverty County ofﬁcials have committed the equivalent
of $17,350 forever to operate and maintain the toll management software
...
Since the capitalized cost represents the
total present worth of ﬁnancing and maintaining a given alternative forever, the alternatives will
automatically be compared for the same number of years (i
...
, inﬁnity)
...
This evaluation is illustrated in
Example 5
...

EXAMPLE 5
...
4) to the point that the life of the seawater option
can be extended to 10 years with a major refurbishment cost after 5 years
...
In $1 million units, the
estimates and PW values (from Figure 5–3) are as follows:
Seawater: PS ϭ $Ϫ20; AOCS ϭ $Ϫ1
...
05(20) ϭ $1
...
31
Groundwater: PG ϭ $Ϫ22; AOCG ϭ $Ϫ2
...
10(22) ϭ $2
...
16
If we assume that the UPW (ultrapure water) requirement will continue for the foreseeable
future, a good number to know is the present worth of the long-term options at the selected
MARR of 12% per year
...
Select the option with the lower CC
...

5
...
31(0
...
43
CCS ϭ Ϫ6
...
12 ϭ $Ϫ53
...
16(0
...
87
CCG ϭ Ϫ5
...
12 ϭ $Ϫ48
...

Comment
If the seawater-life extension is not considered a viable option, the original alternative of
5 years could be used in this analysis
...

AS,5 years ϭ Ϫ20(A͞P,12%,5) Ϫ 1
...
05(20)(A͞F,12%,5)
ϭ $Ϫ7
...
33/0
...
08
Now, the economic advantage of the groundwater option is even larger
...
To determine capitalized cost for the ﬁnite life alternative,
calculate the equivalent A value for one life cycle and divide by the interest rate (Equation [5
...

This procedure is illustrated in Example 5
...

EXAMPLE 5
...
The goal is zero landﬁll by 2020
...
The interest rate for state-mandated projects
is 5% per year
...
No contract period is stated; thus the contract and services are offered for as long as the State needs them
...
Expected life of the equipment is 5 years
with no salvage value
...

(b) Determine the maximum number of sites at which the equipment can be purchased and
still have a capitalized cost less than that of the contractor option
...
The contract, as proposed, has a long life
...
The annual charge of A ϭ $25,000
is divided by i ϭ 0
...
Summing the two values results in
CCC ϭ $Ϫ8
...

For the ﬁnite, 5-year purchase alternative, column B shows the ﬁrst cost ($Ϫ275,000 per
site), AOC ($Ϫ12,000), and equivalent A value of $−755,181, which is determined via
the PMT function (cell tag)
...
1 million
...

(b) A quick way to ﬁnd the maximum number of sites for which CCP Ͻ CCC is to use Excel’s Goal Seek tool, introduced in Chapter 2, Example 2
...
(See Appendix A for details
on how to use this tool
...
The result, shown in column C, indicates that 5
...
Since the number of sites

141

142

Chapter 5

Present Worth Analysis

ϭ PMT($B$1,B13,ϪB12*B4) ϩ B14*B4

Figure 5–6
Spreadsheet solution of Example 5
...

must be an integer, 5 or fewer sites will favor purchasing the equipment and 6 or more
sites will favor contracting the separation services
...
By the way, another way to determine the number of sites is by trial
and error
...

CHAPTER SUMMARY
The present worth method of comparing alternatives involves converting all cash ﬂows to present
dollars at the MARR
...
When the alternatives have different lives, the comparison must be made for equal-service
periods
...
Both approaches compare alternatives in accordance with the equal-service requirement
...

If the life of the alternatives is considered to be very long or inﬁnite, capitalized cost is the
comparison method
...

PROBLEMS
Types of Projects
5
...
2 (a)
(b)

What is meant by the do-nothing alternative?
When is the do-nothing alternative not an
option?

5
...

5
...
11

5
...
6 What two approaches can be used to satisfy the
equal-service requirement?
Alternative Comparison—Equal Lives
5
...
Which one should be selected on the
basis of a present worth analysis at 10% per year?
In-house
First cost, $
Annual cost, $ per year
Annual income, $ per year
Salvage value, $
Life, years

Contract

Ϫ30
Ϫ5
14
2
5

0
Ϫ2
3
...
8 The manager of a canned food processing plant
must decide between two different labeling machines
...
Machine B will cost $51,000 to buy
and will have an annual operating cost of $17,000
during its 4-year life
...
9 A metallurgical engineer is considering two materials for use in a space vehicle
...
(a) Which should be selected on the basis of
a present worth comparison at an interest rate of
12% per year? (b) At what ﬁrst cost for the material not selected above will it become the more
economic alternative?
Material X
First cost, $
Maintenance cost, $ per year
Salvage value, $
Life, years

Material Y

Ϫ15,000
Ϫ9,000
2,000
5

Ϫ35,000
Ϫ7,000
20,000
5

5
...
In one particular year, the company offered 1000 shares of either
class A or class B stock
...
Class B stock was
selling for $20 per share, but its price was expected
to increase by 12% per year
...
Plan A involves remodeling the ﬁre stations
on Alameda Avenue and Trowbridge Boulevard that
are 57 and 61 years old, respectively
...
) The
cost for remodeling the Alameda station is estimated at $952,000 while the cost of redoing the
Trowbridge station is $1
...
Plan B calls for
buying 5 acres of land somewhere between the two
stations, building a new ﬁre station, and selling the
land and structures at the previous sites
...
The size of the new ﬁre station would be 9000
square feet with a construction cost of $151
...
Contractor fees for overhead, proﬁt,
etc
...
(Assume all of the costs for plan B
occur at time 0
...
Under plan B, the sale of the old sites is anticipated to net a positive $500,000 ﬁve years in the
future
...

5
...
The company needs to analyze the economic feasibility of rainwater drains in a 60-acre
area that it plans to develop
...
If
no drains are installed, the cost of reﬁlling and
grading the washed out area is expected to be
$1500 per thunderstorm
...
The cost of the
pipe will be $3 per foot for the total length of
7000 feet required
...

Assuming that thunderstorms occur regularly at
3-month intervals, starting 3 months from now,
which alternative should be selected on the basis of
a present worth comparison using an interest rate of
4% per quarter?
5
...
A
250-mm line will have an initial cost of $155,000,
whereas a 300-mm line will cost $210,000
...
If the lines are

144

Chapter 5

Present Worth Analysis

expected to last for 30 years, which size should be
selected on the basis of a present worth analysis
using an interest rate of 10% per year?
5
...
If gaseous chlorine is added, a chlorinator
will be required that has an initial cost of $8000
and a useful life of 5 years
...
Alternatively, dry chlorine can be added manually at a cost of $1000 per year for chlorine and
$1900 per year for labor
...
15 Anion, an environmental engineering consulting
ﬁrm, is trying to be eco-friendly in acquiring an automobile for general ofﬁce use
...
The hybrid under consideration
is GM’s Volt, which will cost $35,000 and have a
range of 40 miles on the electric battery and several
hundred more miles when the gasoline engine
kicks in
...
The Leaf’s relatively limited range creates a psychological effect known as range anxiety
...
The Leaf
could be leased for $349 per month (end-of-month
payments) for 5 years after an initial $1500 down
payment for “account activation
...
75% per month? Assume the operating cost will be the same for both vehicles
...
16 A pipeline engineer working in Kuwait for the oil
giant BP wants to perform a present worth analysis on
alternative pipeline routings—the ﬁrst predominately
by land and the second primarily undersea
...
Perform the analysis for the engineer at 15% per year
...
17 An electric switch manufacturing company has to
choose one of three different assembly methods
...
Method B will cost $80,000 to buy and
will have an annual operating cost of $6000 over
its 4-year service life
...
Methods A and B will have no
salvage value, but method C will have some equipment worth an estimated $12,000
...

5
...
The company recently settled a lawsuit by
agreeing to pay $60 million in mitigation costs related to acid rain
...

The question of how to distribute the money over
time has been posed
...
5 million
in each of years 1 through 10)
...
Determine which plan is more economical on the
basis of a present worth analysis over a 10-year
period at an interest rate of 10% per year
...
19 Machines that have the following costs are under
consideration for a robotized welding process
...
Show (a) hand and
(b) spreadsheet solutions
...
20 Water for Semiconductor Manufacturing Case

PE

Throughout the present worth analyses, the decision between seawater and groundwater switched
multiple times in Examples 5
...
4
...

145

Problems
Seawater (S)
Life n,
years
10
5

5
(study
period)

Groundwater (G)

PW at
12%, $

Selected

Life n,
years

Ϫ20

Ϫ30
...
31

Ϫ20

Ϫ26
...
16

No

No

10

Ϫ22

Ϫ33
...
32

No

The confusion about the recommended source for
UPW has not gone unnoticed by the general manager
...
The study period is set by the manager
as 10 years, simply because that is the time period on
the lease agreement for the building where the fab
will be located
...
What is the
maximum ﬁrst cost that Angular Enterprises should
pay for the seawater option?
5
...
In a ﬁeld application, physical constraints compromise the pipe layout, so the engineer
is considering installing the airﬂow probes in an
elbow, knowing that ﬂow measurement will be less
accurate but good enough for process control
...

This plan will have a ﬁrst cost of $26,000 with an
annual maintenance cost estimated at $5000
...
The stainless
steel probe can be installed in a drop pipe with the
transmitter located in a waterproof enclosure on the
handrail
...
Its maintenance cost is estimated to be $1400 per year plus
$2500 in year 3 for replacement of signal processing software
...
At an interest rate of 10% per year, which
one should be selected on the basis of a present
worth comparison?
5
...
A plastic liner
will cost $0
...
This removal will cost $500,000
...
20 per square foot
...
23

A sports mortgage is the brainchild of Stadium Capital Financing Group, a company headquartered in
Chicago, Illinois
...
In California, the locked-in price
period is 50 years
...
(a) Which fan made
the better deal if the interest rate is 8% per year?
(b) What should fan X be willing to pay up front for
the mortgage to make the two plans exactly equivalent economically? (Assume he has no reason to
give extra money to UCLA at this point
...
24 A chemical processing corporation is considering
three methods to dispose of a non-hazardous chemical sludge: land application, ﬂuidized-bed incineration, and private disposal contract
...
Determine which has the
least cost on the basis of a present worth comparison at 10% per year for the following scenarios:
(a) The estimates as shown
(b) The contract award cost increases by 20%
every 2-year renewal
Land Application Incineration
First cost, $
Annual operating
cost, $ per year
Salvage value, $
Life, years

Contract

Ϫ130,000
Ϫ95,000

Ϫ900,000
Ϫ60,000

0
Ϫ120,000

25,000
3

300,000
6

0
2

146

Chapter 5

Present Worth Analysis

5
...
The results
in the table use a MARR of 14% per year
...

I
Life n, years
PW over n years, $
PW over 6 years, $
PW over 12 years, $

3
16
...
94
39
...
12
15
...
45

K

L

12
6
Ϫ257
...
46
Ϫ653
...
46
Ϫ257
...
46

Future Worth Comparison
5
...
Robot X will have a ﬁrst cost of $80,000, an
annual maintenance and operation (M&O) cost of
$30,000, and a $40,000 salvage value
...

Which should be selected on the basis of a future
worth comparison at an interest rate of 15% per
year? Use a 3-year study period
...
27 Two processes can be used for producing a polymer that reduces friction loss in engines
...
Process W will have a ﬁrst cost
of $1,350,000, an operating cost of $25,000 per
year, and a $120,000 salvage value after its 4-year
life
...
Which process
should be selected on the basis of a future worth
analysis at an interest rate of 12% per year?
5
...

P
First cost, $
Annual operating cost, $ per year
Salvage value, $
Life, years

Q

Ϫ23,000 Ϫ30,000
Ϫ4,000 Ϫ2,500
3,000
1,000
3
6

5
...
St
...
System K will have a ﬁrst cost of
$1,600,000, an operating cost of $70,000 per year,
and a salvage value of $400,000 after its 4-year life
...
Which system
should be selected on the basis of a future worth
analysis at an interest rate of 12% per year?
5
...
7 million over 10 years
...
However, due to a real estate–induced recession in the
United States, the developer sought and was granted
a new contract
...
Assume that the cost for razing the existing
buildings is $1
...
Determine the difference in
the future worth cost in year 7 of the two contracts
at an interest rate of 10% per year
...
31 A wealthy businessman wants to start a permanent
fund for supporting research directed toward sustainability
...
e
...
If the fund earns interest at a rate of 8% per
year, how much money must be donated each time?
5
...
Determine the capitalized cost of $10,000
every 5 years forever, starting 5 years from now
at an interest rate of (a) 3% and (b) 8% per year
...

5
...
Use an interest rate
of 12% per year
...
34 Water for Semiconductor Manufacturing Case

PE

It is anticipated that the needs for UPW (ultrapure
water) at the new Angular Enterprises site will
continue for a long time, as long as 50 years
...
These
costs were determined (Example 5
...
58 million and CCG ϭ $Ϫ48
...

Groundwater is the clear economic choice
...
It would mean a dependence upon a contractor to supply the water, but the
equipment, treatment, and other costly activities to
obtain UPW on-site would be eliminated
...

(b) The annual cost starts at $5 million for the ﬁrst
year only, and then it increases 2% per year
...
)

5
...

Calculate the capitalized cost of the maintenance
using an interest rate of 10% per year
...
35 Compare the alternatives shown on the basis of
their capitalized costs using an interest rate of
10% per year
...
38 A patriotic group of ﬁreﬁghters is raising money
to erect a permanent (i
...
, inﬁnite life) monument
in New York City to honor those killed in the line
of duty
...
There will be an additional one-time cost
of $20,000 in 2 years to add names of those who
were missed initially
...
37 Because you are thankful for what you learned in
engineering economy, you plan to start a permanent scholarship fund in the name of the professor
who taught the course
...
The interest that is accumulated between now and year 12 is to be added to
the principal of the endowment
...
If you want the amount of the
scholarships to be $40,000 per year, how much
must you donate now if the fund earns interest at a
rate of 8% per year?

ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
5
...

(b) The alternatives will be used only through
the life of the longest-lived alternative
...

(d) At least one of the alternatives will have a
ﬁnite life
...
40 When only one alternative can be selected from
two or more, the alternatives are said to be:
(a) Mutually exclusive
(b) Independent alternatives
(c) Cost alternatives
(d) Revenue alternatives

5
...
42 The alternatives shown are to be compared on the
basis of their present worth values
...
43 The value of the future worth for alternative P at an
interest rate of 8% per year is closest to:
P
Ϫ23,000
Ϫ4,000

(a)
(b)
(c)
(d)

Ϫ30,000
Ϫ2,500

3,000
3

First cost, $
Annual operating cost,
$ per year
Salvage value, $
Life, years

Q

1,000
6

FWP ϭ $Ϫ88,036
FWP ϭ $Ϫ86,026
FWP ϭ $Ϫ81,274
FWP ϭ $Ϫ70,178

5
...
45 A donor (you) wishes to start an endowment that
will provide scholarship money of $40,000 per year
beginning in year 5 and continuing indeﬁnitely
...
46 through 5
...

Alternative I
Initial cost, $
Annual income, $ per year
Annual expenses, $ per year
Salvage value, $
Life, years

Alternative J

Ϫ150,000
20,000
Ϫ9,000
25,000
3

Ϫ250,000
40,000
Ϫ14,000
35,000
6

The interest rate is 15% per year
...
46 In comparing alternatives I and J by the present
worth method, the value of n that must be used in
11,000(P͞A,i,n) for alternative I is:
(a) 3
(b) 6
(c) 18
(d) 36
5
...
48 In comparing alternatives I and J by the present
worth method, the equation that yields the present
worth of alternative I is:
(a) PWI ϭ Ϫ150,000 ϩ 11,000(P͞A,15%,3) ϩ
25,000(P͞F,15%,3)
(b) PWI ϭ Ϫ150,000 ϩ 11,000(P͞A,15%,6) ϩ
25,000(P͞F,15%,6)
(c) PWI ϭ Ϫ150,000 ϩ 11,000(P͞A,15%,6) ϩ
175,000(P͞F,15%,3) ϩ 25,000(P͞F,15%,6)
(d) PWI ϭ Ϫ150,000 ϩ 11,000(P͞A,15%,6) Ϫ
125,000(P͞F,15%,3) ϩ 25,000(P͞F,15%,6)
Problems 5
...
50 are based on the following
information
...

5
...
50 In comparing the machines on a present worth
basis, the present worth of machine Y is closest to:
(a) $Ϫ112,320
(b) $Ϫ122,060
(c) $Ϫ163,040
(d) $Ϫ175,980

Case Study

149

5
...
51 The capitalized cost of $10,000 every 5 years
forever, starting now at an interest rate of 10% per
year, is closest to:
(a) $Ϫ13,520
(b) $Ϫ16,380
(c) $Ϫ26,380
(d) $Ϫ32,590

CASE STUDY
COMPARING SOCIAL SECURITY BENEFITS
Background
When Sheryl graduated from Northeastern University in
2000 and went to work for BAE Systems, she did not pay
much attention to the monthly payroll deduction for social
security
...
However, this was so far in the future that she
fully expected this government retirement beneﬁt system to
be broke and gone by the time she could reap any beneﬁts
from her years of contributions
...
Recently, they both received notices from the
Social Security Administration of their potential retirement
amounts, were they to retire and start social security beneﬁts at preset ages
...

Information
They found that their projected beneﬁts are substantially the
same, which makes sense since their salaries are very close to
each other
...

At age 62, your payment would
be about
At you full retirement age (67
years), your payment would be
about
At age 70, your payment would
be about

$1400 per month

$2000 per month
$2480 per month

These numbers represent a reduction of 30% for early retirement (age 62) and an increase of 24% for delayed retirement (age 70)
...
In other words, if Sheryl starts her $2000 beneﬁt at age 67, Brad can receive a beneﬁt equal to 50% of hers
...
In the meantime, his beneﬁts will have increased by 24%
...

All these options led them to deﬁne four alternative plans
...

B: Each takes full beneﬁts at full retirement age of 67
and receives $2000 per month
...

D: One person takes full beneﬁts of $2000 per month at
age 67, and the other person receives spousal beneﬁts
($1000 per month at age 67) and switches to delayed
beneﬁts of $2480 at age 70
...

Case Study Exercises
Brad and Sheryl are the same age
...
With
this as the interest rate, the analysis for the four alternatives is
possible
...
Can you please help
them? (Do the analysis for one person at a time, not the couple, and stop at the age of 85
...
How much in total (without the time value of money considered) will each plan A through D pay through age 85?
2
...
Plot the future worth values for all four plans on one
spreadsheet graph
...
Economically, what is the best combination of plans for
Brad and Sheryl, assuming they both live to be 85 years
old?
5
...
Answer the question
...

SECTION

TOPIC

LEARNING OUTCOME

6
...

6
...

6
...

6
...

6
...

I

n this chapter, we add to our repertoire of alternative comparison tools
...
Here we learn the equivalent annual worth, or
AW, method
...

Annual worth is also known by other titles
...
The alternative selected by the AW method will always be the same as that selected
by the PW method, and all other alternative evaluation methods, provided they are performed correctly
...
This
method considers all costs of a product, process, or system from concept to phaseout
...
1 Advantages and Uses of Annual Worth Analysis
For many engineering economic studies, the AW method is the best to use, when compared to
PW, FW, and rate of return (Chapters 7 and 8)
...
The AW value, which has the same interpretation as A used thus far, is
the economic equivalent of the PW and FW values at the MARR for n years
...
1]

The n in the factors is the number of years for equal-service comparison
...

When all cash ﬂow estimates are converted to an AW value, this value applies for every year
of the life cycle and for each additional life cycle
...
The AW value determined over
one life cycle is the AW for all future life cycles
...

As with the PW method, there are three fundamental assumptions of the AW method that should
be understood
...
The services provided are needed for at least the LCM of the lives of the alternatives
...
The selected alternative will be repeated for succeeding life cycles in exactly the same manner as for the ﬁrst life cycle
...
All cash ﬂows will have the same estimated values in every life cycle
...
If, in a particular evaluation, the ﬁrst two assumptions are not reasonable, a study period must be established for the analysis
...
In the third assumption, all
cash ﬂows are expected to change exactly with the inﬂation (or deﬂation) rate
...
AW analysis for a stated study period is discussed in Section 6
...

EXAMPLE 6
...
3, National Homebuilders, Inc
...
The PW analysis used the LCM of 18 years
...
The diagram in Figure 6–1 shows the cash ﬂows for all three life
cycles (ﬁrst cost $Ϫ15,000; annual M&O costs $Ϫ3500; salvage value $1000)
...
In Example 5
...

Equal-service
requirement and LCM

152
Figure 6–1

Chapter 6

Annual Worth Analysis

PW = $45,036

PW and AW values
for three life cycles,
Example 6
...

3 life cycles

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

$1000
0

1

2

3

4

5

6
Life cycle 1

i = 15%

$1000

$3500
0

1

2

3

4

5

6

$15,000

Life cycle 2

$1000

$3500
0

1

2

3

4

5

6

$15,000

Life cycle 3

$3500

$15,000
0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20
•••
continues

AW = $7349

Solution
Calculate the equivalent uniform annual worth value for all cash ﬂows in the ﬁrst life cycle
...
Now Equation [6
...

AW ϭ Ϫ45,036(A͞P,15%,18) ϭ $Ϫ7349
The one-life-cycle AW value and the AW value based on 18 years are equal
...
The AW method is especially useful in certain types of studies: asset replacement and
retention time studies to minimize overall annual costs (both covered in Chapter 11), breakeven
studies and make-or-buy decisions (Chapter 13), and all studies dealing with production or manufacturing costs where a cost /unit or proﬁt /unit measure is the focus
...
2

Calculation of Capital Recovery and AW Values

153

If income taxes are considered, a slightly different approach to the AW method is used by some
large corporations and ﬁnancial institutions
...
This approach,
covered in Chapter 17, concentrates upon the wealth-increasing potential that an alternative offers a
corporation
...

6
...
This is the total ﬁrst cost of all assets and services required to initiate
the alternative
...
Use this amount as P
...
This is the terminal estimated value of assets at the end of their useful life
...
For study periods shorter than the useful life, S is the estimated market value or
trade-in value at the end of the study period
...
This is the equivalent annual amount (costs only for cost alternatives;
costs and receipts for revenue alternatives)
...

Salvage/market value

The annual worth (AW) value for an alternative is comprised of two components: capital
recovery for the initial investment P at a stated interest rate (usually the MARR) and the equivalent
annual amount A
...
In equation form,
AW ‫ ؍‬CR ؉ A

[6
...
The total annual amount A is determined from uniform recurring
costs (and possibly receipts) and nonrecurring amounts
...
2]
...
)
The recovery of an amount of capital P committed to an asset, plus the time value of the
capital at a particular interest rate, is a fundamental principle of economic analysis
...
Any expected salvage value is considered in the computation of CR
...
If there is some anticipated
positive salvage value S at the end of the asset’s useful life, its equivalent annual value is recovered using the A͞F factor
...
Accordingly, CR is calculated as
CR ‫؊ ؍‬P(A͞P,i,n) ؉ S(A͞F,i,n)

[6
...
2
Lockheed Martin is increasing its booster thrust power in order to win more satellite launch contracts from European companies interested in opening up new global communications markets
...
Annual operating costs for the system are expected to start the ﬁrst year and continue at
$0
...
The useful life of the tracker is 8 years with a salvage value of $0
...

Calculate the CR and AW values for the system, if the corporate MARR is 12% per year
...
3] to calculate the capital recovery
...
46

Capital recovery

154

Chapter 6

Annual Worth Analysis

CR ϭ Ϫ12
...
5(A͞F,12%,8)
ϭ Ϫ12
...
20130) ϩ 0
...
08130)
ϭ $Ϫ2
...
It means that
each and every year for 8 years, the equivalent total net revenue from the tracker must be at
least $2,470,000 just to recover the initial present worth investment plus the required return of
12% per year
...
9 million each year
...
Since CR ϭ $Ϫ2
...

AW ϭ Ϫ2
...
9 ϭ $Ϫ3
...

Figure 6–2

$0
...
9

6

7 8

0

1

2

3

4

5

6

AW = ?

7

8

(a) Cash ﬂow diagram
for satellite tracker
costs, and (b) conversion to an equivalent
AW (in $1 million),
Example 6
...

$5
...
0
(a)

(b)

There is a second, equally correct way to determine CR
...
There is a relation between the A͞P and A͞F factors
...
3]
...
Subtracting S from the initial investment P before applying
the A͞P factor recognizes that the salvage value will be recovered
...
However, the fact that S is not recovered until year n of ownership is
compensated for by charging the annual interest S(i) against the CR
...
The ﬁrst method,
Equation [6
...

For solution by spreadsheet, use the PMT function to determine CR only in a single spreadsheet cell
...
The format is
‫ ؍‬PMT(i%,n,P,؊S)

[6
...
2
...
46 million
...
46,Ϫ0
...
The answer of $Ϫ2
...

As we learned in Section 3
...

In the case of Example 6
...
The one-cell
PMT function, with the PV function embedded (in bold), can be written as ϭ PMT(12%,8,
8؉PV(12%,1,؊5),Ϫ0
...
47
...
3

155

Evaluating Alternatives by Annual Worth Analysis

6
...
The AW is calculated over the respective life of each alternative, and the selection guidelines are the same as those used for the PW method
...

Two or more alternatives: Select the alternative with the AW that is numerically largest,
that is, less negative or more positive
...

If any of the three assumptions in Section 6
...
Then the cash ﬂow estimates over the study period are converted to AW
amounts
...

EXAMPLE 6
...
Many students at the area universities and community colleges work part-time
delivering orders made via the web
...

The systems provide a link between the web order-placement software and the On-Star system
for satellite-generated directions to any address in the area
...

Each system costs $4600, has a 5-year useful life, and may be salvaged for an estimated
$300
...
The MARR is 10%
...
Perform the solution by hand and by spreadsheet
...
Is this project
ﬁnancially viable at the MARR?
(c) Based on the answer in part (b), determine how much new net income Heavenly Pizza
must have to economically justify the project
...

Solution by Hand
(a) The capital recovery amount calculated by Equation [6
...

CR ϭ Ϫ5[4600(A͞P,10%,5)] ϩ 5[300(A͞F,10%,5)]
ϭ Ϫ5[4600(0
...
16380)]
ϭ $Ϫ5822
The ﬁve systems must generate an equivalent annual new revenue of $5822 to recover the
initial investment plus a 10% per year return
...
The annual operating cost series,
combined with the estimated $6000 annual income, forms an arithmetic gradient series
with a base amount of $5000 and G ϭ $Ϫ100
...
Apply Equation [6
...

AW ϭ CR ϩ A ϭ Ϫ5822 ϩ 5000 Ϫ 100(A͞G,10%,5)
ϭ $Ϫ1003
The system is not ﬁnancially justiﬁed at the net income level of $6000 per year
...
3
...

0 ϭ Ϫ5822 ϩ (R Ϫ 1000) Ϫ 100(A͞G,10%,5)
R ϭ Ϫ5822 Ϫ 1000 Ϫ 100(1
...

Figure 6–4
Spreadsheet solution
of Example 6
...

(a) Capital recovery
in cell B16, (b) AW in
cell E17, and (c) Goal
Seek template and
outcome in cell B5
...
4

AW of a Permanent Investment

157

Cell references are used in the spreadsheet functions to accommodate changes in estimated values
...

(b) The annual worth AW ϭ $–1003 is displayed in column E using the PMT function
shown
...

(c) The minimum required income is determined in the lower part of Figure 6–4
...

EXAMPLE 6
...
Since the meals are
prepared in one central location and distributed by trucks throughout the city, the equipment
that keeps food and drink cold and hot is very important
...
Use the cost estimates below to select the more
economic unit at a MARR of 8% per year
...

AWH ϭ annual equivalent of P Ϫ annual M&O ϩ annual equivalent of S
ϭ Ϫ15,000(A͞P,8%,4) Ϫ 6000 ϩ 0
...
30192) Ϫ 6000 ϩ 3000(0
...
4(20,000)(A͞F,8%,12)
ϭ Ϫ20,000(0
...
7350 ϩ 0
...
13270) ϩ 8000(0
...

If the projects are independent, the AW at the MARR is calculated
...

Independent project
selection

6
...
5
...
For this type of analysis,
the annual worth (and capital recovery amount) of the initial investment is the perpetual annual
interest on the initial investment, that is, A ϭ Pi ϭ (CC) i
...
2]
...
This
automatically annualizes them for each succeeding life cycle
...
2]
...
5
The U
...
Bureau of Reclamation is considering three proposals for increasing the capacity of the
main drainage canal in an agricultural region of Nebraska
...

The capacity of the canal will have to be maintained in the future near its design peak ﬂow because of increased water demand
...
The equipment is expected to have a 10-year life with a $17,000
salvage value
...
To control weeds in the
canal itself and along the banks, environmentally safe herbicides will be sprayed during the irrigation season
...

Proposal B is to line the canal with concrete at an initial cost of $4 million
...
In addition, lining repairs will have to be made every 5 years at a cost of $30,000
...
Estimates are an initial cost
of $6 million, annual maintenance of $3000 for right-of-way, and a life of 50 years
...

Solution
Since this is an investment for a permanent project, compute the AW for one cycle of all recurring costs
...
3], with nA ϭ 10
and nC ϭ 50, respectively
...

Proposal A
CR of dredging equipment:
Ϫ650,000(A͞P,5%,10) ϩ 17,000(A͞F,5%,10)
Annual cost of dredging
Annual cost of weed control
Proposal B
CR of initial investment: Ϫ4,000,000(0
...

Comment
Note the use of the A͞F factor for the lining repair cost in proposal B
...

If the 50-year life of proposal C is considered inﬁnite, CR ϭ P(i) ϭ $Ϫ300,000, instead of
$Ϫ328,680 for n ϭ 50
...
How long lives of 40 or more
years are treated economically is a matter of “local” practice
...
4

159

AW of a Permanent Investment

EXAMPLE 6
...
Bart just received his
bonus in the amount of $8530
...
Bart’s long-term plans are to quit the Coop job some years in the future when he is
still young enough to start his own business
...

(a) Use a spreadsheet to determine the amount of annual year-end withdrawal that he can anticipate (starting 1 year after he quits) that will continue forever
...

(b) Determine the amount Bart must accumulate after 15 and 20 years to generate $3000 per
year forever
...
The accumulated amount after n ϭ 15 years is indicated
as Fafter 15 ϭ ? and the withdrawal series starts at the end of year 16
...

The spreadsheet in Figure 6–6 shows the functions and answers for n ϭ 15 years in
columns C and D
...
The perpetual withdrawal is determined by viewing this accumulated amount
as a P value and by applying the formula
A ϭ P(i) ϭ 23,535(0
...

Answers for n ϭ 20 years are displayed in column E
...

A=?

i = 7%
0

1

2

3

4

13

14

15
∞
16

17

18

19

20

21

22 …

Figure 6–5
P = $8530
Fafter 15 = ?

Diagram for a perpetual
series starting after 15
years of accumulation,
Example 6
...

Figure 6–6
Spreadsheet solution,
Example 6
...

160

Chapter 6

Annual Worth Analysis

(b) To obtain a perpetual annual withdrawal of $3000, it is necessary to determine how much
must be accumulated 1 year before the ﬁrst withdrawal of $3000
...
07
This P value is independent of how long Bart works at the Coop, because he must accumulate
this amount to achieve his goal
...
Note that
the number of years n does not enter into the function ϭ 3000͞B13
...
The row 15 function indicates that Bart will have to work at the Coop for just under 24 additional years
...
5 Life-Cycle Cost Analysis
The PW and AW analysis techniques discussed thus far have concentrated on estimates for ﬁrst
cost P, annual operating and maintenance costs (AOC or M&O), salvage value S, and predictable
periodic repair and upgrade costs, plus any revenue estimates that may favor one alternative over
another
...
A life-cycle cost analysis includes these additional estimates to the extent that they
can be reliably determined
...
Estimates will cover the entire life span from the early
conceptual stage, through the design and development stages, throughout the operating stage, and
even the phaseout and disposal stages
...

Some typical LCC applications are life-span analysis for military and commercial aircraft,
new manufacturing plants, new automobile models, new and expanded product lines, and government systems at federal and state levels
...
S
...

Most commonly the LCC analysis includes costs, and the AW method is used for the analysis,
especially if only one alternative is evaluated
...
Public sector projects are usually
evaluated using a beneﬁt/cost analysis (Chapter 9), rather than LCC analysis, because estimates
to the citizenry are difﬁcult to make with much accuracy
...
Some examples of indirect cost components are taxes, management, legal,
warranty, quality, human resources, insurance, software, purchasing, etc
...

LCC analysis is most effectively applied when a substantial percentage of the life span (postpurchase) costs, relative to the initial investment, will be expended in direct and indirect operating, maintenance, and similar costs once the system is operational
...
However, let’s assume that
Exxon-Mobil wants to evaluate the design, construction, operation, and support of a new type
and style of tanker that can transport oil over long distances of ocean
...

To understand how a LCC analysis works, ﬁrst we must understand the phases and stages of
systems engineering or systems development
...
Generally, the LCC estimates may be categorized into a simpliﬁed

6
...

Acquisition phase: all activities prior to the delivery of products and services
...

• Preliminary design stage—Includes feasibility study, conceptual, and early-stage plans;
ﬁnal go–no go decision is probably made here
...
; there is some acquisition of assets, if economically justiﬁable
...

• Construction and implementation stage—Includes purchases, construction, and implementation of system components; testing; preparation, etc
...

Phaseout and disposal phase: covers all activities to transition to a new system; removal/
recycling/disposal of old system
...
7
In the 1860s, General Mills Inc
...
both started in the ﬂour business in the Twin
Cities of Minneapolis–St
...
In the decade of 2000 to 2010, General Mills purchased Pillsbury for a combination cash and stock deal worth more than $10 billion and integrated the product lines
...
At this point only cost
estimates have been addressed—no revenues or proﬁts
...
Use LCC analysis at the industry MARR of 18% to determine the size of the commitment in AW terms
...
Since all estimates are for costs, they are not preceded by
a minus sign
...
5 million
0
...
5 million
1
...
0 million

Equipment acquisition (years 1 and 2)
Current equipment upgrades (year 2)
New equipment purchases (years 4 and 8)

$2
...
75 million
2
...
0 million
5
...
2 million
per year thereafter
3
...
0 million each year

161

162

Chapter 6

Annual Worth Analysis

Solution
LCC analysis can get complicated rapidly due to the number of elements involved
...
Values are in
$1 million units
...
5
Preliminary design: product and equipment
PW ϭ 1
...
187
Detailed design: product and test marketing, and equipment
PW ϭ 1
...
0(P͞F,18%,2) ϭ $3
...
0(P͞A,18%,2) ϩ 1
...
0(P͞F,18%,4) ϩ 2
...
04 8
1 Ϫ ——
1
...
2 ————— (P͞F,18%,2) ϭ $6
...
14
Use: marketing
PW ϭ 8
...
0(P͞A,18%,8) Ϫ 0
...
0(P͞F,18%,5)
ϭ $20
...
0 million in year 3

΄

(

)

΅

1
...
18
PW ϭ 4
...
412
0
...
0(P͞A,18%,2)(P͞F,18%,8) ϭ $0
...
238 million
...

AW ‫ 832
...
01$ ؍‬million per year
This is the LCC estimate of the equivalent annual commitment to the two proposed
products
...
For example, if one alternative will produce 20 million units per year and a second alternative will operate at 35 million per year, the AW values should be compared on a currency unit/
unit produced basis, such as dollar/unit or euro/hour operated
...
For
some systems, typically defense systems, operating and maintenance costs rise fast after acquisition and remain high until phaseout occurs
...
It is not unusual
to have 75% to 85% of the entire life span LCC committed during the preliminary and detail

Life-Cycle Cost Analysis

Costs

6
...

D
%

%

B

B

Reduced
total LCC
Committed
Cumulative LCC

Cumulative LCC

Committed
Actual

#1

#1

C

E

F

#2

A

A
Time
Acquisition
phase

Operation
phase

Time
Acquisition
phase

(a)

Operation
phase
(b)

Figure 6–8
LCC envelopes for committed and actual costs: (a) design 1, (b) improved design 2
...
As shown in Figure 6–8a, the actual or observed LCC (bottom curve AB) will trail
the committed LCC throughout the life span (unless some major design ﬂaw increases the total
LCC of design #1 above point B)
...
A
more effective design and more efﬁcient equipment can reposition the envelope to design #2 in
Figure 6–8b
...
It is this lower envelope #2 we seek
...

Even though an effective LCC envelope may be established early in the acquisition phase,
it is not uncommon that unplanned cost-saving measures are introduced during the acquisition
phase and early operation phase
...
This style of ad hoc cost savings, often imposed by management early in the design stage and/or construction stage, can substantially increase costs later,
especially in the after-sale portion of the use stage
...

163

164

Chapter 6

Annual Worth Analysis

CHAPTER SUMMARY
The annual worth method of comparing alternatives is often preferred to the present worth method,
because the AW comparison is performed for only one life cycle
...
AW for the ﬁrst life cycle is the AW for the second, third, and
all succeeding life cycles, under certain assumptions
...

For inﬁnite-life (perpetual) alternatives, the initial cost is annualized simply by multiplying P
by i
...

Life-cycle cost analysis is appropriate for systems that have a large percentage of costs in
operating and maintenance
...

PROBLEMS
after its 4-year life
...

Annual Worth Calculations
6
...
6 A sports mortgage is an innovative way to ﬁnance
cash-strapped sports programs by allowing fans to
sign up to pay a “mortgage” for the right to buy
good seats at football games for several decades
with season tickets locked in at current prices
...

If a fan pays a $130,000 “mortgage” fee now (i
...
,
in year 0) when season tickets are selling for $290
each, what is the equivalent annual cost of the
football tickets over the 50-year period at an interest rate of 8% per year?

6
...

6
...
4 James developed the two cash ﬂow diagrams shown
at the bottom of this page
...
Calculate the annual worth value of each over the
respective life cycles to demonstrate that they are
the same
...

6
...
5 An asset with a ﬁrst cost of $20,000 has an annual
operating cost of $12,000 and a $4000 salvage value
Alternative A

Alternative B

$1000

0

1

2

$25

$25

3

i = 10% per year

Year

0

$1000

2

3

4

5

$25

$25

1

$25

$25

$25

$25

$4000
$5000

$5000

6

$25

Year

Problems

6
...
The company
sold it today for $45,000
...
A complete
overhaul at the end of year 4 cost an extra $3600
...

6
...
The cost of
periodic maintenance has been $800 every 2 years
...

Capital Recovery
6
...

He anticipated a salvage value of $50,000 after 10
years
...
(a) Did he recover his investment
and a 12% per year return? (b) If the annual M&O
cost was $10,000 the ﬁrst year and increased by a
constant $1000 per year, was the AW positive or
negative at 12% per year? Assume the $50,000 salvage was realized
...
11

Sylvia has received a $500,000 inheritance from
her favorite, recently deceased aunt in Hawaii
...
She hopes to make 8%
per year on this purchase over an ownership period
of 20 years
...
No annual M&O costs are considered in the analysis
...
12 Humana Hospital Corporation installed a new MRI
machine at a cost of $750,000 this year in its medical professional clinic in Cedar Park
...
Humana uses a return
requirement of 24% per year for all of its medical
diagnostic equipment
...
Also, you are asked to
draw two cash ﬂow diagrams, one showing the
MRI purchase and sale cash ﬂow and a second depicting the required capital recovery each year
...
13 Polypropylene wall caps, used for covering exterior vents for kitchen cooktops, bathroom fans,
dryers, and other building air exhausts, can be
made by two different methods
...
Method Y will have a ﬁrst cost of $140,000,
an operating cost of $24,000 per year, and a
$19,000 salvage value after its 4-year life
...
14 Nissan’s all-electric car, the Leaf, has a base price
of $32,780 in the United States, but it is eligible
for a $7500 federal tax credit
...
The
cost for leasing the vehicle will be $4200 per year
(payable at the end of each year) after an initialization charge of $2500 paid now
...
If the company expects
to be able to sell the car and charging station for
40% of the base price of the car alone at the end of
3 years, should the company purchase or lease the
car? Use an interest rate of 10% per year and annual worth analysis
...
15 A new structural design software package is available for analyzing and designing three-sided guyed
towers and three- and four-sided self-supporting
towers
...
A site license has a one-time cost of $22,000
...
Determine which
strategy should be adopted at an interest rate of
10% per year for a 4-year planning period using
the annual worth method of evaluation
...
16 The city council in a certain southwestern city is
considering whether to construct permanent restrooms in 22 of its smaller parks (i
...
, parks of less

166

Chapter 6

Annual Worth Analysis

than 12 acres) or pay for portable toilets on a yearround basis
...
8 million
...
The service life of a permanent restroom is 20 years
...

6
...
Solar cells will cost $16,600 to install and
will have a useful life of 5 years with no salvage
value
...
,
are expected to be $2400
...
Since the air sampling project
will end in 5 years, the salvage value of the line is
considered to be zero
...
18 The cash ﬂows for two small raw water treatment
systems are shown
...

MF
First cost, $
Annual cost, $ per year
Salvage value, $
Life, years

UF

–33,000
–8,000
4,000
3

–51,000
–3,500
11,000
6

6
...
Two
equivalent robot instruments are available
...
Robot Watcheye will have a
ﬁrst cost of $125,000, annual M&O costs of
$27,000, and a $33,000 salvage value after its
5-year life
...

(a) Which robot is the better economic option?
(b) Using the spreadsheet Goal Seek tool, determine the ﬁrst cost of the robot not selected in
(a) so that it will be the economic selection
...
20 TT Racing and Performance Motor Corporation
wishes to evaluate two alternative CNC machines

for NHRA engine building
...

Machine R
First cost, $
Annual operating cost,
$ per year
Salvage value, $
Life, years

Machine S

Ϫ250,000
Ϫ40,000

Ϫ370,500
–50,000

20,000
3

30,000
5

6
...

For a loader that has a ﬁrst cost of $39,000 and
ﬁrst-year M&O cost of $17,000, compare the
equivalent annual worth of a loader kept for 4 years
with one kept for 5 years at an interest rate of
12% per year
...

6
...
A manager
asked you to determine which of the following
two machines will have the lower (a) capital recovery and (b) equivalent annual total cost
...
Machine Auto1 has a ﬁrst cost of $62,000 and an
operating cost of $21,000 in year 1, increasing
by 8% per year through year 5, after which time
it will have a scavenge value of $2000
...

Permanent Investments
6
...
The ﬁrst cost is $200,000
now, and an update budget of $100,000 every
7 years forever is requested
...

6
...
Use an interest rate of 10% per
year
...
25 A Pennsylvania coal mining operation has installed an in-shaft monitoring system for oxygen
tank and gear readiness for emergencies
...

Maintenance costs are expected to be $150,000 in
year 3, $175,000 in year 4, and amounts increasing by $25,000 per year through year 6 and remain constant thereafter for the expected 10-year
life of the system
...

6
...

Condi
Ϫ25,000
Ϫ9,000
3,000
3

First cost, $
Annual cost, $ per year
Salvage value, $
Life, years

Torro
Ϫ130,000
Ϫ2,500
150,000
ϱ

in usage for crop irrigation, is considering the
purchase of one of the Blanton systems
...
The estimated costs and associated cash ﬂow diagrams over a 10-year period are summarized below and on the next page,
respectively, for each of the ﬁve alternatives
...
For alternatives A and B, there is an extra cost of $15,000 per
installation year to maintain the manual system in
place now
...
Alternative C
is a retroﬁt of the current manual system with no
design or development costs, and there is no level
1 option
...

Alternative

X
First cost, $
Annual cost,
$ per year
Overhaul every
10 years, $
Salvage value, $
Life, years

Y

Z

Ϫ90,000
Ϫ40,000

Ϫ400,000
Ϫ20,000

Ϫ650,000
Ϫ13,000

—

—

Ϫ80,000

7,000
3

25,000
10

200,000
ϱ

Life-Cycle Cost
6
...
A
large farming corporation in India, where depletion is occurring at an alarming rate of 1
...
27 A new bridge across the Allegheny River in
Pittsburgh is expected to be permanent and will
have an initial cost of $30 million
...

The annual inspection and operating costs are estimated to be $50,000
...

6
...

(a) Determine the alternative that is economically best
...
Use a
spreadsheet to answer this question
...
30 The Pentagon asked a defense contractor to estimate the life-cycle cost for a proposed light-duty
support vehicle
...
Use the cost estimates (shown in $1 million) for the 20-year life
cycle to calculate the annual LCC at an interest rate
of 7% per year
...
5
3
...
5
0
...
1

5
...
5
10
...
3
3
...
2
6
...
2

0
...
4
1
...
7
3
...
5
3
...
3
7
...
5

Po͞D

2
...
29

6
...
Reports
on the disposition of each service will also be
entered by ﬁeld personnel, then ﬁled and archived by the system
...
The system is expected to be widely used

over time for other aircraft maintenance scheduling
...
The engineer, who must
make a presentation next week of the best estimates of costs over a 20-year life period, has decided to use the life-cycle cost approach of cost
estimations
...

169

Additional Problems and FE Exam Review Questions
Cost in Year ($ millions)
Cost Category

1

2

3

4

5

6 on 10 18

Field study
0
...
1 1
...
5
Software design
0
...
9
Hardware purchases
5
...
1 0
...
1 0
...
2 0
...
06
development
System implementation
1
...
7
Field hardware
0
...
0 2
...
3 2
...
5 0
...
6 3
...
7

6
...
S
...
Proposal A involves an off-the-shelf
“bare-bones” design and standard grade construction
of walls, windows, doors, and other features
...
The initial cost for
A will be $750,000
...
Minor remodeling will
be required in years 5, 10, and 15 at a cost of $150,000
each time in order to render the units usable for
20 years
...

Proposal B will include tailored design and
construction costs of $1
...

There will be no salvage value at the end of the
20-year life
...
33 A medium-size municipality plans to develop a
software system to assist in project selection during the next 10 years
...
There are three alternatives
under consideration, identiﬁed as M, N, and O
...
Use an annual
life-cycle cost approach to identify the best alternative at 8% per year
...
34 All of the following are fundamental assumptions
for the annual worth method of analysis except:
(a) The alternatives will be needed for only one
life cycle
...

(c) The selected alternative will be repeated for
the succeeding life cycles in exactly the same
manner as for the ﬁrst life cycle
...

6
...

(b) Find the AW of each over the life of the
shortest-lived alternative
...

(d) Find the AW of each alternative over its life
without considering the life of the other
alternatives
...
36 The annual worth of an alternative can be calculated from the alternative’s:
(a) Present worth by multiplying by (A/P,i,n)
(b) Future worth by multiplying by (F/A,i,n)
(c) Either (a) or (b)
(d) Neither (a) nor (b)
6
...
At an interest rate of 10%
per year, the values of n that you could use in the
(A͞P,i,n) factors to make a correct comparison by
the annual worth method are:
A
First cost, $
Annual cost, $ per year
Salvage value, $
Life, years

(a)
(b)
(c)
(d)

B

Ϫ50,000
Ϫ10,000
13,000
3

Ϫ90,000
Ϫ4,000
15,000
6

n ϭ 3 years for A and 3 years for B
n ϭ 3 years for A and 6 years for B
Either (a) or (b)
Neither (a) nor (b)

170

Chapter 6

Annual Worth Analysis

6
...
e
...
At an interest rate of 10% per year, the
equation that represents the perpetual AW of X1 is:
X1
First cost, $
Annual cost, $ per year
Salvage value, $
Life, years

(a)
(b)
(c)

(d)

Y1

Ϫ50,000
Ϫ10,000
13,000
3

Ϫ90,000
Ϫ4,000
15,000
6

AWX1 ϭ Ϫ50,000(0
...
10)
AWX1 ϭ Ϫ50,000(0
...
10) Ϫ 10,000
Ϫ 37,000(P/F,10%,3)(0
...
10)
AWX1 ϭ Ϫ50,000(A/P,10%,3) Ϫ 10,000
ϩ 13,000(A/F,10%,3)

6
...

(b) Multiply $10,000 by (A/F,i,10)
...

(d) Multiply $10,000 by (A/F,i,n) and then
multiply by i
...
40 through 6
...

The alternatives are mutually exclusive and the MARR is
6% per year
...
40 The annual worth of vendor 2 cash ﬂow estimates
is closest to:
(a) $Ϫ63,370
(b) $43,370
(c) $Ϫ43,370
(d) $63,370
6
...
42 The AW values for the alternatives are listed
below
...
43 The capital recovery amount for vendor 3 is:
(a) $40,000 per year
(b) $60,000 per year
(c) $43,370 per year
(d) $100,000 per year
6
...

(b) The estimates are wrong somewhere
...

(d) The alternative should have a longer life so
revenues will exceed costs
...
45 Estimates for one of two process upgrades are as
follows: ﬁrst cost of $40,000, annual cost of $5000
per year, market value that decreases by $2000 per
year to the salvage value of $20,000 after the expected life of 10 years
...
46 The perpetual annual worth of investing $50,000
now and $20,000 per year starting in year 16 and
continuing forever at 12% per year is closest to:
(a) $Ϫ4200
(b) $Ϫ8650
(c) $Ϫ9655
(d) $Ϫ10,655

Case Study

6
...

(b) A monetary estimate of new capital funds required each year for the life of the alternative
...

Does not consider the salvage value, since it
is returned at the end of the alternative’s life
...
The estimates used and the
annual worth analysis at MARR ϭ 15% are summarized
below
...

PowrUp
Cost and installation, $
Annual maintenance cost,
$ per year
Salvage value, $
Equipment repair savings, $
Useful life, years

Lloyd’s

Ϫ26,000
Ϫ800

Ϫ36,000
Ϫ300

2,000
25,000
6

3,000
35,000
10

The spreadsheet in Figure 6–9 is the one Harry used to make
the decision
...
The Lloyd’s protectors were installed
...
In fact, the maintenance contract cost (which includes quarterly inspection) is going from $300 to $1200 per
year next year and will then increase 10% per year for the next
10 years
...
He believes savings will decrease by $2000 per year
hereafter
...

Case Study Exercises
1
...

2
...
If these
estimates had been made 3 years ago, would Lloyd’s
still have been the economic choice?
3
...

CHAPTER 7

Rate of Return
Analysis: One
Project

L E A R N I N G

O U T C O M E S

Purpose: Understand the meaning of rate of return and perform an ROR evaluation of a single project
...
1

Deﬁnition

• State and understand the meaning of rate of
return
...
2

Calculate ROR

• Use a PW or AW relation to determine the
ROR of a series of cash ﬂows
...
3

Cautions

• State the difﬁculties of using the ROR method,
relative to the PW and AW methods
...
4

Multiple RORs

• Determine the maximum number of possible
ROR values and their values for a speciﬁc cash
ﬂow series
...
5

Calculate EROR

• Determine the external rate of return using
the techniques of modiﬁed ROR and return on
invested capital
...
6

Bonds

• Calculate the nominal and effective rate of
return for a bond investment
...
Whether it is an engineering project with cash
ﬂow estimates or an investment in a stock or bond, the rate of return is a wellaccepted way of determining if the project or investment is economically acceptable
...
Correct procedures to calculate a rate of return using a PW or
AW relation are explained here, as are some cautions necessary when the ROR technique is
applied to a single project’s cash ﬂows
...
We will discuss the computation of
ROI in the latter part of this chapter
...
This chapter describes how to recognize this possibility and an approach to ﬁnd the multiple values
...
Two of the techniques are covered: the modiﬁed ROR technique and the ROIC (return on invested capital) technique
...
Finally, the rate of return for a bond investment is discussed
...
1 Interpretation of a Rate of Return Value
From the perspective of someone who has borrowed money, the interest rate is applied to the
unpaid balance so that the total loan amount and interest are paid in full exactly with the last loan
payment
...
The interest rate is the return on this unrecovered balance so that the total amount lent and
the interest are recovered exactly with the last receipt
...

Rate of return (ROR) is the rate paid on the unpaid balance of borrowed money, or the rate
earned on the unrecovered balance of an investment, so that the ﬁnal payment or receipt
brings the balance to exactly zero with interest considered
...
It is
stated as a positive percentage; the fact that interest paid on a loan is actually a negative rate of
return from the borrower’s perspective is not considered
...
In terms of an investment, a return of i ϭ Ϫ100%
means the entire amount is lost
...
Example 7
...

EXAMPLE 7
...
From the lender’s
perspective, the investment in this young engineer is expected to produce an equivalent net
cash ﬂow of $315
...

A ϭ $1000(A͞P,10%,4) ϭ $315
...
Compute the amount
of the unrecovered investment for each of the 4 years using (a) the rate of return on the unrecovered balance (the correct basis) and (b) the return on the initial $1000 investment
...

Solution
(a) Table 7–1 shows the unrecovered balance at the end of each year in column 6 using the
10% rate on the unrecovered balance at the beginning of the year
...

Rate of return

Chapter 7

Rate of Return Analysis: One Project

TA BLE 7–1

Unrecovered Balances Using a Rate of Return of 10% on the Unrecovered
Balance

(1)

(2)

(3) ‫)2( ؋ 01
...
00
Ϫ784
...
51
Ϫ286
...
00
78
...
75
28
...
88

$Ϫ1000
...
47
ϩ315
...
47
ϩ315
...
47
237
...
72
286
...
00

TA BLE 7–2

$Ϫ1000
...
53
Ϫ547
...
79
0

Unrecovered Balances Using a 10% Return on the Initial Amount

(1)

(2)

(3) ‫)2( ؋ 01
...
00
Ϫ784
...
06
Ϫ353
...
00
ϩ315
...
47
ϩ315
...
47

—
$215
...
47
215
...
47
$861
...
00
Ϫ784
...
06
Ϫ353
...
12

(b) Table 7–2 shows the unrecovered balance if the 10% return is always ﬁgured on the initial $1000
...
12,
because only $861
...

(c) As shown in column 3, a total of $400 in interest must be earned if the 10% return each year is
based on the initial amount of $1000
...
88 in interest must be earned if a
10% return on the unrecovered balance is used
...
Figure 7–1 illustrates the correct interpretation of rate of return in Table 7–1
...
00
1000
...
45

$215
...
53
Loan balance, $

174

$237
...
75

547
...
72
$28
...
79

$286
...

Year

7
...
47 receipt represents 10% interest on the unrecovered balance in column 2
plus the recovered amount in column 5
...
Clearly, an interest rate applied only to the principal represents a higher rate than is stated
...
This is sometimes referred to as the installment ﬁnancing problem
...
One popular
example is a “no-interest program” offered by retail stores on the sale of major appliances, audio
and video equipment, furniture, and other consumer items
...
Further,
the program’s ﬁne print may stipulate that the purchaser use a credit card issued by the retail
company, which often has a higher interest rate than that of a regular credit card, for example,
24% per year compared to 15% per year
...
Usually, the correct deﬁnition of i as interest on
the unpaid balance does not apply directly; i has often been manipulated to the ﬁnancial disadvantage of the purchaser
...
4 using the Credit Card Case in
Chapter 4
...
2 Rate of Return Calculation Using a PW
or AW Relation
The ROR value is determined in a generically different way compared to the PW or AW value for
a series of cash ﬂows
...
Using the MARR, which is established independent of any particular project’s cash ﬂows,
a mathematical relation determines the PW value in actual monetary units, say, dollars or euros
...
Therefore, ROR may be
considered a relative measure, while PW and AW are absolute measures
...
Another deﬁnition of rate of return is
based on our previous interpretations of PW and AW
...

To determine the rate of return, develop the ROR equation using either a PW or AW relation, set
it equal to 0, and solve for the interest rate
...
That is, solve for i using either of the relations
0 ‫ ؍‬PW
or

[7
...

0 ‫ ؍‬AW
or

[7
...
It is the root of the ROR
relation
...

Rate of return

176

Chapter 7

Rate of Return Analysis: One Project

Figure 7–2

$1500

Cash ﬂow for which
a value of i is to be
determined
...

If i* Ͻ MARR, the project is not economically viable
...
In rate of return calculations, the objective is to ﬁnd the interest rate i* at which
the cash ﬂows are equivalent
...
For example, if you deposit $1000 now and are promised payments of $500 three years from now and $1500 ﬁve years from now, the rate of return relation
using PW factors and Equation [7
...
3]

The value of i* that makes the equality correct is to be determined (see Figure 7–2)
...
3], we have the form 0 ϭ PW
...
9% by hand using trial and error or using a spreadsheet function
...
Using i* ϭ 16
...
It will show that the unrecovered balances each
year, starting with $Ϫ1000 in year 1, are exactly recovered by the $500 and $1500 receipts in
years 3 and 5
...
That is, if the above interest rate is known to be 16
...
9%,3) ϩ 1500(P͞F,16
...
The only differences are what is given and what is sought
...
The
spreadsheet is faster; the ﬁrst helps in understanding how ROR computations work
...
2
...

i* Using Trial and Error

The general procedure of using a PW-based equation is as follows:

1
...

2
...
1]
...
Select values of i by trial and error until the equation is balanced
...
If the cash ﬂows are combined in such a manner
that the income and disbursements can be represented by a single factor such as P͞F or P͞A, it is

7
...
The problem, then, is to combine the cash ﬂows into the format of only one of the factors
...
Convert all disbursements into either single amounts (P or F) or uniform amounts (A) by
neglecting the time value of money
...
The scheme selected for movement
of cash ﬂows should be the one that minimizes the error caused by neglecting the time value
of money
...

2
...

3
...
The rate obtained is a good estimate for the ﬁrst trial
...
The procedure is illustrated in Example 7
...

i* by Spreadsheet The fastest way to determine an i* value when there is a series of equal
cash ﬂows (A series) is to apply the RATE function
...
The format is
‫ ؍‬RATE(n,A,P,F)

[7
...
The format is
‫ ؍‬IRR(ﬁrst_cell:last_cell,guess)

[7
...

The PW-based procedure for sensitivity analysis and a graphical estimation of the i* value is
as follows:
1
...

3
...

5
...

Set up the ROR relation in the form of Equation [7
...

Enter the cash ﬂows onto the spreadsheet in contiguous cells
...

Use the NPV function to develop a PW graph (PW versus i values)
...

EXAMPLE 7
...
Engineers with Monarch Paints have recommended to management an investment of $200,000 now in novel methods that will reduce the amount of wastewater, packaging
materials, and other solid waste in their consumer paint manufacturing facility
...
Determine the rate of return
using hand and spreadsheet solutions
...

1
...

2
...
1] format for the ROR equation
...
6]

177

178

Chapter 7

Rate of Return Analysis: One Project
$300,000

i* = ?

$15,000

0

1

2

3

4

5

6

7

8

9

10

$200,000

Figure 7–3
Cash ﬂow diagram, Example 7
...

3
...
All income will be regarded
as a single F in year 10 so that the P͞F factor can be used
...
Only for the ﬁrst
estimate of i, deﬁne P ϭ $200,000, n ϭ 10, and F ϭ 10(15,000) ϩ 300,000 ϭ $450,000
...
444
The roughly estimated i is between 8% and 9%
...

0 ϭ Ϫ200,000 ϩ 15,000(P͞A,9%,10) ϩ 300,000(P͞F,9%,10)
0 Ͻ $22,986
The result is positive, indicating that the return is more than 9%
...

0 ϭ Ϫ200,000 ϩ 15,000(P͞A,11%,10) ϩ 300,000(P͞F,11%,10)
0 Ͼ $Ϫ6002
Since the interest rate of 11% is too high, linearly interpolate between 9% and 11%
...
00 ϩ ———————— (2
...
00 ϩ 1
...
58%

Solution by Spreadsheet
The fastest way to ﬁnd i* is to use the RATE function (Equation [7
...
The entry
ϭ RATE(10,15000,Ϫ200000,300000) displays i* ϭ 10
...
It is equally correct to
use the IRR function
...

For a complete spreadsheet analysis, use the procedure outlined above
...

2
...

4
...

Figure 7–3 shows cash ﬂows
...
6] is the ROR relation
...

The IRR function in cell B14 displays i* ϭ 10
...

To graphically observe i * ϭ 10
...
The NPV function is used repeatedly to calculate PW for the xy scatter
chart
...
3

Special Considerations When Using the ROR Method

ϭ NPV(C4,$B$3:$B$12) + $B$2

ϭ IRR(B2:B12)

Figure 7–4
Spreadsheet to determine i* and develop a PW graph, Example 7
...

Just as i* can be found using a PW equation, it may equivalently be determined using an AW
relation
...
Solution by
hand is the same as the procedure for a PW-based relation, except Equation [7
...
In the
case of Example 7
...
55% is determined using the AW-based relation
...
Internally, IRR calculates the NPV function at different i values until NPV ϭ 0
...
)

7
...
As mentioned earlier, an ROR analysis is performed using a different basis than PW and AW analyses
...
As a result, there are some assumptions and special considerations with ROR analysis that must be made when calculating i* and in interpreting its real-world meaning
...

• Multiple i* values
...

This possibility is discussed in Section 7
...

• Reinvestment at i*
...
e
...
However, the ROR method assumes reinvestment at the i* rate
...
g
...

In such cases, the i* value is not a good basis for decision making
...
5
...
To correctly use the ROR method
to choose from two or more mutually exclusive alternatives requires an incremental analysis
procedure that is signiﬁcantly more involved than PW and AW analysis
...

If possible, from an engineering economic study perspective, the AW or PW method at a
stated MARR should be used in lieu of the ROR method
...
And it is easy to
compare a proposed project’s return with that of in-place projects
...

As an illustration, if a project is evaluated at MARR ϭ 15% and has PW Ͻ 0, there is no need to
calculate i*, because i* Ͻ 15%
...

7
...
2 a unique rate of return i* was determined
...
This is called a conventional (or simple) cash ﬂow series
...
Such a series is called nonconventional (nonsimple)
...
Relatively large net cash ﬂow (NCF) changes in amount and sign can
occur in projects that require signiﬁcant spending at the end of the expected life
...
The cash ﬂow diagram will appear
similar to Figure 7–5a
...

TABLE 7–3 Examples of Conventional and Nonconventional Net Cash Flow for
a 6-year Project
Sign on Net Cash Flow by Year
Type of Series

0

1

2

3

4

5

6

Conventional
Conventional
Conventional
Nonconventional
Nonconventional
Nonconventional

Ϫ
Ϫ
ϩ
Ϫ
ϩ
Ϫ

ϩ
Ϫ
ϩ
ϩ
ϩ
ϩ

ϩ
Ϫ
ϩ
ϩ
Ϫ
Ϫ

ϩ
ϩ
ϩ
ϩ
Ϫ
Ϫ

ϩ
ϩ
ϩ
Ϫ
Ϫ
ϩ

ϩ
ϩ
Ϫ
Ϫ
ϩ
ϩ

ϩ
ϩ
Ϫ
Ϫ
ϩ
ϩ

Positive
NCF

0

1

Number of
Sign Changes

Positive
NCF

Positive
NCF

nϪ1 n

2

Year

0

1

1
1
1
2
2
3

nϪ1

2

n

i* = ?
Phaseout
costs
Initial
investment

Initial
investment
(a)

Midlife
investments

i* = ?

(b)

Figure 7–5
Typical cash ﬂow diagrams for projects with (a) large restoration or remediation costs, and (b) upgrade or refurbishment costs
...
4

Multiple Rate of Return Values

When there is more than one sign change in the net cash ﬂows, it is possible that there will be
multiple i* values in the Ϫ100% to plus inﬁnity range
...

Test 1: (Descartes’) rule of signs states that the total number of real-number roots is always less
than or equal to the number of sign changes in the series
...
1] or [7
...
(It is possible that imaginary values or inﬁnity may also satisfy the
equation
...

Zero values in the series are neglected when applying Norstrom’s criterion
...
There may be
negative roots that satisfy the ROR relation, but these are not useful i* values
...
, Sn
...

With the results of these two tests, the ROR relation is solved for either the unique i* or the
multiple i* values, using trial and error by hand, using a programmable calculator, or by spreadsheet using an IRR function that incorporates the “guess” option
...
Examples 7
...
4 illustrate the tests
and solution for i*
...
3
Sept-Îles Aluminum Company operates a bauxite mine to supply its aluminum smelter located about 2 km from the current open pit
...
The
lease for the land will cost $400,000 immediately
...
This
is expected to cost $300,000
...
Perform an ROR analysis that will provide the following information:
(a) Type of cash ﬂow series and possible number of ROR values
(b) PW graph showing all i* values
(c) Actual i* values determined using the ROR relation and spreadsheet function
(d) Conclusions that can be drawn about the correct rate of return from this analysis

Solution
(a) The net cash ﬂows will appear like those in Figure 7–5a with an initial investment of
$Ϫ400,000, annual net cash ﬂow (NCF) of $75,000 for years 1 through 10, and a phaseout cost of $Ϫ300,000 in year 10
...
The
series is nonconventional based on the sign changes throughout the series
...

Test #2: There is one sign change in the cumulative NCF series, which indicates a unique
positive root or one positive i* value
...
3
...
There are two times that
the parabolic-shaped curve crosses the PW ϭ 0 line; these are approximately i1* ϭ Ϫ18%
and i2*ϭ 5%
...
7]

i* values by hand If hand solution is chosen, the same procedure used in Example 7
...
However, the technique to estimate the initial i value will not work
as well in this case since the majority of the cash ﬂows do not ﬁt either the P͞F or the
F͞P factor
...
25%
...
7] with various i values will approximate the
correct answer of about 4
...
This complies with the test results of one positive i* value
...
Entering different values in the
optional “guess” ﬁeld will force the function to ﬁnd multiple i* values, if they exist
...

i1* ϭ Ϫ18
...
53%

This result does not conﬂict with test results, as there is one positive value, but a negative
value also balances the ROR equation
...
53% is accepted as the correct internal rate of return (IROR) for the
project
...

EXAMPLE 7
...
does contract-based work for
automobile manufacturers throughout the world
...

Year

0

1

2

3

Cash Flow ($1000)

ϩ2000

Ϫ500

Ϫ8100

ϩ6800

7
...

(b) Write the PW equation and approximate the i* value(s) by plotting PW vs i
...
Since there are two
sign changes in the cash ﬂow sequence, the rule of signs indicates a maximum of two i*
values
...
As many as two i* values can be found
...
4
Year

Cash Flow
($1000)

Sequence
Number

0
1
2
3

ϩ2000
Ϫ500
Ϫ8100
ϩ6800

Cumulative Cash Flow
($1000)

S0
S1
S2
S3

ϩ2000
ϩ1500
Ϫ6600
ϩ200

(b) The PW relation is
PW ϭ 2000 Ϫ 500(P͞F,i,1) Ϫ 8100(P͞F,i,2) ϩ 6800(P͞F,i,3)
The PW values are shown below and plotted in Figure 7–7 for several i values
...

i%
PW ($1000)

5

10

20

30

40

50

ϩ51
...
55

Ϫ106
...
01

Ϫ11
...
85

Figure 7–7

100

Present worth of cash
ﬂows at several interest
rates, Example 7
...

75

PW (ϫ $1000)

50
25
i%
0
10

20

30 40

50

Ϫ25
Ϫ50
Ϫ75
Ϫ100
Ϫ125

Figure 7–8 presents the spreadsheet PW graph with the PW curve crossing the x axis at
PW ϭ 0 two times
...
The values are
i1* ϭ 7
...
35%

(c) Since both i* values are positive, they are not of much value, because neither can be
considered the true ROR of the cash ﬂow series
...
4
...
This problem is a good example of when an
approach discussed in the next section should be taken
...
The function will ﬁnd the one ROR closest to 10% that satisﬁes the PW relation
...
Often,
the results are unbelievable or unacceptable values that are rejected
...
Assume there are two i* values for a particular cash ﬂow series
...

Both i* Ͼ 0

Discard both values
...

If both i* values are discarded, proceed to the approach discussed in the next section to determine
one rate of return value for the project
...

Always determine the PW or AW at the MARR ﬁrst for a reliable measure of economic justiﬁcation
...

This recommendation is not to dissuade you from using the ROR method
...

7
...
5

Techniques to Remove Multiple Rates of Return

• More than one positive i* value or all negative i* values are obtained when the PW graph and
IRR function are developed, and
• A single, reliable rate of return value is required by management or engineers to make a clear
economic decision
...
The selected approach depends
upon what estimates are the most reliable for the project being evaluated
...

The result of follow-up analysis to obtain a single ROR value when multiple, nonuseful i* values
are present does not determine the internal rate of return (IROR) for nonconventional net
cash ﬂow series
...

We will refer to the resulting value as the external rate of return (EROR) as a reminder that
it is different from the IROR obtained in all previous sections
...
Take the following view: You are
the project manager and the project generates cash ﬂows each year
...
We will call this
the investment rate ii
...
Other years, the net cash ﬂow
will be negative and you must borrow funds from some source to continue
...
Each year, you must consider the time value of money, which must utilize either the
investment rate or the borrowing rate, depending upon the sign on the NCF of the preceding year
...
The resulting ROR value will not be the same for each method, because slightly
different additional information is necessary and the cash ﬂows are treated in slightly different
fashions from the time value of money viewpoint
...
However, the investment and borrowing rates must be reliably estimated, since the results may be quite sensitive to them
...

Return on Invested Capital (ROIC) Approach Though more mathematically rigorous,
this technique provides a more reliable estimate of the EROR and it requires only the investment
rate ii
...

Before covering the techniques, it would be good to review the material in Section 7
...
1
...

Modiﬁed ROR Approach
The technique requires that two rates external to the project net cash ﬂows be estimated
...
This applies to all positive annual NCF
...

• Borrowing rate ib is the rate at which funds are borrowed from an external source to provide
funds to the project
...
The weighted average cost of
capital (WACC) can be used for this rate
...
However, this is not
a good idea as it implies that the company is willing to borrow funds and invest in projects at the
same rate
...
Commonly MARR Ͼ WACC, so usually ii Ͼ ib
...
9 for a quick review
of MARR and WACC and Chapter 10 for a more detailed discussion of WACC
...

Figure 7–9 is a reference diagram that has multiple i* values, since the net cash ﬂows change
sign multiple times
...

1
...

2
...

3
...

FWn ϭ PW0 (F͞P,iЈ%,n)

[7
...
9]

4
...

If iЈ Ͼ MARR, the project is economically justiﬁed
...

As in other situations, on the rare occasion that iЈ ϭ MARR, there is indifference to the project's economic acceptability; however, acceptance is the usual decision
...
5
The cash ﬂows experienced by Honda Motors in Example 7
...
There are
two positive i* values that satisfy the PW relation, 7
...
35% per year
...
Studies indicate that Honda has a WACC of
8
...
Due to the nature of this contract business, any excess funds generated are expected
to earn at a rate of 12% per year
...
5

Techniques to Remove Multiple Rates of Return

Solution by Spreadsheet
Using the information in the problem statement, the rate estimates are as follows:
MARR:
Investment rate, ii:
Borrowing rate, ib:

9% per year
12% per year
8
...
Figure 7–10 shows the result of i' ϭ 9
...
Since 9
...

ϭ MIRR(B2:B5,E2,E3)

Figure 7–10
Spreadsheet application of MIRR function, Example 7
...

It is vital that the interpretation be correct
...
39% is not the internal rate of return
(IROR); it is the external ROR (EROR) based on the two external rates for investing and borrowing money
...

Step 1
...
5%
...
5%,1) Ϫ 8100(P͞F,8
...
Find FW3 of all positive NCF at ii ϭ 12%
...
Find the rate i' at which the PW and FW are equivalent
...
939 (9
...
Since i' Ͼ MARR of 9%, the project is economically justiﬁed using this EROR
approach
...

Return on invested capital (ROIC) is a rate-of-return measure of how effectively a project utilizes
the funds invested in it, that is, funds that remain internal to the project
...

187

188

Chapter 7

Rate of Return Analysis: One Project

The technique requires that the investment rate ii be estimated for excess funds generated in any
year that they are not needed by the project
...
It involves developing a series of future
worth (F) relations moving forward 1 year at a time
...
Usually, ii is set equal to the MARR
...
The ROIC method uses the following procedure to determine a single external rate of return iЉ and to evaluate the economic viability of the
project
...

1
...
, n years)
...
10]
where Ft ‫ ؍‬future worth in year t based on previous year and time value of money
NCFt ‫ ؍‬net cash ﬂow in year t
i
if Ft؊1 Ͼ 0
(extra funds available)
k‫ ؍‬i
iЉ
if Ft؊1 Ͻ 0
(project uses all available funds)

{

2
...
The iЉ value is the ROIC for the speciﬁed investment rate ii
...
Fortunately, the Goal Seek spreadsheet tool can assist in the determination of iЉ because there is only one unknown in the Fn relation and the target value is zero
...
6
...
The guideline for economic decision making is the same as above, namely,
If ROIC Ն MARR, the project is economically justiﬁed
...

It is important to remember that the ROIC is an external rate of return dependent upon
the investment rate choice
...
This is a separate technique to ﬁnd a single rate for the project
...
6
Once again, we will use the cash ﬂows experienced by Honda Motors in Example 7
...
Use the ROIC method to determine the EROR value
...

Year

0

Net Cash Flow ($1000) ϩ2000

1
Ϫ500

2
Ϫ8100

3
ϩ6800

Solution by Hand

The hand solution is presented ﬁrst to provide the logic of the ROIC method
...
Figure 7–11
details the cash ﬂows and tracks the progress as each Ft is developed
...
10] is applied to develop each Ft
...
Year 0:
Year 1:

F0 ϭ $ϩ2000
Since F0 Ͼ 0, externally invest in year 1 at ii ϭ 12%
...
12) Ϫ 500 ϭ $ϩ1740
Since F1 Ͼ 0, use ii ϭ 12% for year 2
...
5

189

Techniques to Remove Multiple Rates of Return
$6800

$6800

F0 ‫0002$ ؍‬

$6800

F1 ‫0471$ ؍‬
F3 = 0
at i؆ = ?

0

1
$500

2

3

0

1

2

3

0

1

2

3

0

1

2

3

Year

F2 ‫1516؊$ ؍‬
$8100

$8100

(a)

(b)

(c)

(d)

Figure 7–11
Application of ROIC method at ii ϭ 12% per year: (a) original cash ﬂow; equivalent form in (b) year 1, (c) year 2, and (d) year 3
...
12) Ϫ 8100 ϭ $Ϫ6151
Now F2 Ͻ 0, use iЉ for year 3, according to Equation [7
...

Year 3:
F3 ϭ Ϫ6151(1 ϩ iЉ) ϩ 6800
This is the last year
...

Go to step 2
...
Solve for iЉ ϭ ROIC from F3 ϭ 0
...
1055 (10
...
Since ROIC > MARR ϭ 9%, the project is economically justiﬁed
...
The future worth values F1 through F3 are determined by the conditional IF statements in rows 3 through 5
...
In each year, Equation [7
...
If there are surplus funds generated by the project,
Ft–1 Ͼ 0 and the investment rate ii (in cell E7) is used to ﬁnd Ft
...

Figure 7–12
Spreadsheet application of ROIC method using Goal Seek, Example 7
...

190

Chapter 7

Rate of Return Analysis: One Project

The Goal Seek template sets the F3 value to zero by changing the ROIC value (cell E8)
...
55% per year
...
55% Ͼ 9%, the MARR, the project
is economically justiﬁed
...
55%) is different than the MIRR rate (9
...
Also
these are both different than the multiple rates determined earlier (7
...
35%)
...

Now that we have learned two techniques to remove multiple i* values, there are some connections between the multiple i* values, the external rate estimates, and the resulting external
rates (iЈ and iЉ) obtained by the two methods
...
That is, all four parameters have the same value
...

Finally, it is very important to remember the following fact
...
When the MARR is established, this is, in effect, ﬁxing the i* value
...

7
...
One very common form of IOU is a bond—a longterm note issued by a corporation or a government entity (the borrower) to ﬁnance major projects
...
Bonds are usually issued in face value amounts of $1000, $5000,
or $10,000
...
The bond dividend is paid c times per
year
...
The amount of interest is
determined using the stated dividend or interest rate, called the bond coupon rate b
...
11]

There are many types or classiﬁcations of bonds
...
For example, Treasury securities are issued in different monetary amounts ($1000 and
up) with varying periods of time to the maturity date (Bills up to 1 year; Notes for 2 to 10 years)
...
S
...
” The safe investment rate indicated
in Figure 1Ϫ6 as the lowest level for establishing a MARR is the coupon rate on a U
...
Treasury
security
...

7
...
7
General Electric just released $10 million worth of $10,000 ten-year bonds
...
(a) Determine the amount a purchaser will receive each
6 months and after 10 years
...
What are the dividend amounts and the ﬁnal payment amount at the maturity date?

Solution
(a) Use Equation [7
...

10,000 (0
...

(b) Purchasing the bond at a discount from face value does not change the dividend or ﬁnal repayment amounts
...

The cash ﬂow series for a bond investment is conventional and has one unique i*, which is
best determined by solving a PW-based rate of return equation in the form of Equation [7
...

EXAMPLE 7
...

It is offering small-denomination bonds at a discount price of $800 for a 4% $1000 bond
that matures in 20 years with a dividend payable semiannually
...
The PW-based equation for calculating the rate of return is
0 ϭ Ϫ800 ϩ 20(P͞A,i*,40) ϩ 1000(P͞F,i*,40)
Solve by the IRR function or by hand to obtain i* ϭ 2
...
The nominal interest rate per year is computed by multiplying i* by 2
...
8435)(2) ϭ 5
...
5], the effective annual rate is
ia ϭ (1
...
7678%

191

192

Chapter 7

Rate of Return Analysis: One Project

EXAMPLE 7
...
He took a ﬁnancial risk and bought a bond from a corporation that
had defaulted on its interest payments
...
The bond paid no interest for the ﬁrst 3 years after Gerry bought it
...
Perform hand and spreadsheet analysis
...
08)
I ϭ —————— ϭ $200 per quarter
4
The effective rate of return per quarter can be determined by solving the PW equation developed on a per quarter basis
...
1% per quarter, which is a nominal 16
...

Solution by Spreadsheet

Once all the cash ﬂows are entered into contiguous cells, the function ϭ IRR(B2:B42) is used
in Figure 7–13, row 43, to display the answer of a nominal rate of return of 4
...

(Note that many of the row entries have been hidden to conserve space
...
10%(4) ϭ 16
...

Figure 7–13
Spreadsheet solution for
a bond investment,
Example 7
...

If a bond investment is being considered and a required rate of return is stated, the same
PW-based relation used to find i* can be used to determine the maximum amount to pay for
the bond now to ensure that the rate is realized
...
As an illustration, in the last
example, if 12% per year, compounded quarterly, is the target MARR, the PW relation is
used to find the maximum that Gerry should pay now; P is determined to be $6004
...

0 ϭ ϪP ϩ 200(P͞A,3%,28)(P͞F,3%,12) ϩ 11,000(P͞F,3%,40)
P ϭ $6004

Problems

193

CHAPTER SUMMARY
The rate of return of a cash ﬂow series is determined by setting a PW-based or AW-based relation
equal to zero and solving for the value of i*
...
Most people, however, can have considerable difﬁculty in calculating a rate of return
correctly for anything other than a conventional cash ﬂow series
...
The maximum number of i* values is equal to the number of
changes in the sign of the net cash ﬂow series (Descartes’ rule of signs)
...

When multiple i* values are indicated, either of the two techniques covered in this chapter can
be applied to ﬁnd a single, reliable rate for the nonconventional net cash ﬂow series
...
Usually, the investment rate is set
equal to the MARR, and the borrowing rate takes on the historical WACC rate
...

If an exact ROR is not necessary, it is strongly recommended that the PW or AW method at
the MARR be used to decide upon economic justiﬁcation
...
1 Under what circumstances would the rate of return
be (a) Ϫ100%, and (b) inﬁnite?
7
...
How much extra interest did
the company pay?
7
...
6 In 2010, the city of Houston, Texas, collected
$24,112,054 in ﬁnes from motorists because of
trafﬁc violations caught by red-light cameras
...
The
net proﬁt, that is, proﬁt after operating costs, is
split equally (that is, 50% each) between the city
and the operator of the camera system
...
4 Assume you borrow $50,000 at 10% per year interest and you agree to repay the loan in ﬁve equal
annual payments
...
7 P&G sold its prescription drug business to
Warner-Chilcott, Ltd
...
1 billion
...
5 International Potash got a $50 million loan amortized over a 10-year period at 10% per year interest
...

(a) What is the amount of each payment?
(b) What is the total amount of interest paid?
How does the total interest paid compare
with the principal of the loan?

7
...
As a result of the claimant payouts,
insurance companies raised homeowners' insurance rates by an average of $59 per year for each
of the 160,000 households in the affected city
...
9 Determine the rate of return for the cash ﬂows shown
in the diagram
...
)
$7000

i=?

0

1

2

3

4

$200

$200

5

6

7

$90

$90

8

$200

Year

$90

$3000

to 1,694,247 in 2015
...
14 U
...
Census Bureau statistics show that the annual
earnings for persons with a high school diploma
are $35,220 versus $57,925 for someone with a
bachelor’s degree
...
(Hint: The investment in
years 1 through 4 is the cost of college plus the
foregone earnings, and the income in years 5
through 35 is the difference in income between a
high school diploma and a bachelor’s degree
...
10 The Ofﬁce of Naval Research sponsors a contest
for college students to build underwater robots that
can perform a series of tasks without human intervention
...
If the
team spent $2000 for parts (at time 0) and the project took 2 years, what annual rate of return did the
team make?

7
...

The bonds carried a 5
...
The U
...
economy was
in a recession at that time, so as part of the federal
stimulus program, the Utility gets a 35% reimbursement on the dividend it pays
...
11 For the cash ﬂows shown, determine the rate of
(b) What is the total dollar amount the Utility will
return
...
12 In an effort to avoid foreclosure proceedings on
struggling mortgage customers, Bank of America
proposed an allowance that a jobless customer
make no payment on their mortgage for up to
9 months
...
The bank would give them
$2000 for moving expenses
...
If the bank
saved $40,000 in foreclosure costs, what rate of
return per month did the bank make on the allowance? Assume the ﬁrst payment that was skipped
was due at the end of month 1 and the $40,000
foreclosure savings and $2000 moving expense
occurred at the end of the 9-month forbearance
period
...
13 The Closing the Gaps initiative by the Texas
Higher Education Coordinating Board established
the goal of increasing the number of students in
higher education in Texas from 1,064,247 in 2000

7
...
The contract also required the company to provide buyout packages
for 400 workers
...

7
...
Rubbersidewalks, Inc
...
The District of Columbia spent $60,000 for
a rubber sidewalk to replace broken concrete in a
residential neighborhood lined with towering willow oaks
...
18 Efﬁcient light jets (ELJs) are smaller aircraft that
may revolutionize the way people travel by plane
...
5 and $3 million, seat 5 to 7
people, and can ﬂy up to 1100 miles at cruising
speeds approaching 425 mph
...
The company invested $500 million
(time 0) and began taking orders 2 years later
...
8 million, what rate of return
will the company make over a 10-year planning
period? Assume 500 of the planes are delivered
each year in years 6 through 10 and that the company’s M&O costs average $10 million per year in
years 1 through 10
...
)
7
...
E-sports entertainment in
New York City purchased ﬁve machines for $6000
each and took in an average of $600 total per week
in sales
...

7
...
The project
took 10 years of planning and cost $4 million
...
What rate of return does the
venture represent, if increased ﬁshing and recreation activities are valued at $270,000 per year
beginning in year 11 and they continue in perpetuity? (If assigned by your instructor, show both
hand and spreadsheet solutions
...
24 According to Descartes’ rule of signs, what is the
maximum number of real-number values that will
balance a rate of return equation?
7
...
26 According to Norstrom’s criterion, there are two
requirements regarding the cumulative cash ﬂows
that must be satisﬁed to ensure that there is only
one positive root in a rate of return equation
...
27 According to Descartes’ rule of signs, how many
possible i* values are there for the cash ﬂows
shown?
Year

1

2

3

4

5

6

Net Cash ϩ4100 Ϫ2000 Ϫ7000 ϩ12,000 Ϫ700 ϩ800
Flow, $

7
...
29 According to Descartes’ rule and Norstrom’s criterion, how many i* values are possible for the cash
ﬂow (CF ) sequence shown?
Year

1

2

3

4

5

Net Cash
ϩ16,000 Ϫ32,000 Ϫ25,000 ϩ50,000 Ϫ8,000
Flow, $
Cumulative ϩ16,000 Ϫ16,000 Ϫ41,000 ϩ9,000 ϩ1,000
CF, $

7
...

Multiple ROR Values

Year

7
...
22 Explain at least three types of projects in which
large net cash ﬂow changes may cause sign changes
during the life of the project, thus indicating the
possible presence of multiple ROR values
...
31 Stan-Rite Corp of Manitowoc, Wisconsin, is a B to
B company that manufactures many types of industrial products, including portable measuring
arms with absolute encoders, designed to perform
3D inspections of industrial parts
...

7
...

0

1

2

3

4

25,000 15,000
4,000
18,000
Ϫ30,000 Ϫ7,000 Ϫ6,000 Ϫ12,000

196

Chapter 7

Rate of Return Analysis: One Project

Year

Expenses

Revenues

0
1
2
3
4
5
6

$Ϫ30
Ϫ20
Ϫ25
Ϫ15
Ϫ22
Ϫ20
Ϫ30

$0
18
19
36
52
38
70

7
...
At the end of February, she spent this
$50 and an additional $150 to buy clothes
...
Her
conclusion was that over the 4 months, she had received $25 more than she spent
...
If so, determine the multiple rates and comment on their validity
...
33 Veggie Burger Boy sells franchises to individuals
who want to start small in the sandwiches-forvegeterians business and grow in net cash ﬂow
over the years
...
He was allowed to borrow at
the end of his ﬁrst year from the corporation’s capital incentive fund with a promise to repay the loan
in addition to the annual share that the corporation
contractually receives from annual sales
...

Year

0

1

NCF, $ 5000 Ϫ10,100

2
500

3

4

5

6

2000 2000 2000 2000

The corporate chief ﬁnancial ofﬁcer (CFO) has
some questions concerning this NCF series
...

(a) Plot the PW versus i graph and estimate the
rate of return for this franchise
...

(c) Basing your conclusions on Descartes’ and
Norstrom’s rules, provide the CFO with
some advice on what ROR value is the most
reliable for this franchise over the 6-year period
...

based on an older technology to produce meat
products
...
In 2012, prior to the sale of the
facility and property, Vaught spent $1 million to
make the site environmentally acceptable to a potential buyer
...
Use a spreadsheet to do
the following
...

(b) Find all rates that are real numbers between
Ϫ25% and ϩ50%, and calculate the PW
value for interest rates in this range
...

Year

Ϫ5

Ϫ6 Ϫ10

7
...
The service was not well received
after the ﬁrst year and was removed from the market
...
Now, in year
5, VistaCare has spent a large sum on research to
broaden the application of this service
...
NCF values
are in $1 million units
...
36 In calculating the external rate of return by the
modiﬁed rate of return approach, it is necessary to
use two different rates of return, the investment rate
ii and the borrowing rate ib
...
37 In the modiﬁed rate of return approach for determining a single interest rate from net cash ﬂows,
state which interest rate is usually higher, the investment rate ii or the borrowing rate ib
...

7
...

Year

0

1

2

3

Net Cash Flow, $ ϩ16,000 Ϫ32,000 Ϫ25,000 ϩ70,000

7
...
34 In 2011, Vaught Industries closed its plant in
Marionsville following labor, environmental, and
safety problems
...

The cash ﬂows shown are those he recorded for the
ﬁrst 6 years as his own boss
...
(After using the procedure, use the MIRR
function to conﬁrm your answer
...
40 Samara, an engineer working for GE, invested her
bonus money each year in company stock
...
e
...
At the end
of year 7, she sold the stock for $52,000 to buy a
condo; she purchased no stock that year
...
Samara
sold all of the remaining stock for $28,000 immediately after the investment at the end of year 10
...

(b) Determine the external rate of return by
hand, using the modiﬁed rate of return approach with an investment rate of 12% per
year and a borrowing rate of 8%
...

(d) Enter the cash ﬂows into a spreadsheet, and use
the IRR function to ﬁnd the i* value
...
Explain why this is so, given that the investment
rate is 12% per year
...
)
7
...
of Solon, Ohio, makes variable area
ﬂowmeters (VAFs) that measure liquid and gas ﬂow
rates by means of a tapered tube and ﬂoat
...
The revenue was $160,000 per year in
years 1 through 10
...

7
...

(a)
(b)

Cash Flow, $1000

0
1
2
3
4

Calculate the external rate of return using the
return on invested capital (ROIC) approach
with an investment rate of 15% per year
...
)

6

NCF, $ Ϫ9000 ϩ4100 Ϫ2000 Ϫ7000 ϩ12,000 ϩ700 ϩ800

Year

(c)

Ϫ65
30
84
Ϫ10
Ϫ12

Determine the number of positive roots to the
rate of return relation
...

7
...

Year

Cash Flow, $

0
1
2
3
4

3000
Ϫ2000
1000
Ϫ6000
3800

7
...
The
product did poorly after only 1 year on the market
...
New development funds have been expended this year
(year 5) at a cost of $1
...
Determine the
external rate of return using the ROIC approach
and an investment rate of 15% per year
...
1% per year
...
45 What is the bond coupon rate on a $25,000 mortgage bond that has semiannual interest payments
of $1250 and a 20-year maturity date?
7
...
What are the amount and frequency of the dividend payments?
7
...
48 What is the present worth of a $50,000 debenture
bond that has a bond coupon rate of 8% per year,
payable quarterly? The bond matures in 15 years
...

198

Chapter 7

Rate of Return Analysis: One Project

7
...
If the
bond maturity date is 20 years from the date they
were issued and the interest rate in the marketplace
is now 12% per year, compounded semiannually,
what is the present worth (now) of one bond?

issued 30-year bonds with a face value of $25 million
...
Because the market interest
rate increased immediately before the bonds were
sold, the city received only $23
...
What was the semiannual interest rate
when the bonds were sold?

7
...
125 million to improve the Van
Buren dam in central El Paso and to ﬁnance three
other drainage projects
...
If the bond
dividend rate would have been 4% per year, payable quarterly, with a bond maturity date 18 years
after issuance, what is the present worth of the
dividend savings to EPWU rate payers? Assume
the market interest rate is 6% per year
...
55 An investor who purchased a $10,000 mortgage
bond today paid only $6000 for it
...

Because the bond is in default, it will pay no dividend for the next 2 years
...
51 A recently issued industrial bond with a face value
of $10,000 has a coupon rate of 8% per year, payable
annually
...
Jeremy is interested in buying one bond
...
52 Due to a signiﬁcant troop buildup at the local military base, a school district issued $10,000,000 in
bonds to build new schools
...
If an investor is able to purchase one of the bonds that has a face value of
$5000 for $4800, what rate of return per 6 months
will the investor realize? Assume the bond is kept
to maturity
...
53 As the name implies, a zero-coupon bond pays no
dividend, only the face value when it matures
...
54 To provide infrastructure in the outlying areas of
Morgantown, West Virginia, the city council

7
...
The bonds have a call date of this year if
GSI decides to take advantage of it
...
If the company buys
the bonds back now for $11 million, determine the
rate of return that the company will make (a) per
quarter and (b) per year (nominal)
...
)
7
...
Market interest rates
dropped, and the company called the bonds (i
...
,
paid them off in advance) at a 10% premium on the
face value
...
5 million to retire the bonds
...
58 All of the following mean the same as rate of
return except:
(a) Internal rate of return
(b) Time for return of capital
(c) Interest rate
(d) Return on investment

7
...
60 The internal rate of return on an investment refers
to the interest rate earned on the:
(a) Initial investment
(b) Unrecovered balance of the investment
(c) Money recovered from an investment
(d) Income from an investment
7
...

(b) The interest rate you get is a simple interest
rate
...

(d) The total of the cumulative cash ﬂows is
equal to 0
...
62 According to Descartes’ rule of signs, for a net
cash ﬂow sequence of Ϫ ϪϩϩϪϩ, the number of
possible i values is:
(a) 2
(b) 3
(c) 4
(d) 5
7
...

(b) The cumulative cash ﬂow must start out
negatively
...

(d) The net cash ﬂow must start out positively
...
64 According to Descartes’ rule and Norstrom’s criterion, the number of positive i* values for the
following cash ﬂow sequence is:
Year

1

Revenue, $
Cost, $

(a)
(b)
(c)
(d)

2

3

4

25,000
30,000

15,000
7,000

4,000
6,000

18,000
12,000

1
2
3
4

7
...
66 A company that uses a minimum attractive rate of
return of 10% per year is evaluating new processes
to improve operational efﬁciency
...

Alternative I

Alternative J

Ϫ40,000
Ϫ15,000
5,000
3

First cost, $
Annual cost, $ per year
Salvage value, $
Life, years

Ϫ50,000
Ϫ12,000
5,000
6

The statement that is most correct is:
(a) The alternatives are revenue alternatives
...

(c) The alternatives are revenue alternatives and
DN is an option
...

7
...
uses a MARR of 8% per
year
...
The estimate associated with the process follows
...
68 When one is using the modiﬁed ROR method to remove multiple ROR values, an additional estimate
needed besides the cash ﬂows and their timings is:
(a) The ROIC value
(b) External rate of return
(c) Investment rate
(d) Internal rate of return
7
...
The correct
computation for the present worth in year 0 is:
Year

1

NCF, $

(a)
(b)

2

3

Ϫ10,000

0

0

4

5

Ϫ19,000 ϩ25,000

Ϫ10,000 Ϫ 19,000(P͞F,12%,4)
Ϫ10,000 Ϫ 19,000(P͞F,12%,4) ϩ
25,000(P͞F,10%,5)

200

Chapter 7

(c)
(d)

25,000(P͞F,10%,5)
Ϫ10,000 Ϫ 19,000(P͞F,10%,4)

7
...
If the future worth computation in year t is
Ft Ͻ 0, the ROIC rate iЉ is used
...

(b) The resulting external rate of return will be
positive
...

(d) The sequence has nonremovable negative
ROR values
...
71 The meaning of return on invested capital for a
corporation is best stated as:
(a) A rate-of-return measure that equates the internal and external ROR
(b) A measure of how effectively the corporation
uses capital funds invested in it
(c) The value at which borrowing ROR and investing ROR are equal
(d) The external rate of return value is based on
total capital invested
7
...
The bond matures
20 years from now
...
73 A $20,000 mortgage bond that is due in 1 year
pays interest of $500 every 3 months
...
5% per year, payable quarterly
(b) 5% per year, payable quarterly
(c) 5% per year, payable semiannually
(d) 10% per year, payable quarterly
7
...
If an investor purchases the bond now for $9000 and holds it to maturity, the rate of return received can be determined
by the following equation:
(a) 0 ϭ Ϫ9000 ϩ 400(P͞A,i,10)
ϩ 10,000(P͞F,i,10)
(b) 0 ϭ Ϫ9000 ϩ 400(P͞A,i,20)
ϩ 10,000(P͞F,i,20)
(c) 0 ϭ Ϫ10,000 ϩ 400(P͞A,i,20)
ϩ 10,000(P͞F,i,20)
(d) 0 ϭ Ϫ9000 ϩ 800(P͞A,i,10)
ϩ 10,000(P͞F,i,10)
7
...
The bond is for sale
now for $4500
...
Over time they decided to work on solar energy production ideas
...
For residential applications, the collector could be mounted along side a
TV dish and be programmed to track the sun
...

The system serves as a supplement to the electricity provided
by the local power company
...
This was great news for lowincome dwellers on government subsidy that are required to
pay their own utility bills
...
Net cash ﬂow after all expenses, loan repayment,
and taxes for the ﬁrst 4 years was acceptable; $55,000 at the

Case Study

end of the ﬁrst year, increasing by 5% each year thereafter
...
However, after serious discussion replaced the initial
excitement of the sales offer, the trio decided to not sell at this
time
...

During the next year, the ﬁfth year of the partnership, the
engineer who had received the patents upon which the collector and generator designs were based became very displeased
with the partnering arrangements and left the trio to go into
partnership with an international ﬁrm in the energy business
...

Net cash ﬂow dropped to $40,000 in year 5 and continued to
decrease by $5000 per year
...
This
was considered too much of a loss, so the two owners did not
accept
...

It is now 12 years since the system was publicly launched
...

201

Case Study Exercises
It is now 12 years after the products were developed, and the
engineers invested most of their savings in an innovative
idea
...
To help with the analysis, determine the following:
1
...

2
...

3
...

4
...
Is there
any indication that multiple rates of return may be present? If so, use the spreadsheet already developed to
search for ROR values in the range Ϯ100% other than
the one determined in exercise 3 above
...
Assume you are an investor with a large amount of
ready cash, looking for an innovative solar energy product
...
Explain your logic for
offering this amount
...

SECTION

TOPIC

LEARNING OUTCOME

8
...

8
...

8
...

8
...

8
...

8
...

8
...

T

his chapter presents the methods by which two or more alternatives can be evaluated using a rate of return (ROR) comparison based on the methods of the previous
chapter
...
The ROR analysis evaluates the increments between two alternatives in pairwise
comparisons
...

8
...
As we have learned, the PW and AW techniques can be used to do so, and are the recommended methods
...

Let’s assume that a company uses a MARR of 16% per year, that the company has $90,000
available for investment, and that two alternatives (A and B) are being evaluated
...
AlA
ternative B requires $85,000 and has an i* of 29% per year
...
However, this is not
necessarily so
...
What happens to the investment capital that is
left over? It is generally assumed that excess funds will be invested at the company’s MARR,
as we learned in previous chapters
...
If alternative A is selected, $50,000 will return
35% per year
...
The rate
of return on the total capital available, then, will be the weighted average
...
35) ϩ 40,000(0
...
6%
90,000
If alternative B is selected, $85,000 will be invested at 29% per year, and the remaining $5000
will earn 16% per year
...
29) ϩ 5000(0
...
3%
90,000
These calculations show that even though the i* for alternative A is higher, alternative B presents
the better overall ROR for the $90,000
...

This simple example illustrates a major fact about the rate of return method for ranking and
comparing alternatives:
Under some circumstances, project ROR values do not provide the same ranking of alternatives
as do PW and AW analyses
...

When independent projects are evaluated, no incremental analysis is necessary between
projects
...

Therefore, the only comparison is with the do-nothing alternative for each project
...

8
...
Based upon the equivalence relations (PW and AW), ROR
evaluation makes the equal-service assumption
...
Therefore,
the LCM (least common multiple) of lives for each pairwise comparison must be used
...

A format for hand or spreadsheet solutions is helpful (Table 8–1)
...
At the end of each life cycle, the salvage value and initial investment for the next cycle
must be included for the LCM case
...
All
incremental cash ﬂows outside the period are neglected
...

Only for the purpose of simpliﬁcation, use the convention that between two alternatives, the
one with the larger initial investment will be regarded as alternative B
...
1]

The initial investment and annual cash ﬂows for each alternative (excluding the salvage value)
are one of the types identiﬁed in Chapter 5:
Revenue alternative, where there are both negative and positive cash ﬂows
Cost alternative, where all cash ﬂow estimates are negative
Revenue or cost
alternative

In either case, Equation [8
...

EXAMPLE 8
...
The company has the opportunity to buy a
slightly used machine for $15,000 or a new one for $21,000
...
Each machine is expected to have a 25-year life
with a 5% salvage value
...

Solution
Incremental cash ﬂow is tabulated in Table 8–2
...
The salvage values in year 25 are separated
from ordinary cash ﬂow for clarity
...
However, remember that several years were combined when performing
the analysis
...

8
...
1

Year

Used Press

New Press

Incremental
Cash Flow
(New – Used)

0
1–25
25

$Ϫ15,000
Ϫ8,200
ϩ750

$Ϫ21,000
Ϫ7,000
ϩ1,050

$Ϫ6,000
ϩ1,200
ϩ300

Cash Flow

EXAMPLE 8
...
Type A has an initial cost of $70,000 and
a life of 8 years
...

The annual operating cost for type A is expected to be $9000, while the AOC for type B is
expected to be $7000
...

Solution by Hand
The LCM of 8 and 12 is 24 years
...

Solution by Spreadsheet
Figure 8–1 shows the incremental cash ﬂows for the LCM of 24 years
...
The incremental
values in column D are the result of subtractions of column B from C
...
The total incremental cash ﬂow should
agree in both the column D total and the subtraction C29 Ϫ B29
...
This possible dilemma is discussed later in the chapter
...
2

Type A

Type B

Incremental
Cash Flow
(B ؊ A)

$ Ϫ70,000
Ϫ9,000
Ϫ70,000
Ϫ9,000
ϩ5,000
Ϫ9,000

$ Ϫ95,000
–7,000

$Ϫ25,000
ϩ2,000

–7,000

ϩ67,000

–7,000
–95,000
–7,000
ϩ10,000
–7,000

ϩ2,000
Ϫ83,000

–7,000

ϩ67,000

–7,000
–7,000
ϩ10,000
$Ϫ338,000

ϩ2,000

Cash Flow
Year
0
1–7
8
9–11
12
13–15
16
17–23
24

Ϫ9,000
Ϫ9,000
Ϫ70,000
Ϫ9,000
ϩ5,000
Ϫ9,000
Ϫ9,000
ϩ5,000
$Ϫ411,000

ϩ2,000

ϩ7,000
$ϩ73,000

205

206

Chapter 8

Rate of Return Analysis: Multiple Alternatives

Figure 8–1
Spreadsheet computation of incremental
cash ﬂows for
unequal-life alternatives, Example 8
...

Starting new life cycle for A
ϭ initial cost ϩ AOC ϩ salvage
‫000,5 ؉ 000,9 ؊ 000,07 ؊ ؍‬

Check on summations
Incremental column should
equal difference of columns

8
...
This is important in an incremental
ROR analysis in order to determine the ROR earned on the extra funds expended for the largerinvestment alternative
...
In Example 8
...
If the new machine is purchased, there will be a “savings” of $1200 per year
for 25 years, plus an extra $300 in year 25
...
If the
equivalent worth of the savings is greater than the equivalent worth of the extra investment at the
MARR, the extra investment should be made (i
...
, the larger ﬁrst-cost proposal should
be accepted)
...

It is important to recognize that the rationale for making the selection decision is the same as
if only one alternative were under consideration, that alternative being the one represented by the
incremental cash ﬂow series
...
As further clariﬁcation of this extra investment rationale, consider the following: The
rate of return attainable through the incremental cash ﬂow is an alternative to investing at the
MARR
...
1 states that any excess funds not invested in the alternative are assumed to
be invested at the MARR
...

8
...
Accordingly,
prior to performing an incremental ROR analysis, it is advisable to determine the internal rate of
return i* for each alternative
...
The guideline is as follows:
For multiple revenue alternatives, calculate the internal rate of return i* for each alternative,
and eliminate all alternatives that have an i* Ͻ MARR
...

As an illustration, if the MARR ϭ 15% and two alternatives have i* values of 12% and 21%,
the 12% alternative can be eliminated from further consideration
...
If both alternatives have i* Ͻ MARR, no alternative is
justiﬁed and the do-nothing alternative is the best economically
...
Alternatives that cannot meet the MARR may be eliminated from further
evaluation using this option
...
The IRR function applied to each alternative’s cash ﬂow estimates can quickly indicate unacceptable alternatives, as demonstrated in Section 8
...

When independent projects are evaluated, there is no comparison on the extra investment
...
For example, assume MARR ϭ 10%, and three independent projects are available
with ROR values of

*
iA ϭ 12%

i* ϭ 9%
B

*
iC ϭ 23%

Projects A and C are selected, but B is not because i* Ͻ MARR
...
4 Rate of Return Evaluation Using PW:
Incremental and Breakeven
In this section we discuss the primary approach to making mutually exclusive alternative selections by the incremental ROR method
...
Use hand solution or spreadsheet functions to ﬁnd ⌬i* , the internal
B–A
*
ROR for the series
...
(⌬i* may replace ⌬i* when only two alternatives are present
...
Because of the reinvestment requirement for PW analysis for different-life
assets, the incremental cash ﬂow series may contain several sign changes, indicating multiple ⌬i*
values
...
The correct
approach is to follow one of the techniques of Section 7
...
This means that the single external
ROR (⌬iЈ or ⌬iЈЈ) for the incremental cash ﬂow series is determined
...
As stated earlier, it is always possible, and generally advisable, to use a PW or
AW analysis at an established MARR in lieu of the ROR method when multiple rates are indicated
...
Order the alternatives by initial investment or cost, starting with the smaller one, called
A
...

2
...

3
...

Independent project
selection

208

Chapter 8

Rate of Return Analysis: Multiple Alternatives

4
...
If necessary, use Norstrom’s criterion to determine if a single positive root exists
...
Set up the PW ϭ 0 equation and determine ⌬i*
...
Select the economically better alternative as follows:

ME alternative
selection

If ⌬i* Ͻ MARR, select alternative A
...

BϪA
If ⌬i* is exactly equal to or very near the MARR, noneconomic considerations help in
the selection of the “better” alternative
...
For example, if the MARR is 15% per year and you have established that
⌬i* is in the 15% to 20% range, an exact value is not necessary to accept B since you already
BϪA
know that ⌬i* Ն MARR
...
Multiple guess
values can be input to ﬁnd multiple roots in the range Ϫ100% to ϱ for a nonconventional series,
as illustrated in Example 7
...
If this is not the case, to be correct, the indication of multiple roots
in step 4 requires that one of the techniques of Section 7
...

EXAMPLE 8
...
Ford and its suppliers are seeking additional sources for
light, long-life transmissions
...
Two United States–
based vendors make the required dies
...
Show both hand and spreadsheet solutions
...
Use the procedure described above to
determine ⌬i*
...
Alternatives A and B are correctly ordered with the higher ﬁrst-cost alternative in column 2
of Table 8–4
...
The cash ﬂows for the LCM of 10 years are tabulated
...
3
Year
0
1–5

Cash Flow A
(1)
$ Ϫ8,000
Ϫ3,500

5

—

6–10
10

Ϫ3,500
—
$Ϫ43,000

Cash Flow B
(2)
$Ϫ13,000
Ϫ1,600
ϩ2,000
Ϫ13,000
Ϫ1,600
ϩ2,000
$Ϫ38,000

Incremental
Cash Flow
(3) ‫)1( ؊ )2( ؍‬
$ Ϫ5,000
ϩ1,900
Ϫ11,000
ϩ1,900
ϩ2,000
$ ϩ5,000

8
...
3
...
The incremental cash ﬂow diagram is shown in Figure 8–2
...
There are three sign changes in the incremental cash ﬂow series, indicating as many as
three roots
...

5
...
2]

In order to resolve any multiple-root problem, we can assume that the investment rate ii in the
ROIC technique will equal the ⌬i* found by trial and error
...
2] for the
ﬁrst root discovered results in ⌬i* between 12% and 15%
...
65%
...
Since the rate of return of 12
...

Comment
In step 4, the presence of up to three i* values is indicated
...
65%
...
65%, we assume that any
positive net cash ﬂows are reinvested at 12
...
If this is not a reasonable assumption, the
ROIC or modiﬁed ROR technique (Section 7
...

The other two roots are very large positive and negative numbers, as the IRR function of
Excel reveals
...

Solution by Spreadsheet
Steps 1 through 4 are the same as above
...
Figure 8–3 includes the same incremental net cash ﬂows from Table 8–4 calculated in
column D
...
65% using the IRR function
...
3
...
Since the rate of return on the extra investment is greater than the 12% MARR, the highercost vendor B is selected
...
For
example, row 17 uses the NPV function to verify that the present worth is positive at MARRϭ12%
...

The rate of return determined for the incremental cash ﬂow series or the actual cash ﬂows can
be interpreted as a breakeven rate of return value
...
Equivalently, the breakeven ROR is the i
value, i*, at which the PW (or AW) values of two alternatives’ actual cash ﬂows are exactly
equal to each other
...
For example, if the PW versus ⌬i graph for the incremental cash ﬂows in
Table 8–4 (and spreadsheet Figure 8–3) is plotted for various interest rates, the graph shown in
Figure 8–4 is obtained
...
65%
...
65%, the extra investment for B is justiﬁed
...
65%, the opposite is true—the extra investment in B should not be made,
and vendor A is selected
...
65%, the alternatives are equally attractive
...
3, provides the same results
...
65%

1800
For MARR
in this range,
select B

1600

For MARR
in this range,
select A

1400
1200
PW of incremental cash flows, $

Breakeven ROR

1000
800
600
400
200
0
6

7

8

9

10

11

12

14

15

16 ⌬i%

– 200
– 400
– 600
Vendor B

Vendor A

– 800

Figure 8–4
Plot of present worth of incremental cash ﬂows for Example 8
...

8
...
Now, the same conclusions are reached using the
following logic:
• If MARR Ͻ 12
...

• If MARR Ͼ 12
...

• If MARR is exactly 12
...

Example 8
...
More of breakeven analysis is covered in Chapter 13
...
4
New ﬁltration systems for commercial airliners are available that use an electric ﬁeld to remove up
to 99
...
This is vitally important, as many
of the ﬂu germs, viruses, and other contagious diseases are transmitted through the systems that
recirculate aircraft air many times per hour
...
, can also be sizable
...

• Plot two graphs: PW versus i values for both alternatives’ cash ﬂows and PW versus ⌬i
values for incremental cash ﬂows
...

Air Cleanser
(Filter 1)
Initial cost per aircraft, $
Estimated savings, $ per year
Estimated life, years

Purely Heaven
(Filter 2)

Ϫ1000
375

Ϫ1500
700 in year 1, decreasing by
100 per year thereafter
5

5

Solution by Spreadsheet
Refer to Figure 8–6 as the solution is explained
...

The cash ﬂow sign tests for each ﬁlter indicate no multiple rates
...
The PW values of ﬁlter 1 and ﬁlter 2 cash ﬂows are plotted
on the right side for i values ranging from 0% to 60%
...
The higher-cost ﬁlter 2
(Purely Heaven) is selected
...
As expected, the curve crosses the PW ϭ 0 line at approximately 17%, indicating the same economic conclusion of ﬁlter 2
...
3 cash ﬂows
(not incremental)
...
4
...
5

213

Rate of Return Evaluation Using AW

Figure 8–6 provides an excellent opportunity to see why the ROR method can result in selecting the wrong alternative when only i* values are used to select between two alternatives
...
The inconsistency
occurs when the MARR is set less than the breakeven rate between two revenue alternatives
...
In Figure 8–6 the incremental breakeven
rate is 16
...
The MARR is lower than breakeven; therefore, the incremental ROR analysis results in correctly selecting ﬁlter 2
...
41% > 23
...
This
error occurs because the rate of return method assumes reinvestment at the alternative’s ROR
value, while PW and AW analyses use the MARR as the reinvestment rate
...

8
...
However, for the AW-based technique, there are two equivalent ways to perform the
evaluation: (1) using the incremental cash ﬂows over the LCM of alternative lives, just as for the
PW-based relation (Section 8
...
There is no difference between
the two approaches if the alternative lives are equal
...

Since the ROR method requires comparison for equal service, the incremental cash ﬂows
must be evaluated over the LCM of lives
...
The same six-step procedure of the previous section (for
PW-based calculation) is used, except in step 5 the AW-based relation is developed
...
Whether the lives are equal or unequal, set up the AW relation for the cash ﬂows of each
alternative, form the relation below, and solve for i*
...
3]

For both methods, all equivalent values are on an AW basis, so the i* that results from Equation [8
...
Example 8
...

EXAMPLE 8
...
3, using an AW-based
incremental ROR method and the same MARR of 12% per year
...
2], for the incremental cash ﬂow in
Example 8
...
65%
...
Write an AW-based
relation on the incremental cash ﬂow series over the LCM of 10 years, or write Equation [8
...

For the incremental method, the AW equation is
0 ϭ Ϫ5000(A͞P,⌬i*,10) Ϫ 11,000(P͞F,⌬i*,5)(A͞P,⌬i*,10) ϩ 2000(A͞F,⌬i*,10) ϩ 1900

Equal-service
requirement

214

Chapter 8

Rate of Return Analysis: Multiple Alternatives

It is easy to enter the incremental cash ﬂows onto a spreadsheet, as in Figure 8–3, column D,
and use the ϭ IRR(D4:D14) function to display ⌬i* ϭ 12
...

For the second method, the ROR is found using the actual cash ﬂows and the respective
lives of 10 years for A and 5 years for B
...

0 ϭ Ϫ13,000(A͞P,i*,5) ϩ 2000(A͞F,i*,5) ϩ 8000(A͞P,i*,10) ϩ 1900
Solution again yields i* ϭ 12
...

Comment
It is very important to remember that when an incremental ROR analysis using an AW-based
equation is made on the incremental cash ﬂows, the LCM must be used
...
6 Incremental ROR Analysis of Multiple Alternatives
This section treats selection from multiple alternatives that are mutually exclusive, using the incremental ROR method
...
The analysis is based upon PW (or AW) relations for incremental cash ﬂows between
two alternatives at a time
...
When the i* values on several alternatives exceed the MARR, incremental ROR evaluation
is required
...
) For all alternatives (revenue or cost), the incremental investment must be separately justiﬁed
...
1
...

ME alternative
selection

Select the one alternative
That requires the largest investment, and
Indicates that the extra investment over another acceptable alternative is justiﬁed
...

The incremental ROR evaluation procedure for multiple, equal-life alternatives is summarized below
...
The terms defender and challenger are dynamic in that they refer, respectively, to the alternative that is currently selected (the defender)
and the one that is challenging it for acceptance based on ⌬i*
...
The steps for solution by hand or by spreadsheet are as follows:
1
...
Record the annual cash
ﬂow estimates for each equal-life alternative
...
Revenue alternatives only: Calculate i* for the ﬁrst alternative
...
If i* Ͻ MARR, eliminate the alternative and
go to the next one
...
The next alternative is now the challenger
...
(Note: This is where
solution by spreadsheet can be a quick assist
...
Label it the
defender and go to step 3
...
Determine the incremental cash ﬂow between the challenger and defender, using the relation
Incremental cash ﬂow ϭ challenger cash ﬂow Ϫ defender cash ﬂow
Set up the ROR relation
...
6

Incremental ROR Analysis of Multiple Alternatives

4
...
(PW
is most commonly used
...
If ⌬i* Ն MARR, the challenger becomes the defender and the previous defender is eliminated
...

6
...
It is the selected one
...
It is vital that the correct alternatives be compared, or the wrong alternative may be selected
...
6
Caterpillar Corporation wants to build a spare parts storage facility in the Phoenix, Arizona,
vicinity
...
The initial cost of earthwork and prefab building and the annual net cash ﬂow estimates are detailed in Table 8–5
...
If the MARR is 10%, use incremental ROR analysis to select the one economically best location
...
6
A

Initial cost, $
Annual cash ﬂow, $ per year
Life, years

Ϫ200,000
ϩ22,000
30

B

C

Ϫ275,000
ϩ35,000
30

Ϫ190,000
ϩ19,500
30

D
Ϫ350,000
ϩ42,000
30

Solution
All sites have a 30-year life, and they are revenue alternatives
...

1
...

2
...
The ROR relation includes only the
P͞A factor
...
7436 and
⌬ic* ϭ 9
...
Since 9
...
Now the comparison is A to
*
DN, and column 2 shows that ⌬iA ϭ 10
...
This eliminates the do-nothing alternative;
the defender is now A and the challenger is B
...
6
C
(1)

Initial cost, $
Cash ﬂow, $ per year
Alternatives compared
Incremental cost, $
Incremental cash ﬂow, $
Calculated (P͞A,⌬i*,30)
⌬i*,%
Increment justiﬁed?
Alternative selected

Ϫ190,000
ϩ19,500
C to DN
Ϫ190,000
ϩ19,500
9
...
63
No
DN

A
(2)
Ϫ200,000
ϩ22,000
A to DN
Ϫ200,000
ϩ22,000
9
...
49
Yes
A

B
(3)
Ϫ275,000
ϩ35,000
B to A
Ϫ75,000
ϩ13,000
5
...
28
Yes
B

D
(4)
Ϫ350,000
ϩ42,000
D to B
Ϫ75,000
ϩ7,000
10
...
55
No
B

215

216

Chapter 8

Rate of Return Analysis: Multiple Alternatives

3
...
From the interest tables, look up the P͞A factor at the MARR, which is (P͞A,10%,30) ϭ
9
...
Now, any P͞A value greater than 9
...
The P͞A factor is 5
...
For reference
purposes, ⌬i* ϭ 17
...

5
...

6
...
7143 (⌬i* ϭ 8
...
Location D is eliminated,
and only alternative B remains; it is selected
...
Since C was not justiﬁed in
this example, location A was not compared with C
...

To demonstrate how important it is to apply the ROR method correctly, consider the following
...
63

10
...
35

11
...
Location D is selected
...
In fact,
it will earn only 8
...
This is another example of the ranking inconsistency problem of the
ROR method mentioned in Section 8
...

For cost alternatives, the incremental cash ﬂow is the difference between costs for two alternatives
...
Therefore, the
lowest-investment alternative is the initial defender against the next-lowest investment (challenger)
...
7 using a spreadsheet solution
...
7
The complete failure of an offshore platform and the resulting spillage of up to 800,000 to
1,000,000 gallons per day into the Gulf of Mexico in the spring of 2010 have made major oil producers and transporters very conscious of the harm done to people’s livelihood and all forms of
aquatic life by spills of this magnitude
...
The Sierra Club, Greenpeace, and other international environmental interest groups are
in favor of the initiative
...
Annual cost estimates are expected to be
high to ensure readiness at any time
...
5%
...

8
...
7
Machine 1

First cost, $
Annual operating cost, $
Salvage value, $
Life, years

Machine 2

Machine 3

Machine 4

Ϫ5,000
Ϫ3,500
ϩ500
8

Ϫ6,500
Ϫ3,200
ϩ900
8

Ϫ10,000
Ϫ3,000
ϩ700
8

Ϫ15,000
Ϫ1,400
ϩ1,000
8

Solution by Spreadsheet
Follow the procedure for incremental ROR analysis
...

1
...

2
...

3
...

4
...
57% by applying the IRR function
...
This return exceeds MARR ϭ 13
...

The comparison continues for 3-to-2 in column E, where the return is negative at ⌬i* ϭ
Ϫ18
...
Finally the 4-to-2 comparison has an incremental ROR of 13
...
5%
...

ϭ IRR(D6:D14)

Figure 8–7
Spreadsheet solution to select from multiple cost alternatives, Example 8
...

Comment
As mentioned earlier, it is not possible to generate a PW versus i graph for each cost alternative
because all cash ﬂows are negative
...
The curves will
cross the PW ϭ 0 line at the ⌬i* values determined by the IRR functions
...
This is another application of the principle of equal-service comparison
...

It is always possible to rely on PW or AW analysis of the incremental cash ﬂows at the MARR
to make the selection
...
However, it is still necessary to make the comparison over the LCM
number of years for an incremental analysis to be performed correctly
...
7 All-in-One Spreadsheet Analysis (Optional)
For professors and students who like to pack a spreadsheet, Example 8
...
Now that the IRR, NPV, and PV functions are
mastered, it is possible to perform a wide variety of evaluations for multiple alternatives on a
single spreadsheet
...
A nonconventional cash ﬂow series
for which multiple ROR values may be found, and selection from both mutually exclusive alternatives and independent projects, are included in this example
...
8
In-ﬂight texting, phone, and Internet connections provided at airline passenger seats are an
expected service by many customers
...
Four optional
data handling features that build upon one another are available from the manufacturer, but at
an added cost per unit
...
g
...
All four options are expected to boost annual revenues by varying amounts
...

(a) Using MARR ϭ 15%, perform ROR, PW, and AW evaluations to select the one level of
options that is the most promising economically
...
If no budget limitations are considered at this time, which options
are acceptable if the MARR is increased to 20% when more than one option may be
implemented?

Figure 8–8
Spreadsheet analysis using ROR, PW, and AW methods for unequal-life, revenue alternatives, Example 8
...

Chapter Summary

Solution by Spreadsheet
(a) The spreadsheet (Figure 8–8) is divided into six sections:
Section 1 (rows 1, 2): MARR value and the alternative names (A through D) are in increasing order of initial cost
...
These are
revenue alternatives with unequal lives
...

Section 4 (rows 21, 22): Because these are all revenue alternatives, i* values are determined by the IRR function
...
Columns F and H were inserted to make space for the
incremental evaluations
...

Section 5 (rows 23 to 25): The IRR functions display the ⌬i* values in columns F and H
...
Since ⌬i* ϭ 19
...
The
ﬁnal comparison of D to C over 12 years results in ⌬i* ϭ 11
...
Alternative C is the chosen one
...
The AW value over the life
of each alternative is calculated using the PMT function at the MARR with an embedded
NPV function
...
For both measures, alternative C has the numerically largest value, as expected
...

(b) Since each option is independent of the others, and there is no budget limitation at this
time, each i* value in row 21 of Figure 8–8 is compared to MARR ϭ 20%
...
Of the four, options B and C have
i* Ͼ 20%
...

Comment
In part (a), we should have applied the two multiple-root sign tests to the incremental cash ﬂow
series for the C-to-B comparison
...
Therefore, up to three realnumber roots may exist
...
42%
CϪB
without using a supplemental (Section 7
...
This means that the investment assumption of 19
...
If the MARR ϭ 15%, or some other
earning rate were more appropriate, the ROIC procedure could be applied to determine a single
rate, which would be different from 19
...
Depending upon the investment rate chosen,
alternative C may or may not be incrementally justiﬁed against B
...

CHAPTER SUMMARY
Just as present worth and annual worth methods ﬁnd the best alternative from among several,
incremental rate of return calculations can be used for the same purpose
...
The incremental investment evaluation is conducted between only two
alternatives at a time, beginning with the lowest initial investment alternative
...

Rate of return values have a natural appeal to management, but the ROR analysis is often
more difﬁcult to set up and complete than the PW or AW analysis using an established MARR
...

If there is no budget limitation when independent projects are evaluated, the ROR value of
each project is compared to the MARR
...

219

220

Chapter 8

Rate of Return Analysis: Multiple Alternatives

PROBLEMS
Understanding Incremental ROR
8
...
2 If the rate of return on the incremental cash ﬂow
between two alternatives is less than the minimum
attractive rate of return, which alternative should
be selected, if any?
8
...
There are revenue and
cost cash ﬂow estimates
...
The ﬁrst one is 3
...
2% below the MARR, and
the third is 2
...
Which alternatives, if any, must he include in the incremental
ROR analysis?
8
...
Alternative Y requires
a larger investment than alternative X
...
5 Victoria is comparing two mutually exclusive
alternatives, A and B
...

(a) What is known about the ROR on the increment between A and B?
(b) Which alternative should be selected?
8
...
Only one can be selected
...
A more expensive microwave model will yield a rate of return of 22% per year
...
7 If $80,000 is invested at 30% and another $50,000
is invested at 20% per year, what is the overall rate
of return on the entire $130,000?
8
...
If the overall rate

of return on the $100,000 was 30% and the rate of
return on the $30,000 invested in Z1 was 15%,
what was the rate of return on Z2?
8
...
, which manufactures rigid shaft couplings, has $600,000 to invest
...

Project X
Project Y
Project Z

iX ϭ 24%
iY ϭ 18%
iZ ϭ 30%

The initial investment required for each project is
$100,000, $300,000, and $200,000, respectively
...
10 Two options are available for setting up a wireless
meter scanner and controller
...
A more permanent system has a higher
ﬁrst cost of $73,000, but it has an estimated life of
6 years and a salvage value of $15,000
...
If
the two options are compared using an incremental
rate of return, what are the incremental cash ﬂows
in (a) year 0 and (b) year 2?
8
...

Machine X
First cost, $
Annual operating cost, $ per year
Salvage value, $
Life, years

Machine Y

Ϫ35,000
Ϫ31,600
0
2

Ϫ90,000
Ϫ19,400
8,000
4

8
...

Alternative P Alternative Q
First cost, $
Annual operating cost, $ per year
Annual revenue, $ per year
Salvage value, $
Life, years

Ϫ50,000
Ϫ8,600
22,000
3,000
3

Ϫ85,000
Ϫ2,000
45,000
8,000
6

8
...
Alternative A has a 3-year life and alternative B a 6-year life
...
14 Standby power for pumps at water distribution
booster stations can be provided by either gasolineor diesel-powered engines
...
15 Several high-value parts for NASA’s reusable
space exploration vehicle can be either anodized
or powder-coated
...
Determine (a) the rate of return on the
incremental cash ﬂows and (b) which one should be
selected if the company’s MARR is 25% per year
...

Powder Coat

?
Ϫ21,000
?
3

Ϫ65,000
?
6,000
3

The incremental AW cash ﬂow equation associated with (powder coat – anodize) is
0 ϭ –14,000(A͞P,i,3) ϩ 5000 ϩ 2000(A͞F,i,3)
What is (a) the ﬁrst cost for anodizing, (b) the annual cost for powder coating, and (c) the resale
(salvage) value of the anodized parts?
Incremental ROR Comparison (Two Alternatives)
8
...

The incremental cash ﬂow associated with two alternatives for chemical storage and handling systems

Incremental Cash Flow
(X3 ؊ P3), $1000

0
1–9
10

$Ϫ4600
1100
2000

8
...
This situation can be dealt with by drilling a new well at a cost of $1,000,000 or by installing a tank and self-cleaning screen ahead of the
desalting equipment
...
A new well will have a pump that is
more efﬁcient than the old one, and it will require
almost no maintenance, so its operating cost will be
only $18,000 per year
...

8
...

The costs associated with producing chemicallytreated vinyl rollers and ﬁber-impregnated rubber
rollers are shown below
...

Assume the company’s MARR is 21% per year
...
)
Treated
First cost, $
Annual cost, $ per year
Salvage value, $
Life, years

Impregnated

Ϫ50,000
Ϫ100,000
5,000
3

Ϫ95,000
Ϫ85,000
11,000
6

8
...
Design 2B will
cost $3 million to install and $135,000 per year to
maintain
...
7 million to install and $70,000 per year to maintain
...

222

Chapter 8

Rate of Return Analysis: Multiple Alternatives

8
...
Additive A has a ﬁrst cost of
$110,000 and an annual operating cost of $60,000
...
If the company
uses a 3-year recovery period for paint products
and a MARR of 20% per year, which process is
economically favored? Use an incremental ROR
analysis
...
21 The manager of Liquid Sleeve, Inc
...
The costs
associated with each are shown below
...
Do the following
using a PW-based rate of return analysis and a
spreadsheet:
(a) Determine which nanoparticle type the company should select using the ⌬i* value
...
Indicate
the breakeven i* value and the MARR value
on the plot
...
Is the answer the same as in part (a)?
Type Fe
First cost, $
Annual operating cost, $ per year
Salvage value, $
Life, years

Type Al

Ϫ150,000 Ϫ280,000
Ϫ92,000 Ϫ74,000
30,000
70,000
2
4

8
...
The incremental cash
ﬂows between the two alternatives, J and S, have an
incremental rate of return that is less than 40%,
which is the MARR of the company
...
She believes the company can implement cost
controls to reduce the annual cost of the more
expensive process
...
23 The incremental cash ﬂows for two alternative
electrode setups are shown
...

(a) Determine which should be selected using an
AW-based rate of return analysis
...

Year

Incremental Cash Flow
(Dryloc ؊ NPT), $

0
1Ϫ8
9

Ϫ56,000
ϩ8,900
ϩ12,000

8
...
It must decide between two machines for a ﬁnishing operation
...

The company’s MARR is 18% per year
...
25 A manufacturer of hydraulic equipment is trying to
determine whether it should use monoﬂange double block and bleed (DBB) valves or a multi-valve
system (MVS) for chemical injection
...
Use an AW-based rate of return
analysis and a MARR of 18% per year to determine the better of the two options
...
26 Poly-Chem Plastics is considering two types of injection molding machines—hydraulic and electric
...
Electric machine
technology (EMT) will have a ﬁrst cost of $800,000,
annual costs of $150,000, and a salvage value of
$130,000 after 5 years
...

223

Problems

(b)
(c)

Determine which machine the company
should select, if the MARR ϭ 16% per year
...

8
...
He calculated iA ϭ 34
...
2% and recommended acceptance of A
since its rate of return exceeded the established
MARR of 30% by a greater amount than project B
...
Do the following to help
Eduardo better understand the rate of return method
and what this reduction in MARR means
...

(b) Perform the correct analysis using each
MARR value
...

Alternative A
First cost, $
Annual operating cost,
$ per year
Annual revenue, $ per year
Salvage value, $
Life, years
i*, %

a MARR of 10% per year to determine which
alternative is best using an incremental rate of
return analysis
...
2

45,000
20,000
6
29
...
28 Four mutually exclusive revenue alternatives
are under consideration to automate a baking
and packaging process at Able Bakery Products
...
The rate of
return on each increment of investment was less
than the MARR
...
29 A WiMAX wireless network integrated with a
satellite network can provide connectivity to
any location within 10 km of the base station
...
An independent cable operator is considering three bandwidth alternatives
...
30 Xerox’s iGenX high-speed commercial printers
cost $1
...
The machines cost
$500,000 to $750,000 depending on what options
the client selects
...
The
operating costs and revenues generated are related
to a large extent to the speed and other capabilities
of the copier
...
The company uses a
3-year planning period and a MARR of 15% per
year
...

Copier

Initial
Investment,
$

Operating
Cost,
$ per Year

Annual
Revenue,
$ per Year

Salvage
Value, $

iGen-1
iGen-2
iGen-3
iGen-4

–500,000
–600,000
–650,000
–750,000

–350,000
–300,000
–275,000
–200,000

ϩ450,000
ϩ460,000
ϩ480,000
ϩ510,000

ϩ70,000
ϩ85,000
ϩ95,000
ϩ120,000

Alternative B

Ϫ40,000
Ϫ5,500

–40,000
–46,000
–61,000

Operating
Annual
Cost,
Income,
$1000 per Year $1000 per Year

8
...
has determined that any one of
ﬁve machines can be used in one phase of its chili
canning operation
...
If the minimum attractive
rate of return is 20% per year, determine which
machine should be selected on the basis of a rate of
return analysis
...
32 Five revenue projects are under consideration by
General Dynamics for improving material ﬂow
through an assembly line
...

From the calculations, determine which project, if
any, should be undertaken if the company’s MARR
is (a) 11
...
5% per year
...

Comparison

Incremental Rate of Return, %

B vs DN
A vs B
D vs DN
E vs B
E vs D
E vs A
C vs DN
C vs A
E vs DN
A vs DN
E vs C
D vs C
D vs B

13%
19%
11%
15%
24%
21%
7%
19%
12%
10%
33%
33%
29%

A
B
C
D
E

–25,000
–35,000
–40,000
–60,000
–75,000

(a)

(b)

(c)

9
...
1
13
...
4
20
...
3 9
...
3 25
...
5 24
...
5 27
...
8
—

If the alternatives are mutually exclusive and
the MARR is 26% per year, which alternative should be selected?
If the alternatives are mutually exclusive and
the MARR is 15% per year, which alternative should be selected?
If the alternatives are independent and the
MARR is 15% per year, which alternative(s)
should be selected?

8
...
The initial costs and cash ﬂows of each
project are shown
...
9% per year,
which alternative should be selected?

D

E

Ϫ80,000
Ϫ60,000
Ϫ40,000
Ϫ30,000
Ϫ20,000

Overall Incremental Rate of Return, %
ROR, % A
B
C
D
E
14
16
17
12
8

—
12
11
17
24

12
—
14
23
21

11
14
—
35
29

17
23
35
—
17

24
21
29
17
—

8
...
An engineer performed the following analysis to select the best
machine, all of which have a 10-year life
...
35 The plant manager at Automaton Robotics is looking at the summarized incremental rate of return
information shown below for ﬁve mutually exclusive alternatives, one of which must be chosen
...
Which alternative is best if the minimum attractive rate of return
is (a) 15% per year and (b) 12% per year?

A
B
C
D
E

Incremental
Rate of Return, %

B

Initial cost, $
Ϫ7,000 Ϫ23,000 Ϫ9,000 Ϫ3,000 Ϫ16,000
Cash ﬂow, $ per
1,000
3,500 1,400
500
2,200
year
Rate of return
14
...
2
15
...
7
13
...
33 The ﬁve alternatives shown here are being evaluated by the rate of return method
...
6
23
...
1
20
...
37

2 to 1
3 to 2
4 to 3
Ϫ16,000 Ϫ12,000 Ϫ26,000
ϩ6,000

ϩ3,000

ϩ5,000

35
...
4

14
...
S
...
At a MARR of
7
...

Project ID

First
Cost, $1000

Annual
Income, $1000

Rate of
Return, %

A
B
C
D
E

Ϫ20,000
Ϫ10,000
Ϫ15,000
Ϫ70,000
Ϫ50,000

ϩ2000
ϩ1300
ϩ1000
ϩ4000
ϩ2600

10
...
0
6
...
7
5
...
38 When conducting a rate of return (ROR) analysis
involving multiple mutually exclusive alternatives, the ﬁrst step is to:
(a) Rank the alternatives according to decreasing initial investment cost
(b) Rank the alternatives according to increasing
initial investment cost
(c) Calculate the present worth of each alternative using the MARR
(d) Find the LCM between all of the alternatives
8
...
40 When comparing independent projects by the
ROR method, you should:
(a) Find the ROR of each project and pick the
ones with the highest ROR
(b) Select all projects that have an overall ROR Ն
MARR
(c) Select the project with an overall ROR Ն
MARR that involves the lowest initial
investment cost
(d) Select the project with the largest initial investment that has been incrementally justiﬁed
8
...
The only scenario that
requires an incremental investment analysis to
select an alternative is that:
(a) X has an overall ROR of 22% per year, and Y
has an overall ROR of 24% per year
(b) X has an overall ROR of 19% per year, and Y
has an overall ROR of 23% per year
(c) X has an overall ROR of 18% per year, and Y
has an overall ROR of 19% per year
(d) X has an overall ROR of 28% per year, and Y
has an overall ROR of 26% per year
8
...
43 For these alternatives, the sum of the incremental
cash ﬂows is:
Year

A

B

0
1
2
3
4
5

Ϫ10,000
ϩ2,500
ϩ2,500
ϩ2,500
ϩ2,500
ϩ2,500

Ϫ14,000
ϩ4,000
ϩ4,000
ϩ4,000
ϩ4,000
ϩ4,000

(a)
(b)
(c)
(d)

$2500
$3500
$6000
$8000

8
...
uses a minimum attractive
rate of return of 8% per year, compounded annually
...
The cash ﬂow estimates
associated with each process are shown below
...
45 For the four independent projects shown, the one or
ones to select using a MARR of 14% per year are:
Project
A
B
C
D

(a)
(b)
(c)
(d)

Rate of Return,
% per Year
14
12
15
10

Only C
Only A and C
Only A
Can’t tell; need to conduct incremental
analysis

226

Chapter 8

Rate of Return Analysis: Multiple Alternatives

Problems 8
...
48 are based on the following
information
...

Initial
Overall ROR
Alternative Investment, $ versus DN, %

A
B
C
D
E

Ϫ25,000
Ϫ35,000
Ϫ40,000
Ϫ60,000
Ϫ75,000

9
...
1
13
...
4
20
...
3 9
...
3 25
...
5 24
...
5 27
...
8
—

8
...
48 If the projects are independent, instead of mutually
exclusive, the one or ones to select at an MARR of
18% per year are:
(a) B and C
(b) B, D, and E
(c) D and E
(d) B, C, and E

8
...
It will be able to translate digital versions of
three-dimensional computer models, containing a wide variety
of part shapes with machined and highly ﬁnished (ultrasmooth) surfaces
...

Additionally, Make-to-Specs will build the code for superﬁne
ﬁnishing of surfaces with continuous control of the ﬁnishing
machines
...
The server ﬁrst cost
and estimated contribution to annual net cash ﬂow are
summarized below
...
They have
asked that, at this stage of the project, all analyses be performed
using both life estimates for each system
...
If the MARR ϭ 12%, which server should be selected?
Use the PW or AW method to make the selection
...
Use incremental ROR analysis to decide between the
servers at MARR ϭ 12%
...
Use any method of economic analysis to display on the
spreadsheet the value of the incremental ROR between
server 2 with a life estimate of 5 years and a life estimate
of 8 years
...
But
I don’t think we can keep doing the same thing for many more
years
...
Elmer
was sharing thoughts on Gulf Coast Wholesale Auto Parts, a
company he has owned and operated for 25 years on the southern outskirts of Houston, Texas
...
Additionally, Gulf Coast operates a rebuild shop serving these same
retailers for major automobile components, such as carburetors,
transmissions, and air conditioning compressors
...
Part of his job at
Energcon Industries is to perform basic rate of return and
present worth analyses on energy management proposals
...
John summarized
all the estimates over a 10-year horizon
...

Option 1: Remove rebuild
...
The removal of the rebuild operations and the
switch to an “all-parts house” are expected to cost
$750,000 in the ﬁrst year
...
Expenses are projected
at $0
...

Option 2: Contract rebuild operations
...
If expenses stay the same for 5 years, they will average
$1
...

Elmer thinks revenues under a contract arrangement
can be $1
...

Option 3: Maintain status quo and sell out after 5 years
(Elmer’s personal favorite)
...
Projections are $1
...
15 million per year in revenue
...

Elmer’s wish is to sell out completely after 5 more
years at this price, and to make a deal that the new
owner pay $500,000 per year at the end of year 5 (sale
time) and the same amount for the next 3 years
...
Elmer has a close friend in the
antique auto parts business who is making a “killing,”
so he says, with e-commerce
...
The trade-out would cost an estimated $1 million for Elmer immediately
...
Expenses are
estimated at $3 million per year and revenues at
$3
...

Option 5: Lease arrangement
...
The ﬁrst-cut estimates
for this option are $1
...

Case Study Exercises
Help John with the analysis by doing the following:
1
...

2
...
Find
any multiple rates in the range of 0% to 100%
...
If John’s father insists that he make 25% per year or more
on the selected option over the next 10 years, what should
he do? Use all the methods of economic analysis you
have learned so far (PW, AW, ROR) so John’s father can
understand the recommendation in one way or another
...
Prepare plots of the PW versus i for each of the ﬁve options
...

5
...
Stewart, Consultant, Communications and High Tech Solutions Engineering, Accenture LLP
...

SECTION

TOPIC

LEARNING OUTCOME

9
...

9
...

9
...

9
...

9
...

9
...

T

he evaluation methods of previous chapters are usually applied to alternatives
in the private sector, that is, for-proﬁt and not-for-proﬁt corporations and businesses
...
In the case of public projects, the owners and users (beneﬁciaries) are the citizens and residents of a government unit—city, county, state, province,
or nation
...

Public-private partnerships have become increasingly common, especially for large infrastructure projects such as major highways, power generation plants, water resource developments, and the like
...
The different formats of B/C analysis, and associated disbeneﬁts of an alternative, are discussed here
...
Performed correctly, the beneﬁt/cost method will always select the same alternative as PW, AW, and ROR
analyses
...
Finally, there is a discussion on professional
ethics and ethical dilemmas in the public sector
...
An expectation of
over 100,000 new residents in the next
several years and 500,000 by 2040
prompted the development of the plant
starting in 2012
...
The project is
termed WTF3, and its initial capital investment is $540 million for the treatment plant and two large steel-pipe
transmission mains (84- and 48-inch)
that will be installed via tunneling approximately 100 to 120 feet under suburban areas of the city to reach current
reservoirs
...
Besides the
treatment plant construction on the 95acre site, there must be at least three
large vertical shafts (25 to 50 feet in diameter) bored along each transmission
main to gain underground access for
equipment and debris removal during
the tunneling operations
...

There are major long-term beneﬁts for
the new facility
...

• The new treatment plant is at a
higher elevation than the current
two plants, allowing gravity ﬂow to
replenish reservoirs, thereby using
little or no electric pumping
...

• It will provide a water quality that
is more consistent due to the location of the raw water intakes
...

The disbeneﬁts are mostly short-term
during the construction of WTF3 and
transmission mains
...

• Large amounts of dust and smoke
will enter the atmosphere in a
residential area during the 3½ years
of construction, tunneling, and
transmission main completion
...

• Natural landscape in plant and tunnel shaft sites will be destroyed
...

• There may be delays in ﬁre and
ambulance services in emergencies,
since many neighborhood streets are
country-road width and offer only
single ingress/egress streets for neighborhoods along the indicated routes
...

• Newly generated revenues will be
used to pay off the capital funding
bonds approved for the plant’s construction
...
Public and elected ofﬁcial intervention has now caused some of the
conclusions using the criteria mentioned
above to be questioned by the general
manager of Allen Water Utilities
...
1)
Incremental B/C analysis, two alternatives (Section 9
...
4)

9
...
(Notable exceptions are the long-life alternatives discussed in Chapters 5 (PW) and 6 (AW) where capitalized
cost analysis was applied
...
These are called public sector projects
...
The primary purpose is to provide service to the citizenry for the
public good at no proﬁt
...

Upon reflection, it is surprising how much of what we use on a daily or as-needed basis is
publicly owned and financed to serve us—the citizenry
...
1

Public Sector Projects

Police and ﬁre protection
Courts and prisons
Food stamp and rent relief programs
Job training

Public housing
Emergency relief
Codes and standards

There are signiﬁcant differences in the characteristics of private and public sector alternatives
...

Characteristic

Public sector

Private sector

Size of investment

Large

Some large; more medium to small

Often alternatives developed to serve public needs require large initial investments, possibly
distributed over several years
...

Characteristic

Public sector

Private sector

Life estimates

Longer (30–50؉ years)

Shorter (2–25 years)

The long lives of public projects often prompt the use of the capitalized cost method, where
inﬁnity is used for n and annual costs are calculated as A ϭ P(i)
...
For example, at i ϭ 7%,
there will be a very small difference in 30 and 50 years, because (A/P,7%,30) ϭ 0
...
07246
...
Public sector projects
often have undesirable consequences, as interpreted by some sectors of the public
...
The economic analysis should
consider these consequences in monetary terms to the degree estimable
...
) To perform a beneﬁt/cost economic analysis of public alternatives, the costs (initial and
annual), the beneﬁts, and the disbeneﬁts, if considered, must be estimated as accurately as
possible in monetary units
...

Beneﬁts—advantages to be experienced by the owners, the public
...
Disbeneﬁts may be indirect economic disadvantages of the alternative
...
For example, assume a short bypass around a congested area in town is
recommended
...
Relative to revenue cash ﬂow estimates in the private sector, beneﬁt estimates are much harder to make, and vary more widely around uncertain averages
...
5
...
In fact, the disbeneﬁt itself may not be known at the time the evaluation is performed
...
Taxes are collected from those who are the owners—the citizens (e
...
, federal gasoline
taxes for highways are paid by all gasoline users, and health care costs are covered by insurance
premiums)
...
Bonds are often issued:
U
...
Treasury bonds, municipal bond issues, and special-purpose bonds, such as utility district
bonds
...
Also, private donors may provide funding
for museums, memorials, parks, and garden areas through gifts
...
Government
agencies are exempt from taxes levied by higher-level units
...
(Private corporations and individual citizens do pay taxes
...
This results in interest rates in the 4% to 8% range
...

As a matter of standardization, directives to use a speciﬁc interest rate are beneﬁcial because
different government agencies are able to obtain varying types of funding at different rates
...
Standardized rates tend to increase the consistency of economic decisions and to reduce gamesmanship
...
The public sector interest rate is identiﬁed
as i; however, it is referred to by other names to distinguish it from the private sector rate
...

Characteristic
Alternative selection
criteria

Public sector

Private sector

Multiple criteria

Primarily based on rate
of return

Multiple categories of users, economic as well as noneconomic interests, and special-interest
political and citizen groups make the selection of one alternative over another much more difﬁcult in public sector economics
...
It is important to describe and itemize the criteria and selection
method prior to the analysis
...
Viewpoint is discussed below
...
Elected ofﬁcials commonly assist with the selection, especially when pressure is brought to bear by voters, developers, environmentalists, and
others
...

The viewpoint of the public sector analysis must be determined before cost, beneﬁt, and disbeneﬁt estimates are made and before the evaluation is formulated and performed
...

9
...
In general, the viewpoint of the analysis should be as broadly deﬁned as those
who will bear the costs of the project and reap its beneﬁts
...
1
...
1 Water Treatment Facility #3 Case
The situation with the location and construction of the new WTF3 and associated transmission
mains described in the chapter’s introduction has reached a serious level because of recent
questions posed by some city council members and citizen groups
...
The lead consultant, Joel Whiterson,
took engineering economy as a part of his B
...
education and has previously worked on
economic studies in the government sector, but never as the lead person
...
He realized that
no viewpoint of the study was deﬁned, and, in fact, the estimates were never classiﬁed as costs,
beneﬁts, or disbeneﬁts
...

Joel deﬁned two viewpoints: a citizen of Allen and the Allen Water Utilities budget
...

Please help with this classiﬁcation
...
Cost of water: 10% annual increase to Allen
households

Average of $29
...
Bonds: Annual debt service at 3% per year on
$540 million

$16
...
2 million (year 20)

3
...
Property values: Loss in value, sales price,
and property taxes

$4 million (years 1–5)

5
...
M&O: Annual maintenance and operations
costs

$300,000 plus 4% per year increase (years 1–20)

7
...
(How this classiﬁcation is done will vary depending upon who does the analysis
...
)
Viewpoint 1: Citizen of the city of Allen
...

Costs: 1, 2, 4, 6

Beneﬁts: 5, 7

Disbeneﬁts: 3

Viewpoint 2: Allen Water Utilities budget
...

Costs: 2, 3, 6

Beneﬁts: 1, 5, 7

Disbeneﬁts: 4

Citizens view costs in a different light than a city budget employee does
...
Similarly, the Allen Water Utilities budget interprets estimate
3 (payment for use of land to Parks and Recreation) as a real cost; but a citizen might interpret
this as merely a movement of funds between two municipal budgets—therefore, it is a disbeneﬁt, not a real cost
...
However, agreement on
the disbeneﬁts and their monetary estimates is difﬁcult (to impossible) to develop, often resulting in the exclusion of any disbeneﬁts from the economic analysis
...
e
...

Most of the large public sector projects are developed through public-private partnerships
(PPPs)
...
Full
funding by the government unit may not be possible using traditional means—fees, taxes, and
bonds
...
The government unit cannot make a proﬁt, but the corporation(s)
involved can realize a reasonable proﬁt; in fact, the proﬁt margin is usually written into the contract that governs the design, construction, and operation of the project
...
In these formats, a government unit took responsibility for
funding and possibly some of the design elements, and later all operation activities, while the
contractor did not share in the risks involved—liability, natural disasters, funding shortfalls, etc
...

Commonly these are called design-build contracts, under which contractors take on more and
more of the functions from design to operation
...
The most reliance is placed upon
a contractor or contractors with a DBOMF contract, as described below
...
It requires the contractor(s) to perform all the DBOMF activities with collaboration and approval of the owner (the government unit)
...
Although a
contractor may assist in some instances, the funding (obtaining the capital funds) remains the
government’s responsibility through bonding, commercial loans, taxation, grants, and gifts
...
In virtually all cases, some forms of design-build
arrangements for public projects are made because they offer several advantages to the government and citizens served:
• Cost and time savings in the design, build, and operate phases
• Earlier and more reliable (less variable) cost estimates

9
...
There are, of course, disadvantages to this arrangement
...
Another risk is that a reasonable proﬁt may not be realized
by the private corporation due to low usage of the facility during the operate phase
...
The subsidy may cover costs plus (contractually agreed-to) proﬁt if usage
is lower than a speciﬁed level
...

9
...

The B/C analysis was developed to introduce greater objectivity into public sector economics,
and as one response to the U
...
Congress approving the Flood Control Act of 1936
...
All cost and
beneﬁt estimates must be converted to a common equivalent monetary unit (PW, AW, or FW) at
the discount rate (interest rate)
...
1]

Present worth and annual worth equivalencies are preferred to future worth values
...
Salvage values
and additional revenues to the government, when they are estimated, are subtracted from costs in
the denominator
...

Most commonly, disbeneﬁts are subtracted from beneﬁts and placed in the numerator
...

The decision guideline is simple:
If B/C Ն 1
...

If B/C Ͻ 1
...

If the B/C value is exactly or very near 1
...

The conventional B/C ratio, probably the most widely used, is calculated as follows:
beneﬁts ؊ disbeneﬁts B ؊ D
B/C ‫——— ؍ —————————— ؍‬
costs
C

[9
...
2] disbeneﬁts are subtracted from beneﬁts, not added to costs
...
For example, if the numbers
10, 8, and 5 are used to represent the PW of beneﬁts, disbeneﬁts, and costs, respectively, the
correct procedure results in B/C ϭ (10 Ϫ 8)͞5 ϭ 0
...
The incorrect placement of disbeneﬁts
in the denominator results in B/C ϭ 10͞(8 ϩ 5) ϭ 0
...
40
...
However, regardless of whether disbeneﬁts are (correctly) subtracted from the numerator or (incorrectly) added to costs in the denominator, a B/C ratio of
less than 1
...
0 by the second
method, and vice versa
...

Maintenance and operation (M&O) costs are placed in the numerator and treated in a manner

Project evaluation

236

Chapter 9

Beneﬁt/Cost Analysis and Public Sector Economics

similar to disbeneﬁts
...
Once all amounts are
expressed in PW, AW, or FW terms, the modiﬁed B/C ratio is calculated as
beneﬁts ؊ disbeneﬁts ؊ M&O costs
Modiﬁed B/C ‫———————————————— ؍‬
initial investment

[9
...
The modiﬁed B/C ratio
will obviously yield a different value than the conventional B/C method
...

The beneﬁt and cost difference measure of worth, which does not involve a ratio, is based on
the difference between the PW, AW, or FW of beneﬁts and costs, that is, B − C
...
This method has the advantage of eliminating the discrepancies noted
above when disbeneﬁts are regarded as costs, because B represents net beneﬁts
...

Subtracting disbeneﬁts from beneﬁts:
Adding disbeneﬁts to costs:

B Ϫ C ϭ (10 Ϫ 8) Ϫ 5 ϭ Ϫ3
B Ϫ C ϭ 10 Ϫ (8 ϩ 5) ϭ Ϫ3

Before calculating the B/C ratio by any formula, check whether the alternative with the larger
AW or PW of costs also yields a larger AW or PW of beneﬁts
...

By the very nature of beneﬁts and especially disbeneﬁts, monetary estimates are difﬁcult to
make and will vary over a wide range
...
This approach assists in determining the economic and public acceptance risk associated with a deﬁned project
...

EXAMPLE 9
...
In a
proposal for the foundation’s board of directors to construct a new hospital and medical clinic
complex in a deprived central African country, the project manager has developed some estimates
...

Award amount:
Annual costs:
Beneﬁts:
Disbeneﬁts:

$20 million (end of) ﬁrst year, decreasing by $5 million per year for 3
additional years; local government will fund during the ﬁrst year only
$2 million per year for 10 years, as proposed
Reduction of $8 million per year in health-related expenses for citizens
$0
...
6 million per year for removal of arable land and commercial
districts

Use the conventional and modiﬁed B/C methods to determine if this grant proposal is economically justiﬁed over a 10-year study period
...

Solution
Initially, determine the AW for each parameter over 10 years
...
864 per year
$2 per year
$8 per year
Use $0
...
2

Beneﬁt/Cost Analysis of a Single Project

237

The conventional B/C analysis applies Equation [9
...

8
...
6
B/C ϭ —————— ϭ 0
...
864 ϩ 2
...
3]
...
0 Ϫ 0
...
0
Modiﬁed B/C ϭ ——————— ϭ 0
...
864
The proposal is not justiﬁed economically since both measures are less than 1
...
If the low
disbeneﬁts estimate of $0
...

It is possible to develop a direct formula connection between the B/C of a public sector and
B/C of a private sector project that is a revenue alternative; that is, both revenues and costs are
estimated
...
3] and the PW method we have used repeatedly
...
) Let’s neglect the initial investment in year 0 for a moment, and
concentrate on the cash ﬂows of the project for year 1 through its expected life
...
3] may be written as
PW of (B Ϫ D) Ϫ PW of C
Modiﬁed B/C ϭ ————————————
PW of initial investment
This relation can be slightly rewritten to form the proﬁtability index (PI), which can be used to
evaluate revenue projects in the public or private sector
...
, n,
PW of NCFt
PI ‫——————————— ؍‬
PW of initial investment

[9
...
The PI measure of
worth provides a sense of getting the most for the investment dollar (euro, yen, etc
...
This is a “bang for the buck”
measure
...

The evaluation guideline for a single project using the PI is the same as for the conventional
B/C or modiﬁed B/C
...
0, the project is economically acceptable at the discount rate
...
0, the project is not economically acceptable at the discount rate
...
The PI has another name: the present worth index
(PWI)
...
This application is discussed in Chapter 12, Section 12
...

EXAMPLE 9
...
51-mile toll road on
the outskirts of Atlanta’s suburban area
...
Highway construction is expected to require

Project evaluation

238

Chapter 9

Beneﬁt/Cost Analysis and Public Sector Economics

5 years at an average cost of $3
...
The discount rate is 4% per year, and the
study period is 30 years
...

Initial investment: $88 million distributed over 5 years; $4 million now and in year 5 and
$20 million in each of years 1 through 4
...

Annual revenue/beneﬁts: Include tolls and retail/commercial growth; start at $2 million in
year 1, increasing by a constant $0
...

Estimable disbeneﬁts: Include loss of business income, taxes, and property value in surrounding areas; start at $10 million in year 1, decrease by $0
...

Solution
The PW values in year 0 for all estimates must be developed initially usually by hand, calculator, or spreadsheet computations
...
All values are
positive because of the sign convention for B/C and PI measures
...
89
PW of costs ϭ $26
...
41
PW of disbeneﬁts ϭ $80
...
3]
...
41 Ϫ 80
...
87
Modiﬁed B/C ϭ —————————— ϭ 0
...
89
The toll road proposal is not economically acceptable, since B/C Ͻ 1
...

(b) From the private corporation viewpoint, Young Construction will apply Equation [9
...

167
...
87
PI ϭ ——————— ϭ 1
...
89
The proposal is clearly justiﬁed without the disbeneﬁts, since PI > 1
...
The private project
perspective predicts that every investment dollar will return an equivalent of $1
...

Comment
The obvious question that arises concerns the correct measure to use
...
The public project setting will commonly use some form of the B/C ratio with
the disbeneﬁt considered
...
Then the numerical dilemma presented above should not occur
...
3 Alternative Selection Using Incremental
B/C Analysis
The technique to compare two mutually exclusive alternatives using beneﬁt/cost analysis is virtually the same as that for incremental ROR in Chapter 8
...
The higher-cost
alternative is justiﬁed if ⌬B/C is equal to or larger than 1
...
The selection rule is as follows:
ME alternative
selection

If ⌬B/C Ն 1
...

If ⌬B/C Ͻ 1
...

9
...
This same rule
was used for incremental ROR analysis
...
We already know the ﬁrst, all costs have a positive sign in the B/C
ratio
...

Alternatives are ordered by increasing equivalent total costs, that is, PW or AW of all cost
estimates that will be utilized in the denominator of the B/C ratio
...

If two alternatives, A and B, have equal initial investments and lives, but B has a larger equivalent
annual cost, then B must be incrementally justiﬁed against A
...
4
below
...

Follow these steps to correctly perform a conventional B/C ratio analysis of two alternatives
...

1
...

2
...
Calculate the incremental cost (⌬C) for the larger-cost alternative
...

3
...

Calculate the incremental beneﬁts (⌬B) for the larger-cost alternative
...

4
...
2], (B Ϫ D)/C
...
Use the selection guideline to select the higher-cost alternative if ⌬B/C Ն 1
...

When the B/C ratio is determined for the lower-cost alternative, it is a comparison with the donothing (DN) alternative
...
0, then DN should be selected and compared to the second
alternative
...
In public sector analysis, the DN alternative is usually the current condition
...
4
The city of Garden Ridge, Florida, has received designs for a new patient room wing to the
municipal hospital from two architectural consultants
...
The costs and beneﬁts are the same in most
categories, but the city ﬁnancial manager decided that the estimates below should be considered to determine which design to recommend at the city council meeting next week and to
present to the citizenry in preparation for an upcoming bond referendum next month
...
The discount rate is 5%, and the life of the building is
estimated at 30 years
...

(b) Once the two designs were publicized, the privately owned hospital in the directly adjacent
city of Forest Glen lodged a complaint that design A will reduce its own municipal hospital’s income by an estimated $500,000 per year because some of the day-surgery features
of design A duplicate its services
...
The city
ﬁnancial manager stated that these concerns would be entered into the evaluation as disbeneﬁts of the respective designs
...

Solution
(a) Since most of the cash ﬂows are already annualized, the incremental B/C ratio will use
AW values
...
Follow the steps of the procedure
above:
1
...

AWA ϭ 10,000,000(A/P,5%,30) ϩ 35,000 ϭ $685,500
AWB ϭ 15,000,000(A/P,5%,30) ϩ 55,000 ϭ $1,030,750
2
...
The incremental cost is
⌬C ϭ AWB Ϫ AWA ϭ $345,250 per year
3
...
The beneﬁts for the ⌬B/C analysis are not the estimates themselves, but the difference if design B is selected
...

⌬B ϭ usageA Ϫ usageB ϭ $450,000 Ϫ $200,000 ϭ $250,000 per year
4
...
2]
...
72
$345,250
5
...
0, indicating that the extra costs associated with design B
are not justiﬁed
...

(b) The revenue loss estimates are considered disbeneﬁts
...
Now
$350,000
⌬B͞C ϭ ———— ϭ 1
...
In this case the inclusion of disbeneﬁts has reversed the previous economic decision
...

New disbeneﬁts will surely be claimed in the near future by other special-interest groups
...
Usually, the expected useful life of a public project is long (25 or 30 or more years), so
alternatives generally have equal lives
...
As with ROR analysis of two alternatives, this is an excellent opportunity
to use the AW equivalency of estimated (not incremental) costs and beneﬁts, if the implied assumption that the project could be repeated is reasonable
...

EXAMPLE 9
...
The
two options for constructing this main were open trench (OT) for the entire 6
...
3 miles
...

9
...
He stated the equivalent annual costs in an
internal e-mail some months ago, based on the expected construction periods of 24 and
36 months, respectively, as equivalent to
AWOT ϭ $1
...
37 million per year
This analysis indicated that the open-trench option was economically better, at that time
...
Use the estimates below that Joel has unearthed to perform a correct incremental
B/C analysis and comment on the results
...

Open trench (OT)

Trench-tunnel (TT)

6
...
3
Trench for 2
...
3 miles: 2100
36
175,000

150,000
140,000
20,000

20,000
60,000
5,000

Distance, miles
First cost, $ per foot
Time to complete, months
Construction support costs, $ per month
Ancillary expenses, $ per month:
Environmental
Safety
Community interface

Solution
One of the alternatives must be selected, and the construction lives are unequal
...
However, the study period of 50 years is a reasonable evaluation time frame, since the mains are considered permanent installations
...

PWOT ϭ PW of construction ϩ PW of construction support costs
ϭ 700(6
...
20 million per year
PWTT ϭ [700(2
...
3)](5280) ϩ 175,000(12)(P͞A,3%,3)
ϭ $61,010,460
AWTT ϭ 61,010,460(A͞P,3%,50)
ϭ $2
...
The incremental cost is
⌬C ϭ AWTT Ϫ AWOT ϭ 2
...
20 ϭ $1
...

PWOT-anc ϭ 310,000(12)(P͞A,3%,2)
ϭ $7,118,220
AWOT-anc ϭ 7,118,220(A͞P,3%,50)
ϭ $276,685 per year

241

242

Chapter 9

Beneﬁt/Cost Analysis and Public Sector Economics

PWTT-anc ϭ 85,000(12)(P͞A,3%,3)
ϭ $2,885,172
AWTT-anc ϭ 2,885,172(A͞P,3%,50)
ϭ $112,147 per year
⌬B ϭ AWOT-anc Ϫ AWTT-anc ϭ 276,685 Ϫ 112,147 ϭ $164,538 per year ($0
...

⌬B/C ϭ 0
...
17 ϭ 0
...
0, the trench-tunnel option is not economically justiﬁed
...

9
...
3
...
6
...
0 when this selected alternative has been compared with another justiﬁed alternative
...
The previous two examples (9
...
5) are good illustrations of
the second type of implied beneﬁt estimation
...
At least one alternative must have B/C Ն 1
...
If all alternatives are unacceptable, the DN alternative is the choice
...
6
...
)
As in the previous section when comparing two alternatives, selection from multiple alternatives by incremental B/C ratio utilizes equivalent total costs to initially order alternatives from
smallest to largest
...
Also, remember that all costs are
considered positive in B/C calculations
...
The procedure for incremental B/C analysis of
multiple alternatives is as follows:
Determine the equivalent total cost for all alternatives
...

Order the alternatives by equivalent total cost, smallest ﬁrst
...

Direct beneﬁts estimation only: Calculate the B/C for the ﬁrst ordered alternative
...
0, eliminate it
...
0
...
0 becomes the defender and the next
higher-cost alternative is the challenger in the next step
...
)
5
...

2
...

4
...
5]

9
...
Calculate the ⌬B/C for the ﬁrst challenger compared to the defender
...
6]

If ⌬B/C Ն 1
...
6], the challenger becomes the defender and the previous defender is eliminated
...
0, remove the challenger and the defender
remains against the next challenger
...
Repeat steps 5 and 6 until only one alternative remains
...

In all the steps above, incremental disbeneﬁts may be considered by replacing ⌬B with ⌬(B Ϫ D)
...
6
Schlitterbahn Waterparks of Texas, a very popular water and entertainment park headquartered in New Braunfels, has been asked by four different cities outside of Texas to consider building a park in their area
...
The annual M&O costs
are expected to be the same for all locations
...

Solution
The viewpoint is that of Schlitterbahn, and the beneﬁts are direct estimates
...
The results are presented in
Table 9–1
...
AW of total costs and an example for city 1 are determined in $1 million units
...
5(0
...
5
ϭ $6
...
The four alternatives are correctly ordered by increasing equivalent total cost in
Table 9–1
...
6
City 1

City 2

City 3

City 4

First cost, $ million
Entrance fee costs, $/year
Annual revenue, $ million/year
Initial cash incentive, $
Property tax reduction, $/year
Sales tax sharing, $/year

38
...
0
250,000
25,000
310,000

40
...
2
350,000
35,000
320,000

45
...
0
500,000
50,000
320,000

60
...
4
800,000
80,000
340,000

AW of total costs, $ million/year
AW of total beneﬁts, $ million/year
Overall B/C
Alternatives compared
Incremental costs ⌬C, $/year
Incremental beneﬁts ⌬B, $/year
⌬B/C
Increment justiﬁed?
City selected

6
...
377
1
...
948
7
...
06
Yes
1

7
...
614
0
...
0

8
...
454
1
...
164
3
...
64
Yes
3

10
...
954
1
...
236
0
...
22
No
3

Eliminated

243

244

Chapter 9

Beneﬁt/Cost Analysis and Public Sector Economics

3
...

AW of beneﬁts ϭ revenue ϩ initial incentive(A͞P,7%,8)
ϩ property tax reduction ϩ sales tax sharing
ϭ 7
...
25(0
...
025 ϩ 0
...
377 ($7,377,000 per year)
4
...
1]
...
377͞6
...
06
City 2 is eliminated with B/C2 ϭ 0
...

5
...

6
...
06, using Equation [9
...
City 1 is economically
justiﬁed and becomes the defender
...
Repeat steps 5 and 6
...
112 Ϫ 6
...
164
⌬B ϭ 10
...
377 ϭ 3
...
077͞1
...
64
City 3 is well justiﬁed and becomes the defender against city 4
...
22
for the 4-to-3 comparison
...
Note that the DN alternative could have been selected had no proposal met the B/C or
⌬B/C requirements
...
The only comparison is between
each project separately with the do-nothing alternative
...
0 are accepted
...
When a budget limitation is imposed, the capital budgeting procedure discussed in
Chapter 12 must be applied
...
Equation [5
...
Example 9
...

EXAMPLE 9
...
3 million; however, when it was publicized, inﬂuential people around Allen spoke strongly against
the location
...
Some of the plant design had already been completed
when the general manager announced that this site was not the best choice anyway, and that it
would be sold and a different, better site (location 2) would be purchased for $28
...

This was well over the budget amount of $22
...
As
it turns out, there was a third site (location 3) available for $35
...

In his review and after much resistance from Allen Water Utilities staff, the consultant, Joel,
received a copy of the estimated costs and beneﬁts for the three plant location options
...
Using the assumption of a very long life for
the WTF3 facility and the established discount rate of 3% per year, determine what Joel discovered when he did the B/C analysis
...
4

Incremental B/C Analysis of Multiple, Mutually Exclusive Alternatives
Location 1

Total beneﬁts, $ per year

Location 3

19
...
0

28
...
0

35
...
0

5
12
6

3
10
6

0
8
6

23

Land cost, $ million
Facility ﬁrst cost, $ million
Beneﬁts, $ per year:
Pumping cost savings
Sales to area communities
Added revenue from Allen

Location 2

19

14

Solution
A spreadsheet can be very useful when performing an incremental B/C analysis of three or
more alternatives
...
Figure 9–1b details all the functions
used in the analysis
...
In $1 million units,
AW of costs ϭ A of land cost ϩ A of facility ﬁrst cost
ϭ (19
...
0)(0
...
379 per year
AW of beneﬁts ϭ $23
Location 2:
AW of costs ϭ $14
...
430
AW of beneﬁts ϭ $14
Location 1:

Though the AW of cost values are close to one another, the increasing order is locations 2, 1,
and 3 to determine ⌬B/C values
...
97) is not economically justiﬁed at the outset
...
Location 2 is justiﬁed
against the DN alternative (B/C2 ϭ 1
...
Location 1 is a clear winner with ⌬B/C ϭ 27
...

In conclusion, Joel has learned that location 1 is indeed the best and that, from the economic
perspective, the general manager was incorrect in stating that location 2 was better
...

(a)

(b)

Figure 9–1
Incremental B/C analysis for WTF3 case: (a) numerical results and (b) functions developed for the analysis
...
Location 1 was initially purchased
and planned for WTF3
...

9
...
A large percentage of service sector projects are generated by and dependent upon the private sector (corporations, businesses,
and other for-proﬁt institutions)
...

A service sector project is a process or system that provides services to individuals, businesses,
or government units
...
Manufacturing and construction activities are commonly not considered a service sector project, though
they may support the theme of the service provided
...

The intangible and intellectual work done by engineers and other professionals is often a part of
a service sector project
...
In other
words, undue risk may be introduced into the decision because of poor monetary estimates
...
This is a public and a service project, but its (economic) beneﬁts are quite difﬁcult
to estimate
...
In all but the last case, beneﬁts in monetary
terms will be poor estimates
...

In service and public sector projects, as expected, it is the beneﬁts that are the more difﬁcult
to estimate
...
The CEA approach utilizes a costeffectiveness measure or the cost-effectiveness ratio (CER) as a basis of ranking projects and
selecting the best of independent projects or mutually exclusive alternatives
...
7]

In the red-light camera example, the effectiveness measure (the beneﬁt) may be one of the samples mentioned earlier, accidents averted or deaths prevented
...
(The reciprocal of Equation [9
...
) With costs in the numerator, smaller ratio values
are more desirable for the same value of the denominator, since smaller ratio values indicate a
lower cost for the same level of effectiveness
...
5

Service Sector Projects and Cost-Effectiveness Analysis

247

alternative selection
...
The ordering criteria are as follows:
Independent projects: Initially rank projects by CER value
...

Return again to the public/service project of red-light cameras
...
If the projects are mutually exclusive, “total accidents averted ” is the correct ranking basis
and an incremental analysis is necessary
...
To select some from several (independent) projects, a budget limit, termed b, is inherently necessary once ordering is complete
...
The procedures and examples follow
...
Determine the equivalent total cost C and effectiveness measure E, and calculate the CER
measure for each project
...
Order projects from the smallest to the largest CER value
...
Determine cumulative cost for each project and compare with the budget limit b
...
The selection criterion is to fund all projects such that b is not exceeded
...
8
Recent research indicates that corporations throughout the world need employees who demonstrate creativity and innovation for new processes and products
...
Rollings
Foundation for Innovative Thinking has allocated $1 million in grant funds to award to corporations that enroll their top R&D personnel in a 1- to 2-month professional training program in
their home state that has a historically proven track record over the last 5 years in helping
individuals earn patents
...
Columns 2 and 3
give the proposed number of attendees and cost per person, respectively, and column 4 provides the historical track record of program graduates in patents per year
...

Solution
We assume that across all programs and all patent awards there is equal quality
...

TA BLE 9–2

Data for Programs to Increase Patents Used for CEA

Program
(1)

Total Personnel
(2)

Cost/Person, $
(3)

5-Year History,
Patents/Graduate/Year
(4)

1
2
3
4
5
6

50
35
57
24
12
87

5000
4500
8000
2500
5500
3800

0
...
1
1
...
1
2
...
6

Independent project
selection

248

Chapter 9

Beneﬁt/Cost Analysis and Public Sector Economics

1
...
7], the effectiveness measure E is patents per year, and the CER is
program cost per person C
CER ϭ —————————— ϭ —
E
patents per graduate
The program cost C is a PW value, and the E values are obtained from the proposals
...
The CER values are shown in Table 9–3 in increasing order, column 5
...
Cost per course, column 6, and cumulative costs, column 7, are determined
...
Programs 4, 2, 5, 3, and 6 (68 of the 87 people) are selected to not exceed $1 million
...
8
Program
(1)
4
2
5
3
6
1

Total
Cost/Person
Personnel
C, $
(2)
(3)
24
35
12
57
87
50

2,500
4,500
5,500
8,000
3,800
5,000

Patents
per Year E
(4)

CER, $ per
Patent
(5) ‫)4(/)3( ؍‬

Program
Cost, $
(6) ‫)3()2( ؍‬

Cumulative
Cost, $
(7) ‫ ؍‬⌺(6)

2
...
1
2
...
9
0
...
5

1,190
1,452
1,897
4,211
6,333
10,000

60,000
157,500
66,000
456,000
330,600
250,000

60,000
217,500
283,500
739,500
1,070,100
1,320,100

Comment
This is the ﬁrst time that a budget limit has been imposed for the selection among independent
projects
...

For mutually exclusive alternatives and no budget limit, the alternative with the highest effectiveness measure E is selected without further analysis
...
The analysis is
based on the incremental ratio ⌬C/E, and the procedure is similar to that we have applied for
incremental ROR and B/C, except now the concept of dominance is utilized
...

For mutually exclusive alternatives, the selection procedure is as follows:
1
...
Record the cost for
each alternative
...
Calculate the CER measure for the ﬁrst alternative
...
This CER is a baseline for the next incremental comparison, and the ﬁrst alternative becomes the new defender
...
Calculate incremental costs (⌬C) and effectiveness (⌬E) and the incremental measure ⌬C/E
for the new challenger using the relation
cost of challenger Ϫ cost of defender
⌬C
⌬C/E ϭ ——————————————————————— ϭ ——
effectiveness of challenger Ϫ effectiveness of defender ⌬E
4
...
Otherwise, no dominance is present and both
alternatives are retained for the next incremental evaluation
...
Dominance present: Repeat steps 3 and 4 to compare the next ordered alternative (challenger) and new defender
...

9
...
Repeat steps 3 and 4 to compare the new challenger and new
defender
...

6
...

7
...

EXAMPLE 9
...
8 decided to fund its 50 R&D
personnel to attend one of the innovation and creativity programs at its own expense
...
0 patents
per year
...

TABLE 9–4

Mutually Exclusive Alternatives Evaluated by Cost-Effectiveness
Analysis, Example 9
...
1
2
...
1

1190
1897
1452

125,000
275,000
225,000

Solution
From Table 9–2, three programs—2, 4, and 5—have a historical record of at least two patents
per graduate per year
...
Use the procedure to perform the incremental analysis
...
The alternatives are ranked by increasing patents per year in Table 9–4, column 4
...
The CER measure for program 4 is compared to the DN alternative
...
1
patents per graduate
3
...

⌬C 5500 Ϫ 2500
5-to-4 comparison: ⌬C/E ϭ —— ϭ —————— ϭ 3750
2
...
1
⌬E
4
...
Program 5 is more expensive for more patents; however, clear dominance is not
present; both programs are retained for further evaluation
...
Dominance not present: Program 5 becomes the new defender, and program 2 is the new
challenger
...

⌬C 4500 Ϫ 5500
2-to-5 comparison: ⌬C/E ϭ —— ϭ —————— ϭ Ϫ5000
3
...
9
⌬E
Compared to C/E5 ϭ 1897, this increment is much cheaper—more patents for less money
per person
...

6
...

⌬C 4500 Ϫ 2500
2-to-4 comparison: ⌬C/E ϭ —— ϭ —————— ϭ 2000
3
...
1
⌬E
This does not represent dominance of program 2 over 4
...
This occurs when there is not lower cost and higher effectiveness of one alternative
over another; that is, one alternative does not dominate all the others
...
Now the budget and other considerations (probably noneconomic) are brought to bear to
make the ﬁnal decision
...

249

250

Chapter 9

Beneﬁt/Cost Analysis and Public Sector Economics

Cost-effectiveness analysis is a form of multiattribute decision-making in which economic
and noneconomic dimensions are integrated to evaluate alternatives from several perspectives by
different decision makers
...

9
...
Above these is the long-standing expectation that public servants have integrity
...
Impartiality, consideration of a wide range of circumstances, and the use of realistic assumptions are but three of
the foundation elements upon which engineers should base their recommendations to decision
makers
...

Engineers are routinely involved in two of the major aspects of public sector activities:
Public policy making—the development of strategy for public service, behavior, fairness,
and justice
...
An example is transportation management
...

Public ofﬁcials use these ﬁndings to establish public transportation policy
...
Consider trafﬁc control,
where the use and placement of trafﬁc control signs, signals, speed limits, parking restrictions, etc
...
(In effect, this is
systems engineering, that is, an application of the life-cycle phases and stages explained in
Section 6
...
)
Whether in the arena of policy making or public planning, engineers can ﬁnd ethical compromise
a possibility when working with the public sector
...

• Use of technology Many public projects involve the use of new technology
...
It is common and expected
that engineers make every attempt to apply the latest technology while ensuring that the public
is not exposed to undue risk
...
These restrictions may be based on ﬁnancial
reasons, politically-charged topics, client-favored options, or a wide variety of other reasons
...

• Negative community impact It is inevitable that public projects will adversely affect some
groups of people, or the environment, or businesses
...
Engineers who ﬁnd (stumble onto) such negative impacts may be pressured by clients, managers, or public ﬁgures to overlook them, though

Chapter Summary

TABLE 9–5

Some Ethical Considerations When Performing B/C and CEA Analysis

What the Study Includes Ethical Dimension

Example

Audience for study

Is it ethical to select a speciﬁc
group of people affected by the
project and neglect possible
effects on other groups?

Construct children’s health care clinics
for city dwellers, but neglect rural families with poor transportation means
...

Greater good for community as a whole

Vulnerable minority groups, especially economically deprived
ones, may be disproportionally
affected
...

Reliance on economic
measures only

Is it acceptable to reduce all
costs and beneﬁts to monetary
estimates for a decision, then
subjectively impute nonquantiﬁed factors in the ﬁnal decision?

Softening of building codes can improve the ﬁnancial outlook for home
builders; however, increased risks of
ﬁre loss, storm and water damage to
structures, and reduced future resale
values are considered only in passing as
a new subdivision is approved by the
planning and zoning committee
...

the Code of Ethics for Engineers dictates a full and fair analysis and report
...
Considering this outcome in the recommendation to the transportation department should be a goal of the analyzing engineers, yet pressure to bias the results may be quite high
...
As discussed earlier,
estimations for beneﬁts, disbeneﬁts, effectiveness measures, and costs can be difﬁcult and inaccurate, but these analysis tools are often the best available to structure a study
...

CHAPTER SUMMARY
The beneﬁt/cost method is used primarily to evaluate alternatives in the public sector
...
0 for the incremental equivalent total cost to be economically justiﬁed
...
For independent projects, no incremental B/C analysis is necessary
...
0 are selected provided there is no budget limitation
...
The characteristics of public sector projects are substantially different from those of the private sector: initial costs are larger; expected
life is longer; additional sources of capital funds include taxation, user fees, and government
grants; and interest (discount) rates are lower
...
Evaluation by B/C
analysis can be difﬁcult with no good way to make monetary estimates of beneﬁts
...
The concept of dominance is incorporated into the procedure for comparing mutually exclusive alternatives
...
Examples are included
...
1 What is the difference between disbeneﬁts and
costs?
9
...

9
...

9
...

(a) Loss of income to local businesses because
of a new freeway
(b) Less travel time because of a loop bypass
(c) $400,000 annual income to local businesses
because of tourism created by a national park
(d) Cost of ﬁsh from a hatchery to stock a lake at
the state park
(e) Less tire wear because of smoother road
surfaces
(f) Decrease in property values due to the closure of a government research lab
(g) School overcrowding because of a military
base expansion
(h) Revenue to local motels because of an extended weekend holiday

public-private partnership between the sheriff’s
ofﬁce and a private security company
...
The
categories chosen will vary depending upon a person’s viewpoint
...

(a) Select the top two viewpoints (in your opinion) for each of the following individuals as
they would categorize estimates as a cost,
beneﬁt, or disbeneﬁt
...
An industrial plant manager in the county
2
...
County commissioner (elected ofﬁce)
4
...

Project B/C Value
9
...
5 What is a fundamental difference between DBOM
and DBOMF contracts?

9
...
The
road will have to be maintained at a cost of $25,000
per year
...
The
improved accessibility has led to a 150% increase
in the property values along the road
...

9
...
Enforcement is proposed to be a

9
...
Since it has been linked to

253

Problems

cancer of the bladder, kidney, and other internal
organs, the EPA has lowered the arsenic standard
for drinking water from 0
...
010 parts per million (10 parts per billion)
...
If it is estimated that there are 90 million
households in the United States and that the lower
standard can save 50 lives per year valued at
$4,000,000 per life, what is the beneﬁt/cost ratio of
the regulation?
9
...
Use an
interest rate of 8% per year
...
11 A project to extend irrigation canals into an area
that was recently cleared of mesquite trees (a nuisance tree in Texas) and large weeds is projected to
have a capital cost of $2,000,000
...
Annual favorable consequences to the general public of $820,000 per year will be offset to
some extent by annual adverse consequences of
$400,000 to a portion of the general public
...
12 Calculate the B/C ratio for the following cash ﬂow
estimates at a discount rate of 7% per year
...
13 The beneﬁts associated with a nuclear power plant
cooling water ﬁltration project located on the Ohio
River are $10,000 per year forever, starting in
year 1
...
Calculate the B/C ratio at i ϭ
10% per year
...
14 A privately funded wind-based electric power
generation company in the southern part of the
country has developed the following estimates

(in $1000) for a new turbine farm
...
Calculate (a) the proﬁtability index and (b) the modiﬁed B/C ratio
...
15 For the values shown, calculate the conventional
B/C ratio at i ϭ 10% per year
...
16 A proposal to reduce trafﬁc congestion on I-5 has
a B/C ratio of 1
...
The annual worth of beneﬁts
minus disbeneﬁts is $560,000
...
17 Oil spills in the Gulf of Mexico have been known
to cause extensive damage to both public and private oyster grounds along the Louisiana and Mississippi shores
...
This procedure inevitably results in death to
some of the saltwater shellﬁsh while preventing
more widespread destruction to public reefs
...
If the
Fish and Wildlife Service spent $110 million in
year 0 and $50 million in years 1 and 2 to minimize environmental damage from one particular
oil spill, what is the beneﬁt-to-cost ratio provided
the efforts resulted in saving 3000 jobs valued at a
total of $175 million per year? Assume disbeneﬁts
associated with oyster deaths amounted to $30 million in year 0
...

9
...
If the operating
cost is $600,000 per year and the public health
beneﬁts are assumed to be $800,000 per year, what
initial investment in the GFH system is necessary
to guarantee a modiﬁed B/C ratio of at least 1
...

254

Chapter 9

Beneﬁt/Cost Analysis and Public Sector Economics

9
...
3 million
...
Beneﬁts of $340,000 per year and disbeneﬁts of $40,000 per year have also been identiﬁed
...

9
...

To the
People
Beneﬁts:

$300,000 now and
$100,000 per
year thereafter

Disbeneﬁts: $40,000 per year

To the
Government
Costs:

$1
...
21 In 2010, Brazil began construction of the Belo
Monte hydroelectric dam on the Xingu River
(which feeds the Amazon River)
...
It will begin producing
electricity in 2015
...
They say it will
devastate wildlife and the livelihoods of 40,000
people who live in the area to be ﬂooded
...
2 billion per year
...
Assume that the
disbeneﬁts will occur evenly through the 5-year
construction period and anticipated beneﬁts will
begin at the end of 2015 and continue indeﬁnitely
...
0
...
22 In the United States, the average number of airplanes in the sky on an average morning is 4000
...

Aerospace company Rockwell Collins developed
what it calls a digital parachute—a panic-button
technology that will land any plane in a pinch at
the closest airport, no matter what the weather or
geography and without the help of a pilot
...
Assume
that the cost of retroﬁtting 20,000 commercial air-

planes is $100,000 each and the plane stays in service for 15 years
...

9
...
This has required homeowners living in valley areas near the
river to purchase ﬂood insurance costing between
$145 and $2766 per year
...
As a result, 13,000 properties
were freed of the federal mandate to purchase
ﬂood insurance
...
If the average cost of ﬂood
insurance is $460 per household per year, calculate the beneﬁt-to-cost ratio of the levee-raising
project
...

9
...

Beneﬁts:

$20,000 in year 0 and
$30,000 in year 5

Disbeneﬁts:

$7000 in year 3

Savings (to government): $25,000 in years 1–4
Cost:

$100,000 in year 0

Project life:

5 years

9
...

9
...
Determine the proﬁtability index for the
ﬁnancial results listed below using a MARR of
8% per year
...
2 million

Year 5

$Ϫ3
...
2 million per year
$2
...
27 A project had a staged investment distributed over
the 6-year contract period
...

255

Problems
Year

0

Investment, $1000
NCF, $1000 per year

1

2

3

Ϫ25
0

0
5

Ϫ10
7

0
9

4

5

6

selected at an interest rate of 8% per year and a
5-year study period
...
28 In comparing two alternatives by the B/C method,
if the overall B/C ratio for both alternatives is calculated to be exactly 1
...
29 In comparing alternatives X and Y by the B/C
method, if B/CX ϭ 1
...
8, what is
known about the B/C ratio on the increment of investment between X and Y?
9
...
The
mountain site (MS) will use injection wells that cost
$4
...
This site will be able to accommodate
150 million gallons per year
...
At this
site, 890 million gallons can be injected each year
...
00 per 1000 gallons, which alternative, if either, should be selected
according to the B/C ratio method? Use an interest
rate of 8% per year and a 20-year study period
...
31 The estimates shown are for a bridge under consideration for a river crossing in Wheeling, West
Virginia
...

East Location

West Location

11 ϫ 10
100,000
990,000
120,000
ϱ

27 ϫ 106
90,000
2,400,000
100,000
ϱ

Initial cost, $
Annual M&O, $/year
Beneﬁts, $/year
Disbeneﬁts, $/year
Life, years

6

9
...
Use a B/C analysis and an interest
rate of 8% per year
...
33 Conventional and solar alternatives are available
for providing energy at a remote radar site
...
34 The two alternatives shown are under consideration
for improving security at a county jail in Travis
County, New York
...

Extra Cameras
(EC)

New Sensors
(NS)

38,000
49,000
110,000
26,000

87,000
64,000
160,000
—

First cost, $
Annual M&O, $/year
Beneﬁts, $/year
Disbeneﬁts, $/year

9
...
S
...
Separate contractors proposed two methods
...
For this method, the costs will be
$14,100 for concrete, $6000 for metal decking,
$4300 for joists, and $2600 for beams
...
Special
additives will be included in the lightweight concrete
that will improve the heat-transfer properties of the
ﬂoor
...

9
...
Determine
which project should be selected on the basis of a
B/C analysis at i ϭ 8% per year and a 20-year
study period
...
37 Two routes are under consideration for a new interstate highway
...
The short transmountain route

256

Chapter 9

Beneﬁt/Cost Analysis and Public Sector Economics

would be 10 kilometers long and would have an initial cost of $45 million
...
Regardless of
which route is selected, the volume of trafﬁc is expected to be 400,000 vehicles per year
...
30 per kilometer, determine which route should be selected on
the basis of (a) conventional B/C analysis and
(b) modiﬁed B/C analysis
...

9
...

Location E would require an investment of $3 million and $50,000 per year to maintain
...
The operating cost of location W will be
$65,000 per year
...
The disbeneﬁts associated
with each location are $30,000 per year for location
E and $40,000 per year for location W
...
Use an interest
rate of 12% per year to determine which location, if
either, should be selected on the basis of (a) the
B/C method and (b) the modiﬁed B/C method
...
39 Three engineers made the estimates shown below
for two optional methods by which new construction
technology would be implemented at a site for public housing
...
Set up a spreadsheet
for B/C sensitivity analysis and determine if option
1, option 2 or the do-nothing option is selected by
each of the three engineers
...

Engineer Bob

Engineer Judy

Engineer Chen

Alternative

A

B

C

D

E

F

PW of capital, $
PW of beneﬁts, $

80
70

50
55

72
76

43
52

89
85

81
84

9
...
One must be accepted
...

Comparison

⌬B/C
Ratio

A versus B
B versus C
C versus D
A versus C
A versus D
B versus D
C versus E
D versus E

0
...
4
1
...
1
0
...
9
1
...
9

9
...
The future worth of
costs, beneﬁts, disbeneﬁts, and cost savings is
shown
...
Determine which of the
projects, if any, should be selected, if the projects
are (a) independent and (b) mutually exclusive
...
43 From the data shown below for six mutually
exclusive projects, determine which project, if
any, should be selected
...
40 A group of engineers responsible for developing
advanced missile detection and tracking technologies, such as shortwave infrared, thermal infrared
detection, target tracking radar, etc
...
The present worth (in $ billions) of the capital requirements
and beneﬁts is shown for each alternative in the
table
...

Annual cost, $
per year
Annual beneﬁts,
$ per year
B/C ratio (alternative vs
...
23

1
...
87

0
...
71

1
...
07
A versus C ϭ 0
...
02
B versus D ϭ 0
...
00
B versus F ϭ 1
...
06
C versus F ϭ 1
...
44 Four mutually exclusive revenue alternatives are
being compared using the B/C method
...
20
23
...
75
10
...
65
15
...
DN A
A
B
C
D

30
38
52
81

0
...
18
1
...
16

B

C

D

— 2
...
30 1
...
58 1
...
45
—

9
...
One accepted measure of effectiveness of
a program is the percentage of enrollees quitting
...

9
...
Louis, Missouri, is considering various proposals regarding the disposal of used tires
...
An incremental B/C
analysis was initiated but never completed
...

(b) Determine which alternative should be selected
...
05
1
...
34

Incremental
B/C When
Compared with
Alternative

J

K

L

M

— ? ? ?
— ? ?
— ?
—

The Cancer Society provides annual cost-offset
funding to cancer patients so more people can afford
these programs
...
Louis has the
capacity to treat each year the number of people
shown
...

(b) Offer programs using as many techniques as
possible to treat up to 1300 people per year
using the most cost-effective techniques
...
46 In an effort to improve productivity in a large
semiconductor manufacturing plant, the plant
manager decided to undertake on a trial basis a series of actions directed toward improving employee morale
...
Periodically, the company surveyed the employees to measure the change in morale
...
The per-employee cost of each strategy
(identiﬁed as A through F) and the resultant measurement score are shown in the next column
...
Determine which strategies are the best to implement
...
)

Cost, $/
Enrollee

Technique
Acupuncture
Subliminal message
Aversion therapy
Outpatient clinic
In-patient clinic
Nicotine replacement
therapy (NRT)

% Quitting

Treatment
Capacity per Year

700
150
1700
2500
1800
1300

9
1
10
39
41
20

250
500
200
400
550
100

9
...
(a) Calculate the cost effectiveness ratio
for each alternative, and (b) use the CER to identify the best alternative
...
49 An engineering student has only 30 minutes before
the ﬁnal exam in statics and dynamics
...
There is time for using only
one method of assistance before the exam; he must
select well
...
The method and estimates follow
...
50 During the design and speciﬁcations development
stages of a remote meter reading system for residential electricity use (a system that allows monthly
usage to be transmitted via phone lines with no
need to physically view meters), the two engineers
working on the project for the city of Forest Ridge
noted something different from what they expected
...
The second designer, an
industrial/systems engineer, further noted that all
the hardware speciﬁcations provided to them by

this same liaison came from the same distributor,
namely, Delsey Enterprises
...
Upon review,
they learned that Don is the son-in-law of the city
liaison and Susan is his stepdaughter
...
51 Explain the difference between public policy making and public planning
...
52 Since transportation via automobile was introduced,
drivers throughout the country of Yalturia in eastern
Europe have driven on the left side of the road
...
This is a major policy change for the country and will require signiﬁcant public planning and
project development to implement successfully and
safely
...

Identify six of the projects you deem necessary
...

ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
9
...
54 All of the following are usually associated with
public sector projects except:
(a) Funding from taxes
(b) Proﬁt
(c) Disbeneﬁts
(d) Inﬁnite life
9
...
com company
(c) Dam with irrigation canals
(d) Mass transit system
9
...
57 In a conventional B/C ratio:
(a) Disbeneﬁts and M&O costs are subtracted
from beneﬁts
...

(c) Disbeneﬁts and M&O costs are added to
costs
...

9
...

(b) Beneﬁts are subtracted from costs
...

(d) M&O costs are put in the numerator
...
59 If two mutually exclusive alternatives have B/C
ratios of 1
...
5 for the lower- and higher-cost
alternatives, respectively, the following is correct:

Case Study

(a)
(b)
(c)
(d)
9
...
5
...
4 and 1
...

The B/C ratio on the increment between
them is less than 1
...

The higher-cost alternative is the better one
economically
...
1,
0
...
3
...
61 An alternative has the following cash ﬂows:
Beneﬁts of $50,000 per year
Disbeneﬁts of $27,000 per year
Initial cost of $250,000
M&O costs of $10,000 per year
If the alternative has an inﬁnite life and the interest
rate is 10% per year, the B/C ratio is closest to:
(a) 0
...
66
(c) 0
...
16
9
...
65
(b) 0
...
80
(d) 1
...
63 In evaluating three mutually exclusive alternatives
by the B/C method, the alternatives are ranked A,
B, and C, respectively, in terms of increasing cost,
and the following results are obtained for the
overall B/C ratios: 1
...
9, and 1
...
On the basis
of these results, you should:
(a) Select A
(b) Select C

(c)
(d)

259

Select A and C
Compare A and C incrementally

9
...
5
...
The ﬁrst cost of the alternative at an interest
rate of 10% per year is closest to:
(a) $23,300
(b) $85,400
(c) $146,100
(d) $233,000
9
...

(b) CEA expresses outcomes in natural units
rather than in currency units
...

(d) CEA is more time-consuming and resourceintensive
...
66 Several private colleges claim to have programs
that are very effective at teaching enrollees how
to become entrepreneurs
...
If the total cost of
the programs is $25,000 and $33,000, respectively, the incremental cost-effectiveness ratio is
closest to:
(a) 6250
(b) 5500
(c) 4000
(d) 1333
9
...
68 Of the following, the word not related to ethics is:
(a) Virtuous
(b) Honest
(c) Lucrative
(d) Proper
9
...

Poor highway lighting may be one reason that proportionately more trafﬁc accidents occur at night
...

For example, an accident with a fatality is valued at

260

Chapter 9

Beneﬁt/Cost Analysis and Public Sector Economics

approximately $4 million, while an accident in which there
is property damage (to the car and contents) is valued at
$6000
...
Observed reductions in accidents seemingly caused by too low
lighting can be translated into either monetary estimates of
the beneﬁts B of lighting or used as the effectiveness
measure E of lighting
...
The
property damage category is commonly the largest based on
the accident rate
...

Number of Accidents Recorded1
Unlighted

Lighted

Accident Type

Day

Night

Day

Night

Property damage

379

199

2069

839

The ratios of night to day accidents involving property damage for the unlighted and lighted freeway sections are 199/379 ϭ
0
...
406, respectively
...
To quantify the beneﬁt, the
accident rate ratio from the unlighted section will be applied to
the lighted section
...
Thus, there would have been (2069)(0
...
This is a difference of 247 accidents
...

To determine the cost of the lighting, it will be assumed that
the light poles are center poles 67 meters apart with 2 bulbs
each
...
Since these data were collected over 87
...
8
Installation cost ϭ $3500 ———
0
...
8/0
...
10 per kWh
...
4 kilowatt/bulb)
ϫ (12 hours/day)(365 days/year)
ϫ ($0
...
Therefore, the
annualized cost C at i ϭ 6% per year is
Total annual cost ϭ $4,585,000(A/P,6%,5)
ϩ 459,024
ϭ $1,547,503
If a beneﬁt/cost analysis is the basis for a decision on additional lighting, the B/C ratio is
1,482,000
B/C ϭ ————— ϭ 0
...
0, the lighting is not justiﬁed
...
If a cost-effectiveness analysis (CEA) is
applied, due to a judgment that the monetary estimates for
lighting’s beneﬁt is not accurate, the C/E ratio is

1,547,503
C/E ϭ ————— ϭ 6265
247
This can serve as a base ratio for comparison when an incremental CEA is performed for additional accident reduction proposals
...

Install poles at twice the distance apart (134 meters)
...

Install cheaper poles and surrounding safety guards,
plus slightly lowered lumen bulbs (350 watts) at a
cost of $2500 per pole; place the poles 67 meters
apart
...

Install cheaper equipment for $2500 per pole with
350-watt lightbulbs and place them 134 meters apart
...

Case Study Exercises
Determine if a deﬁnitive decision on lighting can be determined by doing the following:
1
...

2
...

From an understanding viewpoint, consider the following:
3
...
What would the lighted, night-to-day accident ratio
have to be to make alternative Z economically justiﬁed
by the B/C ratio?
5
...
Does
one seem more appropriate in this type of situation than
the other? Why? Can you think of other bases that might
be better for decisions for public projects such as this one?

Portion of data reported in Michael Grifﬁn, “Comparison of the Safety of Lighting on Urban Freeways,” Public Roads, vol
...
8–15, 1994
...
Any
method—PW, AW, FW, ROR, or B/C—can be used to select one alternative from two or more and
obtain the same, correct answer
...
Yet different information about an alternative is available with each different method
...

Table LS2–1 gives a recommended evaluation method for different situations, if it is not speciﬁed
by the instructor in a course or by corporate practice in professional work
...
Interpretation of the entries in each
column follows
...
Public sector projects are
commonly evaluated using the B/C ratio and usually have long lives that may be considered inﬁnite for economic computation purposes
...
For cost alternatives, the revenue cash ﬂow series is assumed to be equal for all alternatives
...
Service sector projects for which beneﬁts are estimated using a nonmonetary effectiveness measure are usually evaluated with a method such as cost-effectiveness analysis
...

Recommended method: Whether an analysis is performed by hand, calculator, or spreadsheet, the
method(s) recommended in Table LS2–1 will correctly select one alternative from two or more as
TABLE LS2–1

Recommended Method to Compare Mutually Exclusive Alternatives,
Provided the Method Is Not Preselected

Evaluation Period

Type of Alternatives

Recommended Method

Series to
Evaluate

Equal lives of
alternatives

Revenue or cost

AW or PW

Cash ﬂows

Public sector

B/C, based on AW or PW

Incremental
cash ﬂows

Revenue or cost

AW

Cash ﬂows

Public sector

B/C, based on AW

Incremental
cash ﬂows

Revenue or cost

AW or PW

Public sector

B/C, based on AW or PW

Updated cash
ﬂows
Updated incremental
cash ﬂows

Revenue or cost
Public sector

AW or PW
B/C, based on AW

Unequal lives of
alternatives

Study period

Long to inﬁnite

Cash ﬂows
Incremental
cash ﬂows

262

Chapter 9

Beneﬁt/Cost Analysis and Public Sector Economics

rapidly as possible
...
For example, if lives are unequal and the
rate of return is needed, it is best to ﬁrst apply the AW method at the MARR and then determine the selected alternative’s i* using the same AW relation with i as the unknown
...
For spreadsheet analyses, this means that the NPV or PV functions (for present
worth) or the PMT function (for annual worth) is applied
...

Once the evaluation method is selected, a speciﬁc procedure must be followed
...
Table LS2–2 summarizes the important

TABLE LS2–2

Characteristics of an Economic Analysis of Mutually Exclusive Alternatives Once the
Evaluation Method Is Determined
Series to
Evaluate

Rate of
Return;
Interest Rate

Decision
Guideline:
Select1

Equal

Lives

Cash ﬂows

MARR

Unequal

LCM

Cash ﬂows

MARR

PW

Study period

Study period

Long to
inﬁnite

Inﬁnity

Updated
cash ﬂows
Cash ﬂows

MARR

CC
Future worth

Time
Period for
Analysis

PW
Present worth

Lives of
Alternatives

PW

Evaluation
Method

Numerically
largest PW
Numerically
largest PW
Numerically
largest PW
Numerically
largest CC

Equivalence
Relation

FW
AW

Annual worth

AW

MARR

Same as present worth for equal lives, unequal lives,
and study period
Equal or
unequal
Study period

Lives

Cash ﬂows

MARR

Study period

Updated
cash ﬂows
Cash ﬂows

MARR

Incremental
cash ﬂows
Incremental
cash ﬂows
Cash ﬂows

Find ⌬i*

AW

Equal

Lives

PW or AW

Unequal

LCM of pair

AW

Unequal

Lives

PW or AW

Study period

Study period

Updated
incremental
cash ﬂows

Find ⌬i*

PW
Beneﬁt/cost

Inﬁnity

PW or AW

Rate of return

Long to
inﬁnite

Equal or
unequal
Equal or
unequal
Long to
inﬁnite

LCM of pairs

Incremental
cash ﬂows
Incremental
cash ﬂows
Incremental
cash ﬂows

Discount rate

AW
AW or PW

1

Lowest equivalent cost or largest equivalent income
...
0
Last ⌬B/C Ն
1
...
0

Learning Stage 2: Epilogue

elements of the procedure for each method—PW, AW, ROR, and B/C
...
The meaning of the entries in Table LS2–2 follows
...
The capitalized cost (CC) relation is a PW relation for inﬁnite life, and
the FW relation is likely determined from the PW equivalent value
...

Lives of alternatives and time period for analysis The length of time for an evaluation (the
n value) will always be one of the following: equal lives of the alternatives, LCM of unequal
lives, speciﬁed study period, or inﬁnity because the lives are very long
...

• Incremental ROR and B/C methods require the LCM of the two alternatives being compared
...

• CC analysis has an inﬁnite time line and uses the relation P ϭ A͞i
...
The LCM of the two alternatives compared must be used
...
Both
approaches ﬁnd the incremental rate of return ⌬i*
...

Rate of return (interest rate) The MARR value must be stated to complete the PW, FW,
or AW method
...
The ROR method requires that the incremental rate be found in order
to select one alternative
...

Decision guideline The selection of one alternative is accomplished using the general guideline in the rightmost column
...
This is correct for both revenue and cost alternatives
...
This
means that the ⌬i* exceeds MARR, or the ⌬B/C exceeds 1
...

EXAMPLE LS2–1
Read through the problem statement of the following examples, neglecting the evaluation
method used in the example
...
Is this the method used in the example? (a) 8
...
5, (c) 5
...
4
...

(a) Example 8
...
Use the AW or PW value
at the MARR of 10%
...

(b) Example 6
...
The B/C ratio of AW values is the best
choice
...

263

264

Chapter 9

Beneﬁt/Cost Analysis and Public Sector Economics

(c) Since Example 5
...
Since one life is long, capitalized cost, based on P ϭ A/i, is best in this
case
...

(d) Example 5
...
It involves 5-year and 10-year cost
alternatives
...
The PW method for the LCM
of 10 years and a study period of 5 years were both presented in the example
...
The
chapters in this stage introduce information-gathering and
techniques that make decisions better
...

The future is certainly not exact
...

After completing these chapters, you will be able to go beyond
the basic alternative analysis tools of the previous chapters
...

Important note: If asset depreciation and taxes are to be considered by an after-tax analysis, Chapters 16 and 17 should be covered
before or in conjunction with these chapters
...

SECTION

TOPIC

LEARNING OUTCOME

10
...

10
...

10
...

10
...

10
...

10
...

10
...

T

his chapter discusses the different ways to ﬁnance a project through debt and
equity sources and explains how the MARR is established
...
Some of
the parameters speciﬁed earlier are unspeciﬁed here, and in future chapters
...

Until now, only one dimension—the economic one—has been the basis for judging the
economic viability of one project, or the selection basis from two or more alternatives
...

10
...
In Chapter 1 the MARR was described relative to the weighted costs of debt and
equity capital
...

To form the basis for a realistic MARR, the types and cost of each source of project ﬁnancing
should be understood and estimated
...
There are several terms and relationships important to the understanding of project ﬁnancing and the MARR that is speciﬁed to evaluate projects using PW, AW,
FW, or B/C methods
...
9 will complement the following material
...

The MARR is then set relative to the cost of capital
...
MARR values change
over time due to changing circumstances
...
This rate is determined by ﬁnding the ROR (i*) value of the
project cash ﬂows
...

Before we discuss cost of capital, we review the two primary sources of capital
...
Debt ﬁnancing includes borrowing via
bonds, loans, and mortgages
...

The amount of outstanding debt ﬁnancing is indicated in the liabilities section of the corporate
balance sheet
...

Owners’ funds are further classiﬁed as common and preferred stock proceeds or owners’ capital for a private (non-stock-issuing) company
...
The amount of equity is indicated in the net worth
section of the corporate balance sheet
...
Therefore, the cost of debt capital
is 8% as shown in Figure 10–1
...
Management may increase
this MARR in increments that reﬂect its desire for added return and its perception of risk
...
Suppose this
amount is 2%
...
Also, if the risk associated with the investment is considered substantial enough to warrant an additional 1% return requirement, the ﬁnal MARR is 11%
...

The recommended approach does not follow the logic presented above
...
Then the i* value is determined from the
estimated net cash ﬂows
...
Now, additional return requirements and risk factors are considered to determine if 3%
above the MARR of 8% is sufﬁcient to justify the capital investment
...
This is the opportunity cost discussed previously—the unfunded project i* has established the effective MARR for emission
control system alternatives at 11%, not 8%
...
The debt and equity
capital mix changes over time and between projects
...
It is altered for different opportunities and types of projects
...

The effective MARR varies from one project to another and through time because of factors
such as the following:
Project risk
...
This is encouraged by the higher cost of debt
capital for projects considered risky
...

Investment opportunity
...
This common reaction to investment opportunity can create havoc when the
guidelines for setting a MARR are too strictly applied
...

Government intervention
...
This may occur through price limits, subsidies,
import tariffs, and limitation on availability
...
Examples are steel imports,
foreign capital investment, car imports, and agricultural product exports
...
, thus tending to move the MARR up or down
...
If corporate taxes are rising (due to increased proﬁts, capital gains, local
taxes, etc
...
Use of after-tax analysis may assist in
eliminating this reason for a ﬂuctuating MARR, since accompanying business expenses will
tend to decrease taxes and after-tax costs
...
As debt and equity capital become limited, the MARR is increased
...
The
opportunity cost has a large role in determining the MARR actually used
...
2

Debt-Equity Mix and Weighted Average Cost of Capital

Market rates at other corporations
...
These variations are
often based on changes in interest rates for loans, which directly impact the cost of capital
...
1]

The total or effective tax rate, including federal, state, and local taxes, for most corporations is in
the range of 30% to 50%
...
35) ϭ 15
...

EXAMPLE 10
...
Carl, an
architect, has worked in home design with Bulte Homes since graduation
...
They both reside in
Richmond, Virginia
...
Carl and Christy want to expand into a regional e-business corporation
...
Identify
some factors that might cause the loan rate to vary when BA provides the quote
...

Solution
In all cases the direction of the loan rate and the MARR will be the same
...

Investment opportunity: The rate could increase if other companies offering similar services have already applied for a loan at other BA branches regionally or nationwide
...
The intervention may
be designed to boost the housing economic sector in an effort to offset a signiﬁcant
slowdown in new home construction
...

Capital limitation: Assume the computer equipment and software rights held by Carl and
Christy were bought with their own funds and there are no outstanding loans
...

Market loan rates: The local BA branch probably obtains its development loan money
from a large national pool
...
”

10
...
A company with a 40–60 D-E mix has 40% of its capital originating from debt capital
sources (bonds, loans, and mortgages) and 60% derived from equity sources (stocks and retained

269

270

Chapter 10

Project Financing and Noneconomic Attributes

Figure 10–2
General shape of different
cost of capital curves
...
Most projects are funded with a combination of debt and equity capital made available
speciﬁcally for the project or taken from a corporate pool of capital
...

If known exactly, these fractions are used to estimate WACC; otherwise the historical fractions
for each source are used in the relation
WACC ‫( ؍‬equity fraction)(cost of equity capital)
؉ (debt fraction)(cost of debt capital)

[10
...

Since virtually all corporations have a mixture of capital sources, the WACC is a value between the debt and equity costs of capital
...
2] is expanded
...
3]

Figure 10–2 indicates the usual shape of cost of capital curves
...
There is virtually always a mixture of capital sources involved for any capitalization program
...

Most ﬁrms operate over a range of D-E mixes
...
However, another company may be considered “risky” with only 20% debt capital
...

EXAMPLE 10
...
In an effort
to develop sustainable and renewable vegetable sources, a new commercial vertical crop technology is being installed through a public-private partnership with Valcent Products
...
,” www
...
net, June 16, 2010 news release
...
3

Determination of the Cost of Debt Capital

illustration purposes, assume that the present worth of the total system cost is $20 million with
ﬁnancing sources and costs as follows
...
8% per year
$4 million at 5
...
9% per year

There are three existing international vertical farming projects with capitalization and WACC
values as follows:
Project 1:
Project 2:
Project 3:

$5 million with WACC1 ϭ 7
...
2%
$7 million with WACC3 ϭ 4
...

Solution
To apply Equation [10
...
These are 0
...
2 for retained earnings, and 0
...

WACCHK ϭ 0
...
9%) ϩ 0
...
2%) ϩ 0
...
8%) ϭ 6
...
119; project 2 has
0
...
167
...

WACCW ϭ 0
...
9%) ϩ 0
...
2%) ϩ 0
...
8%) ϭ 9
...

The WACC value can be computed using before-tax or after-tax values for cost of capital
...
3 below
...
4]

The effective tax rate is a combination of federal, state, and local tax rates
...
Equation [10
...
2] for an after-tax WACC rate
...

10
...
(We learned about bonds in
Section 7
...
) In most industrialized countries, bond dividends and loan interest payments are taxdeductible as a corporate expense
...
The cost of debt capital is, therefore, reduced
because there is an annual tax savings of the expense cash ﬂow times the effective tax rate Te
...
In formula form,
Tax savings ‫( ؍‬expenses) (effective tax rate) ‫ ؍‬expenses (Te)
Net cash ﬂow ‫ ؍‬expenses ؊ tax savings ‫ ؍‬expenses (1 ؊ Te)

[10
...
6]

To ﬁnd the cost of debt capital, develop a PW- or AW-based relation of the net cash ﬂow
(NCF) series with i* as the unknown
...
This is the cost of debt capital used in the WACC computation, Equation [10
...

EXAMPLE 10
...
If the effective tax rate of the company is 30% and the bonds are discounted
2%, compute the cost of debt capital (a) before taxes and (b) after taxes from the company
perspective
...

Solution by Hand

(a) The annual bond dividend is $1000(0
...
Using the company perspective, ﬁnd the i* in the PW relation
0 ϭ 980 Ϫ 80(P/A, i*,10) Ϫ 1000(P/F, i*,10)
i* ϭ 8
...
3%, which is slightly higher than the
8% bond interest rate, because of the 2% sales discount
...
5]
shows a tax savings of $80(0
...
The bond dividend amount for the PW
relation is now $80 Ϫ 24 ϭ $56
...
87%
...
The after-tax net cash ﬂow is calculated using Equation [10
...
3
...
08

Bond dividend after taxes
ϭ (Ϫ1000*0
...
3)
ϭ IRR(C3:C13)

Figure 10–3
Use of IRR function to determine cost of debt capital before taxes and after taxes, Example10
...

EXAMPLE 10
...
Company managers
have decided to put $10,000 down now from retained earnings and borrow $10,000 at an
interest rate of 6%
...
(a) What is the after-tax cost of debt capital if
the effective tax rate is 42%? (b) How are the interest rate and cost of debt capital used to
calculate WACC?

10
...
42) ϭ $348 by Equation [10
...
The loan repayment is $10,000 in year 10
...
48%
...
Nor is 3
...
The
cost of the $10,000 equity capital is needed to determine the WACC
...
4 Determination of the Cost of Equity
Capital and the MARR
Equity capital is usually obtained from the following sources:
Sale of preferred stock
Sale of common stock
Use of retained earnings
Use of owner’s private capital
The cost of each type of ﬁnancing is estimated separately and entered into the WACC computation
...
One additional method for estimating the cost of equity capital via common stock
is presented
...

Issuance of preferred stock carries with it a commitment to pay a stated dividend annually
...
Preferred stock may be sold at a discount to speed the sale, in
which case the actual proceeds from the stock should be used as the denominator
...
53%
...
The dividends paid
are not a true indication of what the stock issue will actually cost in the future
...
If Re is the cost of equity capital (in
decimal form),
ﬁrst-year dividend
Re ‫ ؉ ———————— ؍‬expected dividend growth rate
price of stock
DV1
‫ ؉ —— ؍‬g
P

[10
...

Stated another way, it is the compound growth rate on dividends that the company believes is
required to attract stockholders
...
S
...
If a 5% or $1 dividend is planned for the ﬁrst year and an
appreciation of 4% per year is anticipated for future dividends, the cost of capital for this common stock issue from Equation [10
...

1
Re ϭ —— ϩ 0
...
09
20
The retained earnings and owner’s funds cost of equity capital is usually set equal to the
common stock cost, since it is the shareholders and owners who will realize any returns from
projects in which these funds are invested
...

Market
security
line

Re

␤Ͼ1
Premium
increases
for more
risky securities
Rm – Rf
Premium

Rm

Rf

Selected
market
portfolio
0

1
...
3]
...
Because of the ﬂuctuations in stock prices and the higher return demanded
by some corporations’ stocks compared to others, this valuation technique is commonly applied
...
8]

where ␤ ϭ volatility of a company’s stock relative to other stocks in the market (␤ ϭ 1
...
S
...
The coefﬁcient ␤
(beta) indicates how the stock is expected to vary compared to a selected portfolio of stocks in
the same general market area, often the Standard and Poor’s 500 stock index
...
0, the
stock is less volatile, so the resulting premium can be smaller; when ␤ Ͼ 1
...

Security is a word that identiﬁes a stock, bond, or any other instrument used to develop capital
...
This is a plot of a market security line, which is a linear ﬁt by regression analysis to indicate the expected return for different
␤ values
...
As ␤ increases, the
premium return requirement grows
...
Once complete, this estimated cost of common stock equity capital can be included
in the WACC computation in Equation [10
...

EXAMPLE 10
...
It is envisioned that processes for prepared meats can be completed more safely and faster using this
automated control software
...
SafeSoft, which has a historical beta value of 1
...
The security
market line indicates that a 5% premium above the risk-free rate is desirable
...
S
...

10
...
8]
...
0 ϩ 1
...
0) ϭ 7
...

In theory, a correctly performed engineering economy study uses a MARR equal to the cost
of the capital committed to the speciﬁc alternatives in the study
...
For a combination of debt and equity capital, the calculated WACC sets the minimum for
the MARR
...
The risks associated with an alternative should be treated separately
from the MARR determination, as stated earlier
...
Unfortunately, the MARR is often set above the WACC because management does want to account for risk by increasing the MARR
...
6
The Engineering Products Division of 4M Corporation has two mutually exclusive alternatives
A and B with ROR values of i* ϭ 9
...
9%
...
5% and the remainder from the same equity funds mentioned above
...
Make the economic decision on alternative A versus B under each ﬁnancing
scenario
...

Solution
The capital is available for one of the two mutually exclusive alternatives
...
Only alternative A is acceptable; alternative B is not since the estimated return of 5
...

Under ﬁnancing plan 2, with a D-E mix of 25–75,
WACC ϭ 0
...
5) ϩ 0
...
0) ϭ 9
...
625%
...

10
...
2
...

The leverage offered by larger debt capital percentages increases the riskiness of projects undertaken by the company
...
This is sometimes referred to as a highly
leveraged corporation
...
Thus, a reasonable balance between debt and equity ﬁnancing is important for

275

276

Chapter 10

Project Financing and Noneconomic Attributes

the ﬁnancial health of a corporation
...
7 illustrates the disadvantages of unbalanced
D-E mixes
...
7
Three auto parts manufacturing companies have the following debt and equity capital amounts
and D-E mixes
...

Amount of Capital
Company

Debt
($ in Millions)

Equity
($ in Millions)

D-E Mix (%–%)

A
B
C

10
20
40

40
20
10

20−80
50−50
80−20

Assume the annual revenue is $15 million for each one and that after interest on debt is considered, the net incomes are $14
...
4, and $10
...
Compute the return on
common stock for each company, and comment on the return relative to the D-E mixes
...
In
million dollars,
14
...
36 (36%)
40
13
...
67 (67%)
ReturnB ϭ ——
20
10
...
00 (100%)
ReturnC ϭ ——
10
As expected, the return is by far the largest for highly leveraged C, where only 20% of the
company is in the hands of the ownership
...

The use of large percentages of debt ﬁnancing greatly increases the risk taken by lenders and
stock owners
...

The leverage of large D-E mixes does increase the return on equity capital, as shown in previous examples; but it can also work against the owners and investors
...
Example 10
...

EXAMPLE 10
...
S
...
As a consequence, the D-E mixes of the so-called traditional companies (American, United, Delta, and others) have become larger on the debt side than is historically acceptable
...
In an effort to reduce costs, assume that
three airlines joined forces to cooperate on a range of services (baggage handling, onboard
food preparation, ticket services, and software development) by forming a new company called
FullServe, Inc
...

Table 10–1 summarizes the D-E mixes and the total equity capitalization for each airline
after its share of $5 B was removed from available equity funds
...
For

10
...
8
Airline
Company

Corporate D-E
Mix, %

Amount Borrowed,
$B

Equity Capital
Available, $ B

National
Global
PanAm

30–70
65–35
91–9

1
...
25
4
...
0
3
...
7

example, National had 30% of its capitalization in debt capital; therefore, 30% of $5 B was
borrowed, and 70% was provided from National’s equity fund
...
0 billion, only 20% of its original value
...
0 billion in equity capital was returned to each airline
...
Assuming the loan and equity amounts are the
same as shown in Table 10–1, determine the resulting equity capital situation for each airline
after it pays off the loan from its own equity funds
...

Solution
Determine the level of post-FullServe equity capital using the following relation, in $ billions
...
0 ϩ 1
...
50 ϭ $4
...
7 ϩ 1
...
25 ϭ $1
...
7 ϩ 1
...
55 ϭ $3
...
The debt capital to fund the failed FullServe effort has affected National
airlines the least, in large part due to its low D-E mix of 30%–70%
...

The same principles discussed above for corporations are applicable to individuals
...
As an example, assume two engineers each have a take-home amount of
$40,000 after all income tax, social security, and insurance premiums are deducted from their
annual salaries
...
If Jamal has a total debt of $25,000 and Barry owes $100,000, the remaining amount of
the annual take-home pay may be calculated as follows:
Amount Paid, $ per Year
Person

Total
Debt, $

Cost of
Debt at 15%, $

Repayment of Debt, $

Amount Remaining
from $40,000, $

Jamal
Barry

25,000
100,000

3,750
15,000

1,250
5,000

35,000
20,000

Jamal has 87
...

277

278

Chapter 10

Project Financing and Noneconomic Attributes

10
...
The
decision-making process explained in that chapter (Figure 1–1) included the seven steps listed on
the right side of Figure 10–5
...
In all prior evaluations, only one attribute—the economic
one—has been identiﬁed and used to select the best alternative
...
As we are all aware, most evaluations do and should take into account multiple attributes in
decision making
...
However,
these noneconomic dimensions tend to be intangible and often difﬁcult, if not impossible, to
quantify with economic and other scales
...
This section and the next describe some of the techniques that accommodate
multiple attributes in an engineering study
...
Public and service sector projects are excellent examples of multiple-attribute problem solving
...

High levels of complexity are introduced into the selection process by the multiple attributes thought
to be important in selecting an alternative for the dam’s location, design, environmental impact, etc
...
The discussion below concentrates on the expanded step 4 and the next section focuses on the evaluation
measure and alternative selection of step 5
...
To seek input from individuals other than the analyst is
important; it helps focus the study on key attributes
...

•
•
•
•

Comparison with similar studies that include multiple attributes
Input from experts with relevant past experience
Surveys of constituencies (customers, employees, managers) impacted by the alternatives
Small group discussions using approaches such as focus groups, brainstorming, or nominal
group technique
• Delphi method, which is a progressive procedure to develop reasoned consensus from different
perspectives and opinions
Figure 10–5

Consider multiple attributes

Expansion of the decisionmaking process to include
multiple attributes
...
Understand the problem; deﬁne the objective
...
Collect relevant information; deﬁne alternatives
...
Make estimates
...
Identify the attributes for decision
making
...
Determine the relative
importance (weights) of attributes
...
For each alternative, determine
each attribute’s value rating
...
Evaluate each alternative using
a multiple-attribute technique
...

4
...

5
...

6
...

7
...

10
...
There are approximately 8000 options for each plane
that must be decided upon by Delta’s engineering, purchasing, maintenance, and marketing personnel before the order to Boeing is placed
...
An economic study based on the equivalent AW of the estimated passenger income per trip has determined that 150 of these options are
clearly advantageous
...
A Delphi study was performed using input from 25 individuals
...
From these two studies it was determined that there are 10 key attributes for options selection
...

• Safety: mean time to failure (MTTF) of ﬂight-critical components
...
(Basically, this is the attribute evaluated by
the economic study already performed
...

The economic attribute of extra revenue may be considered an indirect measure of customer satisfaction, one that is more quantitative than customer opinion/satisfaction survey results
...
However, the point is that the economic study
may directly address only one or a few of the key attributes vital to alternative decision making
...

Risk is a possible variation in a parameter from an expected, desired, or predicted value that
may be detrimental to observing the intended outcome(s) of the product, process, or system
...
Risk is present when there are two or
more observable values of a parameter and it is possible to assume or estimate the chance that
each value may occur
...
Considerations of variation, probabilistic estimates, etc
...
Formalized sensitivity analysis, expected values, simulation,
and decision trees are some of the techniques useful in handling risk
...
The
weight, a number between 0 and 1, is based upon the experienced opinion of one individual or a
group of persons familiar with the attributes, and possibly the alternatives
...
Otherwise,
some averaging technique must be applied to arrive at one weight value for each attribute
...
Weights Wi for each attribute are entered on the left side
...

Attribute weights are usually normalized such that their sum over all the alternatives is 1
...
This
normalizing implies that each attribute’s importance score is divided by the sum S over all attributes
...
, m) are
m

Normalized weights:

͚ W ϭ 1
...
9]

iϭ1

importance scorei
importance score
Weight calculation: Wi ‫———————— ؍ ————————— ؍‬i [10
...

n

Wm

Value ratings Vij

Of the many procedures developed to assign weights to an attribute, an analyst is likely to rely
upon one that is relatively simple, such as equal weighting, rank order, or weighted rank order
...
Each is brieﬂy presented below
...
This is the default approach
...
10]
...
In this case, the ﬁnal evaluation measure for an alternative will be the sum over
all attributes
...
By Equation [10
...
, m͞S
...

Weighted Rank Order The m attributes are again placed in the order of increasing importance
...
The most important attribute is assigned a score, usually 100, and all other attributes are scored relative to it between 100
and 0
...
10] takes the form
si
Wi ‫—— ؍‬
m

[10
...
11] automatically normalizes the weights
...
If repair time is only one-half as important as safety, and the
last two attributes are each one-half as important as repair time, the scores and weights are
as follows
...
50
50/200 ϭ 0
...
125
25/200 ϭ 0
...
000

Pairwise Comparison Each attribute is compared to each other attribute in a pairwise fashion using a rating scale that indicates the importance of one attribute over the other
...
6

TABLE 10–3
Attribute i
Cost
Constructability
Environment
Sum of scores
Weight Wi

Multiple Attribute Analysis: Identiﬁcation and Importance of Each Attribute
Pairwise Comparison of Three Attributes to Determine Weights
1 ‫ ؍‬Cost

2 ‫ ؍‬Constructability

3 ‫ ؍‬Environment

—
2
1

0
—
1

1
1
—

3
0
...
167

2
0
...
Deﬁne the importance comparison scale as follows:
0 if attribute is less important than one compared to
1 if attribute is equally important as one compared to
2 if attribute is more important than one compared to
Set up a table listing attributes across the top and down the side, and perform the pairwise comparison for each column attribute with each row attribute
...
The arrow to the right of the table indicates the direction of comparison, i
...
, column with row attribute
...
The complement score of 0 is placed in the reverse comparison of
constructability with cost
...
11], where the sum for each column is si
...

s1 ϭ 3
si ϭ 3 ϩ 1 ϩ 2 ϭ 6
Cost weight W1 ϭ 3͞6 ϭ 0
...
167 and W3 ϭ 2͞6 ϭ 0
...

There are other attribute weighting techniques, especially for group processes, such as utility
functions, and the Dunn-Rankin procedure
...
If this consistency is important in that
several decision makers with diverse opinions about attribute importance are involved in a study, a
more sophisticated technique may be warranted
...

4-3 Value Rating of Each Alternative by Attribute This is the ﬁnal step prior to calculating the evaluation measure
...

These are the entries within the cells in Table 10–2
...

The scale for the value rating can vary depending upon what is easiest to understand for those
who do the valuation
...
However,
the most popular is a scale of 4 or 5 gradations about the perceived ability of an alternative to
accomplish the intent of the attribute
...
g
...
The last two scales can give a negative impact to the evaluation
measure for poor alternatives
...

281

282

Chapter 10

Project Financing and Noneconomic Attributes

TA BLE 10–4

Completed Layout for Four Attributes and Three
Alternatives for Multiple Attribute Evaluation
Alternatives

Attributes

Weights

1

2

3

Safety
Repair
Crew needs
Economic

0
...
25
0
...
125

6
9
5
5

4
3
6
9

8
1
6
7

If we now build upon the aircraft purchase illustration to include value ratings, the cells are
ﬁlled with ratings awarded by a decision maker
...
Initially, there will be one such table for each decision maker
...
Determination of this evaluation measure is discussed below
...
7 Evaluation Measure for Multiple Attributes
The need for an evaluation measure that accommodates multiple attributes is indicated in step 5
of Figure 10–5
...
This section introduces a single-dimension measure
that is widely accepted
...
The resulting evaluation measure is a formula that calculates an aggregated measure for use in selecting from two or more
alternatives
...

This reduction process removes much of the complexity of trying to balance the different attributes;
however, it also eliminates much of the robust information captured by the process of ranking attributes for their importance and rating each alternative’s performance against each attribute
...
The most used additive model is the weighted attribute method,
also called the additive weight technique
...
12]

i‫1؍‬

The Wi numbers are the attribute importance weights, and Vij is the value rating by attribute i for
each alternative j
...
10]
...
(If an equal weight of Wi ϭ 1
...
)
The selection guideline is as follows:

ME alternative selection

Choose the alternative with the largest Rj value
...

Sensitivity analysis for any score, weight, or value rating is used to determine sensitivity of the decision to it
...

Chapter Summary

EXAMPLE 10
...
niueisland
...
The spreadsheet in Figure 10–6, left two columns, presents the attributes and normalized
weights Wi published in the RFP for use in selecting one of the tenders presenting proposals
...
The next four columns (C through F ) include value ratings
between 0 and 100 developed by a group of decision makers when the details of each proposal were
evaluated against each attribute
...
Use these weights and ratings to determine which proposal to pursue ﬁrst
...
Equation [10
...
As an illustration, for proposal 3,
R3 ϭ 0
...
20(60) ϩ 0
...
35(85) ϩ 0
...
5 ϩ 12
...
5 ϩ 29
...
0
ϭ 84
...

Comment
Any economic measure can be incorporated into a multiple attribute evaluation using this method
...

Figure 10–6
Attributes, weights, ratings, and evaluation measure for Niue workboat proposals, Example 10
...

1

Used with permission of Government of Niue, Infrastructure Department, “Request for Proposal: Supply
of Workboat,” released March 30, 2010, www
...
nu/Documents/workboattender3022
...

CHAPTER SUMMARY
The interest rate at which the MARR is established depends principally upon the cost of capital
and the mix between debt and equity ﬁnancing
...
Risk, proﬁt, and other factors can be considered after the AW,
PW, or ROR analysis is completed and prior to ﬁnal alternative selection
...

If multiple attributes, which include more than the economic dimension of a study, are to be
considered in making the alternative decision, ﬁrst the attributes must be identiﬁed and their
relative importance assessed
...
The
evaluation measure is determined using a model such as the weighted attribute method, where the
measure is calculated by Equation [10
...
The largest value indicates the best alternative
...
Use the results below to determine the opportunity cost for each measure
...
1 List at least three factors that affect the MARR,
and discuss how each one affects it
...
2 State whether each of the following involves debt
ﬁnancing or equity ﬁnancing
...
3 million
(c) Short-term loan of $75,000 from a local bank
(d ) Issuance of $3 million worth of 20-year
bonds
(e) Del Engineering buyback of $8 million of its
own stock using internal funds
10
...
uses an after-tax MARR of
12% per year
...

10
...
He
estimates that his cost to complete the project will
be $7
...
He wants to bid an amount that
will give him an after-tax rate of return of 15% per
year if he gets the job, but he doesn’t know how
much he should bid on a before-tax basis
...

(a) The expression for determining the overall
effective tax rate is
state rate ϩ (1 Ϫ state rate)(federal rate)
What should his before-tax MARR be in
order for him to make an after-tax MARR
of 15% per year?
(b) How much should he bid on the job?
10
...
Immediately
after the investment was made, another investment opportunity came up for which the investors
didn’t have enough capital
...
If the group’s effective tax
rate is 32%, what after-tax rate of return would
the forgone project have yielded?
10
...
5
7,138
12
...
4
1,051
9
...
0
936

8,000
15,000
8,000
8,000
5,000

A
E
C
B
D

8,000
13,000
21,000
36,000
44,000

A
C
E
D
B

8,000
16,000
21,000
29,000
44,000

10
...
He expects a return of 4% per year on all of his investments
...
Tom
has determined that the proposal’s “risk factor”
will require an additional 3% per year return for
him to accept it
...

(b) Determine the effective MARR for his business if Tom turns down the proposal
...
8 Electrical generators produce not only electricity,
but also heat from conductor resistance and from
friction losses in bearings
...
If $18 million came from mortgages and
bond sales, what was the total amount of the
ﬁnancing?
10
...

10
...
The business student has
$30,000 in student loans that come due at graduation
...
The
engineering senior owes $50,000, 50% from his
parents with no interest due and 50% from a

285

Problems

credit union loan
...

(a) What is the D-E mix for each student?
(b) If their grandparents pay the loans in full at
graduation, what are the amounts on the
checks they write for each graduate?
(c) When grandparents pay the full amount at
graduation, what percent of the principal
does the interest represent?

dividends at a rate of 5% per year, and the remaining 60% from retained earnings, which currently
earn 9% per year
...

10
...
The balance sheet for First Engineering indicates a total
debt of $87 million, and that of Midwest Development indicates a net worth of $62 million
...

10
...
1% for the year in its
report to stockholders
...
The annual report also mentions that projects within the corporation are 75%
funded by its own capital
...

10
...
invested $50 million
...
What is the
return on the company’s equity, if the net income is
$5 million on a revenue base of $6 million?
10
...
The
ﬁnancing proﬁle, with interest rates, is as follows:
$3 million in stock sales at 15% per year, $4 million in bonds at 9%, and $6 million in retained
earnings at 7% per year
...
14 Growth Transgenics Enterprises (GTE) is contemplating the purchase of its rival
...
He learned there are two
plans being considered
...
Plan 2 requires only 20% equity
funds with the balance borrowed at a higher rate of
8% per year
...
2% will not be exceeded,
what is the maximum cost of debt capital allowed for each plan? Are these rates higher
or lower than the current estimates?
10
...
The current plan is 60%
equity capital and 40% debt ﬁnancing
...
17 BASF will invest $14 million this year to upgrade
its ethylene glycol processes
...
Equity capital costs 14
...
Debt capital costs 10%
per year before taxes
...

(a) Determine the amount of annual revenue
after taxes that is consumed in covering the
interest on the project’s initial cost
...
Determine the
amount of annual revenue needed to cover
the interest with this plan, and explain the effect it may have on the corporation’s ability
to borrow in the future
...
18 A couple planning for their child’s college education can fund part of or all the expected $100,000
tuition cost from their own funds (through an education IRA) or borrow all or part of it
...
Use a spreadsheet to
generate a plot of the WACC curve with the estimated loan interest rates below and determine the
best D-E mix for the couple
...
0
6
...
0
9
...
0
13
...
19 Over the last few years, Carol’s Fashion Store, a
statewide franchise, has experienced the D-E
mixes and costs of debt and equity capital on several projects summarized below
...

(b) Determine what mix of debt and equity capital provided the lowest WACC
...
5%
13
...
0
11
...
9
12
...
8%
7
...
9
9
...
5
12
...
20 For Problem 10
...
0% increases
to 14
...

Cost of Debt Capital
10
...
If the company’s effective tax rate is 33%, determine the company’s cost
of debt capital (a) before taxes and (b) after taxes
...
22 A company that makes several different types of
skateboards, Jennings Outdoors, incurred interest
expenses of $1,200,000 per year from various
types of debt ﬁnancing
...
If the company’s effective tax rate is
29%, what was the company’s cost of debt capital
(a) before taxes and (b) after taxes?
10
...
, a manufacturer of cable assemblies for
polycrystalline photovoltaic solar modules, requires $3
...
The company
plans to sell 15-year bonds that carry a dividend of
6% per year, payable semiannually
...
Determine
(a) the nominal annual after-tax cost of debt capital and (b) the effective annual after-tax cost of
debt capital
...
24 Tri-States Gas Producers expects to borrow
$800,000 for ﬁeld engineering improvements
...
The company will pay an effective 8% per year to the bank
for 8 years
...
The
bond issue will be for 800 10-year bonds of $1000
each that require a 6% per year dividend payment
...
25 The Sullivan Family Partnership plans to purchase
a refurbished condo in their hometown for investment purposes
...
5%
per year after all relevant income taxes are paid
...
If the effective
tax rate is 22% per year, based only on these data,
answer the following
...
26 Determine the cost of equity capital to Hy-Lok
USA if the company sells 500,000 shares of its
preferred stock at a 5% discount from its price of
$130
...

10
...
92 per share
...
2% per year, determine the cost of equity capital
for the stock offering
...
28 The cost of debt capital is lower after taxes than
before taxes
...
Why
are the after-tax rates not the same for both types
of ﬁnancing?
10
...
The beta value for its stock
is 1
...
Use the capital asset pricing model and a
3
...
2%
...
30 Management at Hirschman Engineering has asked
you to determine the cost of equity capital based
on the company’s common stock
...
Last year, the ﬁrst year for
dividends, the stock paid $0
...
50 on the New York Stock Exchange
...
Hirschman Engineering stock
has a volatility that is higher than the norm at 1
...

If safe investments are returning 5
...

10
...
93 per share on an average price of $18
...
The company expects
to grow the dividend rate at a maximum of 1
...
The stock volatility is 1
...
95% per year dividend
...
S Treasury bills are
returning 4
...
Determine the company’s cost of
equity capital last year using (a) the dividend
method and (b) the CAPM
...
32 Last year a Japanese engineering materials corporation, Yamachi Inc
...
S
...

Now, Euro bonds are being purchased with a realized average return of 3
...
The volatility
factor of Yamachi stock last year was 1
...
18
...
1% dividends per year
...

10
...
She
asked the ﬁnance manager for the corporate
MARR
...

Funds Source

Amount, $ Average Cost, %

Retained earnings
Stock sales
Long-term loans

4,000,000
6,000,000
5,000,000

7
...
8
9
...

The study is after taxes and part (a) provided
the before-tax MARR
...

Different D-E Mixes
10
...
35 In a leveraged buyout of one company by another,
the purchasing company usually obtains borrowed
money and inserts as little of its own equity as possible into the purchase
...

10
...
Financing will be equally split
between common stock ($250,000) and a loan
with an 8% after-tax interest rate
...
The effective tax rate is 50%
...
Recent analysis shows that it
has a volatility rating of 0
...
Nationally, the safest investment is currently paying 3% per year
...
37 Fairmont Industries primarily relies on 100% equity ﬁnancing to fund projects
...

The Fairmont owner can supply the money from
personal investments that currently earn an average of 7
...
The annual net cash ﬂow
from the project is estimated at $30,000 for the
next 15 years
...
If the MARR is the WACC, determine which
plan, if either, should be undertaken
...

10
...
has an opportunity to invest $10,000,000 in a new engineering remote
control system for offshore drilling platforms
...
Omega’s share of the annual net cash ﬂow
is estimated to be $1
...
Omega is about to initiate CAPM as

288

Chapter 10

Project Financing and Noneconomic Attributes

its common stock evaluation model
...
22 and
is paying a premium of 5% on its common stock
dividend
...
S
...
Is the venture ﬁnancially attractive if the MARR equals (a) the cost of equity
capital and (b) the WACC?
10
...
The phase I
installed price for the dies and machinery is
$2,000,000
...
The WACC over the
last 5 years has averaged 10% per year
...
The ﬁrst requires an investment of
40% equity funds at 9% and a loan for the
balance at an interest rate of 10% per year
...
5% per year
...
With this restriction, what is
the maximum loan interest rate that can be incurred for each of the ﬁnancing alternatives?
10
...
Utilize a
spreadsheet solution to (a) select any combination
of the projects if the MARR is equal to the aftertax WACC and (b) determine if the same projects
should be selected if the risk factors are enough to
require an additional 2% per year for the investment to be made
...
5% per year
...
The effective tax rate is 30% per year
...
41 Two friends each invested $20,000 of their own
(equity) funds
...
Theresa, being a risk taker, leveraged the
$20,000 and purchased a $100,000 condo for rental

property
...

(a) Determine the year-end values of their equity
funds if there was a 10% increase in the
value of the stocks and the condo
...

(c) Use your results to explain why leverage can
be ﬁnancially risky
...
42 In multiple attribute analysis, if three different alternatives are to be evaluated on the basis of eight
attributes that are considered of equal importance,
what is the weight of each attribute?
10
...
Determine the weight of each attribute
using the importance scores
...
Safety
2
...
Impact
4
...
Acceptability

Importance Score
60
40
80
30
20

10
...
, and J
...

10
...
Use the statements
to determine the normalized weights if assigned
scores are between 0 and 100
...
Flexibility (F)
2
...
Uptime (U)
4
...
46 Different types and capacities of crawler hoes are
being considered for use in a major excavation on
a pipe-laying project
...
For the information that follows,
determine the weighted rank order, using a 0-to-10
scale and the normalized weights
...
Truck versus hoe height 90% as important as trenching speed
2
...
Type of subsoil
30% as important as trenching speed
4
...
Hoe trenching speed
Most important attribute
6
...
47 John, who works at Swatch, has decided to use the
weighted attribute method to compare three systems for manufacturing a watchband
...
John’s
ratings for each alternative are as follows:
Alternative
Attribute
Economic return Ͼ MARR
High throughput
Low scrap rate

1

2

3

50
100
100

70
60
40

100
30
50

Use the weights below to evaluate the alternatives
...
48 The Athlete’s Shop has evaluated two proposals
for weight lifting and exercise equipment
...
In addition to
this economic measure, three more attributes
were independently assigned a relative importance score from 0 to 100 by the shop manager
and the lead trainer
...
0 as shown in the following table
...

Attribute
Economics
Durability
Flexibility
Maintainability

Proposal A

Proposal B

1
...
35
1
...
25

0
...
00
0
...
00

Select the better proposal using each of the following methods
...
49 The term opportunity cost refers to:
(a) The ﬁrst cost of an alternative that has been
accepted for funding
(b) The total cost of an alternative that has been
accepted for funding
(c) The rate of return or proﬁt available on the
next-best alternative that had to be forgone
due to lack of capital funds
(d) The cost of an alternative that was not recognized as an alternative that actually represented a good opportunity
10
...
51 All of the following are examples of debt capital
except:
(a) Retained earnings
(b) Long-term bonds
(c) Loan from a local bank
(d) Purchase of equipment using a credit card
10
...
53 If a public utility expands its capacity to generate
electricity by obtaining $41 million from retained
earnings and $30 million from municipal bond
sales, the utilities’ debt-to-equity mix is closest to:
(a) 58% debt and 42% equity
(b) 73% debt and 27% equity
(c) 27% debt and 73% equity
(d) 42% debt and 58% equity

If Medzyme does not want to exceed its historical
weighted average cost of capital (WACC), and it is
forced to go to a D-E mix of 75–25, the maximum
acceptable cost of equity capital is closest to:
(a) 7
...
2%
(c) 9
...
9%

10
...
ﬁnanced a new product as follows:
$5 million in stock sales at 13
...
9% per year, and
$3 million through convertible bonds at 7
...
The company’s WACC is closest to:
(a) 9% per year
(b) 10% per year
(c) 11% per year
(d) 12% per year

10
...
The weight to assign to attribute
1 is:
(a) 0
...
20
(c) 0
...
25

10
...
2% and the corporate effective tax rate is
39%, the approximated before-tax rate of return is
closest to:
(a) 6
...
4%
(c) 18
...
7%
10
...
Equity capital
has cost 11%; however, debt capital that historically cost 9% has now increased by 20% per year
...
58 For eight attributes rank-ordered in terms of increasing importance, the weighting of the sixth attribute is closest to:
(a) 0
...
14
(c) 0
...
03

CASE STUDY
WHICH IS BETTER—DEBT OR EQUITY FINANCING?
Background

Information

Pizza Hut Corporation has decided to enter the catering business in three states within its Southeastern U
...
Division,
using the name Pizza Hut At-Your-Place
...
5 million
...

A feasibility study completed last year indicated that the
At-Your-Place business venture could realize an estimated
annual net cash ﬂow of $300,000 before taxes in the three
states
...

An engineer with Pizza Hut’s Distribution Division has
worked with the corporate ﬁnance ofﬁce to determine how to
best develop the $1
...
There are two viable ﬁnancing plans
...
(A simplifying assumption that $75,000 of the principal is repaid with each
annual payment can be made
...
The ﬁnancial manager informed the
engineer that stock is paying $0
...
This dividend pattern is expected to continue,
based on the current ﬁnancial environment
...
What values of MARR should the engineer use to determine the better ﬁnancing plan?

Case Study

2
...
He does not know
how to consider all the tax angles for the debt ﬁnancing
in plan A
...
Is A or B
the better plan?
3
...
Plot the WACC curve and compare
its shape with that of Figure 10–2
...

SECTION

TOPIC

LEARNING OUTCOME

11
...

11
...

11
...

11
...

11
...

11
...

O

ne of the most common and important issues in industrial practice is that of replacement or retention of an asset, process, or system that is currently installed
...
The fundamental question answered by a replacement study (also called a replacement͞retention
study) about a currently installed system is, Should it be replaced now or later? When an
asset is currently in use and its function is needed in the future, it will be replaced at some
time
...

A replacement study is usually designed to ﬁrst make the economic decision to retain or
replace now
...
If the decision is to retain,
the cost estimates and decision can be revisited each year to ensure that the decision to
retain is still economically correct
...

A replacement study is an application of the AW method of comparing unequal-life alternatives, ﬁrst introduced in Chapter 6
...

If a study period is speciﬁed, the replacement study procedure is different from that used
when no study period is set
...
After-tax
replacement analysis is included in Chapter 17
...

The products are sold to a wide range of
industries from the nuclear and solar power
industry to sports equipment manufacturers
of specialty golf and tennis gear, where kiln
temperatures up to approximately 1700°C
are needed
...
Two are in use
currently at plant locations on each coast
of the country; one kiln is 10 years old, and
the second was purchased only 2 years ago
and serves, primarily, the ceramics industry
needs on the west coast
...

During the last two or three quarterly
maintenance visits, the Harper team
leader and the head of B&T quality have
discussed the ceramic and metal industry
needs for higher temperatures
...
These
may ﬁnd use in hypersonic vehicles,
engines, plasma arc electrodes, cutting
tools, and high-temperature shielding
...
This unit will have lower
operating costs and signiﬁcantly greater
furnace efﬁciency in heat time, transit,
and other crucial parameters
...

For identiﬁcation, let
PT identify the currently installed
pusher-plate tunnel kiln (defender)
GH identify the proposed new graphite hearth kiln (challenger)
Relevant estimates follow in $ millions
for monetary units
...
2; starts at $3
...
4
increases
10%͞year

Life, years

6 (remaining)

Heating
element,
$M

—

$38; with no
trade-in

12 (estimated)
$2
...
2)
Replacement study (Section 11
...
5)
Replacement value (Section 11
...
18 and 11
...
1 Basics of a Replacement Study
The need for a replacement study can develop from several sources:
Reduced performance
...
This usually
results in increased costs of operation, higher scrap and rework costs, lost sales, reduced quality, diminished safety, and larger maintenance expenses
...
New requirements of accuracy, speed, or other speciﬁcations cannot
be met by the existing equipment or system
...

Obsolescence
...
The ever-decreasing development cycle time to bring new products to market is
often the reason for premature replacement studies, that is, studies performed before the
estimated useful or economic life is reached
...

Defender and challenger are the names for two mutually exclusive alternatives
...
A
replacement study compares these two alternatives
...
(This is the
same terminology used earlier for incremental ROR and B͞C analysis, but both alternatives
were new)
...
Also called trade-in value, this estimate is obtained from professional appraisers, resellers, or liquidators familiar with the industry
...
In replacement analysis,
the salvage value at the end of one year is used as the market value at the beginning of the
next year
...
The term equivalent uniform annual cost (EUAC) may be used in lieu of
AW, because often only costs are included in the evaluation; revenues generated by the
defender or challenger are assumed to be equal
...
) Therefore, all values will be negative when only
costs are involved
...

Economic service life

Economic service life (ESL) for an alternative is the number of years at which the lowest AW
of cost occurs
...
The next section
explains how to ﬁnd the ESL
...
The current
market value (MV) is the correct estimate to use for P for the defender in a replacement
study
...
It is
incorrect to use the following as MV for the defender ﬁrst cost: trade-in value that does
not represent a fair market value, or the depreciated book value taken from accounting
records
...
), this cost is added to the MV to obtain the estimated
defender ﬁrst cost
...
This amount is almost always equal to P, the ﬁrst cost
of the challenger
...
1

295

Basics of a Replacement Study

If an unrealistically high trade-in value is offered for the defender compared to its fair
market value, the net cash ﬂow required for the challenger is reduced, and this fact should be
considered in the analysis
...
In equation form, this is P Ϫ (TIV Ϫ MV)
...
e
...
e
...
Of course, when the trade-in and market values are the same, the challenger P
value is used in all computations
...

Sometimes, an analyst or manager will attempt to increase this ﬁrst cost by an amount equal to the
unrecovered capital remaining in the defender, as shown on the accounting records for the asset
...
This leads us to identify two additional characteristics of
replacement analysis, in fact, of any economic analysis: sunk costs and nonowner’s viewpoint
...
The replacement alternative for an asset, system, or process that has
incurred a nonrecoverable cost should not include this cost in any direct fashion; sunk costs
should be handled in a realistic way using tax laws and write-off allowances
...
For example, assume an asset costing $100,000
two years ago has a depreciated value of $80,000 on the corporate books
...
If the replacement alternative (challenger) has
a ﬁrst cost of $150,000, the $80,000 from the current asset is a sunk cost were the challenger
purchased
...

The second characteristic is the perspective taken when conducting a replacement study
...

The nonowner’s viewpoint, also called the outsider’s viewpoint or consultant’s viewpoint, provides the greatest objectivity in a replacement study
...
Additionally, it assumes the services provided by the defender can be purchased now by making an “initial
investment” equal to the market value of the defender
...

As mentioned in the introduction, a replacement study is an application of the annual worth
method
...
If the planning horizon is unlimited, that is, a study period is not speciﬁed, the assumptions are as follows:
1
...

2
...

When this challenger replaces the defender (now or later), it will be repeated for succeeding
life cycles
...
Cost estimates for every life cycle of the defender and challenger will be the same as in their
ﬁrst cycle
...
We discussed this previously for the
AW method (and the PW method)
...
The replacement procedure discussed in Section 11
...
When the
planning horizon is limited to a speciﬁed study period, the assumptions above do not hold
...
5 discusses replacement analysis over a ﬁxed study period
...
1
Only 2 years ago, Techtron purchased for $275,000 a fully loaded SCADA (supervisory control and data acquisition) system including hardware and software for a processing plant operating on the Houston ship channel
...
Actual M&O costs have been $25,000 per year, and the book value
is $187,000
...
Given these
factors, the system is likely worth nothing if kept in use for the ﬁnal 3 years of its anticipated
useful life
...
A 5-year life, salvage
value of 15% of stated ﬁrst cost or $60,000, and an M&O cost of $15,000 per year are good
estimates for the new system
...

Using the above values as the best possible today, state the correct defender and challenger
estimates for P, M&O, S, and n in a replacement study to be performed today
...
All
others—original cost of $275,000, book value of $187,000, and trade-in value of
$100,000—are irrelevant to a replacement study conducted today
...

First cost
P ϭ $Ϫ400,000
M&O cost
A ϭ $Ϫ15,000 per year
Expected life
n ϭ 5 years
Salvage value
S ϭ $60,000

Comment
If the replacement study is conducted next week when estimates will have changed, the defender’s ﬁrst cost will be $80,000, the new market value according to the appraiser
...

11
...
In reality, the best life
estimate to use in the economic analysis is not known initially
...
The best life estimate is called the economic service life
...

11
...

To

W
tal A

AW

of c

o sts

OC
of A

C a p it a l r

e c o v ery

0

Years
Economic
service life

The ESL is also referred to as the economic life or minimum cost life
...
When n years have passed, the ESL indicates that the asset should be replaced
to minimize overall costs
...

The ESL is determined by calculating the total AW of costs if the asset is in service 1 year,
2 years, 3 years, and so on, up to the last year the asset is considered useful
...
1]

The ESL is the n value for the smallest total AW of costs
...
Therefore, $–200 is a lower cost than $Ϫ500
...
The CR component of total
AW decreases, while the AOC component increases, thus forming the concave shape
...

Decreasing cost of capital recovery
...
Capital recovery is calculated by Equation [6
...
The salvage value S, which usually decreases with time, is the estimated
market value (MV) in that year
...
2]

Increasing cost of AW of AOC
...
To calculate the AW of the AOC series for 1, 2, 3,
...

The complete equation for total AW of costs over k years (k ϭ 1, 2, 3,
...
3]

j‫1؍‬

P ϭ initial investment or current market value
Sk ϭ salvage value or market value after k years
AOCj ϭ annual operating cost for year j ( j ϭ 1 to k)

Capital recovery

298

Chapter 11

Replacement and Retention Decisions

The current MV is used for P when the asset is the defender, and the estimated future MV values
are substituted for the S values in years 1, 2, 3,
...

To determine ESL by spreadsheet, the PMT function (with embedded NPV functions as
needed) is used repeatedly for each year to calculate capital recovery and the AW of AOC
...
The PMT function formats for the capital recovery
and AOC components for each year k (k ϭ 1, 2, 3,
...
4]
؊PMT(i%,years,NPV(i%,year_1_AOC:
year_k_AOC)؉0)

When the spreadsheet is developed, it is recommended that the PMT functions in year 1 be developed using cell-reference format; then drag down the function through each column
...
Augmenting the table with an Excel xy
scatter chart graphically displays the cost curves in the general form of Figure 11–1, and the ESL
is easily identiﬁed
...
2 illustrates ESL determination by hand and by spreadsheet
...
2
A 3-year-old backup power system is being considered for early replacement
...
Estimated future market values and annual operating costs for the next
5 years are given in Table 11–1, columns 2 and 3
...

Solution by Hand

Equation [11
...
, 5
...
Column 5
gives the equivalent AW of AOC for k years
...
3] is
Total AW3 ϭ ϪP(A͞P,i,3) ϩ MV3(A͞F,i,3) Ϫ [PW of AOC1,AOC2, and AOC3](A͞P,i,3)
ϭ Ϫ20,000(A͞P,10%,3) ϩ 6000(A͞F,10%,3) Ϫ [5000(P͞F,10%,1)
ϩ 6500(P͞F,10%,2) ϩ 8000(P͞F,10%,3)](A͞P,10%,3)
ϭ Ϫ6230 Ϫ 6405 ϭ $Ϫ12,635
A similar computation is performed for each year 1 through 5
...
Therefore, the defender ESL is n ϭ 3 years,
and the AW value is $Ϫ12,635
...

Solution by Spreadsheet

See Figure 11–2 for the spreadsheet screen shot and chart that shows the ESL is n ϭ 3 years
and AW ϭ $Ϫ12,634
...
2

Economic Service Life

Current market value

ϭ PMT($B$1,$A9,$B$2,Ϫ$B9)

ϭ ϪPMT($B$1,$A9,NPV($B$1,$C$5:$C9)ϩ0)

ESL of defender
Total AW is minimum
at n ϭ 3 years

Total AW
curve

Capital
recovery
curve

Figure 11–2
Determination of ESL and plot of curves, Example 11
...
) Contents of columns D and E are described below
...
4]
...
The $ symbols are included for absolute cell referencing, needed
when the entry is dragged down through the column
...
For example, in actual numbers, the
cell-reference PMT function in year 5 shown on the spreadsheet reads ϭ PMT (10%,5,20000,
Ϫ0), resulting in $Ϫ5276
...

Column E: The NPV function embedded in the PMT function obtains the present worth in
year 0 of all AOC estimates through year k
...
For example, in year 5, the PMT in numbers is ϭ ϪPMT(10%,5,NPV
(10%,C5:C9)ϩ0)
...
The graph plots the AW of
AOC curve, which constantly increases in cost because the AOC estimates increase
each year
...
If the same MV were estimated for each
year, the curve would appear like Figure 11–1
...
This indicates that the ESL is relatively insensitive to costs
...
Previously we had a speciﬁc life estimated to be n years
with associated other estimates: ﬁrst cost in year 0, possibly a salvage value in year n, and an
AOC that remained constant or varied each year
...
This is the economic service life when n
is ﬁxed
...
Therefore,
we can conclude the following:
When the expected life n is known and speciﬁed for the challenger or defender, no ESL computations are necessary
...
This AW value is the correct one
to use in the replacement study
...
First the market/salvage series is needed
...
For example, an asset with a
ﬁrst cost of P can lose market value of, say, 20% per year, so the market value series for years
0, 1, 2,
...
8P, 0
...
, respectively
...

Marginal costs (MC) are year-by-year estimates of the costs to own and operate an asset for that
year
...
The
sum of the AW values of the ﬁrst two of these components is the capital recovery amount
...
3]
...

AW of marginal costs ‫ ؍‬total AW of costs

[11
...
The ESL analysis presented in Example 11
...
This is demonstrated in Example 11
...

EXAMPLE 11
...
When the current tunnel kiln was purchased 2 years ago for $25 million, an ESL study indicated that the minimum cost life was between 3 and 5 years of the expected 8-year life
...
Now, the same type of question arises for the proposed graphite hearth
model that costs $38 million new: What are the ESL and the estimated total AW of costs? The
Manager of Critical Equipment at B&T estimates that the market value after only 1 year will
drop to $25 million and then retain 75% of the previous year’s value over the 12-year expected
life
...

Solution
Figure 11–3 is a spreadsheet screen shot of the two analyses in $ million units
...

A brief description of each analysis follows
...
2

Economic Service Life

ESL analysis

Two AW series
are identical

Marginal cost analysis

Figure 11–3
Comparison of annual worth series resulting from ESL analysis and marginal cost analysis, Example 11
...

ESL analysis: Equation [11
...
, 12 years (columns C,
D, and E) in the top of Figure 11–3
...

The result in column F is the total AW series that is of interest now
...
Row 33 details the functions
for year 12
...

The two AW series are identical, thus demonstrating that Equation [11
...
Therefore, either an ESL or a marginal cost analysis will provide the same information for a replacement study
...
32 million at its full 12-year life
...
These conclusions are based on the extent to which detailed annual estimates are
made for the market value
...
Year-by-year market value estimates are made
...
These are the best n and AW
values for the replacement study
...
Yearly market value estimates are not available
...
Use it to calculate the AW over n years
...

301

302

Chapter 11

Replacement and Retention Decisions

Upon completion of the ESL analysis (item 1 above), the replacement study procedure in Section 11
...
3 Performing a Replacement Study
Replacement studies are performed in one of two ways: without a study period speciﬁed or with
one deﬁned
...
The procedure discussed in this section applies when no study period (planning horizon) is speciﬁed
...
5 is applied
...
The complete study is ﬁnished if the challenger (C) is selected to replace the defender (D) now
...
Use the annual worth and life
values for C and D determined in the ESL analysis in the following procedure
...

The replacement study procedure is:
New replacement study:
1
...
When the
challenger is selected, replace the defender now, and expect to keep the challenger for nC
years
...
If the defender is selected, plan to retain it for up
to nD more years
...
) Next year, perform the
following steps
...
Determine if all estimates are still current for both alternatives, especially ﬁrst cost, market
value, and AOC
...
If yes and this is year nD, replace the defender
...
This step
may be repeated several times
...
Whenever the estimates have changed, update them and determine new AWC and AWD values
...

Figure 11–4

Replacement study

Overview of replacement
study approaches
...
3

Performing a Replacement Study

If the defender is selected initially (step 1), estimates may need updating after 1 year of retention (step 2)
...
Either signiﬁcant
changes in defender estimates or availability of a new challenger indicates that a new replacement study is to be performed
...

Example 11
...
The planning horizon is unspeciﬁed in this
example
...
4
Two years ago, Toshiba Electronics made a $15 million investment in new assembly line
machinery
...
The equipment sorts, tests, and performs insertion-order kitting on
electronic components in preparation for special-purpose circuit boards
...
Due to the new standards, coupled with rapidly changing technology, a
new system is challenging the retention of these 2-year-old machines
...
The i is 10% and the
estimates are below
...

(b) Perform the replacement study now
...
The challenger is making large
inroads to the market for electronic components assembly equipment, especially with the
new international standards features built in
...
Also, this prematurely outdated
equipment is more costly to keep serviced, so the estimated AOC next year has been increased from $8000 to $12,000 and to $16,000 two years out
...

Solution
(a) The results of the ESL analysis, shown in Figure 11–5, include all the MV and AOC estimates in columns B and C
...
The total AW of costs is for each year, should the challenger be placed into service
for that number of years
...
3], where the A͞G factor accommodates the arithmetic gradient
series in the AOC
...
4
...
4]
...

The lowest AW cost (numerically largest) values for the replacement study are as follows:
Challenger:
Defender:

AWC ϭ $Ϫ19,123
AWD ϭ $Ϫ17,307

for nC ϭ 4 years
for nD ϭ 3 years

The challenger total AW of cost curve (Figure 11–5) is classically shaped and relatively ﬂat
between years 3 and 6; there is virtually no difference in the total AW for years 4 and 5
...

(b) To perform the replacement study now, apply only the ﬁrst step of the procedure
...
Prepare to perform the one-year-later analysis 1 year from now
...
Apply the steps for the one-year-later analysis:
2
...
Go to step 3 to
perform a new ESL analysis for the defender
...
The defender estimates in Figure 11–5 are updated below for the ESL analysis
...
3]
...

Year k

Market
Value, $

AOC, $

Total AW
If Retained k More Years, $

0
1
2

12,000
2,000
0

—
Ϫ12,000
Ϫ16,000

—
Ϫ23,200
Ϫ20,819

11
...
Therefore, replace the defender
now, not 2 years from now
...

PE

EXAMPLE 11
...
A
marketing study revealed that the improving business activity on the west coast implies that
the revenue proﬁle between the installed kiln (PT) and the proposed new one (GH) would be
the same, with the new kiln possibly bringing in new revenue within the next couple of years
...
Assume you are the lead
engineer and that you previously completed the ESL analysis on the challenger (Example 11
...
It indicates that for the GH system the ESL is its expected useful life
...
21؊$ ؍‬million
The president asked you to complete the replacement study, stipulating that, due to the rapidly
rising annual operating costs (AOC), the defender would be retained a maximum of 6 years
...

Solution
After some data collection, you have good evidence that the market value for the PT system
will stay high, but that the increasing AOC is expected to continue rising about $1
...
The best estimates for the next 6 years in $ million units are these:
Year

1

2

3

4

5

6

Market Value, $ M
AOC, $ M per year

22
...
2

22
...
4

22
...
6

20
...
8

18
...
0

18
...
2

You developed a spreadsheet and performed the analysis in Figure 11–6
...
0(A͞P,15%,3) ϩ 22
...
2(P͞F,15%,1) ϩ 6
...
6(P͞F,15%,3)](A͞P,15%,3)
ϭ –9
...
34 Ϫ[14
...
43798)
ϭ $Ϫ9
...

Defender: ESL nPT ‫ 1 ؍‬year with total equivalent annual cost AWPT ‫ 05
...
5
...
Since AWPT ϭ
$Ϫ8
...
32 million, you should recommend keeping the current kiln only 1 more year and doing another study during the year
to determine if the current estimates are still reliable
...
A comparison of Figure 11–3 (top), column F, and Figure 11–6, column F, shows us that the largest
total AW of the current system ($–11
...
32 M for 12 years)
...
It would take some signiﬁcant estimate
changes to justify the challenger
...
4 Additional Considerations in a
Replacement Study
There are several additional aspects of a replacement study that may be introduced
...

• Future-year replacement decisions at the time of the initial replacement study
• Opportunity cost versus cash ﬂow approaches to alternative comparison
• Anticipation of improved future challengers
In most cases when management initiates a replacement study, the question is best framed as,
“Replace now, 1 year from now, 2 years from now, etc
...
In other words,
at the time it is performed, step 1 of the procedure does answer the replacement question for
multiple years
...

The ﬁrst costs (P values) for the challenger and defender have been correctly taken as the
initial investment for the challenger C and current market value for the defender D
...
This approach, also called the conventional
approach, is correct for every replacement study
...
Use of
the cash ﬂow approach is strongly discouraged for at least two reasons: possible violation of the
equal-service requirement and incorrect capital recovery value for C
...
Therefore, the cash ﬂow approach can work only when challenger and defender lives are exactly equal
...
If this equalservice comparison reason is not enough to avoid the cash ﬂow approach, consider what happens
to the challenger’s capital recovery amount when its ﬁrst cost is decreased by the market value of
the defender
...
3] will decrease, resulting in a
falsely low value of CR for the challenger, were it selected
...
The conclusion is simple:
Use the initial investment of C and the market value of D as the ﬁrst costs in the ESL analysis and
in the replacement study
...
The expectation of ever-improving challengers can offer strong encouragement to retain the defender until
some situational elements—technology, costs, market ﬂuctuations, contract negotiations, etc
...
5

307

Replacement Study over a Speciﬁed Study Period

stabilize
...
A large expenditure on equipment when
the standards changed soon after purchase forced an early replacement consideration and a large loss
of invested capital
...
It is
important to understand trends, new advances, and competitive pressures that can complement the
economic outcome of a good replacement study
...
Adding needed features to a currently installed defender may prolong its useful life and productivity until challenger choices are more appealing
...
If taxes should be considered, proceed now, or after the next section, to Chapter 17
and the after-tax replacement analysis in Section 17
...

11
...

The AW values for the challenger and for the remaining life of the defender are not based on the
economic service life; the AW is calculated over the study period only
...

This means that the defender or challenger is not needed beyond the study period
...
1—service needed for indefinite future, best challenger available now, and estimates will be identical for future life cycles
...
This is especially important for
the defender
...

When the defender’s remaining life is shorter than the study period, the cost of providing
the defender’s services from the end of its expected remaining life to the end of the study
period must be estimated as accurately as possible and included in the replacement study
...

1
...
Develop all the viable ways to use the defender and challenger during the study period
...
The AW values for the challenger and
defender cash ﬂows are used to build the equivalent cash ﬂow values for each option
...
Selection of the best option
...
Select the option with the lowest cost, or highest income if revenues are estimated
...
)
The following examples use this procedure and illustrate the importance of making cost estimates for the defender alternative when its remaining life is less than the study period
...
6
Claudia works with Lockheed-Martin (LMCO) in the aircraft maintenance division
...
S
...
A key piece of equipment for maintenance
operations is an avionics circuit diagnostics system
...
It has no capital recovery costs remaining, and the following are
reliable estimates: current market value ϭ $70,000, remaining life of 3 more years, no salvage
value, and AOC ϭ $30,000 per year
...

Claudia has found that there is only one good challenger system
...

Study period

308

Chapter 11

Replacement and Retention Decisions

Realizing the importance of accurate defender alternative cost estimates, Claudia asked the
division chief what system would be a logical follow-on to the current one 3 years hence, if
LMCO wins the contract
...
The company would keep it
for the entire 10 additional years for use on an extension of this contract or some other application that could recover the remaining 3 years of invested capital
...
Claudia’s estimate of the ﬁrst cost of this same system 3 years from now is
$900,000
...

The division chief mentioned any study had to be conducted using the interest rate of 10%,
as mandated by the U
...
Ofﬁce of Management and Budget (OMB)
...

Solution
The study period is ﬁxed at 10 years, so the intent of the replacement study assumptions is not
present
...

Further, any analyses to determine the ESL values are unnecessary since alternative lives are
already set and no projected annual market values are available
...
Since the defender will be replaced now or in
3 years, there are only two options:
1
...

2
...

Cash ﬂows are diagrammed in Figure 11–7
...

Equation [11
...
6
...
5

Replacement Study over a Speciﬁed Study Period

The second option has more complex cost estimates
...
Added to this is the capital recovery for the defender follow-on for the
next 7 years
...
(It is not
unusual for the recovery of invested capital to be moved between projects, especially for contract
work
...
The ﬁnal cash ﬂows are shown in Figure 11–7b
...

The CR and AW for all 10 years are
CRDF ϭ Ϫ900,000(A͞P,10%,10) ϭ $Ϫ146,475
AWDF ϭ (Ϫ146,475 Ϫ 50,000)(F͞A,10%,7)(A͞F,10%,10) ϭ $Ϫ116,966

[11
...
This is the AW
for option 2
...
Retain the defender now and expect
to purchase the follow-on system 3 years hence
...
If this assumption were not made, its capital recovery cost would
be calculated over 7 years, not 10, in Equation [11
...
This raises the
annual worth to AWD ϭ $Ϫ163,357
...

EXAMPLE 11
...
Because of ﬂight increases, new ﬁre-ﬁghting capacity is needed once again
...
Estimates are presented below
...

Presently Owned
First cost P, $
AOC, $
Market value, $
Salvage value, $
Life, years

New Purchase

Double Capacity

Ϫ151,000 (3 years ago)
Ϫ1,500
70,000
10% of P
12

Ϫ175,000
Ϫ1,500
—
12% of P
12

Ϫ190,000
Ϫ2,500
—
10% of P
12

Solution
Identify option 1 as retention of the presently owned truck and augmentation with a new samecapacity vehicle
...

Option 1

Option 2

Presently Owned
P, $
AOC, $
S, $
n, years

Augmentation

Double Capacity

Ϫ70,000
Ϫ1,500
15,100
9

Ϫ175,000
Ϫ1,500
21,000
12

Ϫ190,000
Ϫ2,500
19,000
12

309

310

Chapter 11

Replacement and Retention Decisions

(a) For a full-life 12-year study period of option 1,
AW1 ϭ (AW of presently owned) ϩ (AW of augmentation)
ϭ [Ϫ70,000(A͞P,12%,9) ϩ 15,100(A͞F,12%,9) Ϫ 1500]
ϩ [Ϫ175,000(A͞P,12%,12) ϩ 21,000(A͞F,12%,12) Ϫ 1500]
ϭ Ϫ13,616 Ϫ 28,882
ϭ $Ϫ42,498
This computation assumes the equivalent services provided by the current ﬁre truck can be
purchased at $−13,616 per year for years 10 through 12
...

(b) The analysis for an abbreviated 9-year study period is identical, except that n ϭ 9 in each
factor; that is, 3 fewer years are allowed for the augmentation and double-capacity trucks
to recover the capital investment plus a 12% per year return
...

AW1 ϭ $Ϫ46,539

AW2 ϭ $Ϫ36,873

Option 2 is again selected
...
It involves the capital recovery amount for the challenger, when the strict deﬁnition of a study period is applied
...
Highly abbreviated study periods tend to disadvantage the challenger
because no consideration of time beyond the end of the study period is made in calculating the
challenger’s capital recovery amount
...
For example, if the study period is 5 years and
the defender will remain in service 1 year, or 2 years, or 3 years, cost estimates must be made to
determine AW values for each defender retention period
...

Option

Defender
Retained, Years

Challenger
Serves, Years

W
X
Y
Z

3
2
1
0

2
3
4
5

The respective AW values for defender retention and challenger use deﬁne the cash ﬂows for
each option
...
8 illustrates the procedure using the progressive example
...
8 Keep or Replace the Kiln Case

PE

We have progressed to the point that the replacement study between the defender PT and
challenger GH was completed (Example 11
...
The defender was the clear choice with a
much smaller AW value ($Ϫ8
...
32 M)
...
They know the current tunnel kiln is much cheaper
than the new graphite hearth, but the prospect of future new business should not be dismissed
...
5

Replacement Study over a Speciﬁed Study Period

TA BLE 11–2

Replacement Study Options and Total AW Values, Example 11
...
50
Ϫ9
...
59
Ϫ10
...
16
Ϫ11
...
21
Ϫ15
...
31
Ϫ18
...
10
—

The president asked, “Is it possible to determine when it is economically the cheapest to
purchase the new kiln, provided the current one is kept at least 1 year, but no more than 6
years, its remaining expected life?” The chief ﬁnancial ofﬁcer answered, yes, of course
...
(b) Discuss the next step in the analysis based on
the conclusion reached here
...
We know the MARR is 15% per year, the study period has been established at 6 years, and the defender PT will stay in place between 1 and 6 years
...
The total AW values
were determined for the defender in Example 11
...
3 (Figure 11–3)
...
Use the
procedure for a replacement study with a ﬁxed study period
...
There are six options in this case; the defender
is retained from 1 to 6 years while the challenger is installed from 0 to 5 years
...
Figure 11–8 presents the options and the AW series for each option
from Table 11–2
...

Step 2: Selection of the best option
...
The conclusion is clearly to keep the
defender in place for 6 more years
...
If the analysis is to be carried further, the possibility of increased revenue based on
services of the challenger’s high-temperature and operating efﬁciency should be considered next
...
A revenue increase for the challenger will reduce its AW of costs and possibly make it more
economically viable
...
8
...
6 Replacement Value
Often it is helpful to know the minimum market value of the defender necessary to make the
challenger economically attractive
...

This is a breakeven value between AWC and AWD; it is referred to as the replacement value
(RV)
...
The AWC is known, so RV can be determined
...

Determination of the RV for a defender is an excellent opportunity to utilize the Goal Seek tool
in Excel
...
Example 11
...

EXAMPLE 11
...
This is based
on the ESL analysis that concluded the following (Examples 11
...
5):
Defender:
Challenger:

ESL nPT ϭ 1 year with AWPT ϭ $Ϫ8
...
32 million

The original defender price was $25 million, and a current market value of $22 million was
estimated earlier (Figure 11–6)
...
What will you discover RV to be? The MARR is 15% per year
...
32
and solve for RV
...
20
MV ϭ $22
...
32 ϭ ϪRV(A͞P,15%,1) ϩ 22
...
20
1
...
32 ϩ 22
...
20
RV ϭ $25
...

Comment
To ﬁnd RV using a spreadsheet, return to Figure 11–6
...
50), and the required value is $Ϫ12
...
The “changing” cell is the current market value (cell F2), currently $22
...

When “OK” is touched, $25
...
This is the RV
...
Best
(economic) challenger is described as the one with the lowest annual worth (AW) of costs for
some number of years
...
However, if reasonable estimates of the expected market value (MV) and AOC

Problems

313

for each year of ownership can be made, these year-by-year (marginal) costs help determine the
best challenger
...
The resulting nC and AWC values are used
in the replacement study
...

Replacement studies in which no study period (planning horizon) is speciﬁed utilize the annual worth method of comparing two unequal-life alternatives
...

When a study period is speciﬁed for the replacement study, it is vital that the market value and
cost estimates for the defender be as accurate as possible
...
All the viable options for using the defender and challenger are enumerated, and their AW
equivalent cash ﬂows are determined
...
This option determines how long the defender is retained before replacement
...
1 In a replacement study, what is meant by “taking
the nonowner’s viewpoint”?
11
...
The company thought the asset would last
5 years and that its book value would decrease by
$20,000 each year and, therefore, be worthless at
the end of year 5
...
If the replacement is
purchased immediately at a ﬁrst cost of $75,000
and if it will have a lower annual worth, what is the
amount of the sunk cost? Assume the company’s
MARR is 15% per year
...
3 As a muscle car aﬁcionado, a friend of yours likes
to restore cars of the 60s and 70s and sell them for
a proﬁt
...
Another opportunity has
come up (a 1969 Dodge Charger) that he is thinking of buying because he believes he could sell it
for a proﬁt of $60,000 after it is completely restored
...
He thought that the
completely restored Shelby would be worth
$195,000, resulting in a tidy proﬁt of $22,000, but
in its half-restored condition, the most he could get
now is $115,000
...

(a) What is wrong with this thinking?
(b) What is his sunk cost in the Shelby?

11
...

11
...
At that time, it was expected to be used for 10 years and then traded in
for its salvage value of $10,000
...
The company estimates that the old crane
can be used, if necessary, for another 3 years, at
which time it would have a $23,000 estimated
market value
...
Determine the values of
P, n, S, and AOC that should be used for the existing crane in a replacement analysis
...
6 Equipment that was purchased by Newport Corporation for making pneumatic vibration isolators
cost $90,000 two years ago
...

Experience with this type of equipment has shown
that the operating cost for the ﬁrst 4 years is
$65,000 per year, after which it increases by $6300
per year
...
Determine the values of P, S, and AOC
if a replacement study is done (a) now and
(b) 1 year from now
...
7 A piece of equipment that was purchased 2 years
ago by Toshiba Imaging for $50,000 was expected
to have a useful life of 5 years with a $5000 salvage

314

Chapter 11

Replacement and Retention Decisions

value
...
Increased demand now requires that the equipment
be upgraded again for another $17,000 so that it
can be used for 3 more years
...
Alternatively,
it can be replaced with new equipment priced at
$65,000 with operating costs of $14,000 per year
and a salvage value of $23,000 after 6 years
...
Determine the values of P, S, AOC, and n
for the defender in a replacement study
...
8 For equipment that has a ﬁrst cost of $10,000 and
the estimated operating costs and year-end salvage
values shown below, determine the economic service life at i ϭ 10% per year
...
9 To improve package tracking at a UPS transfer facility, conveyor equipment was upgraded with
RFID sensors at a cost of $345,000
...
The
salvage value of the equipment is expected to be
$140,000 for the ﬁrst 3 years, but due to obsolescence, it won’t have a signiﬁcant value after that
...
10 Economic service life calculations for an asset are
shown below
...

Years
Retained

AW of First
Cost, $

AW of Operating
Cost, $ per Year

AW of
Salvage Value, $

1
2
3
4
5

Ϫ51,700
Ϫ27,091
Ϫ18,899
Ϫ14,827
Ϫ12,398

Ϫ15,000
Ϫ17,000
Ϫ19,000
Ϫ21,000
Ϫ23,000

35,000
13,810
6,648
4,309
2,457

11
...
Maintenance can be
done at 1-, 2-, 3-, or 4-year intervals, but the longer
the interval between servicing, the higher the cost
...
What interval should
be scheduled for maintenance to minimize the
overall equivalent annual cost? The interest rate is
8% per year
...
12 A construction company bought a 180,000 metric
ton earth sifter at a cost of $65,000
...
The operating cost is expected to follow the
series described by 40,000 ϩ 10,000k, where k is
the number of years since it was purchased (k ϭ 1,
2,
...
The salvage value is estimated to be
$30,000 for years 1 and 2 and $20,000 for years 3
through 7
...

11
...
[Note that the numbers are annual worth values associated with
various years of retention; that is, if the equipment is kept for, say, 3 years, the AW (years 1
through 3) of the ﬁrst cost is $32,169, the AW of
the operating cost is $51,000, and the AW of the
salvage value is $6042
...
From the information available, determine the following:
(a) The interest rate used in the ESL calculations
...
Use the interest rate determined
in part (a)
...
14 A large, standby electricity generator in a hospital
operating room has a ﬁrst cost of $70,000 and may
be used for a maximum of 6 years
...
15)n,
where n is the number of years after purchase
...
At an interest rate of 12% per
year, what are the economic service life and the
associated AW value?

315

Problems

11
...
The salvage value cannot go below zero
...
The interest rate is 15% per year
...

11
...
Use an interest rate of
14% per year and hand solution
...
17 Use the annual marginal costs to ﬁnd the economic
service life for Problem 11
...
Assume the salvage values are the best estimates of future market value
...

11
...
3, the market value (salvage value)
series of the proposed $38 million replacement kiln
(GH) dropped to $25 million in only 1 year and then
retained 75% of the previous year’s market value
through the remainder of its 12-year expected life
...
Additionally, the heating element replacement in year 6 will probably cost $4 million, not
$2 million
...
Starting in
year 5, the AOC is expected to increase by 25% per
year, not 10% as predicted earlier
...
3)
...

(b) In percentage changes, estimate how much
these new cost estimates may affect the
minimum-cost life and AW of cost estimate
...
19 In a one-year-later analysis, what action should be
taken if (a) all estimates are still current and the
year is nD, (b) all estimates are still current and the
year is not nD, and (c) the estimates have changed?
11
...

If Retained This
Number of Years

AW Value,
$ per Year

1
2
3
4
5

Ϫ62,000
Ϫ51,000
Ϫ49,000
Ϫ53,000
Ϫ70,000

A challenger has ESL ϭ 2 years and AWC ϭ
$Ϫ48,000 per year
...

(b) When should the next replacement evaluation
take place?
11
...
The presently owned ones were purchased 4
years ago for $250,000
...

Alternatively, new controlled-environment rooms
could be purchased at a cost of $270,000
...
Determine whether the
company should upgrade or replace
...

11
...

Increased demand necessitated an upgrade costing
$30,000 one year ago
...
Its annual operating cost will be $47,000, and
it will have a $22,000 salvage after 3 years
...
If replaced now, the existing equipment will
be sold for $9000
...

316

Chapter 11

Replacement and Retention Decisions

11
...
A new car will cost $26,000 and have
annual operation and maintenance costs of $1200 per
year with an $8000 salvage value in 5 years (which is
its estimated economic service life)
...
Its operating cost is expected to be $1900
this year, with costs increasing by $200 per year
...
Assuming used cars like
the one presently owned will always be available,
should the presently owned car be sold now, 1 year
from now, 2 years from now, or 3 years from now?
Use annual worth calculations at i ϭ 10% per year
and show your work
...
24 A pulp and paper company is evaluating whether it
should retain the current bleaching process that
uses chlorine dioxide or replace it with a proprietary “oxypure” process
...
Use an interest rate of
15% per year to perform the replacement study
...
27 The data associated with operating and maintaining an asset are shown below
...
e
...
At an interest rate of 10%
per year, estimate the AW of keeping the machine
from year 1 to year 2
...
26 A crushing machine that is a basic component of a
metal recycling operation is wearing out faster than
expected
...
At that time, the buyer thought the
machine would serve its needs for at least 5 years,
at which time the machine would be sold to a
smaller independent recycler for $80,000
...
If it is
kept, the machine’s operating cost will be $37,000
per year for the next 2 years, after which it will be
scrapped for $1000
...
Alternatively, the company can outsource the process now
for a ﬁxed cost of $56,000 per year
...
25 A machine that is critical to the Phelps-Dodge copper reﬁning operation was purchased 7 years ago
for $160,000
...
The situation has changed
...
If kept in
service, it can be minimally upgraded at a cost of
$43,000, which will make it usable for up to
2 more years
...

Alternatively, the company can purchase a new
system that will have an equivalent annual worth
of $Ϫ47,063 per year over its ESL
...
Calculate the relevant annual worth values, and determine when the
company should replace the machine
...
28 A machine that cost $120,000 three years ago can
be sold now for $54,000
...
Its
operating cost was $18,000 for the ﬁrst 3 years of
its life, but the M&O cost is expected to be
$23,000 for the next 2 years
...
At an interest rate of 10%
per year, determine if the presently owned machine should be replaced now, 1 year from now, or
2 years from now
...
29 The projected market value and M&O costs associated with a presently owned machine are shown
(next page)
...
If the presently
owned machine is replaced now, the cost of the
ﬁxed-price contract will be $33,000 for each of the
next 3 years
...
Determine if
and when the defender should be replaced with the
outside vendor using an interest rate of 10% per
year
...

Year

Market Value, $

M&O Cost,
$ per Year

0
1
2
3
4

30,000
25,000
14,000
10,000
8,000

—
Ϫ24,000
Ϫ25,000
Ϫ26,000
—

11
...
It is appraised at a current market value of only $50,000
...
The challenger,
which can be purchased for $300,000, has an expected life of 10 years and a $50,000 salvage
value
...
Assume the AOC estimates are the same for both
alternatives
...
31 For the estimates in Problem 11
...
Is this a maximum or minimum for the upgrade, if the current system is to be retained?
11
...
It is expected to have the market values and
annual operating costs shown below for its remaining useful life of up to 3 years
...

Richter should retain the present plotter
...

11
...
34 State what is meant by the cash ﬂow approach to
replacement analysis, and list two reasons why it is
not a good idea to use this method
...
35 ABB Communications is considering replacing
equipment that had a ﬁrst cost of $300,000 ﬁve
years ago
...
Since the present equipment or the proposed
equipment can be used for any or all of the 3-year
period, one of the company’s industrial engineers
produced AW cost information for the defender
and challenger as shown below
...
Determine
when the defender should be replaced to minimize
the cost to ABB for the 3-year study period using
an interest rate of 10% per year
...
36 The table below shows present worth calculations
of the costs associated with using a presently
owned machine (defender) and a possible replacement (challenger) for different numbers of years
...
Show solutions (a) by hand and
(b) by spreadsheet
...
At an interest rate
of 15% per year, determine how many more years

11
...
Accordingly, a senior
engineer has recommended that a 2-year-old piece

318

Chapter 11

Replacement and Retention Decisions

of precision measurement equipment be replaced
immediately
...

(a) Perform the replacement analysis using the
annual worth method for a speciﬁed 3-year
study period
...
Comment on the effect
made by the 3-year study period
...
38 An industrial engineer at a ﬁber-optic manufacturing company is considering two robots to reduce
costs in a production line
...
Robot Y will have a ﬁrst cost
of $97,000, an annual M&O cost of $27,000, and
salvage values of $60,000, $51,000, and $42,000
after 1, 2, and 3 years, respectively
...
39 A 3-year-old machine purchased for $140,000 is not
able to meet today’s market demands
...
The current machine will have an annual operating cost of $85,000
per year and a $30,000 salvage value in 3 years
...
The replacement machine,
which will serve the company now and for at least
8 years, will cost $220,000
...
It will have an estimated operating cost of $65,000 per year
...

(a) Should the company replace the presently
owned machine now, or do it 3 years from now?
(b) Compare the capital recovery requirements for
the replacement machine (challenger) over the
study period and an expected life of 8 years
...
40 Two processes can be used for producing a polymer that reduces friction loss in engines
...
Process L, the challenger, will have a
ﬁrst cost of $230,000, an operating cost of $65,000
per year, and salvage values of $100,000 after 1
year, $70,000 after 2 years, $45,000 after 3 years,
and $26,000 after its maximum expected 4-year
life
...
You
have been asked to determine which process to select when (a) a 2-year study period is used and
(b) a 3-year study period is used
...
41 Keep or Replace the Kiln Case

In Example 11
...
This is a signiﬁcantly shortened
period compared to the expected 12-year life of
the challenger
...

11
...
The plant
manager, who is dedicated to cutting costs but not
sacriﬁcing quality and hygiene, has the projected
data shown in the table below if the current system
were retained for up to its maximum expected life
of 5 years
...
0 million per year
if Nabisco signs on for 4 to 10 years, and $5
...

Years Retained

AW, $ per Year

Close-Down Expense, $

0
1
2
3
4
5

—
Ϫ2,300,000
Ϫ2,300,000
Ϫ3,000,000
Ϫ3,000,000
Ϫ3,500,000

Ϫ3,000,000
Ϫ2,500,000
Ϫ2,000,000
Ϫ1,000,000
Ϫ1,000,000
Ϫ500,000

(a)

At a MARR ϭ 8% per year, perform a replacement study for the plant manager with a
ﬁxed study period of 5 years, when it is anticipated that the plant will be shut down due
to the age of the facility and projected technological obsolescence
...
(Hint:
Calculate AW values for all combinations of
defender͞challenger options
...
43 In 2008, Amphenol Industrial purchased a new
quality inspection system for $550,000
...

Currently the expected remaining life is 3 years
with an AOC of $27,000 per year and an estimated
salvage value of $30,000
...
If the
MARR for the corporation is 12% per year, ﬁnd
the minimum trade-in value now necessary to
make the president’s replacement economically
advantageous
...
44 A CNC milling machine purchased by Proto Tool
and Die 10 years ago for $75,000 can be used for 3
more years
...
A challenger will cost $130,000 with an economic life of
6 years and an operating cost of $32,000 per year
...
On the basis of

319

these estimates, what market value for the existing
asset will render the challenger equally attractive?
Use an interest rate of 12% per year
...
45 Hydrochloric acid, which fumes at room temperatures, creates a very corrosive work environment,
causing steel tools to rust and equipment to fail
prematurely
...
Its operating cost is $75,000 per year
...
It is expected to have a $30,000 salvage
value after its 6-year ESL
...
46 Engine oil puriﬁer machines can effectively remove acid, pitch, particles, water, and gas from
used oil
...
Its operating cost is higher than expected,
so if it is not replaced now, it will likely be used for
only 3 more years
...
A more efﬁcient challenger, puriﬁer B, will cost $150,000
with a $50,000 salvage value after its 8-year ESL
...

What is the market value for A that will make the
two puriﬁers equally attractive at an interest rate of
12% per year?

ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
11
...
48 A sunk cost is the difference between:
(a) The ﬁrst cost and the salvage value
(b) The market value and the salvage value
(c) The ﬁrst cost and the market value
(d) The book value and the market value
11
...
The operating costs
are $9000 per year, and it is expected to last 4 more
years with a $5000 salvage value
...
The value that should
be used as P for the presently owned vehicle in a
replacement study is:
(a) $45,000
(b) $5000
(c) $50,000
(d) $24,000
11
...
51 In looking for ways to cut costs and increase
proﬁt (to make the company’s stock go up), one
of the company’s industrial engineers (IEs) determined that the equivalent annual worth of an
existing machine over its remaining useful life of
3 years will be $–70,000 per year
...
If the engineer uses a
3-year study period and an interest rate of 15%
per year, she should recommend that the existing
machine be:
(a) Replaced now
(b) Replaced 1 year from now
(c) Replaced 2 years from now
(d) Not replaced
11
...
It can be replaced now or later with a
machine that will have an AW of $–90,000 per
year if it is kept for 2 years or less, –$65,000 if it
is kept between 3 and 5 years, and $–110,000 if it
is kept for 6 to 8 years
...
The replacement must be made now or 3 years from
now, according to the department supervisor
...
53 The cost characteristics of a CO testing machine at
Dytran Instruments are shown below
...
The equation for determining the AW of keeping the tester for 2 years is:
(a) AW ϭ Ϫ100,000(A͞P,i,8) Ϫ [42,000(P͞F,i,1)
ϩ 47,000(P͞F,i,2)](A͞P,i,8) ϩ 40,000(A͞F,i,8)
(b) AW ϭ Ϫ100,000(A͞P,i,2) Ϫ [42,000(P͞F,i,1)
ϩ 47,000(P͞F,i,2)](A͞P,i,2) ϩ 40,000(A͞F,i,2)

(c)
(d)

AW ϭ Ϫ100,000(A͞P,i,2) Ϫ 47,000
ϩ 40,000(A͞F,i,2)
AW ϭ Ϫ100,000(A͞P,i,2) Ϫ 42,000
ϩ 40,000(A͞F,i,2)

Machine Age,
Years

M&O Costs,
$ per Year

Salvage Value
at End of Year, $

1
2
3
4
5
6
7
8

Ϫ42,000
Ϫ47,000
Ϫ49,000
Ϫ50,000
Ϫ52,000
Ϫ54,000
Ϫ63,000
Ϫ67,000

60,000
40,000
31,000
24,000
15,000
10,000
10,000
10,000

11
...

Retention
Period, Years

AW Value,
$ per Year

1
2
3
4
5

Ϫ92,000
Ϫ81,000
Ϫ87,000
Ϫ89,000
Ϫ95,000

A challenger has an economic service life of
7 years with an AW of $–86,000 per year
...
If all future costs remain as estimated for the analysis, the company should purchase the challenger:
(a) Now
(b) After 2 years
(c) After 3 years
(d) Never
11
...
The defender should be replaced:
(a) Now
(b) 1 year from now
(c) 2 years from now
(d) 3 years from now
AW Value, $ per Year

Number of
Years Retained

Defender

Challenger

1
2
3
4
5

Ϫ14,000
Ϫ13,700
Ϫ16,900
Ϫ17,000
Ϫ18,000

Ϫ21,000
Ϫ18,000
Ϫ13,100
Ϫ15,600
Ϫ17,500

Case Study

321

CASE STUDY
WILL THE CORRECT ESL PLEASE STAND?
Background
New pumper system equipment is under consideration by a
Gulf Coast chemical processing plant
...
Because of the variable quality of the raw
chemical and the high pressures imposed on the pump chassis
and impellers, a close log is maintained on the number of
hours per year that the pump operates
...
As currently planned, rebuild and
M&O cost estimates are increased accordingly when cumulative operating time reaches the 6000-hour mark
...
Estimates made for
this pump are as follows:

First cost:
Rebuild cost:

M&O costs:

MARR:

$Ϫ800,000
$Ϫ150,000 whenever 6000 cumulative hours are logged
...
A maximum of
3 rebuilds is allowed
...
Determine the economic service life of the pump
...
The plant superintendent told you, the safety engineer,
that only one rebuild should be planned for, because
these types of pumps usually have their minimum-cost
life before the second rebuild
...

Comment on the practicality of ESL ϭ 6 years, given
the MV calculated
...
In a separate conversation, the line manager told you
to not plan for a rebuild after 6000 hours, because
the pump will be replaced after a total of 10,000
hours of operation
...
He also told you to assume now that the 15%
growth rate applies from year 1 forward
...
What do you think of these suggestions from the plant
superintendent and the line manager?

CHAPTER 12

Independent
Projects with
Budget
Limitation

L E A R N I N G

O U T C O M E S

Purpose: Select independent projects for funding when there is a limitation on the amount of capital available for investment
...
1

Capital rationing

• Explain how a capital budgeting problem is
approached
...
2

Equal-life projects

• Use PW-based capital budgeting to select from
several equal-life independent projects
...
3

Unequal-life projects

• Use PW-based capital budgeting to select from
several unequal-life independent projects
...
4

Linear programming

• Set up the linear programming model and use the
Solver spreadsheet tool to select projects
...
5

Ranking options

• Use the internal rate of return (IROR) and
proﬁtability index (PI) to rank and select from
independent projects
...
If the projects are not mutually exclusive,
they are categorized as independent of one another, as discussed at the beginning of Chapter 5
...
It
is possible to select any number of projects from none (do nothing) to all viable projects
...
This limit is considered as each independent project is economically
evaluated
...
They determine the economically best rationing of initial investment
capital among independent projects based upon different measures, such as PW, ROR, and
the proﬁtability index
...

12
...
When a corporation has
several options for placing investment capital, a “reject or accept” decision must be made for
each project
...
Selection of one project does not impact the selection decision for any other project
...

The term project is used to identify each independent option
...
The term mutually exclusive alternative continues
to identify a project when only one may be selected from several
...
Two examples of contingent projects A and
B are as follows: A cannot be accepted unless B is accepted; and A can be accepted in lieu of B,
but both are not needed
...
For example, B must be accepted if both A and C are accepted
...

A capital budgeting study has the following characteristics:
• Several independent projects are identiﬁed, and net cash ﬂow estimates are available
...

• A stated budgetary constraint restricts the total amount available for investment
...
This investment limit
is identiﬁed by the symbol b
...

By nature, independent projects are usually quite different from one another
...
In the private sector, sample projects may
be a new warehousing facility, expanded product base, improved quality program, upgraded information system, automation, and acquisition of another ﬁrm
...
The typical
capital budgeting problem is illustrated in Figure 12–1
...

Present worth analysis using the capital budgeting process is the recommended method to
select projects
...

Independent project
selection
Limited budget

324

Chapter 12

Figure 12–1

Independent Projects with Budget Limitation

Independent
projects

Basic characteristics of a
capital budgeting study
...

Life

Capital
investment
limit

Investment

B
...

Life

Select 0 to all 3 projects

Objective: Maximize PW
value of selection
within capital limit

Equal-service requirement

This guideline is not different from that used for selection in previous chapters for independent
projects
...
The primary difference now is that the amount of
money available to invest is limited, thus the title capital budgeting or rationing
...

Previously, PW analysis had the requirement of equal service between alternatives
...
Rather, the selection guideline has the following implied assumption
...

Opportunity cost

This fundamental assumption is demonstrated to be correct at the end of Section 12
...

Another dilemma of capital rationing among independent projects concerns the ﬂexibility of the
capital investment limit b
...
For example, assume project A has a positive PW value at the MARR
...
However, in the examples here, we will not exceed a stated investment limit
...
9 and 10
...
The lack of capital to fund the next project deﬁnes the ROR level that
is forgone
...

12
...
Two are the internal rate of return (IROR), discussed in Chapter 7, and the
proﬁtability index (PI), also called the present worth index (PWI), introduced in Chapter 9
...
The capital budgeting process,
covered in the next 3 sections, does ﬁnd the optimal solution for PW values
...
Application of these two measures is presented in Section 12
...

12
...
Each
feasible bundle must have a total investment that does not exceed b
...
The total number of bundles for m projects is 2m
...
For m ϭ 4, there are 24 ϭ 16 bundles, and for m ϭ 16, 216 ϭ 65,536 bundles
...
The bundle with the largest PW value
is selected
...

Project

Initial Investment, $

A
B
C
D

Ϫ10,000
Ϫ5,000
Ϫ8,000
Ϫ15,000

If the investment limit is b ϭ $25,000, of the 16 bundles, there are 12 feasible ones to evaluate
...
The
viable bundles are shown below
...
Develop all mutually exclusive bundles with a total initial investment that does not exceed
the capital limit b
...
Sum the net cash ﬂows NCFjt for all projects in each bundle j ( j ϭ 1, 2,
...
, nj)
...

3
...

PWj ‫ ؍‬PW of bundle net cash ﬂows ؊ initial investment
t‫؍‬nj

‫؍‬

͚ NCF (P͞F,i,t) ؊ NCF
jt

j0

[12
...
Select the bundle with the (numerically) largest PWj value
...

Any bundle with PWj Ͻ 0 is discarded, because it does not produce a return of at least the MARR
...
1
The projects review committee of Microsoft has $20 million to allocate next year to new software product development
...
All
amounts are in $1000 units
...
Select the project(s)
if a 15% return is expected
...

Remember the units are in $1000
...
There are 25 ϭ 32 possible bundles
...
The $21,000 investment for E eliminates it from all bundles
...
The bundle net cash flows, column 4, are the sum of individual project net cash
flows
...
Use Equation [12
...
Since the annual NCF
and life estimates are the same for a bundle, PWj reduces to
PWj ϭ NCFj(P͞A,15%,9) Ϫ NCFj0
4
...
Bundle 2 does not return
15%, since PW2 Ͻ 0
...
This leaves $6 million uncommitted
...

12
...
3 Capital Rationing Using PW Analysis
of Unequal-Life Projects
Usually independent projects do not have the same expected life
...
1, the
PW method for solution of the capital budgeting problem assumes that each project will last for
the period of the longest-lived project nL
...
Therefore, use of the LCM of lives is not
necessary, and it is correct to use Equation [12
...

EXAMPLE 12
...

Solve by hand and by spreadsheet
...
Of the 24 ϭ 16 bundles, 8 are economically feasible
...
1] are summarized in Table 12–3
...

TA BLE 12–3

Present Worth Analysis for Unequal-Life Independent Projects,
Example 12
...
It is necessary
to initially develop the mutually exclusive bundles manually and total net cash ﬂows each year
using each project’s NCF
...
Bundle 5 (projects A and C) has the largest PW value (row 16)
...
2
...
1] is correct
...
Refer to
Figure 12–3, which uses the general layout of a two-project bundle
...
The P͞A factor is used for PW computation
...
At the end of the shorter-lived project, the bundle has a total future worth
of NCFj(F͞A,MARR,nj) as determined for each project
...
The assumption of the return at the MARR
is important; this PW approach does not necessarily select the correct projects if the return is not
at the MARR
...
Finally,

Figure 12–3
Representative cash ﬂows
used to compute PW for a
bundle of two independent unequal-life projects
by Equation [12
...

FWB
PWB

nB = nL

Project B
Investment
for B

FWA
PWA
Future worth
Period of
reinvestment
at MARR
nA

Project A
Investment
for A

Bundle PW = PWA + PWB

nL

12
...
2
...
This is the bundle PW ϭ PWA ϩ PWB
...
2]

Substitute the symbol i for the MARR, and use the factor formulas to simplify
...
3]

ϭ NCFj(P͞A,i,nj)
Since the bracketed expression in Equation [12
...

To demonstrate numerically, consider bundle j ϭ 7 in Example 12
...
The evaluation is in
Table 12–3, and the net cash ﬂow is pictured in Figure 12–4
...

FW ϭ 5220(F͞A,15%,4)(F͞P,15%,5) ϩ 2680(F͞P,15%,4) ϭ $57,111
The present worth at the initial investment time is
PW ϭ Ϫ16,000 ϩ 57,111(P͞F,15%,9) ϭ $235
The PW value is the same as PW7 in Table 12–3 and Figure 12–2
...
If this assumption is not realistic, the PW analysis
must be conducted using the LCM of all project lives
...
4 Capital Budgeting Problem Formulation
Using Linear Programming
The procedure discussed above requires the development of mutually exclusive bundles one
project at a time, two projects at a time, etc
...
As the number of independent projects increases, this process
becomes prohibitively cumbersome and unworkable
...
The problem is formulated using the
integer linear programming (ILP) model, which means simply that all relations are linear and that
the variable x can take on only integer values
...
The formulation in
words follows
...

Constraints:
• Capital investment constraint is that the sum of initial investments must not exceed a speciﬁed limit
...

For the math formulation, deﬁne b as the capital investment limit, and let xk (k ϭ 1 to m projects)
be the variables to be determined
...
Note that the subscript k represents each independent project, not a mutually
exclusive bundle
...
4]

Յb

kϭ1

xk ϭ 0 or 1

for k ϭ 1, 2,
...
1] at MARR ϭ i
...
5]

t‫1؍‬

Computer solution is accomplished by a linear programming software package which treats the ILP
model
...
The Solver tool is similar to Goal Seek with signiﬁcantly more capabilities
...
This means that the
function Z in Equation [12
...
Also, multiple changing cells can be identiﬁed, so
the 0 or 1 value of the unknowns can be determined
...
4] can be
accommodated
...
3 illustrates its use
...
3
Review Example 12
...
(a) Formulate the capital budgeting problem using the math programming model presented in Equation [12
...
(b) Select the projects using Solver
...
The capital investment limit is b ϭ $20,000 in Equation [12
...

kϭ4

͚ PW x

Maximize:

k k

ϭZ

kϭ1

kϭ4

͚ NCF

Constraints:

k0 xk

Յ 20,000

kϭ1

xk ϭ 0 or 1

for k ϭ 1, 2, 3, 4

Now, substitute the PWk and NCFk0 values from Table 12–3 into the model
...
We have the complete 0-or-1 ILP formulation
...
2 is written as
x1 ϭ 1

x2 ϭ 0

x3 ϭ 1

x4 ϭ 0

12
...
3
...
The spreadsheet
template can be expanded in either direction if needed
...
The descriptions below and the cell tag identify the contents of the rows and cells
in Figure 12–5, and their linkage to Solver parameters
...
Cell I5 is the expression for Z, the sum of the PW values for
the projects
...

Rows 6 to 18: These are initial investments and net cash ﬂow estimates for each
project
...

Row 19: The entry in each cell is 1 for a selected project and 0 if not selected
...
Since each entry must be 0 or 1, a binary constraint is placed on all row 19 cells in Solver, as shown in Figure 12–5
...
Solver will ﬁnd the solution to maximize Z
...
The
NPV functions are developed for any project with a life up to 12 years at the
MARR entered in cell B1
...

Row 22: This row shows the initial investment for the selected projects
...
This cell has the budget limitation placed on it by the constraint in Solver
...

To solve the example, set all values in row 19 to 0, set up the Solver parameters as described
above, and click on Solve
...
) If needed, further directions on saving the solution,
making changes, etc
...
5, and on the Excel help function
...

331

332

Chapter 12

Independent Projects with Budget Limitation

12
...
2 to 2
...
However, it is
very common in industrial, professional, and government settings to learn that the rate of return
is the basis for ranking projects
...
2, is determined by setting a PW or AW relation equal to zero and solving for i* — the
IROR
...
6]

tϭ1

This is the same as Equation [12
...
The spreadsheet
function RATE or IRR will provide the same answer
...

If there is no budget limit, select all projects that have IROR Ն MARR
...
This can occur because IROR ranking maximizes the overall rate of return, not
necessarily the PW value
...

Another common ranking method is the proﬁtability index (PI) that we learned in
Section 9
...
This is a “bang for the buck” measure that provides a sense of getting the most for
the investment dollar over the life of the project
...
2 for more details
...
The PI measure is deﬁned as
t‫؍‬nj

͚ NCF (P͞F, i, t)
jt

PW of net cash ﬂows
t‫1؍‬
——————————— ‫———————— ؍‬
PW of initial investment

ͦ NCFj0 ͦ

[12
...
The numerator has only cash ﬂows that result from the project for years 1 through its life
nj
...
Similar to the
previous case, the selection guideline is as follows:

Independent project
selection

Once the project ranking by PI is complete, select all projects in order without exceeding the
investment limit b
...
0
...
Example 12
...
None
of these results are incorrect; they simply maximize different measures, as you will see
...
Additionally, greater complexity is introduced when dependent and contingent projects are involved
...
4
Georgia works as a ﬁnancial analyst in the Management Science Group of General Electronics
...
5

333

Additional Project Ranking Measures
IROR, PI, and PW Values for Five Projects, Example 12
...
5
1
...
8
0
...
4
1
...
6
0
...
0
1
...
She has conﬁrmed the computations and is ready to do the ranking and make the
selection
...

Use the PI measure to rank and select projects
...

Compare the selected projects by the three methods and determine which one will maximize the overall ROR value of the $18 million budget
...

(a) Ranking by overall IROR values indicates that projects 1 and 5 should be selected with
$13 million of the $18 million budget expended
...

(b) As an example, the PI for project 1 is calculated using Equation [12
...

4000(P͞A,15%,6)
PI1 ϭ ————————
|Ϫ8000|
ϭ 15,138͞8,000
ϭ 1
...
Again, the remaining $5 million is assumed to generate a return of MARR ϭ 15% per year
...

Projects 1 and 3, rather than 1 and 5, are selected for a total PW ϭ $8
...
The remaining $2 million is assumed to earn 15% per year
...
PW ranking results in the
selection of projects 1 and 3
...
In $1000 units,
Projects 1 and 5

NCF, year 0:
NCF, years 1–3:
NCF, years 4–6:

$Ϫ13,000
$6,600
$4,000

0 ϭ Ϫ13,000 ϩ 6600(P͞A,i,3) ϩ 4000(P͞A,i,3)(P͞F,i,3)
Ranking of Projects by Different Measures, Example 12
...
5
26
...
89
1
...
4
12
...
6

3
2
4

21,000

1
...
92
0
...
1%
...
1(13,000) ϩ 15
...
4%
Projects 1 and 3

NCF, year 0:
NCF, years 1–5:
NCF, year 6:

$Ϫ16,000
$6,700
$4,000

0 ϭ Ϫ16,000 ϩ 6700(P͞A,i,5) ϩ 4000(P͞F,i,6)
By IRR function, the rate of return is 33
...
The overall return on the entire budget is
ROR ϭ [33
...
0(2000)]͞18,000
ϭ 31
...
4%
...
189 (i
...
, 7
...
051) million, as determined
from Table 12–5, column 7
...
Capital budgeting involves proposed projects, each
with an initial investment and net cash ﬂows estimated over the life of the project
...

•
•
•
•

Selection is made from among independent projects
...

Maximizing the present worth of the net cash ﬂows is the objective
...

The present worth method is used for evaluation
...

There are a maximum of 2m bundles for m projects
...
Reinvestment of net positive cash ﬂows at the
MARR is assumed for all projects with lives shorter than that of the longest-lived project
...
Excel’s Solver tool solves this problem by
spreadsheet
...
Two measures are the
internal rate of return (IROR) and the proﬁtability index (PI), also called the PW index
...
When there are a large number of projects, the IROR basis is commonly applied in
industrial settings
...
1 Deﬁne the following terms: bundle, contingent
project, dependent project
...
2 State two assumptions made when doing capital rationing using a PW analysis for unequal-life projects
...
3 For independent projects identiﬁed as A, B, C, D,
E, F, and G, how many mutually exclusive bundles
can be formed?

12
...

Projects X and Y perform the same function with
different processes; both should not be selected
...
5 Five projects have been identiﬁed for possible
implementation by a company that makes dry ice
blasters—machines that propel tiny dry ice pellets at supersonic speeds so they ﬂash-freeze and
then lift grime, paint, rust, mold, asphalt, and

335

Problems

other contaminants off in-place machines and a
wide range of surfaces
...
Determine which bundles are possible,
provided the budget limitation is (a) $34,000 and
(b) $45,000
...
The project costs and 18% per year PW
values are as shown
...
7 Develop all acceptable mutually exclusive bundles for the four independent projects described
below if the investment limit is $400 and the following project selection restriction applies: Project 1 can be selected only if both projects 3 and 4
are selected
...
6 Four independent projects (1, 2, 3, and 4) are proposed for investment by Perfect Manufacturing,
Inc
...

Projects 1 and 4 should not both be selected;
they are essentially duplicates
...
10 The capital fund for research project investment at
SummaCorp is limited to $100,000 for next year
...

Initial
Project Investment, $
I
II
III

Ϫ25,000
Ϫ30,000
Ϫ50,000

A
B
C

12
...
Select the best bundle if the capital budget limit is $45,000 and the MARR is the cost of
capital, which is 9% per year
...
9 The general manager for Woodslome Appliance
Company Plant #A14 in Mexico City has four
independent projects that she can fund this year to

6,000
9,000
15,000

Life, Salvage
Years Value, $
4
4
4

4,000
Ϫ1,000
20,000

12
...
Use (a) hand and (b) spreadsheet-based
PW analysis and a 15% per year return requirement
to help this engineer make the best decision from a
purely economic perspective
...
12 Dwayne has four independent vendor proposals to
contract the nationwide oil recycling services for
Ford Corporation manufacturing plants
...
Revenue sharing of
recycled oil sales with Ford is a part of the requirement
...
The corporate MARR
is 10% per year
...

(b) A larger budget of $5
...

(c) There is no limit on spending
...
5 million
Ϫ3
...
8 million
Ϫ2
...
14 Charlie’s Garage has $70,000 to spend on new
equipment that may increase revenue for his car
repair shop
...
All are expected to last 3 years
...
17 The independent project estimates below have
been developed by the engineering and ﬁnance
managers
...
Select the economically best projects using the PW
method and (a) hand solution and (b) spreadsheet
solution
...
5
Ϫ3
...
8
Ϫ2
...
18 Use the PW method to evaluate four independent
projects
...
The MARR is 12% per year, and up to
$16,000 in capital investment funds are available
...
13 Use the PW method at 8% per year to select up to
three projects from the four available ones if no
more than $20,000 can be invested
...

Project

12
...
Use the PW method to
evaluate mutually exclusive bundles to
make the selection
...
)

2

3

4

Ϫ5000
5

Ϫ8,000
5

Ϫ9,000
3

Ϫ10,000
4

Year
1
2
3
4
5

NCF Estimates, $ per Year
1000
1700
2400
3000
3800

500
500
500
500
10,500

5000
5000
2000

0
0
0
17,000

12
...
18 using a spreadsheet
...

12
...

12
...
5% per year
...
0
million can be invested
...

(b) If the life of project 3 can be increased from
5 to 10 years for the same $1 million investment, use Goal Seek to determine the NCF in
year 1 for project 3 alone to have the same
PW as the best bundle in part (a)
...
With these new
NCF and life estimates, what are the best
projects for investment?

337

Problems
Estimated NCF, $ per Year
Project

Investment,
$M

Life,
Years

Year 1

Ϫ0
...
1
Ϫ1
...
21 Formulate the linear programming model, develop
a spreadsheet, and solve the capital rationing problem in Example 12
...

12
...
17
...
25 Using the estimates in Problem 12
...

Other Ranking Measures
12
...

(a) Determine the IROR, PI, and PW values if
the MARR is 15% per year
...
27 An engineer at Delphi Systems is considering the
projects below, all of which can be considered to
last indeﬁnitely
...

(a) Determine which projects should be selected
on the basis of IROR if the budget limitation
is $39,000
...
0
19
...
3
14
...
0

12
...
The company’s MARR is 15% per year
...
(Solve by hand or
spreadsheet as instructed
...

12
...
20(a), using the linear programming model and
a spreadsheet
...
24 Johnson and Johnson is expanding its ﬁrst-aid
products line for individuals allergic to latex
...
18
...
The MARR is 12%
per year, and the budget limit is $16 million
...

First Cost, $

A
B
C
D
E

Gradient
after Year 1

1
2
3

Project

Annual
Project
Project First Cost, $ Income, $ per Year Life, Years
L
A
N
D
T

Ϫ30,000
Ϫ15,000
Ϫ45,000
Ϫ70,000
Ϫ40,000

9,000
4,900
11,100
9,000
10,000

10
10
10
10
10

12
...
The American Society of Civil Engineers (ASCE) has teamed
with municipalities, counties, and several excavation
companies to develop robots that can travel through
mains, detect leaks, and repair many of them immediately
...
There is a $100 million limit on capital funding,
and the MARR is established at 12% per year
...
Solve by spreadsheet, unless instructed to use hand solution
...
Is it economically justiﬁed?
(c) Determine the overall rate of return for the
$100 million with the projects selected in
part (a)
...

Project

First
Cost, $ M

Estimated Annual
Savings, $M per Year

Project
Life, Years

W
X
Y
Z

Ϫ12
Ϫ25
Ϫ45
Ϫ60

5
...
3
12
...
0

3
4
6
8

338

Chapter 12

Independent Projects with Budget Limitation

12
...

12
...
All projects have a 10-year life
...

(c) Are different projects selected using the two
methods?
Project

First Cost, $

Net Income,
$ per Year

A
B
C
D
E

Ϫ18,000
Ϫ15,000
Ϫ35,000
Ϫ60,000
Ϫ50,000

4,000
2,800
12,600
13,000
8,000

12
...
The company always has more projects to engage in than it has
capital to fund projects
...
Since all projects are considered long-term ventures, the company uses an inﬁnite period for their life
...

Project

IROR, %
18
...
3
34
...
3
9
...
33 A project that has a condition associated with its
acceptance or rejection is known as:
(a) A mutually exclusive alternative
(b) A contingent project
(c) A dependent project
(d) Both (b) and (c)
12
...
35 All of the following are correct when a capital
budgeting problem is solved using the 0-1 integer
linear programming model except:
(a) Partial investment in a project is acceptable
...

(c) Budget constraints may be present for the
ﬁrst year only or for several years
...

12
...

A bundle may consist of only one project
...

A bundle may include contingent and dependent projects
...
37 When there are 5 projects involved in a capital
budgeting study, the maximum number of bundles
that can be formulated is:
(a) 6
(b) 10
(c) 31
(d) 32
12
...
39 The independent projects shown below are under
consideration for possible implementation by
Renishaw Inc
...

If the company’s MARR is 14% per year and it
uses the IROR method of capital budgeting, the
projects it should select under a budget limitation
of $105,000 are:
(a) A, B, and C
(b) A, B, and D
(c) B, C, and D
(d) A, C, and D
Project

First Cost, $

Annual
Income, $ per Year

Rate of
Return, %

A
B
C
D
E

Ϫ20,000
Ϫ10,000
Ϫ15,000
Ϫ70,000
Ϫ50,000

4,000
1,900
2,600
10,000
6,000

20
...
0
17
...
3
12
...
40 For a project that requires an initial investment
of $26,000 and yields $10,000 per year for
4 years, the PI at an interest rate of 10% per year
is closest to:
(a) 1
...
22
(c) 1
...
56

CHAPTER 13

Breakeven and
Payback
Analysis

L E A R N I N G

O U T C O M E S

Purpose: Determine the breakeven for one or two alternatives and calculate the payback period with and without a
return required
...
1

Breakeven point

• Determine the breakeven point for one
parameter
...
2

Two alternatives

• Calculate the breakeven point of a parameter
and use it to select between two alternatives
...
3

Payback period

• Determine the payback period of a project at
i ϭ 0% and i Ͼ 0%
...

13
...

B

reakeven analysis is performed to determine the value of a variable or parameter of a project or alternative that makes two elements equal, for example, the
sales volume that will equate revenues and costs
...
Breakeven
analysis is commonly applied in make-or-buy decisions when a decision is needed about the
source for manufactured components, services, etc
...
There are two types of payback: return (i Ͼ 0%) and no return
(i ϭ 0%)
...

These aspects are discussed in depth in this chapter
...
If the variable of interest is allowed to vary, the
approaches of sensitivity analysis (Chapter 18) should be used
...

13
...
This form of breakeven analysis has been used many times so far
...
Methods
used to determine the quantity include
Direct solution by hand if only one factor is present (say, P͞A) or only single amounts are
estimated (for example, P and F)
Trial and error by hand or calculator when multiple factors are present
Spreadsheet when cash ﬂow and other estimates are entered into cells and used in resident
functions (PV, FV, RATE, IRR, NPV, PMT, and NPER) or tools (Goal Seek and Solver)
...
For example, the variable may be a design element to minimize cost or the
production level needed to realize revenues that exceed costs by 10%
...
The breakeven point QBE is determined from mathematical relations, e
...
, product revenue and costs or
materials supply and demand or other parameters that involve the parameter Q
...

The unit of the parameter Q may vary widely: units per year, cost per kilogram, hours per month,
percentage of full plant capacity, etc
...
A linear revenue
relation is commonly assumed, but a nonlinear relation is often more realistic
...

Costs, which may be linear or nonlinear, usually include two components—ﬁxed and
variable—as indicated in Figure 13–1b
...
These include costs such as buildings, insurance, ﬁxed overhead, some
minimum level of labor, equipment capital recovery, and information systems
...
These include costs such as direct labor, materials, indirect costs, contractors, marketing, advertisement, and warranty
...
Even if no
units are produced, ﬁxed costs are incurred at some threshold level
...

R, revenue per year

Nonlinear
(2)

Linear

Q, units per year
(a) Revenue relations—(1) increasing and
(2) decreasing revenue per unit
TC ϭ FC ϩ VC

Total
cost, TC
Variable, VC

Cost per year

Cost per year

TC ϭ FC ϩ VC

TC

VC

FC

Fixed, FC
Q, units per year
(b) Linear cost relations

Q, units per year
(c) Nonlinear cost relations

cannot last long before the plant must shut down to reduce ﬁxed costs
...

Variable costs change with production level, workforce size, and other parameters
...

When FC and VC are added, they form the total cost relation TC
...
Figure 13–1c shows a general TC curve for a
nonlinear VC in which unit variable costs decrease as the quantity level rises
...
If Q Ͼ QBE, there is a
predictable proﬁt; but if Q Ͻ QBE, there is a loss
...
Proﬁt is calculated as
Proﬁt ‫ ؍‬revenue ؊ total cost
‫ ؍‬R ؊ TC
‫ ؍‬R ؊ (FC ؉ VC)

[13
...

R ϭ TC
rQ ϭ FC ϩ vQ
where

r ϭ revenue per unit
v ϭ variable cost per unit

Solve for the breakeven quantity Q ϭ QBE for linear R and TC functions
...
2]

13
...

$
TC

R

Profit
maximized

Profit range

Loss

Loss
QBE

QP

QBE

Q, units per year

Figure 13–3
Breakeven points and maximum proﬁt point for a nonlinear analysis
...
For example, if the variable cost per unit is reduced, then the
TC line has a smaller slope (Figure13–2) and the breakeven point will decrease
...
A similar
analysis is possible for ﬁxed VC and increased levels of production, as shown in the next example
...
Figure 13–3
presents this situation for two breakeven points
...

343

344

Chapter 13

Breakeven and Payback Analysis

Of course, no static R and TC relations—linear or nonlinear—are able to estimate exactly the
revenue and cost amounts over an extended period of time
...

EXAMPLE 13
...
Normal production level is 60 diverter systems per month, but
due to signiﬁcantly improved economic conditions in Asia, production is at 72 per month
...

Fixed costs
Variable cost per unit
Revenue per unit

FC ϭ $2
...
2] to determine the breakeven number of units
...

FC
QBE ϭ ———
rϪv
2400
ϭ ———— ϭ 60 units per month
75 Ϫ 35
Figure 13–4 is a plot of R and TC lines
...
The
increased production level of 72 units is above the breakeven value
...
1
...
2

345

Breakeven Analysis Between Two Alternatives

(b) To estimate proﬁt (in $1000 units) at Q ϭ 72 units per month, use Equation [13
...

Proﬁt ϭ R Ϫ TC ϭ rQ Ϫ (FC ϩ vQ)
ϭ (r Ϫ v)Q Ϫ FC
ϭ (75 Ϫ 35)72 Ϫ 2400
ϭ $480

[13
...

(c) To determine the required difference r Ϫ v, use Equation [13
...
4 million
...
33 per unit
45
The spread between r and v must be $53,330
...
e
...

In some circumstances, breakeven analysis performed on a per unit basis is more meaningful
...
2], but the TC relation is divided by Q to
obtain an expression for cost per unit, also termed average cost per unit Cu
...
4]
Cu ϭ —— ϭ ———— ϭ —— ϩ v
Q
Q
Q
At the breakeven quantity Q ϭ QBE, the revenue per unit is exactly equal to the cost per unit
...
4] takes on the shape of a hyperbola
...
This is the same as setting proﬁt equal to zero in Equation [13
...

It may be necessary to perform some dimensional analysis initially to obtain the correct revenue
and total cost relations in order to use the same dimension for both relations, for example, $ per
unit, miles per month, or units per year
...
2 Breakeven Analysis Between Two Alternatives
Now we consider breakeven analysis between two mutually exclusive alternatives
...
Equating the two PW or AW relations determines the breakeven point
...

The parameter can be the interest rate i, ﬁrst cost P, annual operating cost (AOC), or any parameter
...

For example, the incremental ROR value (⌬i*) is the breakeven rate between alternatives
...
In
Section 11
...
If the market value is
larger than RV, the decision should favor the challenger
...
Figure 13–5 illustrates
the breakeven concept for two alternatives with linear cost relations
...
However, alternative 2 has a smaller variable cost,
as indicated by its lower slope
...
Thus, if the number of units of the common
variable is greater than the breakeven amount, alternative 2 is selected, since the total cost
will be lower
...

Breakeven

346

Chapter 13

Breakeven and Payback Analysis

Figure 13–5

Alt
...

Alt
...
1 FC

Alt
...
The AW is preferred when the variable units are
expressed on a yearly basis, and AW calculations are simpler for alternatives with unequal lives
...

1
...

2
...

3
...

Selection between alternatives is based on this guideline:
If the anticipated level of the common variable is below the breakeven value, select the alternative with the higher variable cost (larger slope)
...

(Refer to Figure 13–5
...
2
A small aerospace company is evaluating two alternatives: the purchase of an automatic feed
machine and a manual feed machine for a ﬁnishing process
...

One person will operate the machine at a rate of $12 per hour
...
Annual maintenance and operating cost is expected to be $3500
...
However, three workers will be required at $8 per
hour each
...
All
projects are expected to generate a return of 10% per year
...

1
...

2
...
5x
The VC is developed in dollars per year
...
5x
ϭ $–6992 Ϫ 1
...
2

Breakeven Analysis Between Two Alternatives

Similarly, the annual variable cost and AW for the manual feed machine are
$8
1 hour x tons
Annual VC ϭ —— (3 operators) ——— ———
hour
6 tons year
ϭ 4x
AWmanual ϭ Ϫ8000(A͞P,10%,5) Ϫ 1500 Ϫ 4x
ϭ $Ϫ3610 Ϫ 4x
3
...

AWauto ϭ AWmanual
Ϫ6992 Ϫ 1
...
5 is smaller than the manual feed VC slope of 4
...
This means
the company contracts to buy the product or service from the outside, or makes it within the company
...
Where the two cost relations cross is the make-buy decision quantity
...

EXAMPLE 13
...
It is faced with
a make-or-buy decision
...
The steel arm of the lift can be purchased internationally for $3
...
If manufactured on site, two machines will be required
...
Machine A will require an overhaul after 3 years costing $3000
...
A total of four
operators will be required for the two machines at a rate of $12
...
In a
normal 8-hour period, the operators and two machines can produce parts sufﬁcient to manufacture 1000 units
...

(a) Number of units to manufacture each year to justify the in-house (make) option
...
The company expects to produce 10,000 units
per year
...

1
...

2
...

Annual VC ϭ (cost per unit)(units per year)
4 operators $12
...
4x
The annual ﬁxed costs for machines A and B are the AW amounts
...

347

Chapter 13

Breakeven and Payback Analysis

3
...
50x) and the make option yields
Ϫ3
...
4x
Ϫ3
...
5]

A minimum of 6565 lifts must be produced each year to justify the make option, which
has the lower variable cost of 0
...

(b) Substitute 10,000 for x and PA for the to-be-determined ﬁrst cost of machine A (currently
$18,000) in Equation [13
...
Solution yields PA ϭ $58,295
...

Even though the preceding examples treat only two alternatives, the same type of analysis can
be performed for three or more alternatives
...
The results are the ranges through which each alternative is
more economical
...
Between 40 and 60, alternative 2 is more economical; and above 60,
alternative 3 is favored
...
If the costs increase
or decrease uniformly, mathematical expressions that allow direct determination of the breakeven point can be developed
...
3 Payback Analysis
Payback analysis is another use of the present worth technique
...
Payback is allied with breakeven analysis; this is illustrated later in the section
...

Alternative 1
Alternative 2

Alternative 3
Total cost, $/year

348

Breakeven
points

40
60
Output, units/hour

Figure 13–6
Breakeven points for three alternatives
...
3

349

Payback Analysis

The payback period np is an estimated time for the revenues, savings, and any other monetary beneﬁts to completely recover the initial investment plus a stated rate of return i
...

No return; iϭ0%: Also called simple payback, this is the recovery of only the initial investment
...

An example application of payback may be a corporate senior manager who insists that every
proposal return the initial cost and some stated return within 3 years
...
The payback period should be determined using a required i Ͼ 0%
...
After the formulas are presented, a couple of cautions about payback usage are provided
...
For both types, the terminology is P for the initial investment in the asset, project, contract, etc
...
Using Equation [1
...

Note that np is usually not an integer
...
, np,
t‫؍‬np

No return, i ‫ ;%0 ؍‬NCFt varies annually:

0 ‫؊ ؍‬P ؉

͚ NCF

t

[13
...
7]
t‫؍‬np

Discounted, i Ͼ 0%; NCFt varies annually:

0 ‫؊ ؍‬P ؉

͚ NCF (P͞F, i, t)
t

[13
...
9]

After np years, the cash ﬂows will recover the investment in year 0 plus the required return of
i%
...
If the estimated life is less than np years, there is not enough time to recover the
investment and i% return
...
Consequently, it is preferable to use payback as an
initial screening method or supplemental tool rather than as the primary means to select an
alternative
...

• Either type of payback disregards all cash ﬂows occurring after the payback period
...

Payback analysis utilizes a signiﬁcantly different approach to alternative evaluation than the
primary methods of PW, AW, ROR, and B͞C
...
However, the information obtained from discounted payback analysis performed at an appropriate i Ͼ 0% can be very useful in that a sense of the risk
involved in undertaking an alternative is provided
...
Even here, the 6-year payback is considered supplemental information and does not
replace a complete economic analysis
...
4
The board of directors of Halliburton International has just approved an $18 million worldwide engineering construction design contract
...
The contract has a potentially lucrative repayment clause

Payback period

350

Chapter 13

Breakeven and Payback Analysis

to Halliburton of $3 million at any time that the contract is canceled by either party during the
10 years of the contract period
...
(b) Determine
the no-return payback period and compare it with the answer for i ϭ 15%
...
Show both hand and spreadsheet solutions
...
The single $3 million payment (call it CV for
cancellation value) could be received at any time within the 10-year contract period
...
9] is altered to include CV
...
3 years, found by trail and error
...

(b) If Halliburton requires absolutely no return on its $18 million investment, Equation [13
...

0 ϭ Ϫ18 ϩ 5(3) ϩ 3
There is a very signiﬁcant difference in np for 15% and 0%
...
3 years, while the no-return payback period requires only 5 years
...

Solution by Spreadsheet

Enter the function ϭ NPER(15%,3,Ϫ18,3) to display 15
...
Change the rate from 15% to
0% to display the no-return payback period of 5 years
...
When cash ﬂows that occur
after np are neglected, it is possible to favor short-lived assets even when longer-lived assets
produce a higher return
...
Comparison of short- and long-lived assets in Example 13
...

EXAMPLE 13
...
Machine 2 is expected to be versatile and technologically advanced enough
to provide net income longer than machine 1
...
8] and [13
...
57 years at i ϭ 15%
...

13
...
57

$12,000
Cash flows neglected
by payback analysis

$3000 per year
$1000 per year

0

1

2

3

4

5

6

7

Machine 2

8

9

10

11

12

13

14

np = 9
...
5
...
57 years, which is less than the 7-year life
...
52 years, which is less than the 14-year life
...

Now, use a 15% PW analysis to compare the machines and comment on any difference in the
recommendation
...
Compare them over the LCM of 14 years
...

This result is the opposite of the payback period decision
...
As illustrated in Figure 13–7 (for one life
cycle for each machine), payback analysis neglects all cash ﬂow amounts that may occur after
the payback time has been reached
...
Often a shorter-lived alternative evaluated by payback analysis may
appear to be more attractive, when the longer-lived alternative has cash ﬂows later in its life
that make it more economically attractive
...

They can be used in conjunction to determine the payback period when a desired level of breakeven is speciﬁed
...
By working together in
this fashion, better economic decisions can be made
...
6 illustrates the second of the
situations mentioned above
...
6
The president of a local company expects a product to have a proﬁtable life of between 1 and
5 years
...
The cost and revenue estimates are as follows:
Fixed costs: Initial investment of $80,000 with $1000 annual operating cost
...

Revenue: Twice the variable cost for the ﬁrst 5 years and 50% of the variable cost thereafter
...
Since values of XBE are
sought for np ϭ 1, 2, 3, 4, 5, solve for breakeven by substituting each payback period
...

80,000
——— ϩ 1000
Fixed cost, FC
np
Revenue per unit, r

$16

Variable cost per unit, v

(years 1 through 5 only)

$8

The breakeven relation from Equation [13
...

[13
...
Equation [13
...
The breakeven values are the same as those above,
e
...
, sell 5125 units per year to pay back in 2 years
...

-

Figure 13–8
Breakeven number of units for different payback periods, Example 13
...

13
...
Examples 13
...
8 demonstrate the use of Goal Seek for both types of problems
...
4

More Breakeven and Payback Analysis on Spreadsheets

EXAMPLE 13
...
autoblog
...
The design is based on the fact that a person naturally steps downward on his or her foot when surprised, shocked, or struck with a medical
emergency
...
Assume that for the manufacture of pedal components, two equally qualiﬁed machines have been identiﬁed and estimates made
...
However, the automated controls, safety features, and ergonomic design of machine 2 make it a better choice for
the plant in the opinion of the project engineer
...

The parameters to concentrate on are (a) ﬁrst cost, (b) net cash ﬂow, and (c) life of machine 2,
if all other estimates remain the same
...
7
...

(a) Figure 13–10a: By forcing the AW for machine 2 to equal $193, Goal Seek ﬁnds a breakeven of $96,669
...

(b) Figure 13–10b: (Remember to reset the ﬁrst cost to $ –110,000 on the spreadsheet
...
Therefore, if the NCF estimate can realistically be
increased from $22,000 to $25,061, again machine 2 will be economically equivalent
...
The easiest approach is to use the NPER function to ﬁnd the payback period
...
13 years
...

353

354

Chapter 13

Breakeven and Payback Analysis

(a) First cost

(b) NCF

Figure 13–10
Breakeven values for (a) ﬁrst cost and (b) annual net cash ﬂow using Goal Seek, Example 13
...

EXAMPLE 13
...
The estimated costs each year
for repairs, insurance, etc
...

At an expected 8% per year return, use spreadsheet analysis to determine the payback period
if the building is (a) kept for 2 years and sold for $290,000 sometime beyond year 2 or (b) kept
for 3 years and sold for $370,000 sometime beyond 3 years
...
The NPV function is applied (columns D and F) to
determine when the PW changes sign from plus to minus
...
When PW Ͼ 0, the 8% return is exceeded
...
If the building is sold
after exactly 3 years for $290,000, the payback period was not exceeded; but after 4 years
it is exceeded
...
If the building is sold after 4 or 5 years, the payback is not exceeded; however, a
sale after 6 years is beyond the 8%-return payback period
...
8

355

Problems

CHAPTER SUMMARY
The breakeven point for a variable for one project is expressed in terms such as units per year or
hours per month
...
Use the following decision guideline:
Single Project

(Refer to Figure 13–2
...
Use the
following guideline to select an alternative:
Two Alternatives

(Refer to Figure 13–5
...
This is a supplemental analysis technique used primarily for initial screening prior to a full evaluation by PW or some other method
...

PROBLEMS
Breakeven Analysis for a Project
13
...
Banner Engineering’s QT50R radar-based
sensor features frequency-modulated technology to
accurately monitor or detect objects up to 15 miles
away while resisting rain, wind, humidity, and extreme temperatures
...

(a) What could the company’s ﬁxed cost per
year be in order for Banner to break even
with sales of 9000 units per year?
(b) If Banner’s ﬁxed cost is actually $750,000
per year, what is the proﬁt at a sales level of
7000 units per year?
13
...
The ﬁxed costs associated
with manufacturing are $800,000 per year
...
3 A metallurgical engineer has estimated that the
capital investment cost for recovering valuable

metals (nickel, silver, platinum, gold, etc
...
The equipment will have a useful life
of 15 years with no salvage value
...
9,
where E is the efﬁciency of the metal recovery operation (in decimal form)
...
What must the
average selling price per pound be for the precious metals that are recovered and sold in order
for the company to break even at its MARR of
15% per year?
Problems 13
...
7 are based on the following
information
...
Use the following cost
and revenue ﬁgures, quoted in U
...
dollars per hundredweight (cwt), recorded for this year to calculate the answers for each plant
...
50
2
...
4 Determine the breakeven point for each plant
...
5 Estimate the minimum revenue per hundredweight
required for next year if breakeven values and
variable costs remain constant, but ﬁxed costs
increase by 10%
...
6 During this year, the French plant sold 950 units in
Europe, and the U
...
plant sold 850 units
...

13
...
Determine the decreases in dollar
amounts and percentages in variable cost necessary to meet this goal, if the number of units sold is
the same as this year
...
8 The National Highway Trafﬁc Safety Administration raised the average fuel efﬁciency standard to
35
...
The rules will cost consumers an average of $926 extra per vehicle in the
2016 model year
...
5 mpg and keeps it
for 5 years
...
75% per month?
13
...
S
...
K
...
The ﬁxed cost of the center is $775,000
with an average variable cost of $1 and revenue of
$2
...

(a) Find the percentage of the call capacity that
must be placed each year to break even
...
This is expected to
increase the center’s ﬁxed cost to $900,000 of
which 50% will be allocated to the new product line
...

How does this required revenue compare with
the current center revenue of $2
...
10 The addition of a turbocharger to a small V-6 engine that gets 18 miles per gallon of gasoline can
boost its power to that of a V-8 engine and increase
fuel efﬁciency at the same time
...
25 per gallon and the interest rate is 1%
per month
...
11 Transporting extremely heavy patients (people
who weigh more than 500 pounds) is much more
difﬁcult than transporting normal-weight patients
...
The extra fees
are justiﬁed by the ambulance companies on the
basis of the specialty equipment required and the
extra personnel involved
...

13
...
A number of companies make devices
that they claim will signiﬁcantly increase a vehicle’s fuel efﬁciency
...
Assuming the device works as claimed, for a
vehicle that currently gets 20 miles per gallon
(mpg), how many miles would the owner have to
drive each year to break even in 5 years? Assume
the cost of gasoline is $3
...

13
...
Assume you had been buying gasoline for
$2
...
98 per
gallon at the station where you usually go
...

13
...
The initial cost for equipment conversion
will be $200 million with a 20% salvage value
anytime within a 5-year period
...
The
production capacity for the ﬁrst year will be
4000 units
...
15 For the last 2 years, The Health Company has
experienced a ﬁxed cost of $850,000 per year
and an (r Ϫ v) value of $1
...
International
competition has become severe enough that
some ﬁnancial changes must be made to keep
market share at the current level
...

(b) If ﬁxed costs and revenue per unit remain at
their current values, what type of change
must take place to make the breakeven point
go down?
13
...
15) Expand
the analysis performed in Problem 13
...
The ﬁnancial manager estimates that ﬁxed costs will fall to $750,000
when the required production rate to break even is
at or below 600,000 units
...
17 Providing restrooms at parks, zoos, and other cityowned recreation facilities is a considerable expense for municipal governments
...
The
cost of renting and servicing a portable restroom
is $7500 per year
...
He remarked that the rather high cost
is due to the necessity to use expensive materials
and construction techniques that are tailored to
minimize damage from vandalism that often occurs in unattended public facilities
...

13
...
A geosynthetic bentonite clay liner
(GCL) will cost $1
...
Alternatively,

357

a high-density polyethylene (HDPE) geomembrane can be installed that will have a useful life of
12 years
...
19 An irrigation canal contractor wants to determine
whether he should purchase a used Caterpillar
mini excavator or a Toro powered rotary tiller for
servicing irrigation ditches in an agricultural area
of California
...

Fixed costs for insurance, license, etc
...
The excavator will require
one operator at $15 per hour and maintenance at
$1 per hour
...
15 mile of ditch can be
prepared
...
The tiller costs $1200 and has a useful life of
5 years with no salvage value
...
20 per hour, and with the tiller,
the two workers can prepare 0
...
The contractor’s MARR is 10% per year
...

13
...

Such a system involves recirculation of the partially treated water back into the feed tank, causing the water to heat up
...

The single-pass system, good for 3 years, requires
a small chiller costing $920 plus stainless steel
tubing, connectors, valves, etc
...
The
cost of water, treatment charges, electricity, etc
...
10 per hour
...
28 per hour to operate
...

13
...
A
high-use component (expected usage is 5000 units
per year) can be purchased for $25 per unit with
delivery promised within a week
...
Labor

358

Chapter 13

and other operating costs are estimated to be
$35,000 per year over the study period of 5 years
...
Neglect the element of availability (a) to determine the breakeven quantity
and (b) to recommend making or buying at the expected usage level
...
22 A partner in a medium-size A/E (architectural͞engineering) design ﬁrm is evaluating two alternatives
for improving the exterior appearance of the building they occupy
...
The paint is expected to
remain attractive for 4 years, at which time repainting will be necessary
...
Alternatively, the building can be sandblasted now and every 6 years at a cost 40% greater
than the previous time
...
23 A junior mechanical engineering student is cooping this semester at Regency Aircraft, which
customizes the interiors of private and corporate
jets
...
The ﬁrst cost is
not easy to estimate due to many options, but the
annual revenue and M&O costs should net out at
$ϩ15,000 per year over a 10-year life
...
Determine the
breakeven ﬁrst cost of the machine to just recover
its ﬁrst cost and a return of 8% per year under two
scenarios:
I: No outside revenue will be developed by
the machine
...

Solve using (a) hand and (b) spreadsheet solutions
...
24 Ascarate Fishing Club (a nonproﬁt organization
dedicated to teaching kids how to ﬁsh) is considering two options for providing a heavily stocked
pond for kids who have never caught a ﬁsh before
...
The purchase price will be $400
...

Breakeven and Payback Analysis

Option 2 is an in-ground pond that will be excavated by club members at no cost and lined with
fabric that costs $1 per square foot
...
Assume 300 ft2
of liner will be purchased
...
Maintenance inside the
fence is expected to cost $20 per year
...
At an interest rate of 6%
per year, how long would the above-ground pool
have to last to break even?
13
...
Alternative 1 is a gravel base and pavement with an initial cost of $500,000 that will last for 15 years
and has an annual upkeep cost of $100 per mile
...
Annual reapplication of the
mix is required
...
05 mile
...
(b) A drive in a pickup indicates a
total of 12
...
Which is the more
economical alternative?
13
...
When the lagoon is
full, it is necessary to remove the sludge to a site
located 8
...
Currently, when the lagoon is full, the sludge is removed by pump into a tank truck and hauled away
...
The
company pays a contract individual to operate the
pump and oversee environmental and safety factors at a rate of $100 per day, plus the truck and
driver must be rented for $200 per day
...
The pump would have an initial
cost of $1600 and a life of 10 years and will cost
$3 per day to operate
...

(a) If the pipeline will cost $12 per meter to construct and will have a 10-year life, how many
days per year must the lagoon require pumping to justify construction of the pipeline?
(b) If the company expects to pump the lagoon
once per week every week of the year, how
much money can it afford to spend now on
the 10-year life pipeline to just break even?
13
...
(a) Use an AW
relation to determine the minimum number of
hours per year to operate the pumps that will justify the Auto Green system, if the MARR is 10%
per year
...
(c) Write the spreadsheet functions to display the payback period for
both 0% and 3% per month
...
31 (a)

Nutra Jet (N) Auto Green (A)
Initial cost, $
Life, years
Rebuild cost, $
Time before rebuild, annually
or minimum hours
Cost to operate, $ per hour

Ϫ4,000
3
Ϫ1,000
2,000

Ϫ10,300
6
Ϫ2,200
8,000

1
...

If the asset will be in service for 12 years,
should it be purchased?

0
...
28 An engineering practitioner can lease a fully
equipped computer and color printer system for
$800 per month or purchase one for $8500 now and
pay a $75 per month maintenance fee
...

Show both (a) hand and (b) spreadsheet solutions
...
29 The ofﬁce manager of an environmental engineering consulting ﬁrm was instructed to make an ecofriendly decision in acquiring an automobile for
general ofﬁce use
...
The hybrid under consideration is
GM’s Volt, which will cost $35,000, will have a
salvage value of $15,000 after 5 years, and will
have a range of 40 miles on the electric battery,
plus several hundred more miles when the gasoline engine kicks in
...
The Leaf’s relatively limited
range creates a psychological effect known as
range anxiety (RA), which has the company leaning toward purchasing the Volt
...

The accountant for the consulting ﬁrm told the
ofﬁce manager that the Leaf is the better economic
option based on an evaluation she performed earlier
...
75% per month
...
30 How long will you have to sell a product that has
an income of $5000 per month and expenses of

(b)

13
...
15 million
...
Determine the
number of years the equipment must be used to
obtain payback at MARR values of (a) 0% and 8%
per year and (b) 15% and 16% per year
...

13
...

The year index is k ϭ 1, 2, 3,
...

For a preliminary conclusion, should the
equipment be purchased if the actual useful
life is 7 years?

13
...

She is considering the purchase of a three-bedroom
lodge in upper Montana that will cost $250,000
...
If
Clarisa spends an average of $500 per month for
utilities and the investment increases at a rate of
2% per month, how long would it be before she
could sell the property for $100,000 more than she
has invested in it?
13
...
When she completed
her engineering management degree, she sold the
business and her grandparents told her to keep the

360

Chapter 13

Breakeven and Payback Analysis

money as a graduation present
...
36 Buhler Tractor sold a tractor for $45,000 to Tom
Edwards 10 years ago
...
37 National Parcel Service has historically owned and
maintained its own delivery trucks
...
The study period is no
more than 24 months for either alternative
...
Use
the ﬁrst cost and net cash ﬂow estimates to determine the payback in months with a nominal 9%
per year return for the (a) purchase option and
(b) lease option
...
38 Julian Browne, owner of Clear Interior Environments, purchased an air scrubber, HEPA vacuum,
and other equipment for mold removal for $15,000
eight months ago
...
For the last 4 months, a
contract generated a net $6000 per month
...

Determine (a) the no-return payback period and
(b) the nominal 18%-per-year payback period
...
39 Explain why payback analysis may favor an alternative with a shorter payback period when it is not
the better choice economically
...
40 When comparing two alternatives, why is it best to
use no-return payback analysis as a preliminary
screening tool prior to conducting a complete PW
or AW evaluation?
Spreadsheet Problems
13
...
007Q2 ϩ 32Q
TC ϭ 0
...
2Q ϩ 8
(a) Plot R and TC
...

Estimate the amount of proﬁt at this quantity
...
The equations are
Proﬁt ϭ aQ2 ϩ bQ ϩ c
Ϫb
Qp ϭ ——
2a
Ϫb2 ϩ c
Maximum proﬁt ϭ ——
4a
Use these relations to conﬁrm the graphical
estimates you made in (a)
...
)
13
...
Revenue
for the ﬁrst year was $50,000
...

Problems 13
...
44 are based on the following
information
...
The current system has a ﬁxed cost of $300,000
per year and a variable cost of $10 per unit
...
A newly proposed process will add onboard features that allow the revenue to increase to $16 per
unit, but the ﬁxed cost will now be $500,000 per year
...
2 hour required to produce each unit
...
43 Determine the annual breakeven quantity for
(a) the current system and (b) the new system
...
44 Plot the two proﬁt relations and estimate graphically the breakeven quantity between the two
alternatives
...
45 through 13
...

13
...

Mid-Valley Industrial Extension Service, a state-sponsored
agency, provides water quality sampling services to all
business and industrial ﬁrms in a 10-county region
...
Now an outsourcing agency has offered to take over this function on a
per sample basis
...
The MARR for government projects is 5%
per year, and a study period of 8 years is chosen
...
46 Use a spreadsheet to graph the AW curves for both
options for test loads between 0 and 4000 per year
in increments of 1000 tests
...

Sample costs average $25 each
...

Outsourced: Contractors quote sample cost averages
of $100 for the ﬁrst 5 years, increasing to
$125 per sample for years 6 through 8
...
47 The service director has asked the outsource company to reduce the per sample costs by 25% across
the board over the 8-year study period
...
46 before answering
...
48 Assume the Extension Service can reduce its annual salaries from $175,000 to $100,000 per
year and the per sample cost from $25 to $20
...
) What is
the new annual breakeven test quantity?

ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
13
...
52 AW1 ϭ Ϫ23,000(A͞P,10%,10)ϩ4000(A͞F,10%,10)
Ϫ 3000 Ϫ 3x
AW2 ϭ Ϫ8,000(A͞P,10%,4) Ϫ 2000 − 6x
For these two AW relations, the breakeven point x,
in miles per year, is closest to:
(a) 1130
(b) 1224
(c) 1590
(d) 655

13
...
Alternative A will
have ﬁxed costs of $42,000 per year and will require 2 workers at $48 per day each
...
Alternative B will have ﬁxed costs of
$56,000 per year, but with this alternative, 3 workers will generate 200 units of product
...
53 To make an item in-house, equipment costing
$250,000 must be purchased
...
Buying the item externally will cost $100 per unit
...
51 When the variable cost is reduced for linear total
cost and revenue lines, the breakeven point decreases
...

(b) The two lines will now cross at zero
...

(d) The total cost line becomes nonlinear
...
54 A procedure at Mercy Hospital has ﬁxed costs of
$10,000 per year and variable costs of $50 per test
...
The number of tests that must be performed each year for the two operations to break
even is closest to:
(a) 290 (b) 455 (c) 750 (d) Over 800

362

Chapter 13

13
...
Alternative X has ﬁxed
costs of $10,000 per year with a variable cost of
$50 per unit
...

The number of units that must be produced each
year in order for alternative Y to be favored is
closest to:
(a) Y will be favored for any level of production
(b) 125
(c) 375
(d) X will be favored for any level of production
13
...
Material C will cost
$100,000 per mile and last for 10 years
...
Material D will cost $30,000 per mile
and last for 5 years
...
57 A construction company can purchase a piece of
equipment for $50,000 and spend $100 per day in
operating costs
...
Alternatively, the company can lease the equipment for $400 per day
...
58 A tractor has a ﬁrst cost of $40,000, a monthly operating cost of $1500, and a salvage value of
$12,000 in 10 years
...

An identical tractor can be rented for $3200 per
month (operating cost not included)
...
4%,10) Ϫ 1500n
ϩ 12,000(A͞F,11
...
59 An anticorrosive coating for a chemical storage
tank will cost $5000 and last 5 years if touched
up at the end of 3 years at a cost of $1000
...
60 The price of a car is $50,000 today
...
You now
have $25,000 in an investment that is earning 20%
per year
...
61 Process A has a ﬁxed cost of $16,000 per year and
a variable cost of $40 per unit
...

If the company’s MARR is 10% per year, the ﬁxed
cost of process B that will make the two alternatives have the same annual cost at a production
rate of 1000 units per year is closest to:
(a) Less than $10,000
(b) $18,000
(c) $27,000
(d) Over $30,000
13
...
63 Two methods of weed control in an irrigation canal
are under consideration
...
The lining will last 20 years
...
Method B involves spraying
a chemical that costs $40 per gallon
...
In determining the number of
miles per year that would result in breakeven, the
variable cost for method B is closest to:
(a) $5 per mile
(b) $15 per mile
(c) $20 per mile
(d) $40 per mile
13
...
Aeration is used primarily for the physical removal of
gases or volatile compounds, while sludge recirculation can
be beneﬁcial for turbidity removal and hardness reduction
...
With the huge increases in
electricity cost that have occurred in some localities, however, it became necessary to review the cost-effectiveness of
all water treatment processes that consume signiﬁcant
amounts of energy
...

there was neither aeration nor recirculation
...
The reduction was 18% when neither aeration nor
recirculation was used
...
With sludge recirculation alone, the turbidity reduction was only 6%, meaning that
sludge recirculation alone actually resulted in an increase in
turbidity—the difference between 18% and 6%
...
The calculations
are based on the following data:
Aerator motor ϭ 40 hp
Aerator motor efﬁciency ϭ 90%
Sludge recirculation motor ϭ 5 hp
Recirculation pump efﬁciency ϭ 90%

Information
This study was conducted at a 106 m3 per minute watertreatment plant where, under normal operating circumstances, sludge from the secondary clariﬁers is returned to
the aerator and subsequently removed in the primary clariﬁers
...

To evaluate the effect of sludge recirculation, the sludge
pump was turned off, but aeration was continued
...
Finally, both processes were discontinued
...

The results obtained from the four operating modes
showed that the hardness decreased by 4
...
e
...
When only sludge was recirculated, the reduction was
3
...
There was no reduction due to aeration only, or when

Chemical
additions

Electricity cost ϭ 9 ¢͞kWh (previous analysis)
Lime cost ϭ 7
...
62 mg͞L per mg͞L hardness
Coagulant cost ϭ 16
...
5
As a ﬁrst step, the costs associated with aeration and sludge
recirculation were calculated
...

Aeration cost:
40 hp ϫ 0
...
09 $͞kWh ϫ 24 h͞day
Ϭ 0
...
75 kW͞hp ϫ 0
...
90 ϭ $9 per day or $275 per month
Figure 13–12

Flash
mix

Flocculation

Primary
clarifier

Secondary
clarifier

Aerator

Filter

Schematic of water
treatment plant
...

I
...

Alternative
Description

1

Aeration only
Sludge recirculation
only
Neither aeration nor
sludge recirculation

Recirculation
(2)

4

Total Savings
(3) ‫)2( ؉ )1( ؍‬

Hardness
(4)

Turbidity
(5)

Total
Extra Cost
(6) ‫)5( ؉ )4( ؍‬

Net
Savings
(7) ‫)6( ؊ )3( ؍‬

Normal operating condition

Sludge recirculation
and aeration

2
3

Aeration
(1)

Extra Cost for
Removal of

—
2196

275
—

275
2196

1380
262

469
845

1849
1107

Ϫ1574
ϩ1089

2196

275

2471

1380

469

1849

ϩ622

The estimates appear in columns 1 and 2 of the cost summary
in Table 13–1
...
The calculations below are based on a design ﬂow of
53 m3͞minute
...
The extra turbidity reaching the
ﬂocculators could require further additions of the coagulating
chemical
...
Since the average dosage before discontinuation of aeration was 10 mg͞L,
the incremental chemical cost incurred because of the increased turbidity in the clariﬁer efﬂuent would be
(10 ϫ 0
...
165 $͞kg ϫ 60 min͞h
ϫ 24 h͞day ϭ $27
...
e
...

Changes in hardness affect chemical costs by virtue of the
direct effect on the amount of lime required for water softening
...
1 mg͞L (that is, 258 mg͞L ϫ 4
...

However, with sludge recirculation only, the reduction was
9
...
3 mg͞L attributed to
aeration
...
3 mg͞L ϫ 0
...
079 $͞kg
ϫ60 min͞h ϫ 24 h͞day ϭ $8
...

The total savings and total costs associated with changes
in plant operating conditions are tabulated in columns 3 and
6 of Table 13–1, respectively, with the net savings shown in
column 7
...
” This condition would result in
a net savings of $1089 per month, compared to a net savings
of $622 per month when both processes are discontinued and
a net cost of $1574 per month for aeration only
...

In summary, the commonly applied water treatment practices of sludge recirculation and aeration can signiﬁcantly affect the removal of some compounds in the primary clariﬁer
...

Case Study Exercises
1
...
Does a decrease in the efﬁciency of the aerator motor
make the selected alternative of sludge recirculation only
more attractive, less attractive, or the same as before?
3
...
If the efﬁciency of the sludge recirculation pump were
reduced from 90% to 70%, would the net savings difference between alternatives 3 and 4 increase, decrease, or
stay the same?
5
...
If the cost of electricity decreased to 8 ¢͞kWh, which
alternative would be the most cost-effective?
7
...

L E A R N I N G S TA G E 4
Rounding Out the Study

LEARNING STAGE 4

Rounding Out
the Study
CHAPTER

14

Effects of Inﬂation
CHAPTER

15

Cost Estimation and
Indirect Cost
Allocation
CHAPTER

16

Depreciation Methods
CHAPTER

17

After-Tax Economic
Analysis
CHAPTER

18

Sensitivity Analysis
and Staged Decisions
CHAPTER

19

More on Variation
and Decision Making
under Risk

T

his stage includes topics to enhance your ability to perform a
thorough engineering economic study of one project or several alternatives
...
Techniques of cost
estimation to better predict cash ﬂows are treated in order to base
alternative selection on more accurate estimates
...
An expanded version of sensitivity analysis is
developed to examine parameters that vary over a predictable range
of values
...
Finally, the elements of risk and probability
are explicitly considered using expected values, probabilistic analysis,
and spreadsheet-based Monte Carlo simulation
...
Use the chart in the Preface to
determine appropriate points at which to introduce the material in
Learning Stage 4
...

SECTION

TOPIC

LEARNING OUTCOME

14
...

14
...

14
...

14
...

T

his chapter concentrates upon understanding and calculating the effects of inﬂation in time value of money computations
...

The annual inﬂation rate is closely watched and historically analyzed by government
units, businesses, and industrial corporations
...
In the ﬁrst decade of the 21st century, inﬂation has not
been a major concern in the United States or most industrialized nations
...
Factors such as the cost
of energy, interest rates, availability and cost of skilled people, scarcity of materials, political stability, and other, less tangible factors have short-term and long-term impacts on the
inﬂation rate
...
The basic techniques to do so are covered here
...
1 Understanding the Impact of Inﬂation
We are all very well aware that $20 now does not purchase the same amount as $20 did in 2005
and purchases signiﬁcantly less than in 2000
...

Inﬂation is an increase in the amount of money necessary to obtain the same amount of
goods or services before the inﬂated price was present
...
Inﬂation decreases
the purchasing ability of money in that less goods or services can be purchased for the same
one unit of money
...
The
value of money has decreased, and as a result, it takes more money for the same amount of
goods or services
...
To make comparisons between monetary amounts
that occur in different time periods, the different-valued money ﬁrst must be converted to constant-value money in order to represent the same purchasing power over time
...

Money in one period of time t1 can be brought to the same value as money in another period
of time t2 by using the equation
amount in period t2
Amount in period t1 ‫————————————— ؍‬
inﬂation rate between t1 and t2

[14
...
Dollars in period t2 are called future dollars or then-current dollars and have inﬂation
taken into account
...
1] is
future dollars
Constant-value dollars ‫—————— ؍‬
(1 ؉ f )n
Future dollars ‫ ؍‬constant-value dollars(1 ؉ f )n

[14
...
3]

We can express future dollars in terms of constant-value dollars, and vice versa, by applying the
last two equations
...
As an illustration, use the price of a cheese pizza
...
99

March 2011

If inﬂation on food prices averaged 5% during the last year, in constant-value 2010 dollars, this
cost is last year’s equivalent of
$8
...
05 ϭ $8
...
3], is
$8
...
05) ϭ $9
...
44 in 2012 buys exactly the same cheese pizza as $8
...
If inﬂation
averages 5% per year over the next 10 years, Equation [14
...

$8
...
05)10 ϭ $13
...
In some areas of the world, hyperinﬂation may average 50% per
year
...
99 to $518
...

Placed into an industrial or business context, at a reasonably low inﬂation rate averaging 4%
per year, equipment or services with a ﬁrst cost of $209,000 will increase by 48% to $309,000
over a 10-year span
...
Make no mistake: Inﬂation is a formidable
force in our economy
...
Only the ﬁrst two are interest rates
...
This is the rate at which interest is earned when the
effects of changes in the value of currency (inﬂation) have been removed
...
(The equation used to calculate i, with the
inﬂuence of inﬂation removed, is derived later in Section 14
...
) The real rate of return that
generally applies for individuals is approximately 3
...
This is the “safe investment”
rate
...

Inﬂation-adjusted or market interest rate if
...
This is the interest rate we hear everyday
...
It is also known as the inﬂated interest rate
...
The determination
of this value is discussed in Section 14
...

Inﬂation rate f
...

Deflation is the opposite of inflation in that when deflation is present, the purchasing
power of the monetary unit is greater in the future than at present
...
Inflation
occurs much more commonly than deflation, especially at the national economy level
...

Temporary price deﬂation may occur in speciﬁc sectors of the economy due to the introduction of improved products, cheaper technology, or imported materials or products that
force current prices down
...
However, deﬂation over a short time in a speciﬁc sector of an economy can be
orchestrated through dumping
...
The prices will go
down for the consumer, thus forcing domestic manufacturers to reduce their prices in order
to compete for business
...
Prices may then return to normal levels and, in fact, become inﬂated over time, if competition has been signiﬁcantly
reduced
...
However, if deﬂation occurs at a more general

14
...
Another result is that individuals and families have less money to spend due to fewer jobs, less
credit, and fewer loans available; an overall “tighter” money situation prevails
...
In
the extreme case, this can evolve over time into a deﬂationary spiral that disrupts the entire
economy
...

Engineering economy computations that consider deﬂation use the same relations as those for
inﬂation
...
2] and [14
...
For example, if deﬂation is estimated to be 2% per year, an asset that costs $10,000 today would have a ﬁrst cost 5 years from
now determined by Equation [14
...

10,000(1 Ϫ f )n ϭ 10,000(0
...
9039) ϭ $9039

14
...
The calculations involved in this procedure are illustrated in Table14–1, where the inﬂation rate is 4% per year
...
Column 3 shows the cost in future dollars,
and column 4 veriﬁes the cost in constant-value dollars via Equation [14
...
When the future
dollars of column 3 are converted to constant-value dollars (column 4), the cost is always
$5000, the same as the cost at the start
...
The actual cost (in inﬂated dollars) of the
item 4 years from now will be $5849, but in constant-value dollars the cost in 4 years will still
amount to $5000
...

Two conclusions can be drawn
...
And
$5000 four years in the future has a PW of only $3415 now in constant-value dollars at a real
interest rate of 10% per year
...
The effect of compounded inﬂation and interest rates can be large, as you
can see by the shaded area
...
Consider the P͞F
formula, where i is the real interest rate
...
1͞)3( ؍‬n

Present Worth
at Real
i ‫$,%01 ؍‬
(5) ϭ (4)(P͞F,10%,n)

0
1
2
3
4

5000(0
...
04) ϭ 208
5408(0
...
04) ϭ 225

5000
5200
5408
5624
5849

5000
5200͞(1
...
04)2 ϭ 5000
5624͞(1
...
04)4 ϭ 5000

5000
4545
4132
3757
3415

369

370

Chapter 14

Effects of Inﬂation

Figure 14–1
Comparison of constantvalue dollars, future dollars, and their present
worth values
...
2]
...
4]
(1 ϩ i ϩ f ϩ if )n
If the term i ϩ f ϩ if is deﬁned as if, the equation becomes
1
P ‫ ؍‬F ———— ‫ ؍‬F(P͞F,if ,n)
(1 ؉ if )n

[14
...
6]

where i ϭ real interest rate
f ϭ inﬂation rate
For a real interest rate of 10% per year and an inﬂation rate of 4% per year, Equation [14
...
4%
...
10 ϩ 0
...
10(0
...
144
Table 14–2 illustrates the use of if ϭ 14
...
As shown in column 4, the present worth for each year
is the same as column 5 of Table 14–1
...
That is, either i or if is introduced into the P͞A, P͞G, or Pg factors, depending upon whether the cash ﬂow is expressed in constant-value (today’s) dollars or
future dollars, respectively
...

If the cash ﬂow is expressed in future dollars, the PW value is obtained using if
...
2] and then ﬁnd the PW at the real interest rate i
...
2

Present Worth Calculations Adjusted for Inﬂation

TABLE 14–2

Present Worth Calculation Using an Inﬂated Interest Rate

Year
n
(1)

Cost in
Future Dollars, $
(2)

(P͞F,14
...
8741
0
...
6679
0
...
1
Glyphosate is the active ingredient in the herbicide Roundup® marketed by Monsanto Co
...
Contributions to Monsanto’s
revenue have been reduced signiﬁcantly by international dumping of generic glyphosate, as
announced in mid-2010
...
Assume when the price was set at $16 per gallon,
there was a prediction that in 5 years the price would inﬂate to $19 per gallon
...

(a) Determine the annual rate of inﬂation over 5 years to increase the price from $16 to $19
...
Compare this result
with $10 per gallon that Monsanto predicted would be the longer-term price
...

(d) Determine the market interest rate that must be used in economic equivalence
computations, if inﬂation is considered and an 8% per year real return is expected
by Monsanto
...

(a) Solve Equation [14
...

19
16 ϭ 19(P͞F,f,5) ϭ ————
(1 ϩ f )5
1 ϩ f ϭ (1
...
2
f ϭ 0
...
5% per year)

(b) If the price deﬂation rate is 3
...

F ϭ P(F͞P,Ϫ3
...
035)5
ϭ 12(0
...
04
The price will fall to exactly $10 per gallon after 5 years, as Monsanto predicted
...
Kilman and I
...

371

372

Chapter 14

Effects of Inﬂation

(c) Five years in the future, at 3
...
5%,5) ϭ 12(1
...
1877)
ϭ $14
...
25 versus $16 per gallon)
...
5% per year and a real return of 8% per year, Equation [14
...
78% per year
...
08 ϩ 0
...
08)(0
...
1178 (11
...
2
A 15-year $50,000 bond that has a dividend rate of 10% per year, payable semiannually, is currently for sale
...
5% each 6-month period, what is the
bond worth now (a) without an adjustment for inﬂation and (b) when inﬂation is considered?
Show both hand and spreadsheet solutions
...
10)]͞2 ϭ $2500
...

if ϭ 0
...
025 ϩ (0
...
025) ϭ 0
...
6%,30) ϩ 50,000(P͞F,6
...
9244) ϩ 50,000(0
...
Without
an inﬂation adjustment, the PV function is developed at the nominal 4% rate for 30 periods;
with inﬂation considered the rate is if ϭ 6
...

Comment
The $18,985 difference in PW values illustrates the tremendous negative impact made by only
2
...
06% per year)
...
Yet, this is worth only
$39,660 in constant-value dollars
...
2
...
2

Present Worth Calculations Adjusted for Inﬂation

EXAMPLE 14
...
She wishes to calculate a project’s PW
with estimated costs of $35,000 now and $7000 per year for 5 years beginning 1 year from now
with increases of 12% per year thereafter for the next 8 years
...

Solution
(a) Figure 14–3 presents the cash ﬂows
...
34] and [2
...

PW ϭ Ϫ35,000 Ϫ 7000(P͞A,15%,4)
1
...
15
Ϫ ———————— (P͞F,15%,4)
0
...
12

{

[ ( ) ]}

ϭ Ϫ35,000 Ϫ 19,985 Ϫ 28,247
ϭ $Ϫ83,232
In the P͞A factor, n ϭ 4 because the $7000 cost in year 5 is the A1 term in Equation [2
...

PW = ?

01

PWg = ?

2

45

3
0

$7000

i = 15% per year
6

1

2

7
3

8
4

9 10 11 12 13
5

6

7

8

9

Year
Geometric series year

$7840

$35,000

$17,331
12% increase
per year

Figure 14–3
Cash ﬂow diagram, Example 14
...

(b) To adjust for inﬂation, calculate the inﬂated interest rate by Equation [14
...

if ϭ 0
...
11 ϩ (0
...
11) ϭ 0
...
65%,4)
1
...
2765
Ϫ ————————— (P͞F,27
...
2765 Ϫ 0
...
2545) Ϫ 30,945(0
...
The present value of future inﬂated
dollars is signiﬁcantly less when the inﬂation adjustment is included
...

373

374

Chapter 14

Effects of Inﬂation

Examples 14
...
3 above add credence to the “buy now, pay later” philosophy
...
If cash is not readily available at that time, the debts cannot be repaid
...
In the longer term, this
buy now, pay later approach must be tempered with sound ﬁnancial practices now, and in the future
...
3 Future Worth Calculations
Adjusted for Inﬂation
In future worth calculations, a future amount F can have any one of four different interpretations:
Case 1
...

Case 2
...

Case 3
...

Case 4
...

Depending upon which interpretation is intended, the F value is calculated differently, as
described below
...

Case 1: Actual Amount Accumulated It should be clear that F, the actual amount of money
accumulated, is obtained using the inﬂation-adjusted (market) interest rate
...
7]

For example, when we quote a market rate of 10%, the inﬂation rate is included
...

P(1 ؉ if)n P (F͞P, if ,n)
F ‫—————— ؍ ———— ؍‬
(1 ؉ f) n
(1 ؉ f )n

[14
...
The percentage loss in purchasing power is a measure of how much less
...
In 7 years, the purchasing power has risen, but only to $1481
...
3159
(1
...
Therefore, we
conclude that 4% inﬂation over 7 years reduces the purchasing power of money by 24%
...
This real interest rate is the i in Equation [14
...

if ϭ i ϩ f ϩ if
ϭ i(1 ϩ f ) ϩ f
if ؊ f
i ‫——— ؍‬
1؉f

[14
...
3

Future Worth Calculations Adjusted for Inﬂation

The real interest rate i represents the rate at which today’s dollars expand with their same purchasing power into equivalent future dollars
...
The use of this interest rate is appropriate for calculating the future worth of an investment (such as a savings account or money market fund) when
the effect of inﬂation must be removed
...
9]
0
...
04
i ϭ ————— ϭ 0
...
04

(5
...
77%,7) ϭ $1481
The market interest rate of 10% per year has been reduced to a real rate that is less than 6% per
year because of the erosive effects of 4% per year inﬂation
...
Simply put, future dollars are worth less, so more are needed
...
This is the situation if someone asks, “How
much will a car cost in 5 years if its current cost is $20,000 and its price will increase by the inﬂation rate of 6% per year?” (The answer is $26,765
...

To ﬁnd the future cost, substitute f for the interest rate in the F͞P factor
...
10]

Reconsider the $1000 used previously
...
Maintaining purchasing power and earning interest must account for both increasing
prices (case 3) and the time value of money
...
Thus, to make a real
rate of return of 5
...
For the same $1000 amount,
if ϭ 0
...
04 ϩ 0
...
04) ϭ 0
...
77% per year and inﬂation of f ϭ 4% per year
...
The calculations made in this section explain the following:
•
•
•
•

The amount of $1000 now at a market rate of 10% per year will accumulate to $1948 in 7 years
...

An item with a cost of $1000 now will cost $1316 in 7 years at an inﬂation rate of 4% per year
...
77%
with inﬂation considered at 4% per year
...
5% mentioned earlier
...
Deﬁne the symbol MARRf as the inﬂation-adjusted or market
MARR, which is calculated in a fashion similar to if
...
11]

375

376

Chapter 14

TA BLE 14–3

Effects of Inﬂation
Calculation Methods for Various Future Worth Interpretations

Future Worth
Desired
Case 1: Actual dollars
accumulated

Use stated market
rate if in equivalence
formulas

Case 2: Purchasing power
of accumulated dollars in
terms of constant-value
dollars

Example for
P ‫ ,0001$ ؍‬n ‫,7 ؍‬
if ‫ ,%01 ؍‬f ‫%4 ؍‬

Method of
Calculation

Use market rate if in
equivalence and
divide by (1 ϩ f )n
or
Use real i

F ϭ 1000(F͞P,10%,7)

1000 (F͞P,10%,7)
F ϭ ————————
(1
...
77%,7)

Case 3: Dollars required for
same purchasing power

Use f in place of i in
equivalence
formulas

F ϭ 1000(F͞P,4%,7)

Case 4: Future dollars to
maintain purchasing
power and to earn a return

Calculate if and use in
equivalence
formulas

F ϭ 1000(F͞P,10%,7)

The real rate of return i used here is the required rate for the corporation relative to its cost of
capital
...
The
inﬂation-adjusted MARR is calculated by including the inﬂation rate of, say, 4% per year
...
11]
...
13 ϩ 0
...
13(0
...
1752

(17
...
4
Abbott Mining Systems wants to determine whether it should upgrade a piece of equipment
used in deep mining operations in one of its international operations now or later
...
However, if the company selects plan I, the purchase will be deferred for 3 years when the cost is expected to rise
to $300,000
...
The inﬂation rate
in the country has averaged 3% per year
...

Solution

(a) Inﬂation not considered: The real rate, or MARR, is i ϭ 12% per year
...
Calculate the FW value for plan A three years from now and
select the lower cost
...

(b) Inﬂation considered: This is case 4; the real rate (12%), and inﬂation of 3% must be
accounted for
...
11]
...
12 ϩ 0
...
12(0
...
1536

14
...

FWA ϭ Ϫ200,000(F͞P,15
...

The inﬂation rate of 3% per year has raised the equivalent future worth of costs by 9
...
This is the same as an increase of 3% per year, compounded
over 3 years, or (1
...
3%
...
, are present
...
In these cases, the government may take drastic actions: redeﬁne the currency in terms of
the currency of another country, control banks and corporations, and control the ﬂow of capital
into and out of the country in order to decrease inﬂation
...
To appreciate
the disastrous effect of hyperinﬂation on a company’s ability to keep up, we can rework Example 14
...
The FWA amount skyrockets and plan I is a
clear choice
...

Good economic decisions in a hyperinﬂated economy are very difﬁcult to make using traditional engineering economy methods, since the estimated future values are totally unreliable
and the future availability of capital is uncertain
...
4 Capital Recovery Calculations
Adjusted for Inﬂation
It is particularly important in capital recovery (CR) calculations used for AW analysis to include
inﬂation because current capital dollars must be recovered with future inﬂated dollars
...
This suggests the use of the inﬂated interest rate in the
A͞P formula
...
8%,5) ϭ $325
...

This suggests the use of a higher interest rate, that is, the if rate, to produce a lower A value in the
A͞F formula
...
8%,5) ϭ $137
...
80
...
59 versus $163
...

377

378

Chapter 14

Effects of Inﬂation

EXAMPLE 14
...
58 today, if the market interest rate is 10% per year and inﬂation is
8% per year?

Solution
First, ﬁnd the actual number of inﬂated dollars required 5 years in the future that is equivalent
to $680
...
This is case 3; Equation [14
...

F ϭ (present purchasing power)(1 ϩ f )5 ϭ 680
...
08)5 ϭ $1000
The actual amount of the annual deposit is calculated using the market interest rate of 10%
...

A ϭ 1000(A͞F,10%,5) ϭ $163
...
85% as determined using Equation [14
...
To put these calculations into perspective, if the inﬂation rate is zero when the real interest rate is 1
...
58 today is obviously $680
...

Then the annual amount required to accumulate this future amount in 5 years is A ϭ
680
...
85%,5) ϭ $131
...
This is $32
...
80 calculated above for
f ϭ 8%
...
To make up the
purchasing power difference, more higher-value dollars are required
...
63 per year is required
...
People tend to pay off less of their incurred debt at each payment because they
use any excess money to purchase additional items before the price is further inﬂated
...
All this is due to the spiraling effect of increasing inﬂation
...

CHAPTER SUMMARY
Inﬂation, treated computationally as an interest rate, makes the cost of the same product or service increase over time due to the decreased value of money
...
Some important relations are the following:
Inﬂated interest rate: if ϭ i ϩ f ϩ if
Real interest rate: i ϭ (if − f )͞(1 ϩ f)
PW of a future amount with inﬂation considered: P ϭ F(P͞F, if, n)
Future worth in constant-value dollars of a present amount with the same purchasing power:
F ϭ P(F͞P,i,n)
Future amount to cover a current amount with inﬂation only: F ϭ P(F͞P, f, n)
Future amount to cover a current amount with inﬂation and interest: F ϭ P(F͞P,if , n)
Annual equivalent of a future amount: A ϭ F(A͞F,if , n)
Annual equivalent of a present amount in future dollars: A ϭ P(A͞P,if , n)
Hyperinﬂation implies very high f values
...
This can, and usually does, cause a national ﬁnancial disaster when it continues over extended periods of time
...
1 What is the difference between today’s dollars and
constant-value dollars (a) when using today as the
reference point in time and (b) when using 2 years
ago as the reference point?
14
...

14
...
4 Determine the inﬂation-adjusted interest rate for a
growth company that wants to earn a real rate of
return of 20% per year when the inﬂation rate is
5% per year
...
5 For a high-growth company that wants to make a
real rate of return of 30% per year, compounded
monthly, determine the inﬂation-adjusted nominal
interest rate per year
...
5% per month
...
6 A high-tech company whose stock trades on the
NASDAQ stock exchange uses a MARR of 35% per
year
...
7 Calculate the inﬂation-adjusted interest rate per
quarter when the real interest rate is 4% per quarter
and the inﬂation rate is 1% per quarter
...
8 Calculate the real interest rate per month if the
nominal inﬂation-adjusted interest rate per year,
compounded monthly, is 18% and the inﬂation rate
per month is 0
...

14
...
The contract price is ﬁxed at
$45,000 per year for 4 years
...

14
...
11 Assume that you want to retire 30 years from now
with an amount of money that will have the same
value (same purchasing power) as $1
...
If you estimate the inﬂation rate will be 4%
per year, how many future (then-current) dollars
will you need?
Adjusting for Inﬂation
14
...
13 The inﬂation rate in a Central American country is
6% per year
...
14 During periods of hyperinﬂation, prices increase
rapidly over short periods of time
...
6 billion
...
4 billion
...
15 A trust was set up by your grandfather that states
you are to receive $250,000 exactly 5 years from
today
...

14
...

Determine (a) the number of constant-value dollars 5 years in the future that is equivalent to
$30,000 now and (b) the number of future dollars
that will be equivalent to $30,000 now
...
17 The Pell Grant program of the federal government
provides ﬁnancial aid to needy college students
...
18 Ford Motor Company announced that the price of
its F-150 pickup trucks is going to increase by only
the inﬂation rate for the next 3 years
...
1% per year, what is the
expected price of a comparably equipped truck
next year? 3 years from now?
14
...
The cost of the dampeners today is $120,000,
but the company has to wait until a permit is approved for its bidirectional port-to-plant product
pipeline
...

Because of intense foreign competition, the manufacturer plans to increase the price only by the inﬂation rate each year
...
8% per
year and the company’s MARR is 20% per year,
estimate the cost of the dampeners in 2 years in
terms of (a) today’s dollars and (b) future dollars
...
20 A machine currently under consideration by
Holzmann Industries has a cost of $45,000
...
When the purchasing manager
checked the invoice for the machine he purchased
5 years ago, he saw that the price was $29,000
...
21 A report by the National Center for Public Policy
and Higher Education stated that tuition and fees
(T&F) at public colleges and universities increased
by 439% over the last 25 years
...
When the report was written, T&F at a
4-year public university constituted 28% of the
MFI of $52,000 (tuition and fees at a private university constituted 76% of MFI)
...
22 The headline on a Chronicle of Higher Education
article reads “College Costs Rise Faster than Inﬂation
...
(a) What was the average annual percentage increase over that period of time? (b) If the
real increase in tuition (i
...
, without inﬂation) was
5% per year, what was the inﬂation rate per year?
14
...
” At the University of Kansas,
Jayhawks fans can sign up to pay $105,000 over
10 years for the right to buy top seats for football
during the next 30 years
...
Season tickets in tier 1 are currently selling for $350
...
What is the dollar amount of
the savings on the tickets (with no interest considered), if ticket prices rise at a rate of 3% per year
for the next 30 years?
Present Worth Calculations with Inﬂation
14
...
What are they?
14
...
At a real interest rate of
10% per year and inﬂation rate of 4% per year, what
is the present worth of the cost of the equipment?
14
...
The supplier quoted a price of
$125,000 if the unit is purchased within the next
3 years
...
Assuming the tower will not be purchased for 3 years, calculate the present worth at
an interest rate of 10% per year and an inﬂation
rate of 4% per year
...
27 How much can the manufacturer of superconducting magnetic energy storage systems afford to
spend now on new equipment in lieu of spending
$75,000 four years from now? The company’s real
MARR is 12% per year, and the inﬂation rate is
3% per year
...
28 Find the present worth of the cash ﬂows shown
...
Assume a real
interest rate of 8% per year and an inﬂation rate of
6% per year
...
29 A doctor is on contract to a medium-sized oil company to provide medical services at remotely located, widely separated reﬁneries
...
The doctor
can buy a used Lear jet now or wait for a new very
light jet (VLJ) that will be available 3 years from
now
...
9 million, payable when the plane is delivered in 3 years
...
If the
MARR is 15% per year and the inﬂation rate is projected to be 3% per year, what is the present worth of
the VLJ with inﬂation considered?
14
...
The HDD machine will include an innovative pipe loader design and
maneuverable undercarriage system
...
At a real MARR of
10% per year and an inﬂation rate of 5% per year,
determine if the company should buy now or later
(a) without any adjustment for inﬂation and
(b) with inﬂation considered
...
31 An engineer must recommend one of two rapidprototyping machines for integration into an upgraded manufacturing line
...
Salesman
A gave her the estimates in constant-value
(today’s) dollars, while saleswoman B provided
the estimates in future (then-current) dollars
...
Use PW analysis to determine which
machine the engineer should recommend
...
32 A salesman from Industrial Water Services (IWS),
who is trying to get his foot in the door for a large
account in Fremont, offered water chlorination
equipment for $2
...
This is $400,000 more
than the price offered by a competing saleswoman
from AG Enterprises
...
If the equipment has

381

a 2-year warranty, determine which offer is better
...

14
...
A small pipeline
will cost less to purchase (including valves and
other appurtenances) but will have a high head loss
and, therefore, a higher pumping cost
...
Determine present worth values if future
worth values are FWS ϭ $2
...
5 million
...
4% per
month
...

14
...
1 million 5 years from
now, or (3) pay an amount of money 5 years from
now that will have the same purchasing power as
$850,000 now
...
35 If the inﬂation rate is 6% per year and a person
wants to earn a true (real) interest rate of 10% per
year, determine the number of future dollars she
has to receive 10 years from now if the present investment is $10,000
...
36 How many future dollars would you need 5 years
from now just to have the same buying power as
$50,000 now, if the deﬂation rate is 3% per year?
14
...
38 Harmony Corporation plans to set aside $60,000
per year beginning 1 year from now for replacing
equipment 5 years from now
...
39 The strategic plan of a solar energy company that
manufactures high-efﬁciency solar cells includes
an expansion of its physical plant in 4 years
...
If the company sets aside
$7,000,000 now into an account that earns interest
at 7% per year, what will the inﬂation rate have to
be in order for the company to have exactly the
right amount of money for the expansion?
14
...
The account is now worth $25,000
...
41 A Toyota Tundra can be purchased today for
$32,350
...
If the price of the
truck increases exactly in accordance with an estimated inﬂation rate of 3
...

14
...
In part, his will stated: “The
capital shall be invested by my executors in safe
securities and shall constitute a fund, the interest
on which shall be annually distributed in the form
of prizes to those who, during the preceding year,
shall have conferred the greatest beneﬁt on mankind
...
In addition to a
gold medal and a diploma, each recipient receives
a substantial sum of money that depends on the
Foundation’s income that year
...
In 1996, the award was $653,000; it was
$1
...

(a) If the increase between 1996 and 2009 was
strictly due to inﬂation, what was the average
inﬂation rate per year during that 13-year
period?
(b) If the Foundation expects to invest money
with a return of 5% above the inﬂation rate,
how much will a laureate receive in 2020,
provided the inﬂation rate averages 3% per
year between 2009 and 2020?

Effects of Inﬂation

14
...
The company
is planning to add larger-capacity robotic arms to
one of its assembly lines 3 years from now
...
4 million
...
8% per year
...
44 The data below show two patterns of inﬂation that
are exactly the opposite of each other over a
20-year time period
...
45 Factors that increase costs and prices—especially
for materials and manufacturing costs sensitive to
market, technology, and labor availability—can be
considered separately using the real interest rate i,
the inﬂation rate f, and additional increases that
grow at a geometric rate g
...
The
geometric rate is the same one used in the geometric series (Chapter 2)
...
This is over and above the inﬂation rate
...
46 An electric utility is considering two alternatives
for satisfying state regulations regarding pollution control for one of its generating stations
...
S
...
The station is
currently producing excess VOCs and oxides of
nitrogen
...
Plan A involves replacing
the burners and switching from fuel oil to natural
gas
...
Plan B involves going to the foreign city
and running gas lines to many of the “backyard”
brick-making sites that now use wood, tires, and
other combustible waste materials for ﬁring the
bricks
...
S
...
The initial cost of plan B will be
$1
...
Additionally, the electric company would subsidize
the cost of gas for the brick makers to the extent
of $200,000 per year
...
For a 10-year project period
and no salvage value for either plan, which one
should be selected on the basis of an annual
worth analysis at a real interest rate of 7% per
year and an inﬂation rate of 4% per year?
14
...
Her goal is to save enough
money over the next 3 years so that when she begins her trip, the amount she has accumulated will
have the same buying power as $72,000 today
...
48 An entrepreneur engaged in wildcat oil well drilling is seeking investors who will put up $500,000
for an opportunity to reap high returns if the venture is successful
...
How much will the investors have to
receive each year to recover their money if an

inﬂation rate of 5% per year is to be included in
the calculation?
14
...
manufactures in situ calibration
veriﬁcation systems that conﬁrm ﬂow measurement accuracies without removing the meters
...
If the company’s real MARR is 15%
per year, which process has the lower annual cost
when inﬂation of 5% per year is considered?
Process X
First cost, $
Operating cost, $ per year
Salvage value, $
Life, years

Process Y

Ϫ65,000
Ϫ40,000
0
5

Ϫ90,000
Ϫ34,000
10,000
5

14
...
The company is considering a new
annealing-drawing process to reduce costs
...
7 million now, how much
must be saved each year to recover the investment
in 5 years if the company’s MARR is a real 12%
per year and the inﬂation rate is 3% per year?
14
...
Determine the equivalent cost of the system if the real
interest rate is 10% per year and the inﬂation rate
is 4% per year
...
52 Maintenance costs for pollution control equipment
on a pulverized coal cyclone furnace are expected
to be $180,000 now and another $70,000 three
years from now
...
If the company
uses a real interest rate of 9% per year and the inﬂation rate averages 3% per year, what is the
equivalent annual cost of the equipment?
14
...
5 million for 5 years to ﬁnance start-up costs for a new project involving
site reclamation
...
The average
inﬂation rate is 5% per year
...

(b) Determine the capital recovery if the company is satisﬁed with accumulating $2
...

(c) Determine the capital recovery in part (b)
without considering inﬂation
...
54 Inﬂation occurs when:
(a) Productivity increases
(b) The value of the currency decreases
(c) The value of the currency increases
(d) The price of gold decreases
14
...
56 To calculate how much something will cost if you
expect its cost to increase by exactly the inﬂation
rate, you should:
(a) Multiply by (1 ϩ f )n
(b) Multiply by [(1 ϩ f )n͞(1 ϩ i)n]
(c) Divide by (1 ϩ f )n
(d) Multiply by (1 ϩ if )n
14
...
4%
(b) 7%
(c) 9%
(d) 15
...
58 The market interest rate if is 12% per year, compounded semiannually
...
59 Construction equipment has a cost today of
$40,000
...
60 The amount of money that would be accumulated
now from an investment of $1000 25 years ago at

a market rate of 5% per year and an inﬂation rate
averaging 2% per year over that time period is
closest to:
(a) $1640
(b) $3385
(c) $5430
(d) Over $5500
14
...
62 If you expect to receive a gift of $50,000 six years
from now, the present worth of the gift at a real
interest rate of 4% per year and an inﬂation rate of
3% per year is closest to:
(a) $27,600
(b) $29,800
(c) $33,100
(d) $37,200
14
...
For a real interest rate of 5%
per year and an inﬂation rate of 4% per year, the
annual capital recovery requirement for the equipment (in future dollars) is determined by:
(a) AW ϭ Ϫ30,000(A͞P,4%,20) Ϫ 7000
ϩ 5000(A͞F,4%,20)
(b) AW ϭ Ϫ30,000(A͞P,5%,20) Ϫ 7000
ϩ 5000(A͞F,5%,20)
(c) AW ϭ Ϫ30,000(A͞P,9%,20) Ϫ 7000
ϩ 5000(A͞F,9%,20)
(d) AW ϭ Ϫ30,000(A͞P,9
...
2%,20)
14
...
When
inﬂation is moderately high, bonds offer a low return relative
to stocks, because the potential for market growth is not present with bonds
...
However, bonds do offer a steady income that may be
important to an individual, and they serve to preserve the
principal invested in the bond, because the face value is returned at maturity
...
He has a collection of stocks in his
retirement portfolio, but no bonds
...
” He can choose additional
stocks or bonds, but has decided to not split the $50,000 between the two forms of investments
...

He assumes the effects of federal and state income taxes will
be the same for both forms of investment
...

Bond purchase: If he purchased a bond, he would have a
predictable income of 5% per year and the $50,000 face
value after the 12-year maturity period
...
Can you answer them for him for both choices?
1
...
If he decided to sell the stock or bond immediately after
the ﬁfth annual dividend, what is his minimum selling
price to realize a 7% real return? Include an adjustment
of 4% per year for inﬂation
...
If Earl needed some money in the future, say, immediately after the ﬁfth dividend payment, what would be
the minimum selling price in future dollars, if he were
only interested in recovering an amount that maintained
the purchasing power of the original price?
4
...
Earl plans to keep the stocks or bonds for 12 years, that
is, until the bond matures
...
For what amount must he
sell the stocks after 12 years, or buy the bonds now to
ensure he realizes this return? Do these amounts seem
reasonable to you, given your knowledge of the way
that stocks and bonds are bought and sold?

CHAPTER 15

Cost
Estimation and
Indirect Cost
Allocation
L E A R N I N G

O U T C O M E S

Purpose: Make cost estimates using different methods; demonstrate the allocation of indirect costs using traditional and
activity-based costing rates
...
1

Approach

• Explain the bottom-up and design-to-cost (topdown) approaches to cost estimation
...
2

Unit method

• Use the unit method to make a preliminary cost
estimate
...
3

Cost index

• Use a cost index to estimate a present cost based
on historical data
...
4

Cost capacity

• Use a cost-capacity equation to estimate
component, system, or plant costs
...
5

Factor method

• Estimate total plant cost using the factor
method
...
6

Indirect cost rates

• Allocate indirect costs using traditional indirect
cost rates
...
7

ABC allocation

• Use the Activity-Based Costing (ABC) method to
allocate indirect costs
...
8

Ethics and proﬁt

• Describe how biased estimation can become an
ethical dilemma
...
In reality, they are not; they must be estimated
...
Cost estimation is important in all aspects of a project, but especially in the stages of project conception, preliminary design, detailed design, and economic analysis
...
In
engineering practice, the estimation of costs receives much more attention than revenue
estimation; costs are the topic of this chapter
...
Therefore, allocation of indirect costs for functions
such as utilities, safety, management and administration, purchasing, and quality is made
using some rational basis
...

15
...
In general, most cost estimates are developed for either a
project or a system; however, combinations of these are very common
...
A system is usually an operational design that involves processes, services, software, and other nonphysical items
...
Of course, many projects will have major elements that are not physical, so estimates of both types must be developed
...

There would be no operational system if only the costs of computer hardware plus wire and
wireless connectors were estimated; it is equally important to estimate the software, personnel, and maintenance costs
...
In real-world practice, the cash ﬂows for
costs and revenues must be estimated prior to the evaluation of a project or comparison of alternatives
...
Revenue estimates utilized by engineers are usually developed in marketing,
sales, and other departments
...
Normally direct costs are estimated with some detail, then the indirect costs are added using standard rates and factors
...
Accordingly, many industrial settings require some estimating for indirect costs
as well
...
Primarily, direct costs are discussed here
...

•
•
•
•

What cost components must be estimated?
What approach to cost estimation will be applied?
How accurate should the estimates be?
What estimation techniques will be utilized?

Costs to Estimate If a project revolves around a single piece of equipment, for example, an
industrial robot, the cost components will be signiﬁcantly simpler and fewer than the components
for a complete system such as the manufacturing and testing line for a new product
...
Examples of cost
components are the ﬁrst cost P and the annual operating cost (AOC), also called the M&O costs

Direct͞Indirect costs

388

Chapter 15

Cost Estimation and Indirect Cost Allocation

(maintenance and operating) of equipment
...

Listed below are sample elements of the ﬁrst cost and AOC components
...

AOC component (part of the equivalent annual cost A):
Elements: Direct labor cost for operating personnel
Direct materials
Maintenance costs (daily, periodic, repairs, etc
...
When costs for an entire system must be estimated, the number of cost components and elements is likely to be in the hundreds
...

For familiar projects (houses, ofﬁce buildings, highways, and some chemical plants) there are
standard cost estimation software packages available
...
) and estimates costs with time-proven, built-in relations
...
However, there are no “canned” software
packages for a large percentage of industrial, business, service and public sector projects
...
For a simple rendition of this approach, see
Figure 15–1 (left side)
...
The
price is then determined by adding indirect costs and the proﬁt margin, which is usually a percentage of the total cost
...
This approach works well when competition is not a dominant factor in pricing the product or service
...
The competitive price establishes the target cost
...
This approach is useful in encouraging innovation,
new design, manufacturing process improvement, and efﬁciency
...

This approach places greater emphasis on the accuracy of the price estimation activity
...
The designto-cost approach is best applied in the early stages of a new or enhanced product design
...

Usually, the resulting approach is some combination of these two philosophies
...
Historically, the bottom-up
approach was more predominant in Western engineering cultures, especially in North America
...
1

389

Understanding How Cost Estimation Is Accomplished
Figure 15–1

Top-down approach
Required
price

Competitive
price
Desired
profit

ϩ

Allowed
profit

Ϫ

Total cost ϭ

ϭ Target cost
Indirect
costs

ϩ

ϩ

Indirect
costs

Direct
costs
Maintenance
and operations

ϩ

ϩ

Maintenance
and operations

Direct
labor

ϩ

ϩ

Direct
labor

Direct
materials

ϩ

ϩ

Direct
materials

Equipment and
ϩ
capital recovery

ϩ

Equipment
capital recovery

Cost
component
estimates

At design stage

Before design stage

Bottom-up approach

Cost
component
estimates

Design-to-cost approach

globalization of engineering design has speeded the adoption of the design-to-cost approach
worldwide
...
The accuracy required increases as the project progresses from preliminary design to detailed design and on to
economic evaluation
...

When utilized at early and conceptual design stages, estimates are referred to as order-ofmagnitude estimates and generally range within Ϯ20% of actual cost
...
Every project setting has its own characteristics, but a range of Ϯ5% of actual
costs is expected at the detailed design stage
...
Obviously, the desire for better accuracy has to be balanced against the cost of obtaining it
...

Simpliﬁed cost estimation
processes for bottom-up
and top-down approaches
...
The use of the unit method and cost indexes base
the present estimate on past cost experiences, with inﬂation considered
...
They are called cost estimating relationships (CERs)
...

Most cost estimates made in a professional setting are accomplished in part or wholly using
software packages linked to updated databases that contain cost indexes and rates for the locations, products, or processes being studied
...
Corporations usually standardize on one or two packages to ensure consistency over time and projects
...
2 Unit Method
The unit method is a popular preliminary estimation technique applicable to virtually all professions
...

CT ϭ u ϫ N

[15
...
Some sample unit cost factors (and values) are
Total average cost of operating an automobile (52¢ per mile)
Cost to bury ﬁber cable in a suburban area ($30,000 per mile)
Cost to construct a parking space in a parking garage ($4500 per space)
Cost of constructing interstate highway ($6
...
If house construction costs
average $225 per square foot, a preliminary cost estimate for an 1800-square-foot house, using
Equation [15
...
Similarly, a 200-mile trip should cost about $104 for the car only
at 52¢ per mile
...
This is illustrated in Example 15
...

EXAMPLE 15
...
Since a Ϯ20% estimate is acceptable at this preliminary stage, a unit
method estimate is sufﬁcient
...

Materials: 3000 tons at $45
...
1] to each of the ﬁve areas and sum the results to obtain the total cost
estimate of $566,700
...

15
...
1

Resource

Amount N

Unit Cost Factor u, $

Cost Estimate, u ؋ N, $

Materials
Machinery, tooling
Labor, casting
Labor, ﬁnishing
Labor, indirect

3000 tons
1500 hours
3000 hours
1200 hours
400 hours

45
...
3 Cost Indexes
This section explains indexes and their use in cost estimation
...
A preliminary cost estimate is often
based on a cost index
...
The index is
dimensionless and measures relative cost change over time
...
” Timely updating of the index is very important
...
This index includes such items as rent, food, transportation, and certain services
...
Table 15–2 is a listing of some of the more common indexes
...
S
...
S
...
1
381
...
5
389
...
6
394
...
3
395
...
7
444
...
2
499
...
4
575
...
9
555
...
5
1039
...
8
1061
...
3
1089
...
9
1104
...
6
1178
...
5
1302
...
3
1449
...
6
1461
...
2]

where Ct ϭ estimated cost at present time t
C0 ϭ cost at previous time t0
It ϭ index value at time t
I0 ϭ index value at time t0
Generally, the indexes for equipment and materials are made up of a mix of components that
are assigned certain weights, with the components sometimes further subdivided into more basic
items
...
These subcomponents, in turn, are built up from even more basic items such as pressure pipe, black pipe, and galvanized pipe
...
The base period of 1957 to
1959 is assigned a value of 100 for the CEPCI, 1913 ϭ 100 for the ENR index, and 1926 ϭ 100
for the M&S equipment cost index
...
For example, the CE plant cost index is available at www
...
com͞pci
...
construction
...
This latter site offers a comprehensive
series of construction-related resources, including several ENR cost indexes and cost estimation
systems
...
eng-tips
...

EXAMPLE 15
...
The engineer ﬁnds that a project of similar complexity and magnitude was completed 5 years ago at a skilled labor cost of $360,000
...
What is the estimated skilled labor cost for
the new project?

15
...
Using Equation [15
...

The cost index will vary, perhaps with the region of the country, the type of product or service,
and many other factors
...
The development of the cost index requires the actual
cost at different times for a prescribed quantity and quality of the item
...
The index each year (period) is determined as the cost divided by the base-year cost and multiplied by 100 (or 1)
...

EXAMPLE 15
...
He has decided to expand into new
areas and wants to make cost estimates for three of the more signiﬁcant labor costs involved
in making these types of ﬁlms
...

(a) Make 2008 the base year, and determine the cost indexes using a basis of 1
...
Comment
on the trend of each index over the years
...
The cost in 2010 was $78 per hour;
assume a worst-case scenario is that the graphics index continues the same arithmetic trend
it had from 2010 to 2011
...

Solution

(a) For each type of service, calculate It͞I0 where t ϭ 2005, 2006,
...
Table 15–5 presents the indexes
...

Stuntmen labor cost: Rising until 2009, then stable
...

TA BLE 15–4

Average Hourly Costs for Three Services, Example 15
...
3
Index It ͞I0

Type of Service

2005

2006

2007

2008

2009

2010

2011

Graphics
Stuntmen
Actors

0
...
71
0
...
84
0
...
89

0
...
86
1
...
00
1
...
00

1
...
24
0
...
16
1
...
83

1
...
21
0
...
16
...
18; the index value in 2014 will be
1
...
18) ϭ 1
...
Equation [15
...

C2014 ϭ C2010(I2014͞I2010) ϭ 78(1
...
16)
ϭ 78(1
...
4 Cost-Estimating Relationships:
Cost-Capacity Equations
Design variables (speed, weight, thrust, physical size, etc
...
Cost-estimating relationships (CERs) use these
design variables to predict costs
...

One of the most widely used CER models is a cost-capacity equation
...
This is also known
as the power law and sizing model
...
3]

where C1 ϭ cost at capacity Q1
C2 ϭ cost at capacity Q2
x ϭ correlating exponent
The value of the exponent for various components, systems, or entire plants can be obtained or
derived from a number of sources, including Plant Design and Economics for Chemical Engineers, Preliminary Plant Design in Chemical Engineering, Chemical Engineers’ Handbook,
technical journals (especially Chemical Engineering), the U
...
Environmental Protection Agency,
professional or trade organizations, consulting ﬁrms, handbooks, and equipment companies
...
When an exponent value for a particular unit is not known, it is common practice to use the value of x ϭ 0
...
In
fact, in the chemical processing industry, Equation [15
...

The exponent x in the cost-capacity equation is commonly in the range 0 Ͻ x Յ 1
...

If x ϭ 1, a linear relationship is present
...

It is especially powerful to combine the time adjustment of the cost index (It͞I0) from Equation [15
...
If the index is
embedded into the cost-capacity computation in Equation [15
...

C2,t ϭ (cost at time 0 of level 2) ϫ (time adjustment cost index)

[ ( ) ] ( II )

Q2
ϭ C1,0 ——
Q1

x

t
—
0

This is commonly expressed without the time subscripts
...

[15
...
5

TABLE 15–6

Cost-Estimating Relationships: Factor Method
Sample Exponent Values for Cost-Capacity Equations
Size Range

Exponent

1–100 MGD
0
...
1–100 MGD
5–300 hp
200–2100 hp
20–8000 ft3͞min
15–400 ft2
0
...
05–50 MGD
0
...
04–5 MGD
0
...
2 MGD
100–2000 gal

0
...
14
0
...
71
0
...
98
0
...
32
0
...
71
0
...
55
0
...
02
1
...
69
0
...
35
0
...
67

Component͞System͞Plant
Activated sludge plant
Aerobic digester
Blower
Centrifuge
Chlorine plant
Clariﬁer
Compressor, reciprocating (air service)
Compressor
Cyclone separator
Dryer
Filter, sand
Heat exchanger
Hydrogen plant
Laboratory
Lagoon, aerated
Pump, centrifugal
Reactor
Sludge drying beds
Stabilization pond
Tank, stainless

Note: MGD ϭ million gallons per day; hp ϭ horsepower; scfd ϭ standard cubic feet per day
...
4
The total design and construction cost for a digester to handle a ﬂow rate of 0
...
7 million in 2010
...
0 MGD
...
2 to 40 is 0
...
The cost index in 2010
of 131 has been updated to 225 for this year
...
3] can estimate the cost of the larger system in 2010, but it must be updated by
the cost index to today’s dollars
...
4] performs both operations at once
...
0 0
...
5
ϭ 1,700,000(1
...
718) ϭ $3
...
5 Cost-Estimating Relationships:
Factor Method
Another widely used model for preliminary cost estimates of process plants is called the factor
method
...
The method is based on the premise that fairly reliable total plant costs can be
obtained by multiplying the cost of the major equipment by certain factors
...
These factors are commonly referred to as Lang factors after Hans J
...

395

396

Chapter 15

Cost Estimation and Indirect Cost Allocation

In its simplest form, the factor method is expressed in the same form as the unit method
CT ‫ ؍‬hCE

[15
...
This follows the cost estimation approaches presented in Figure 15–1
...
10;
solid-ﬂuid process plants, 3
...
74
...

EXAMPLE 15
...
08 million
...
63, estimate the plant’s total cost
...
5]
...
63(2,080,000)
ϭ $7,550,400

Subsequent reﬁnements of the factor method have led to the development of separate factors
for direct and indirect cost components
...
1 are speciﬁcally
identiﬁable with a product, function, or process
...
Examples of indirect costs are general administration, computer services, quality, safety,
taxes, security, and a variety of support functions
...
1
...

For indirect costs, some of the factors apply to equipment costs only, while others apply to the
total direct cost
...
The overall cost factor h can be
written as
n
hϭ1ϩ

͚f

[15
...
Therefore, Equation [15
...
6 and 15
...

)]

(1 ؉ fI)

[15
...
6

Traditional Indirect Cost Rates and Allocation

EXAMPLE 15
...

If the direct cost factor is 1
...
25, determine the total plant cost
...
6]
...
61 ϩ 0
...
86
The total plant cost is
CT ϭ 2
...
7
A new container-handling crane at the Port of Singapore is expected to have a deliveredequipment cost of $875,000
...
, is 0
...
The construction factor is 0
...
21
...

Solution
(a) Total equipment cost is $875,000
...
6] is
h ϭ 1 ϩ 0
...
53 ϩ 0
...
23
The total cost is
CT ϭ 2
...
7] is used to estimate total
cost
...
49 ϩ 0
...
02
i

iϭ1

CT ϭ [875,000(2
...
21) ϭ $2,138,675

Comment
Note the decrease in estimated cost when the indirect cost is applied to the equipment cost only
in part (a)
...

15
...
For the manufacturing environment, it can be stated generally that the
statement of cost of goods sold (discussed in Appendix B) is one end product of this system
...
All costs incurred in one department
or process line are collected under a cost center title, for example, Department 3X
...
Of course, this in itself is no easy chore, and the cost of the tracking
system may prohibit collection of all direct cost data to the level of detail desired
...

397

398

Chapter 15

Cost Estimation and Indirect Cost Allocation

TA BLE 15–7

Sample Indirect Cost Allocation Bases

Cost Category
Taxes
Heat, light
Power
Receiving, purchasing
Personnel, machine shop
Building maintenance
Software
Quality control

Indirect costs

Possible Allocation Basis
Space occupied
Space, usage, number of outlets
Space, direct labor hours, horsepower-hours, machine hours
Cost of materials, number of orders, number of items
Direct labor hours, direct labor cost
Space occupied, direct labor cost
Number of accesses
Number of inspections

Indirect costs are costs associated with property taxes, service and maintenance departments, personnel, legal, quality, supervision, purchasing, utilities, software development,
etc
...
Detailed collection of these data is
cost-prohibitive and often impossible; thus, allocation schemes are utilized to distribute the
expenses on a reasonable basis
...
Historically, common bases have
been direct labor cost, direct labor hours, machine-hours, number of employees, space, and direct
materials
...

estimated total indirect costs
Indirect cost rate ‫————————————— ؍‬
estimated basis level

[15
...
For example, if a division has
two producing departments, the total indirect cost allocated to a department is used as the numerator in Equation [15
...
Example 15
...

EXAMPLE 15
...
The following information was obtained from last year’s budget for the three machines
...

Cost Source

Allocation Basis

Machine 1
Machine 2
Machine 3

Direct labor cost
Direct labor hours
Direct material cost

Estimated Activity Level
$100,000
2000 hours
$250,000

Solution
Applying Equation [15
...
50 per direct labor dollar
indirect budget
50,000
Machine 2 rate ϭ ———————— ϭ ———
2000
direct labor hours
ϭ $25 per direct labor hour
indirect budget
50,000
Machine 3 rate ϭ ——————— ϭ ————
material cost
250,000
ϭ $0
...
6

Traditional Indirect Cost Rates and Allocation

Now the actual direct labor costs and hours and material costs are determined for this
year, and each dollar of direct labor cost spent on machine 1 implies that $0
...
Similarly, indirect costs are added for machines 2 and 3
...
For example, if direct materials are the basis for allocation to
four separate processing lines, the blanket rate is
total indirect costs
Indirect cost rate ϭ ———————————
total direct materials cost
If $500,000 in indirect costs and $3 million in materials are estimated for the four lines, the blanket indirect rate is 500,000͞3,000,000 ϭ $0
...
Blanket rates are easier
to calculate and apply, but they do not account for differences in the type of activities accomplished in each cost center
...
For example, light machinery may contribute less per hour than heavy, more expensive machinery
...
The use of blanket rates in these cases is not recommended, as the
indirect cost will be incorrectly allocated
...
The approach should be the application of different bases for
different machines, activities, etc
...
8
...
Realization that more
than one basis should be normally used in allocating indirect costs has led to the use of activitybased costing methods, as discussed in the next section
...
This results in the total
cost of production, which is called the cost of goods sold, or factory cost
...

If the total indirect cost budget is correct, the indirect costs charged to all cost centers for the
period of time should equal this budget amount
...
Experience in indirect cost estimation assists in reducing the variance at the
end of the accounting period
...
9
Since the manager determined indirect cost rates for BestWay (Example 15
...
Perform the computations using the data in
Table 15–8
...

Solution
Start with the cost of goods sold (factory cost) relation given by Equation [B
...
8 are applied:
Machine 1 indirect ϭ (labor cost)(rate) ϭ 2500(0
...
00) ϭ $18,750
Machine 3 indirect ϭ (material cost)(rate) ϭ 19,550(0
...

Based on the annual indirect cost budget of $150,000, one month represents 1͞12 of the
total or
150,000
Monthly budget ϭ ————
12
ϭ $12,500
The allocation variance for total indirect cost is
Variance ϭ 12,500 Ϫ 23,910 ϭ $Ϫ11,410
This is a large budget underallocation, since much more was actually charged than allocated
...
3% underallocation of indirect
costs
...

Once estimates of indirect costs are determined, it is possible to perform an economic analysis of
the present operation versus a proposed operation
...
10
...
10
For several years, Cuisinart Corporation has purchased the carafe assembly of its major coffeemaker line at an annual cost of $2
...
The suggestion to make the component in-house
has been made
...
The allocated hours column is the time
necessary to produce the carafes for a year
...
Perform an economic analysis for the make alternative,
assuming that a market rate of 15% per year is the MARR
...
Use the data of Table 15–9 to calculate the indirect cost allocation
...
10
Indirect Costs

Department

Basis,
Hours

Rate
per Hour, $

Allocated
Hours

Material
Cost, $

Direct
Labor
Cost, $

A
B
C

Labor
Machine
Labor

10
5
15

25,000
25,000
10,000

200,000
50,000
50,000

200,000
200,000
100,000

300,000

500,000

15
...

15
...
Where
once as much as 35% to 45% of the final product cost was represented in labor, now the
labor component is commonly 5% to 15% of total manufacturing cost
...
The use of
bases, such as direct labor hours, to allocate indirect cost is not accurate enough for automated and technologically advanced environments
...
8]
...

A product that by traditional methods may have contributed a large portion to proﬁt may actually be a loser when indirect costs are allocated more correctly
...
This may indicate that they are
proﬁtable, when in actuality they are losing money
...
It
is designed to identify cost centers, activities, and cost drivers
...

Cost centers: The ﬁnal products or services of the corporation are called cost centers or cost
pools
...

Activities: These are usually support departments (purchasing, quality, IT, maintenance, engineering, supervision) that generate the indirect costs which are then distributed to the cost
centers
...
Examples are the number of purchase orders, cost of engineering change orders,
number of machine setups, number of safety violations, and the like
...

1
...

2
...

3
...

total cost of activity
ABC indirect cost rate ‫———————————— ؍‬
total volume of cost driver

[15
...
Use the rate to allocate indirect cost to cost centers for each activity
...
The allocation
for costs generated by the purchasing department, for example, is based on the number of purchase orders (step 2) to support laser production
...

401

402

Chapter 15

Cost Estimation and Indirect Cost Allocation

EXAMPLE 15
...
However, accounts such as business travel
have historically been allocated on the basis of the number of employees at the plants in France,
Italy, Germany, and Spain
...
The ABC system is chosen to augment the traditional method to more precisely
allocate travel costs to major product lines at each plant
...
If total employment of 29,100 is
distributed as follows, allocate the $500,000
...
The ABC method will be
applied to allocate travel costs to major product lines
...
5% of $1 million
30% of $500,000

Further, the study indicates that in 1 year a total of 500 travel vouchers were processed by
the management of the major ﬁve product lines produced at the four plants
...

Product lines—1, 3, and 5; vouchers—80 for line 1, 30 for 3, 30 for 5
...

Product line—5; vouchers—140 for line 5
...

Solution
(a) In this case, Equation [15
...

travel budget
Indirect cost rate ϭ ———————
total workforce
$500,000
ϭ ———— ϭ $17
...

Paris
Florence
Hamburg
Barcelona

$17
...
Use the 4-step procedure to allocate travel costs to the ﬁve products
...
The total amount to be allocated is determined from the percentages of each
plant’s support budget devoted to travel
...
05(2,000,000) ϩ
...
30(500,000) ϭ $500,000

15
...
11
Product Line
1

Paris
Florence
Hamburg
Barcelona
Total

2

50
80
100

25

$230

$50

3

4

30
25

5
30

20
$30

$20

140
$170

Total
75
140
145
140
$500

Step 2
...
The allocation will be to
the products directly, not to the plants
...

Step 3
...
9] determines an ABC allocation rate
...
Table 15–10 summarizes the vouchers and allocation by product line and by city
...
Comparison of the by-plant totals in Table 15–10 (far right
column) with the respective totals in part (a) indicates a substantial difference in
the amounts allocated, especially to Paris, Hamburg, and Barcelona
...

Comment
Let’s assume that product 1 has been produced in small lots at the Hamburg plant for a number
of years
...
In the ABC analysis, Hamburg has a total of $145,000 travel dollars
allocated, $100,000 from product 1
...
This indicates to management the need to examine the manufacturing lot size practices at Hamburg and
possibly other plants, especially when a product is currently manufactured at more than one plant
...
This is not a good approach, since ABC is
not a complete cost system
...
The two systems
work well together with the traditional methods allocating costs that have identiﬁable direct
bases, for example, direct labor
...

15
...
Public agencies,
private corporations, and not-for-proﬁt businesses all make economic decisions based on these

403

404

Chapter 15

Cost Estimation and Indirect Cost Allocation

estimates, most of which are made by employees of the organizations or by outside consultants
hired to perform speciﬁc activities under contract
...
The personal morals and adherence to codes of professional ethics discussed in Section 1
...

The NSPE Code of Ethics for Engineers (Appendix C) referenced previously starts with a list
of six Fundamental Canons
...
”
Acts that bias the results from experimental samples, previous cost data, or survey results for the
purposes of personal gain, increased proﬁts, or favoritism are examples of unethical behavior
...

Use accepted theory and techniques in taking statistical samples, building budget elements,
and drawing conclusions that are included in proposals, applications, and recommendations
...

The second case study at the end of this chapter presents an example of some ethical challenges
present when preparing estimates and proposals for contract work
...
There are bottom-up and
top-down approaches; each treats price and cost estimates differently
...
The Consumer Price Index (CPI) is an often-quoted example of cost indexing
...
Two of them are
Cost-capacity equation—good for estimating costs from design variables for equipment, materials, and construction
Factor method—good for estimating total plant cost
Traditional cost allocation uses an indirect cost rate determined for a machine, department, product line, etc
...

With increased automation and information technology, different techniques of indirect cost allocation have been developed
...

The ABC method allocates indirect costs on the rationale that purchase orders, inspections,
machine setups, reworks, etc
...
Improved understanding of how the company
or plant accumulates indirect costs is a major by-product of implementing the ABC method
...
1 Rank the following estimate types in terms of time
spent to carry out the estimate (most time to least
time): partially designed, design 60% to 100%
complete, order of magnitude, scoping͞feasibility,
detailed estimate
...
2 Classify the following cost elements as ﬁrst cost
(FC) components or annual operating cost (AOC)

components for a piece of equipment on the
shop floor: supplies, insurance, equipment cost,
utility cost, installation, delivery charges, labor
cost
...
3 State whether actual (A) or estimated (E) costs are
more likely to be used to carry out the following
activities: calculate taxes, make bids, pay bonuses,
determine proﬁt or loss, predict sales, set prices,

405

Problems

evaluate proposals, distribute resources, plan production, and set goals
...
4 Identify the output and input variables in both the
bottom-up and top-down approaches to cost estimating
...
5 Classify the following costs as typically direct (D)
or indirect (I):
Project staff
Utilities
Raw materials
Project supplies
Administrative staff

Audit and legal
Rent
Equipment training
Labor
Miscellaneous ofﬁce supplies

15
...

15
...
8 Use the unit cost method to determine the preliminary cost of a guardrail for a small bridge if a total
of 120 linear feet will be required at a cost of
$58
...

15
...
Prepare a preliminary
cost estimate for the garage if the cost per parking
space is $4700
...
10 The Ofﬁce of the Undersecretary of Defense periodically releases unit cost data for use in military
construction programs
...
11 The Department of Defense uses area cost factors
(ACFs) to adjust for differences in construction costs
in different parts of the country (and world)
...
70 while the ACF for Rapid City, South Dakota, is
0
...
If a cold storage processing warehouse costs
$1,350,000 in Rapid City, estimate the cost for
Andros Island
...
12 Preliminary cost estimates for jails can be made
using costs based on either unit area (square feet)
or unit volume (cubic feet)
...
13 The unit area and unit volume total project costs
for a library are $114 per square foot and $7
...
Based on these numbers,
what is the average height of library rooms?
15
...
If the vibrators cost $76 per day and the concrete pump costs
$580 per day, estimate (a) the cost of the equipment
per cubic yard of concrete and (b) the equipment
cost for placing 56 cubic yards of concrete
...
15 A labor crew for placing concrete consists of
1 labor foreman at $25
...
60 per hour, 5 laborers at $23
...
45 per
hour
...
Determine
(a) the cost per day of labor for the C20 crew, (b) the
cost of the C20 crew per cubic yard of concrete, and
(c) the cost to place 250 cubic yards of concrete
...
16 Site work activities associated with constructing a
small bridge are shown in the table below
...
Use the data to determine (a) the total cost for structural excavation, (b)
the total cost for the pile-driving rig, and (c) the
total labor cost for the site work
...
35

1
...
31

5
...
78

1
...
13

2
...
57

Pile-driving rig
Piling, steel,
driving

Legend: cy ϭ cubic yard; ls ϭ lump sum; lf ϭ linear foot

Cost Indexes
15
...
If the state index for Texas is
76
...
5, estimate the
total construction cost of a middle school for 800
students in each state
...
18 A consulting engineering ﬁrm is preparing a preliminary cost estimate for a design-construct project of a coal processing plant
...
Use the
values in Table 15–3 to estimate the cost of construction for a similar-size plant in mid-2010
...
19 If the editors at ENR decide to redo the construction cost index so that the year 2000 has a base
value of 100, determine the value for the year (a)
1995 and (b) 2009
...
20 (a) Estimate the value of the ENR construction
cost index by using the average (compounded)
percentage change in its value between 1995
and 2005 to predict the value in 2009
...
21 An engineer who owns a construction company
that specializes in large commercial projects noticed that material costs increased at a rate of 1%
per month over the past 12 months
...

15
...
35
...
38
...
23 Electropneumatic general-purpose pressure transducers convert supply pressure to regulated output
pressure in direct proportion to an electrical input
signal
...
6
...
3?
15
...
40 in February 2007
...
32 and 4874
...
If a general
contractor in Atlanta won construction jobs totaling $54
...

Cost Estimation and Indirect Cost Allocation

15
...
37 when the base year
was 1913 with a value of 100
...
0, the CCI for August 2010
would be 8
...
In this case, what is the CCI
value for 1967 when the base year is 1913?
15
...
51 in August 2010
...
55 per ton
...
27 A contractor purchased equipment costing $40,000
in 2010 when the M&S equipment cost index was
at 1461
...
He remembers purchasing the same
equipment for $21,771 many years ago, but he does
not remember the year that he did so
...
68% per year
over that time period and the equipment increased
in price exactly in proportion to the index, (a) in
what year did he purchase the equipment and (b)
what was the value of the index in that year?
Cost-Capacity and Factor Methods
15
...
29 Use the exponent values in Table 15–6 to estimate
the cost for the following equipment to be placed
on an offshore drilling platform
...

(b) The cost of a 1700-gallon stainless steel tank
if a 900-gallon tank costs $4100
...
30 A high-pressure stainless steel pump (1000 psi)
with a variable-frequency drive (VFD) is installed
in a seawater reverse-osmosis pilot plant that is recovering water from membrane concentrate at a
rate of 4 gallons per minute (gpm)
...
Because of favorable results
from the pilot study, the city utility wants to go with
a full-scale system that will produce 500 gpm
...
37
...
31 A 0
...
Estimate
the cost of a 2-MGD tower if the exponent in the
cost-capacity equation is 0
...

Problems

15
...
Make the best cost estimate possible for
the VFD for a 100-hp motor
...

15
...
52 times the cost of the 30 m2 unit
...
34 Reinforced concrete pipe (RCP) that is 12 inches
in diameter had a cost of $12
...
The cost for 24-inch RCP
was $27
...
If the cross-sectional area of
the pipe is considered the “capacity” in the costcapacity equation, determine the value of the exponent in the cost-capacity equation that exactly
relates the two pipe sizes
...
35 A 100,000 barrel per day (bpd) fractionation tower
cost $1
...
3
...
8, provided the exponent in the
cost-capacity equation is 0
...
36 A mini wind tunnel for calibrating vane or hotwire anemometers cost $3750 in 2002 when the
M&S equipment index value was 1104
...
If the
index value is now 1620
...
The cost-capacity equation
exponent is 0
...

15
...

The engineer used the M&S equipment index for
the years 1998 and 2008 and the cost-capacity
equation with an exponent value of 0
...
If the
original equipment had only one-fourth the capacity of the new equipment, what was the cost of the
original equipment in 1998?
15
...
For example, a SRR of 2
...
942
...
942C1 when Q2͞Q1 ϭ 2
...

15
...
8 million
...
25, what is the total plant
cost expected to be?

407

15
...
)
while maximizing oriﬁce life and machine performance
...
32 million, what is the overall
cost factor for the system?
15
...
If the direct cost factor is 1
...
45 (applies to equipment
only), determine the expected cost of the laboratory
...
42 The equipment cost for a 10 gallon per minute
farm-scale ethanol fuel production plant is
$243,000
...
28 and for installation is 0
...
The indirect cost
factor for licenses, insurance, etc
...
84 (applied
to total direct cost)
...

15
...
3 million
...
35 and the indirect cost factor is
0
...

15
...
If the
direct cost factor is 3
...
38, what is the estimated total plant cost? The
indirect cost factor applies to the total direct cost
...
45 Nicole is an engineer on temporary assignment at
a reﬁnery operation in Seaside
...

The equipment itself is estimated at $250,000 with
a construction cost factor of 0
...
30
...
(a) If the indirect cost factor
should be 0
...
40 indirect
cost factor is used
...
46 The company you work for currently allocates insurance costs on the basis of cost per direct labor
hour
...
If the direct labor hours for

408

Chapter 15

Cost Estimation and Indirect Cost Allocation

departments A, B, and C are expected to be 2000,
8000, and 5000, respectively, this year, determine
the allocation to each department
...
However, the accountant advises the manager to not
be concerned because the allocation rates have
decreased each month
...
47 The director of public works needs to distribute the
indirect cost allocation of $1
...
The recorded amounts
for this year are as follows:

Indirect Cost, $
Month

Branch

Miles Driven

Direct Labor
Hours

275,000
247,000
395,000

38,000
31,000
55,500

North
South
Midtown

Branch

Direct
Labor
Hours

Basis

Indirect Cost
Allocation
Last Year, $

North
South
Midtown

350,000
200,000
500,000

40,000
20,000
64,000

Miles
Labor
Labor

300,000
200,000
450,000

(a)
(b)

Determine the rates for this year for each
branch
...
What percentage of this year’s
budget is now distributed?

15
...
Because of the nature and use of three of
these stations, each is considered a separate cost
center for indirect cost allocation
...
Machine
operating hours are used as the allocation basis for
all machines
...
Use the data collected
this year to determine the indirect cost rate for
each center
...
40
1
...
37
1
...
92

2800
3400
3500
3600
6000

2600
3800
3500

During the evaluation, the following additional
information from departmental and accounting
records is obtained
...
This information follows:
Miles
Driven

Allocated

February
March
April
May
June

Records for This Year

Rate

Direct Labor
Hours

Month
February
March
April
May
June

(a)
(b)

Cost, $

Material
Cost, $

Departmental
Space, ft2

640
640
640
640
800

2560
2560
2560
2720
3320

5400
4600
5700
6300
6500

2000
2000
3500
3500
3500

With this information determine the allocation basis used each month
...

15
...
9 million
this year
...
Your manager asks
that both direct and indirect costs be included
when in-house manufacturing (make alternative)
is evaluated
...
5 million and a life
of 6 years
...
5 million per year
...
Perform the AW evaluation at
MARR ϭ 12% per year over a 6-year study period
...

Department

15
...
The manager has obtained records of allocation rates and actual charges for
the prior 3 months and estimates for this month
(May) and next month (see the table)
...
40 per $
$0
...
51 The municipal water and desalinization utility in a
California city currently allocates some costs for
maintenance shop workers to pumping stations
based on the number of pumps at each station
...
Information about the stations is below
...

(a) Allocate the budget to each station based on
the number of service trips
...

Station ID

No
...
52 Factory Direct manufactures and sells manufactured
homes
...
Each plant builds different models and ﬂoor
plans
...
Because of these advances, the CFO plans to use build-time per unit as
the new basis
...
The data shown represents average costs and times
...

Plant

Texas

Oklahoma

20,000

12,700

18,600

400

415
355

Allocation, $

DFW
YYZ
MEX

55,000
20,833
15,000

6
9
10

330,000
187,500
150,000

The airline’s baggage management director suggests that
an allocation on the basis of baggage trafﬁc, not ﬂights,
will better represent the distribution, primarily based on
the fact that the high fees now charged to passengers to
check luggage have signiﬁcantly changed the number of
bags handled at the major hubs
...

15
...
54 Using the baggage-trafﬁc basis, determine the allocation rate using last year’s total of $667,500,
and distribute this amount to the hubs this year
...
55 What are the percentage changes in allocation at
each hub using the two different bases?
15
...
For this year, in round
numbers, the budgets and allocation of $1 million
advertising indirect costs are as follows:
Site
A
Budget, $
Allocation, $

(a)

B

C

D

2 million
200,000

3 million
300,000

4 million
400,000

1 million
100,000

Determine the allocation if the ABC method is
used with a new basis
...
The cost
driver is the number of guests during the year
...
The average number of lodging
nights for guests at each site is as follows:
Site
A

Problems 15
...
55 use the following
information
...
Last year $667,500 was distributed as follows:

Length of stay, nights

(c)

B

C

D

3
...
5

1
...
75

Comment on the distribution of advertising
costs using the two methods
...

410

Chapter 15

Cost Estimation and Indirect Cost Allocation

ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
15
...

(b) Cost estimates are an output variable
...

(d) Both (a) and (b) are correct
...
58 A ratio of the cost of something today to its cost at
some time in the past is called a:
(a) Cost-capacity index
(b) Cost index
(c) Buyer’s guide
(d) Bluebook index
15
...
60 A 50-hp turbine pump was purchased for $2100
...
76, a 200-hp turbine pump could be expected to cost about:
(a) $6020
(b) $5320
(c) $4890
(d) $4260

100,000 units per day was $3 million, the value of
the exponent in the cost-capacity equation is:
(a) 0
...
39
(c) 0
...
60
15
...
If the overall cost factor
for the complete system is 2
...
65 The delivered-equipment cost for setting up a
production and assembly line for high-sensitivity,
gas-damped accelerometers is $650,000
...
82 and
0
...
61 The city built a recreation park in 1980 for
$500,000
...
17 at that time
...
16, the estimated
cost is closest to:
(a) $695,800
(b) $750,700
(c) $820,300
(d) $910,500

15
...
An allocation basis that may not be
reasonable is:
(a) Miles of toll road monitored
(b) Average number of cars patrolling per hour
(c) Amount of car trafﬁc per section of toll
road
(d) Cost to operate a patrol car

15
...
9
...
32 and the M&S index
value was 1449
...
67 The IT department allocates indirect costs to user
departments on the basis of CPU time at the rate of
$2000 per second
...
If the IT indirect budget for the
year is $8
...
5%
(c) 55%
(d) Not enough information to determine

15
...
If the cost for a plant with a capacity of

15
...

2
...

(a)
(b)
(c)
(d)

15
...

Cost of engineering changes processed
Size of the workforce
Administrative cost to process a change
order
1
2
3
1 and 3

2
...

(a)
(b)
(c)
(d)

It is an excellent replacement for a traditional
cost accounting system
...

It can help explain the economic impact of
management decisions
...
, initiated the manufacture and
sales of a portable sterilization unit (Quik-Sterz) that can be
placed in the hospital room of a patient
...
This new unit
makes the instruments available at the point and time of use
for burn and severe wound patients who are in a regular patient room
...
The standard
version sells for $10
...
75
...

Information
Medical Dynamics has historically used an indirect cost allocation system based upon direct hours to manufacture for

TA BLE 15–11

all its other product lines
...
However, Arnie, the person who performed
the indirect cost analysis and set the sales price, is no longer
at the company, and the detailed analysis is no longer available
...
A search of
these ﬁles revealed the manufacturing and cost information in
Table 15–11
...
73 per unit for the standard model and $27
...

Last year management decided to place the entire plant on
the ABC system of indirect cost allocation
...
Five activities and their cost drivers were identiﬁed for the Medical Dynamics manufacturing operations
(Table 15–12)
...

Historical Records of Direct and Indirect Cost Analyses for Quik-Sterz
Quik-Sterz Direct Cost (DC) Evaluation

Model
Standard
Premium

Direct Labor,
$͞Unit1

Direct Material,
$͞Unit

Direct Labor,
Hours͞Unit

Total Direct
Labor Hours

5
...
00

2
...
75

0
...
50

187,500
125,000

Allocated
IDC, $

Sales,
Units/Year

1
...
33 million

250,000

Quik-Sterz Indirect Cost (IDC) Evaluation
Model

Direct Labor,
Hours͞Unit

Standard

0
...
50

1

Average direct labor rate is $20 per hour
...
The
ﬁrst impression of the production manager is that the new system will show that indirect costs for Quik-Sterz are about the
same as they have been for other products over the last several
years when a standard model and an upgrade (premium) model
were sold
...
Fundamentally, there are two reasons
why the production manager does not like to produce premium
versions: They are less proﬁtable for the company, and they require signiﬁcantly more time and operations to manufacture
...
Use traditional indirect cost allocation to verify Arnie’s
cost and price estimates
...
Use the ABC method to estimate the indirect cost allocation and total cost for each model
...
If the prices and number of units sold are the same
next year (750,000 standard and 250,000 premium),
and all other costs remain constant, compare the
profit from Quik-Sterz under the ABC method with
the profit using the traditional indirect cost allocation method
...
What prices should Medical Dynamics charge next year
based on the ABC method and a 10% markup over cost?
What is the total proﬁt from Quik-Sterz predicted to be
if sales hold steady?
5
...

CASE STUDY
DECEPTIVE ACTS CAN GET YOU IN TROUBLE
Contributed by Dr
...

The law requires that land disturbed by these types of mining
activities be returned to a productive capacity that is as good
as or better than its productive capacity before mining
...
To this end,
mining companies must sample the different soils found in
the areas to be disturbed by mining activities
...

Soils in a low pH range (pH values Ͻ 5) are indicative of
low fertility
...

To meet the baseline for pH, the acreage of the mine soils
with low pH should not exceed the acreage of the unmined
soils with low pH
...
The different soils
within the 600 acres were depicted in the County Soil Survey
where the mining activities were to take place
...
Assessment of the data
indicated that 30% of the area (180 acres) occupied by the
soils in the area to be disturbed had pH values between 4
...
9
...

Six years later, 450 acres had been mined and the terrain
surface had been leveled to reestablish premine slopes
...
0 and
4
...
The president of Yucatan indicated that the company
would submit a revised soil baseline based on new sampling
in the remaining 150 acres of unmined soils because, in his
opinion, the ﬁrst soil baseline was biased
...
The company quickly hired
a consultant to develop the new baseline, and subsequently
Yucatan submitted the ﬁnal report from its consultant to the
State Department of Mining and Reclamation
...
0 and 4
...
Comparative results between the old and new samples can be expressed as follows:
Percent and Acreage of Area
Soil Baselines

Old Soil Baseline

Revised Soil Baseline

pH: 4
...
9

30%
180 acres

45%
270 acres

A rough statistical check between the old and revised soil
baselines indicated that the results were mixed
...
Contained in the submitted new-sample package
was a letter from the Yucatan consultant
...

413

The State Department of Mining and Reclamation staff
concluded that the revised soil pH sample data had been carefully “screened” to reduce the amount of remediation work
that Yucatan Mining would have to complete
...
It was also noted
that should Yucatan Mining disagree with this response, the
case would be ﬁled with the legal staff as a contested case
...

Case Study Questions
1
...
What
actions would you direct your staff to take concerning
this situation?
2
...
What type of
“audit” procedures might you want implemented to
identify these possibly unethical activities?
3
...
When the 30 lowest pH samples were used to establish the new baseline, were the samples still random,
according to experimental design standards? If so, why?
If not, why not?
4
...
You have golfed together
several times, your and his children are on the same soccer team at school, and your families are members of
the same community swimming pool club
...
As a matter of principle and practice, do you believe
there is some amount of data-altering or bias-making
that is allowed before an application (such as the one
described here) should be considered the result of professionally unethical acts? How would you deﬁne such
a threshold limit?

CHAPTER 16

Depreciation
Methods

L E A R N I N G

O U T C O M E S

Purpose: Use depreciation or depletion methods to reduce the book value of a capital investment in an asset and natural
resource
...
1

Terminology

• Deﬁne and use the basic terms of asset
depreciation
...
2

Straight line

• Apply the straight line (SL) method of
depreciation
...
3

Declining balance

• Apply the declining balance (DB) and double
declining balance (DDB) methods of
depreciation
...
4

MACRS

• Apply the modiﬁed accelerated cost recovery
system for tax depreciation purposes for U
...
based corporations
...
5

Recovery period

• Select the asset recovery period for MACRS
depreciation
...
6

Depletion

• Explain depletion; apply cost depletion and
percentage depletion methods
...
1

Historical methods

• Apply the sum-of-years-digits (SYD) and unitof-production (UOP) methods of depreciation
...
2

Switching

• Switch between classical depreciation methods;
explain how MACRS provides for switching
...
3

MACRS and switching

• Calculate MACRS rates using switching between
classical methods and MACRS rules
...
Although the depreciation amount is
not an actual cash ﬂow, the process of depreciating an asset on the books of the corporation accounts for the decrease in an asset’s value because of age, wear, and obsolescence
...
An introduction to depreciation types, terminology, and classical methods is followed by a discussion
of the Modiﬁed Accelerated Cost Recovery System (MACRS), which is the standard in the
United States for tax purposes
...

Why is depreciation important to engineering economy? Depreciation is a tax-allowed
deduction included in tax calculations in virtually all industrialized countries
...

This chapter concludes with an introduction to two methods of depletion, which are used
to recover capital investments in deposits of natural resources such as oil, gas, minerals, ores,
and timber
...
Additionally, the appendix includes an in-depth
derivation of the MACRS depreciation rates from the straight line and declining balance
rates
...

16
...
Most descriptions are applicable to corporations as well as individuals who own depreciable assets
...

The method used to depreciate an asset is a way to account for the decreasing value of the asset
to the owner and to represent the diminishing value (amount) of the capital funds invested in it
...

Though the term amortization is sometimes used interchangeably with the term depreciation,
they are different
...

In addition, the term capital recovery is sometimes used to identify depreciation
...
The term depreciation is used
throughout this book
...

Tax depreciation Used by a corporation or business to determine taxes due based on current tax laws of the government entity (country, state, province, etc
...

The methods applied for these two purposes may or may not utilize the same formulas
...
There are classical, internationally accepted depreciation methods used to determine book depreciation: straight line, declining balance, and the historical

416

Chapter 16

Depreciation Methods

sum-of-years-digits method
...

In most industrialized countries, the annual tax depreciation is tax-deductible; that is, it is subtracted from income when calculating the amount of taxes due each year
...

Tax depreciation may be calculated and referred to differently in countries outside the United
States
...

Where allowed, tax depreciation is usually based on an accelerated method, whereby the
depreciation for the ﬁrst years of use is larger than that for later years
...
In effect, accelerated methods defer
some of the income tax burden to later in the asset’s life; they do not reduce the total tax
burden
...

First cost P or unadjusted basis B is the delivered and installed cost of the asset including
purchase price, delivery and installation fees, and other depreciable direct costs incurred to
prepare the asset for use
...
When the
ﬁrst cost has no added, depreciable costs, the basis is the ﬁrst cost, that is, P ϭ B
...
The book
value is determined at the end of each year t (t ϭ 1, 2,
...

Recovery period n is the depreciable life of the asset in years
...
Both of these values may be different from the asset’s
estimated productive life
...
Because of the structure of depreciation laws,
the book value and market value may be substantially different
...
However, a computer workstation may have a market value much lower
than its book value due to rapidly changing technology
...

The salvage value, expressed as an estimated dollar amount or as a percentage of the ﬁrst cost,
may be positive, zero, or negative due to dismantling and carry-away costs
...
This rate may be the same each year, which is called the straight line rate d, or
different for each year of the recovery period
...

Included is most manufacturing and service industry property—vehicles, manufacturing
equipment, materials handling devices, computers and networking equipment, communications equipment, ofﬁce furniture, reﬁning process equipment, construction assets, and
much more
...
Land itself is considered real property, but it is not depreciable
...
This convention is utilized in
this text and in most U
...
-approved tax depreciation methods
...

16
...

Book value, $

SL

MACRS

DB

Estimated
salvage
value

Time, years

Recovery period

As mentioned before, there are several models for depreciating assets
...
Accelerated models, such as the declining balance (DB) method, decrease the book value to zero (or to the salvage value) more rapidly than
the straight line method, as shown by the general book value curves in Figure 16–1
...
Each function is introduced and illustrated as the method is explained
...
One
that may be of interest to a U
...
-based small or medium-sized business performing an economic
analysis is the Section 179 Deduction
...
Up to a
speciﬁed amount, the entire basis of an asset is treated as a business expense in the year of purchase
...
The limit changes with time; it was
$24,000 in 2002; $102,000 in 2004; $125,000 in 2007; and $250,000 in 2008–2010
...
Investments above these limits must be depreciated using MACRS
...
S
...
In 1981, all classical methods, including straight line, declining balance, and sum-of-yearsdigits depreciation, were disallowed as tax deductible and replaced by the Accelerated Cost Recovery System (ACRS)
...
To this date, the following is the law in the United States
...

MACRS has the DB and SL methods, in slightly different forms, embedded in it, but these two
methods cannot be used directly if the annual depreciation is to be tax deductible
...
S
...
Most other countries still recognize the classical methods of straight line and declining balance for tax or book purposes
...
Appendix Section 16A
...

418

Chapter 16

Depreciation Methods

Tax law revisions occur often, and depreciation rules are changed from time to time in the
United States and other countries
...
S
...
irs
...
Pertinent publications can be downloaded
...
MACRS and most corporate tax depreciation laws are discussed in it
...
2 Straight Line (SL) Depreciation
Straight line depreciation derives its name from the fact that the book value decreases linearly
with time
...

Straight line depreciation is considered the standard against which any depreciation model is
compared
...
For tax depreciation, as mentioned earlier, it is not used directly in the United States, but it is commonly used in most countries for tax purposes
...
S
...
5)
...
In equation form,
Dt ‫( ؍‬B ؊ S)dt
B؊S
‫——— ؍‬
n

[16
...
, n)
Dt ϭ annual depreciation charge
B ϭ ﬁrst cost or unadjusted basis
S ϭ estimated salvage value
n ϭ recovery period
dt ϭ depreciation rate ϭ 1͞n
Book Value

B

Since the asset is depreciated by the same amount each year, the book value after t years of service, denoted by BVt, will be equal to the ﬁrst cost B minus the annual depreciation times t
...
2]

Earlier we deﬁned dt as a depreciation rate for a speciﬁc year t
...
3]
d ϭ dt ϭ —
n
The format for the spreadsheet function to display the annual depreciation Dt in a single-cell
operation is
‫ ؍‬SLN(B, S, n)

[16
...
1
If an asset has a ﬁrst cost of $50,000 with a $10,000 estimated salvage value after 5 years,
(a) calculate the annual depreciation and (b) calculate and plot the book value of the asset after
each year, using straight line depreciation
...
1]
...

16
...
1
...
2]
...
For years 1 and 5, for example,
BV1 ϭ 50,000 Ϫ 1(8000) ϭ $42,000
BV5 ϭ 50,000 Ϫ 5(8000) ϭ $10,000 ϭ S

16
...
5]

In this case the method is called double declining balance (DDB)
...
2; so 20% of the book value is removed annually
...
5͞n
...

Dt ‫( ؍‬d)BVt؊1

[16
...
7]

If BVtϪ1 is not known, the depreciation in year t can be calculated using B and d
...
8]

Book value in year t is determined in one of two ways: by using the rate d and basis B or by
subtracting the current depreciation charge from the previous book value
...
9]
[16
...
Like the
SL method, DB is embedded in the MACRS method, but the DB method itself cannot be used to
determine the annual tax-deductible depreciation in the United States
...

Declining balance is also known as the ﬁxed percentage or uniform percentage method
...
If d ϭ 0
...
Therefore, the depreciation amount decreases each year
...
The implied salvage value after n
years is the BVn amount, that is,
Implied S ϭ BVn ϭ B(1 Ϫ d)n

[16
...
However, if the implied S Ͻ estimated S, it is
necessary to stop charging further depreciation when the book value is at or below the estimated
salvage value
...
(This
guideline is important when the DB method can be used directly for tax depreciation purposes
...
The range for d is 0 Ͻ d Ͻ 2͞n
...
12]

The spreadsheet functions DDB and DB are used to display depreciation amounts for speciﬁc
years
...
For the double declining balance method, the format is
‫ ؍‬DDB(B, S, n, t, d)

[16
...
If omitted, this optional
entry is assumed to be 2 for DDB
...
5 makes the DDB function display 150%
declining balance method amounts
...
No further depreciation is charged when this
occurs
...
11]
...
1]
...
Caution is needed when using this function
...
12]
...
Therefore, if the depreciation rate is known, always use the DDB function to ensure
correct results
...
2 and 16
...

EXAMPLE 16
...
The
equipment will be DDB depreciated over an expected life of 12 years
...
(a) Calculate the depreciation and book value for
years 1 and 4
...

(b) Calculate the implied salvage value after 12 years
...
1667 per year
...
8]
and [16
...

Year 1:
Year 4:

D1 ϭ (0
...
1667)1Ϫ1 ϭ $4167
BV1 ϭ 25,000(1 Ϫ 0
...
1667)(25,000)(1 Ϫ 0
...
1667)4 ϭ $12,054

The DDB functions for D1 and D4 are, respectively, ϭ DDB(25000,2500,12,1) and
ϭ DDB(25000,2500,12,4)
...
3

Declining Balance (DB) and Double Declining Balance (DDB) Depreciation

(b) From Equation [16
...
1667)12 ϭ $2803
Since the estimated S ϭ $2500 is less than $2803, the asset is not fully depreciated when
its 12-year expected life is reached
...
3
Freeport-McMoRan Copper and Gold has purchased a new ore grading unit for $80,000
...
Use the DB and DDB
methods to compare the schedule of depreciation and book values for each year
...

Solution by Hand
An implied DB depreciation rate is determined by Equation [16
...

(

10,000
d ϭ 1 Ϫ ———
80,000

)

1͞10

ϭ 0
...
1877 Ͻ 2͞n ϭ 0
...

Table 16–1 presents the Dt values using Equation [16
...
10] rounded to the nearest dollar
...
1877(64,984) ϭ $12,197
BV2 ϭ 64,984 Ϫ 12,197 ϭ $52,787
Because we round off to even dollars, $2312 is calculated for depreciation in year 10, but
$2318 is deducted to make BV10 ϭ S ϭ $10,000 exactly
...
2 result in the depreciation and book value series in Table 16–1
...
3
Declining Balance, $

Double Declining Balance, $

Year t

Dt

BVt

Dt

BVt

0
1
2
3
4
5
6
7
8
9
10

—
15,016
12,197
9,908
8,048
6,538
5,311
4,314
3,504
2,846
2,318

80,000
64,984
52,787
42,879
34,831
28,293
22,982
18,668
15,164
12,318
10,000

—
16,000
12,800
10,240
8,192
6,554
5,243
4,194
3,355
2,684
737

80,000
64,000
51,200
40,960
32,768
26,214
20,972
16,777
13,422
10,737
10,000

Solution by Spreadsheet
The spreadsheet in Figure 16–3 displays the results for the DB and DDB methods
...
Since the ﬁxed rates are close—0
...
2 for
DDB—the annual depreciation and book value series are approximately the same for the two
methods
...
12], but note in the cell tags
that the DDB function is used in both columns B and D to determine annual depreciation
...
12] and maintains it to only three signiﬁcant digits
...
1

ϭ DDB(B$2,B$3,B$4,$A18,10*$B$5)
ϭ DDB(B$2,B$3,B$4,$A18)

Figure 16–3
Annual depreciation and book value using DB and DDB methods, Example 16
...

were used in column B (Figure 16–3), the ﬁxed rate applied would be 0
...
The resulting
Dt and BVt values for years 8, 9, and 10 would be as follows:
t

D t, $

BVt, $

8
9
10

3,501
2,842
2,308

15,120
12,277
9,969

Also noteworthy is the fact that the DB function uses the implied rate without a check to halt
the book value at the estimated salvage value
...
However, the DDB function uses a relation different from that of the DB function to determine annual depreciation—one that correctly stops depreciating at the estimated
salvage value, as shown in Figure 16–3, cells E17–E18
...
4 Modiﬁed Accelerated Cost Recovery
System (MACRS)
In the 1980s, the United States introduced MACRS as the required tax depreciation method for
all depreciable assets
...
Corporations are free to apply
any of the classical methods for book depreciation
...
1
Many aspects of MACRS deal with the speciﬁc depreciation accounting aspects of tax law
...
Additional information on how the DDB, DB, and SL methods are embedded into MACRS and how
to derive the MACRS depreciation rates is presented and illustrated in the chapter appendix,
Sections 16A
...
3
...
14]

R
...

16
...
As for other methods, the book value
in year t is determined by subtracting the depreciation amount from the previous year’s book
value
BVt ϭ BVtϪ1 Ϫ Dt

[16
...

BVt ‫ ؍‬ﬁrst cost ؊ sum of accumulated depreciation
j‫؍‬t

‫؍‬B؊

͚D

[16
...

The basis B (or ﬁrst cost P) is completely depreciated; salvage is always assumed to be zero,
or S ϭ $0
...
g
...
5 or 39 years
for real property (e
...
, rental property or structures)
Depreciation rates provide accelerated write-off by incorporating switching between classical
methods
...
5 explains how to determine an allowable MACRS recovery period
...
14] are included in Table 16–2
...

The MACRS rates start with the DDB rate or the 150% DB rate and switch when the SL method
offers faster write-off
...
33
44
...
81
7
...
00
32
...
20
11
...
52

14
...
49
17
...
49
8
...
00
18
...
40
11
...
22

5
...
50
8
...
70
6
...
75
7
...
68
6
...
71

5
...
92
8
...
46

7
...
55
6
...
56
6
...
23
5
...
90
5
...
90

5
...
89
4
...
46
4
...
28

5
...
90
5
...
90
5
...
46
4
...
46
4
...
46

2
...
46
4
...
23

6
7
8
9
10
11
12
13
14
15
16
17–20
21

423

424

Chapter 16

Depreciation Methods

For real property, MACRS utilizes the SL method for n ϭ 39 throughout the recovery period
...
02564
...
The MACRS real property rates in percentage amounts are
Year 1
Years 2–39
Year 40

100d1 ϭ 1
...
564%
100d40 ϭ 1
...
5 years, which applies only to residential rental property,
uses the SL method in a similar fashion
...
Also note that the extra-year rate is one-half of the previous year’s rate
...
This convention assumes that all property is placed in service at the midpoint of the tax year of installation
...
This removes some of
the accelerated depreciation advantage and requires that one-half year of depreciation be taken in
year n ϩ 1
...
However, the
variable declining balance (VDB) function, which is used to determine when to switch between
classical methods, can be adapted to display MACRS deprecation for each year
...
2 of this chapter and Appendix A of the text
...
5), MIN(n, t؊0
...
17]

where B ϭ ﬁrst cost
0 ϭ salvage value of S ϭ 0
n ϭ recovery period
if MACRS n ϭ 3, 5, 7, or 10
dϭ 2
1
...

EXAMPLE 16
...
This chemical is a resin used in plastic pipe, retail bags, blow
molding, and injection molding
...
The chief engineer asked the ﬁnance director to provide an analysis of the difference between (1) the DDB method, which is the internal book depreciation and book value method used at the plant, and (2) the required
MACRS tax depreciation and its book value
...
Use hand and spreadsheet
solutions to do the following:
(a) Determine which method offers the larger total depreciation after 2 years
...

Solution by Hand

The basis is B ϭ $400,000 and the estimated S ϭ 0
...
The MACRS rates
for n ϭ 3 are taken from Table 16–2, and the depreciation rate for DDB is dmax ϭ 2͞3 ϭ
0
...
Table 16–3 presents the depreciation and book values
...
6667) ϭ $29,629, except this would make BV3 Ͻ $20,000
...

16
...
4
MACRS

DDB

Year

Rate

Tax
Depreciation, $

Book
Value, $

Book
Depreciation, $

0
1
2
3
4

0
...
4445
0
...
0741

133,320
177,800
59,240
29,640

400,000
266,680
88,880
29,640
0

266,667
88,889
24,444

Book
Value, $
400,000
133,333
44,444
20,000

(a) The 2-year accumulated depreciation values from Table 16–3 are
MACRS:
DDB:

D1 ϩ D2 ϭ $133,320 ϩ 177,800 ϭ $311,120
D1 ϩ D2 ϭ $266,667 ϩ 88,889 ϭ $355,556

The DDB depreciation is larger
...
)
(b) After 2 years the book value for DDB at $44,444 is 50% of the MACRS book value of $88,880
...
This
occurs because MACRS always removes the entire ﬁrst cost, regardless of the estimated salvage value
...
4)
...

(a) The 2-year accumulated depreciation values are
MACRS (add cells B6 ϩ B7):
DDB (add cells D6 ϩ D7):

$133,333 ϩ 177,778 ϭ $311,111
$266,667 ϩ 88,889 ϭ $355,556

(b) Book values after 2 years are
MACRS (cell C7):
DDB (cell E7):

$88,889
$44,444

The book values are plotted in Figure 16–4
...

Comment
It is advisable to set up a spreadsheet template for use with depreciation problems in this and
future chapters
...

Figure 16–4
Spreadsheet screen shot of MACRS and DDB depreciation and book value, Example 16
...

425

426

Chapter 16

Depreciation Methods

MACRS simpliﬁes depreciation computations, but it removes much of the ﬂexibility of
method selection for a business or corporation
...

16
...
For book depreciation the n value should be the
expected useful life
...
There are tables that assist in determining the life and recovery period
for tax purposes
...

The U
...
government requires that all depreciable property be classiﬁed into a property class
which identiﬁes its MACRS-allowed recovery period
...
Virtually any property
considered in an economic analysis has a MACRS n value of 3, 5, 7, 10, 15, or 20 years
...
The ﬁrst is the general depreciation system (GDS) value, which we use in examples and problems
...
The
rates utilize the DDB method or the 150% DB method with a switch to SL depreciation
...

The far right column of Table 16–4 lists the alternative depreciation system (ADS) recovery period range
...
5

Ofﬁce furniture; some manufacturing equipment;
railroad cars, engines, tracks; agricultural machinery; petroleum and natural gas equipment; all
property not in another class

7

10–15

Equipment for water transportation, petroleum reﬁning, agriculture product processing, durable-goods
manufacturing, shipbuilding

10

15–19

Land improvements, docks, roads, drainage, bridges,
landscaping, pipelines, nuclear power production
equipment, telephone distribution

15

20–24

Municipal sewers, farm buildings, telephone switching buildings, power production equipment (steam
and hydraulic), water utilities

20

25–50

Residential rental property (house, mobile home)

27
...
6

Depletion Methods

recovery period than the GDS
...
The use of ADS is generally a choice left to a company, but it is required
for some special asset situations
...
This electable SL option is, however, sometimes chosen
by businesses that are young and do not need the tax beneﬁt of accelerated depreciation during the
ﬁrst years of operation and asset ownership
...

16
...
We now turn to irreplaceable natural resources and the equivalent of depreciation, which
is called depletion
...
The two methods of depletion for book or tax purposes are
used to write off the ﬁrst cost, or value of the estimated quantity, of resources in mines, wells,
quarries, geothermal deposits, forests, and the like
...
Details for
U
...
taxes on depletion are found in IRS Publication 535, Business Expenses
...
Cost depletion may be applied to most types of
natural resources and must be applied to timber production
...

ﬁrst cost
CDt ‫———————— ؍‬
resource capacity

[16
...
The total cost depletion
cannot exceed the ﬁrst cost of the resource
...

Percentage depletion This is a special consideration given for natural resources
...
The depletion amount for year t is calculated as
Percentage depletion t ‫ ؍‬percentage depletion rate
؋ gross income from property
‫ ؍‬PD ؋ GIt

[16
...
The
U
...
government does not generally allow percentage depletion to be applied to oil and gas wells
(except small independent producers)
...
S
...

Deposit
Sulfur, uranium, lead, nickel, zinc,
and some other ores and minerals
Gold, silver, copper, iron ore, and
some oil shale
Oil and natural gas wells (varies)
Coal, lignite, sodium chloride
Gravel, sand, peat, some stones
Most other minerals, metallic ores

Percentage of
Gross Income, PD
22
15
15–22
10
5
14

427

428

Chapter 16

Depreciation Methods

EXAMPLE 16
...
An estimated 350 million board feet of lumber is harvestable
...

(b) After 2 years the total recoverable board feet was reestimated upward to be 450 million from
the time the rights were purchased
...

Solution
(a) Use Equation [16
...

700,000
CDt ϭ ———— ϭ $2000 per million board feet
350
Multiply CDt by the annual harvest to obtain depletion of $30,000 in year 1 and $44,000
in year 2
...

(b) After 2 years, a total of $74,000 has been depleted
...
Additionally, with the
new estimate of 450 million board feet, a total of 450 Ϫ 15 Ϫ 22 ϭ 413 million board feet
remains
...
, the cost depletion factor is
626,000
CDt ϭ ———— ϭ $1516 per million board feet
413

EXAMPLE 16
...
It has an anticipated gross income of $5
...
0 million per year after year 5
...
Compute annual depletion amounts for the mine
...
15
...
15(5
...
15(3
...
75 million is written off in 5 years, and the remaining $6
...
The total number of years is
$6
...
9 ϭ 18
...

In many of the natural resource depletion situations, the tax law allows the larger of the two
depletion amounts to be claimed each year
...
Therefore, it is wise to calculate both depletion
amounts and select the larger
...

CDAt ϭ cost depletion amount
PDAt ϭ percentage depletion amount
TIt ϭ taxable income
The guideline for the tax-allowed depletion amount for year t is

{

Depletion ϭ max[CDAt, PDAt]
max[CDAt, 50% of TIt]

if PDAt Յ 50% of TIt
if PDAt Ͼ 50% of TIt

Chapter Summary

For example, assume a medium-sized quarry owner calculates the following for 1 year
...
For tax purposes, apply the guideline above and use the cost depletion of $275,000, since it is larger than
50% of TI
...
In the United States, the MACRS method is the only
one allowed for tax depreciation
...
Depreciation does not result in cash
ﬂow directly
...
The annual depreciation amount is tax deductible, which can result in actual cash
ﬂow changes
...
Common relations for each method are summarized in Table 16–5
...

• The estimated salvage value is always considered
...

Declining Balance (DB)
• The method accelerates depreciation compared to the straight line method
...

• The most used rate is twice the SL rate, which is called double declining balance (DDB)
...

• It is not an approved tax depreciation method in the United States
...

Modiﬁed Accelerated Cost Recovery System (MACRS)
• It is the only approved tax depreciation system in the United States
...

• It always depreciates to zero; that is, it assumes S ϭ 0
...

• Depreciation rates are tabulated
...

• MACRS straight line depreciation is an option, but recovery periods are longer than those for
regular MACRS
...
The annual
cost depletion factor is applied to the amount of resource removed
...
Percentage depletion, which can recover more
than the initial investment, reduces the investment value by a constant percentage of gross income each year
...
1 Sum-of-Years-Digits (SYD) and Unit-ofProduction (UOP) Depreciation
The SYD method is a historical accelerated depreciation technique that removes much of the
basis in the ﬁrst one-third of the recovery period; however, write-off is not as rapid as for DDB
or MACRS
...

The mechanics of the method involve the sum of the year’s digits from 1 through the recovery
period n
...

depreciable years remaining
Dt ‫( ————————————— ؍‬basis ؊ salvage value)
sum of years digits
n؊t؉1
Dt ‫( ————— ؍‬B ؊ S)
SUM

[16A
...

jϭn

SUM ϭ

n(n ϩ 1)
͚ j ϭ ————
2
jϭ1

The book value for any year t is calculated as
t(n ؊ t͞2 ؉ 0
...
2]

The rate of depreciation decreases each year and equals the multiplier in Equation [16A
...

nϪtϩ1
dt ϭ ————
SUM

[16A
...
The function format is
ϭ SYD(B,S,n,t)

EXAMPLE 16A
...

Solution
The sum of the year’s digits is 36, and the depreciation amount for the second year by Equation [16A
...

Figure 16A–1 is a plot of the book values for an $80,000 asset with S ϭ $10,000 and n ϭ
10 years using the four depreciation methods that we have learned
...

16A
...

A second depreciation method that is not allowed for tax purposes, but useful in some situations is the unit-of-production (UOP) method
...
Suppose a highway contractor
has a series of state highway department contracts that will last several years and that earth moving equipment is purchased for use on all contracts
...
For year t, UOP
deprecation is calculated as
actual usage for year t
Dt ϭ —————————— (basis Ϫ salvage)
total lifetime usage

[16A
...
2
Zachry Contractors purchased an $80,000 mixer for use during the next 10 years for contract
work on IH-10 in San Antonio
...
Use the actual usage per
year shown in Table 16A–1 and the unit-of-production method to determine annual depreciation
...
4] to determine the annual depreciation based on the estimated total lifetime amount of material, 2 million m3 in this
case
...
If the mixer
is continued in service after the 2 million m3 is processed, no further depreciation is allowed
...
2

Year t

Actual
Usage, 1000 m3

Annual
Depreciation Dt , $

Cumulative
Depreciation, $

1
2–8
9–10

400
200
100

16,000
8,000
4,000

16,000
72,000
80,000

Total

2000

80,000

431

432

Chapter 16

Depreciation Methods

16A
...

It also maximizes the present value of accumulated and total depreciation over the recovery period
...
The approach below is an inherent part of MACRS
...
General rules of switching are summarized here
...
Switching is recommended when the depreciation for year t by the currently used method is
less than that for a new method
...

2
...

3
...
When switching from a DB method, the estimated salvage value,
not the DB-implied salvage value, is used to compute the depreciation for the new method;
we assume S ϭ 0 in all cases
...
)
4
...

In all situations, the criterion is to maximize the present worth of the total depreciation PWD
...

t‫؍‬n

PWD ‫؍‬

͚ D (P͞F, i, t)
t

[16A
...

Switching is most advantageous from a rapid write-off method such as DDB to the SL model
...
11] exceeds the salvage value estimated at purchase time; that is, switch if
BVn ϭ B(1 Ϫ d)n Ͼ estimated S

[16A
...
Depending upon the values of d and n, the
switch may be best in the later years or last year of the recovery period, which removes the implied S inherent to the DDB model
...
For each year t, compute the two depreciation charges
...
7]
[16A
...
Select the larger depreciation value
...
9]

3
...
5]
...
This approach is usually not taken, but the switching technique will work
correctly for all depreciation methods
...
Once these are understood, the mechanics of the switching can be speeded up by applying the spreadsheet function

16A
...
This is a quite powerful function that determines the depreciation for 1 year or the total over several years for the DB-to-SL switch
...
10]

Appendix A explains all the ﬁelds in detail, but for simple applications, where the DDB and SL
annual Dt values are needed, the following are correct entries:
start_t is the year (tϪ1)
end_t is year t
d is optional; 2 for DDB is assumed, the same as in the DDB function
no_switch is an optional logical value:
FALSE or omitted—switch to SL occurs, if advantageous
TRUE—DDB or DB method is applied with no switching to SL depreciation considered
...
This is discussed in Example 16A
...
You may notice that the VDB function is the same one used to calculate annual MACRS depreciation
...
3
The Outback Steakhouse main ofﬁce has purchased a $100,000 online document imaging system with an estimated useful life of 8 years and a tax depreciation recovery period of 5 years
...
(d) Perform the DDB-to-SL switch using a spreadsheet and plot
the book values
...

Solution by Hand
The MACRS method is not involved in this solution
...
1] determines the annual SL depreciation
...

PWD ϭ 20,000(P͞A,15%,5) ϭ 20,000(3
...
40
...
The value PWD ϭ
$69,915 exceeds $67,044 for SL depreciation
...

(c) Use the DDB-to-SL switching procedure
...
The DDB values for Dt in Table 16A–2 are repeated in Table 16A–3 for comparison
with the DSL values from Equation [16A
...
The DSL values change each year because
BVtϪ1 is different
...

For illustration, compute DSL values for years 2 and 4
...
The column “Larger Dt” indicates a switch in year 4 with D4 ϭ $10,800
...
Total depreciation
with switching is $100,000 compared to the DDB amount of $92,224
...
With switching, PWD ϭ $73,943, which is an increase over both the SL and DDB
methods
...
3b

Year
t

Dt , $

BVt , $

(P͞F,15%,t)

Present Worth
of Dt , $

0
1
2
3
4
5

40,000
24,000
14,400
8,640
5,184

100,000
60,000
36,000
21,600
12,960
7,776

0
...
7561
0
...
5718
0
...
3c

DDB Method, $
DDDB
BVt

0
1
2
3
4*
5

—
40,000
24,000
14,400
8,640
5,184

Totals

100,000
60,000
36,000
21,600
12,960
7,776

SL Method
DSL, $

Larger
Dt , $

P͞F
Factor

Present
Worth of
Dt, $

20,000
15,000
12,000
10,800
12,960

40,000
24,000
14,400
10,800
10,800

0
...
7561
0
...
5718
0
...

Solution by Spreadsheet
(d) In Figure 16A–2, column D entries are the VDB functions to determine that the DDB-toSL switch should take place in year 4
...
If TRUE were entered, the declining
balance model would be maintained throughout the recovery period, and the annual depreciation amounts would be equal to those in column B
...
The terminal book value in year 5 for the DDB
method is BV5 ϭ $7776, while the DDB-to-SL switch reduces the book value to zero
...
The results here are the
same as in parts (b) and (c) above
...

ϭ DDB($C$3,0,5,$A7)
ϭ VDB($E$3,0,5,$A6,$A7,2,FALSE)

Figure 16A–2
Depreciation for DDB-to-SL switch using the VDB function, Example 16A
...

16A
...
When the switch to SL takes place, which is usually in the last 1 to
3 years of the recovery period, any remaining basis is charged off in year n ϩ 1 so that the book
value reaches zero
...
For recovery periods of 15 and 20 years, 150% DB with the half-year convention and the
switch to SL apply
...
Only the MACRS rates for the GDS recovery periods (Table 16–4) utilize the DDBto-SL switch
...

EXAMPLE 16A
...
3, parts (c) and (d), the DDB-to-SL switching method was applied to a
$100,000, n ϭ 5 years asset resulting in PWD ϭ $73,943 at i ϭ 15%
...

Solution
Table 16A–4 summarizes the computations for depreciation (using Table 16–2 rates), book
value, and present worth of depreciation
...
This is so, in part, because the half-year
convention disallows 50% of the ﬁrst-year DDB depreciation (which amounts to 20% of the
basis)
...

TA BLE 16A–4

Depreciation and Book Value Using MACRS,
Example 16A
...
20
0
...
192
0
...
1152
0
...
000
tϭ6

PWD ϭ

͚ D (P͞F,15%,t) ϭ $69,016
t

tϭ1

16A
...
In the ﬁrst year, some adjustments have been made to compute the
MACRS rate
...
The half-year convention is always imposed, and any remaining book value in year n is removed in year n ϩ 1
...

Since different DB depreciation rates apply for different n values, the following summary may
be used to determine Dt and BVt values
...

435

436

Chapter 16

Depreciation Methods

For n ϭ 3, 5, 7, and 10 Use DDB depreciation with the half-year convention, switching to SL
depreciation in year t when DSL Ն DDB
...
2, and add onehalf year when computing DSL to account for the half-year convention
...

[16A
...
12]

tϭ1

2 n
BVtϪ1
—————
n Ϫ t ϩ 1
...
13]
t ϭ 2, 3,
...

For n ϭ 15 and 20 Use 150% DB with the half-year convention and the switch to SL when
DSL Ն DDB
...
7]
DDB ϭ dt(BVtϪ1)
where
dt ϭ

{

0
...
50
——
n

t ϭ 2, 3,
...
14]

EXAMPLE 16A
...
(a) Use Equations [16A
...
13] to obtain the annual
depreciation and book value
...

Solution

(a) With n ϭ 5 and the half-year convention, use the DDB-to-SL switching procedure to obtain the results in Table 16A–5
...
4(2880) ϭ $1152
2880
DSL ϭ ————— ϭ $1152
5 Ϫ 4 ϩ 1
...
13]
...

(b) The actual rates are computed by dividing the “Larger Dt” column values by the ﬁrst cost
of $10,000
...

t

1

2

3

4

5

6

dt

0
...
32

0
...
1152

0
...
0576

16A
...
5

TABLE 16A–5
Years

DDB

t

dt

DDB, $

SL Depreciation
DSL, $

Larger
Dt, $

BVt, $

0
1
2
3
4
5
6

—
0
...
4
0
...
4
0
...
But the logic behind the MACRS
rates is described here for those interested
...
The subscripts DB and SL have been inserted along with
the year t
...
, t) on d
...
, n are

(

iϭtϪ1

dDB,t ϭ d 1 Ϫ

(

͚d

i

iϭ1

iϭtϪ1

1Ϫ

͚d

i

)

)

[16A
...
5

[16A
...

dSL,nϩ1 ϭ 0
...
17]

The DB and SL rates are compared each year to determine which is larger and when the switch
to SL depreciation should occur
...
6
Verify the MACRS rates in Table 16–2 for a 3-year recovery period
...
33, 44
...
81, and 7
...

Solution

The ﬁxed rate for DDB with n ϭ 3 is d ϭ 2͞3 ϭ 0
...
Using the half-year convention in year 1
and Equations [16A
...
17], the results are as follows:
d1:

dDB,1 ϭ 0
...
5(0
...
3333

d2: Cumulative depreciation rate is 0
...

dDB,2 ϭ 0
...
3333) ϭ 0
...
3333
dSL,2 ϭ ————— ϭ 0
...
5

(larger value)

437

438

Chapter 16

Depreciation Methods

d3: Cumulative depreciation rate is 0
...
4445 ϭ 0
...

dDB,3 ϭ 0
...
7778) ϭ 0
...
7778
dSL,2 ϭ ————— ϭ 0
...
5
Both values are the same; switch to straight line depreciation
...

d4 ϭ 0
...
5(0
...
0741

PROBLEMS
Fundamentals of Depreciation
16
...
2 What is the difference between book value and
market value?
16
...

16
...

16
...

16
...
S
...
irs
...

(a) What is the deﬁnition of depreciation according to the IRS?
(b) What is the description of the term salvage
value?
(c) What are the two depreciation systems within
MACRS, and what are the major differences
between them?
(d) What are the properties listed that cannot be
depreciated under MACRS?
(e) When does depreciation begin and end?
(f ) What is a Section 179 deduction?

Salvage value ϭ 10% of purchase price
Operating cost (with technician) ϭ $185,000
per year
The manager of the department asked the newest
hire to enter the appropriate data in the tax accounting program
...
8 Stahmann Products paid $350,000 for a numerical
controller during the last month of 2007 and had it
installed at a cost of $50,000
...
Stahmann sold
the system at the end of 2011 for $45,000
...

16
...
The annual depreciation was
1͞n using the relevant life value
...

Straight Line Depreciation

16
...

16
...
11 Pneumatics Engineering purchased a machine that
had a ﬁrst cost of $40,000, an expected useful life
of 8 years, a recovery period of 10 years, and a
salvage value of $10,000
...
The
inﬂation rate is 6% per year and the company’s

Problems

MARR is 11% per year
...

16
...

Determine (a) the ﬁrst cost of the asset and (b) the
assumed salvage value
...
13 Lee Company of Westbrook, Connecticut, manufactures pressure relief inserts for thermal relief
and low-ﬂow hydraulic pressure relief applications
where zero leakage is required
...

If the book value at the end of year 3 is $30,000 and
the company assumed that the machine would be
worthless at the end of its 5-year useful life,
(a) what is the book depreciation charge each year
and (b) what was the ﬁrst cost of the machine?
16
...
Write a single-cell spreadsheet function to
display the book value after 5 years of straight line
depreciation
...

16
...
The company
planned to use the machine for 10 years; however,
due to rapid obsolescence it will be retired after
only 4 years in 2012
...

(a) What is the amount of capital investment remaining when the asset is prematurely retired?
(b) If the asset is sold at the end of 4 years for
$175,000, what is the amount of capital investment lost based on straight line depreciation?
(c) If the new-technology machine has an estimated cost of $300,000, how many more
years should the company retain and depreciate the currently owned machine to make its
book value and the ﬁrst cost of the new machine equal to each other?
16
...
Tabulate the values for SL
depreciation, accumulated depreciation, and book
value for each year if (a) S ϭ 0 and (b) S ϭ $16,000
...

439

16
...
S
...
It has B ϭ
$2,000,000 and a salvage value of 20% of B
...
The general managers of the two plants want to know the
difference in (a) the depreciation amount for year
5 and (b) the book value after 5 years
...

Declining Balance Depreciation
16
...
Explain the differences between these rates
...
19 Equipment for immersion cooling of electronic
components has an installed value of $182,000
with an estimated trade-in value of $40,000 after
15 years
...

16
...

However, the station will be book-depreciated to
zero over a recovery period of 30 years
...
(c) What is the implied
salvage value for DDB? (d) Use a spreadsheet to
build the depreciation and book value schedules
for both methods to verify your answers
...
21 A video recording system was purchased 3 years
ago at a cost of $30,000
...
The system is to be replaced this year with
a trade-in value of $5000
...
22 An engineer with Accenture Middle East BV in
Dubai was asked by her client to help him understand
the difference between 150% DB and DDB depreciation
...

(a) What are the book values after 12 years for
both methods?
(b) How do the estimated salvage and these book
values compare in value after 12 years?
(c) Which of the two methods, when calculated
correctly considering S ϭ $30,000, writes off
more of the ﬁrst cost over 12 years?

440

Chapter 16

Depreciation Methods

16
...
For tax
depreciation, the SL method with n ϭ10 years
was used, but for book depreciation, Boyditch applied the DDB method with n ϭ 7 years and neglected the salvage estimate
...

(a) Compare the sales price today with the book
values using the SL and DDB methods
...

16
...
Therefore, he
keeps two sets of books, one for tax purposes
(MACRS) and one for equipment management
purposes (SL)
...

16
...
The instructor asked her to
graphically compare the total percent of ﬁrst cost
depreciated for an asset costing B dollars over a
life of n ϭ 5 years for DDB and 125% DB depreciation
...
Use a spreadsheet
unless otherwise instructed
...
30 The manager of a Glidden Paint manufacturing
plant is aware that MACRS and DDB are both accelerated depreciation methods; however, out of curiosity, she wants to determine which one provides
the faster write-off in the ﬁrst 3 years for a recently
purchased mixer that has a ﬁrst cost of $300,000, a
5-year recovery period, and a $60,000 salvage
value
...
The annual MACRS depreciation rates are 20%, 32%, and
19
...

MACRS Depreciation
16
...

Give an example of each
...
26 What was one of the prime reasons that MACRS
depreciation was initiated in the mid-1980s?
16
...
Since the robot is
unique in its capabilities, the company expects to
be able to sell it in 4 years for $95,000
...

(b) Determine the book value of the robot at the
end of year 2
...
28 Animatics Corp
...
The company purchased an asset 2 years
ago that has a 5-year recovery period
...

(a) What was the ﬁrst cost of the asset?
(b) How much was the depreciation charge in
year 1?
(c) Develop the complete MACRS depreciation
and book value schedule using the VDB
function
...
31 Railroad cars used to transport coal from Wyoming
mines to Texas power plants cost $1
...
Develop the depreciation and book value schedules for
the GDS MACRS method by using two methods on
a spreadsheet—the VDB function and the MACRS
rates
...
32 A 120-metric-ton telescoping crane that cost
$320,000 is owned by Upper State Power
...
(a) Compare book values
for MACRS and standard SL depreciation over a
7-year recovery period
...

16
...
Estimated salvage
is $150,000 for any year after 5 years of use
...

(a) GDS MACRS where a recovery period of
10 years is allowed
(b) Double declining balance with a recovery
period of 15 years
(c) ADS straight line as an alternative to MACRS,
with a recovery period of 15 years
16
...
com has installed $100,000 worth of
depreciable software and equipment that represents the latest in Internet teaming and basket
competition, intended to allow anyone to enjoy

441

Problems

the sport on the Web or in the alley
...
The company can depreciate
using MACRS for a 5-year recovery period or opt
for the ADS alternate system over 10 years using
the straight line method
...

(a) Construct the book value curves for both
methods on one graph
...

(b) After 3 years of use, what percentage of the
$100,000 basis is removed for each method?
Compare the two percentages
...
35 A company has purchased special-purpose equipment for the manufacture of rubber products
(asset class 30
...
In this case, the ADS alternative to
MACRS is required for tax depreciation purposes
...
Using a recovery period of 3 years, except for the ADS alternative, which requires a 4-year recovery with
half-year convention included, prepare a single
graph showing the annual recovery rates (in
percent) for the three methods
...
38 A sand and gravel pit purchased for $900,000 is
expected to yield 50,000 tons of gravel and 80,000
tons of sand per year
...

(a) Determine the depletion charge according to
the percentage depletion method
...

(b) If taxable income is $100,000 for the year, is
this depletion charge allowed? If not, how
much is allowed?
16
...
The cost of the purchase was $9 million
...
If the company
sold 20,000 tons in year 1 and 30,000 tons in year
2, what would the depletion charges be each year
according to the cost depletion method?
16
...
The
Colorado mine has the taxable income and sales
results summarized below
...

Year

Taxable
Income, $

Sales,
Ounces

Average
Sales Price,
$ per Ounce

1
2
3

1,500,000
2,000,000
800,000

1800
5500
2300

1115
1221
1246

Depletion
16
...

During the past 3 years the amount of coal removed was 21,000, 18,000, and 20,000 tons,
respectively
...

Determine (a) the cost depletion allowance for
each year and (b) the percentage of the purchase
price depleted thus far
...
37 NA Forest Resources purchased forest acreage
for $500,000 from which an estimated 200 million board feet of lumber is recoverable
...
10 per board
foot
...

In years 3 to 10, however, the company expects
to remove 20 million board feet per year
...

(a) Determine the depletion amount in year 2 by
the cost depletion method
...

16
...
During the last 5 years, the amount extracted each year was 60,000, 50,000, 58,000
60,000, and 65,000 tons
...
5 million tons of usable stones
and gravel
...
2 million
...

(a) Determine the depletion charge each year,
using the larger of the values for the two depletion methods
...

(b) Compute the percent of the initial cost that
has been written off in these 5 years, using
the depletion charges in part (a)
...
5 million tons remaining,
rework parts (a) and (b)
...
42 All of the following types of real property are depreciable except:
(a) Warehouses
(b) Land
(c) Ofﬁce buildings
(d) Test facilities
16
...

(b) The taxpayer must use the property in an
income-producing activity
...

(d) The property must have a determinable useful life of more than 1 year
...
44 A machine with a 5-year life has a ﬁrst cost of
$20,000 and a $2000 salvage value
...
According to the
classical straight line method, the depreciation
charge in year 2 is nearest to:
(a) $2800
(b) $3600
(c) $4500
(d) $5300
16
...
The machine has a ﬁrst cost of $40,000
with a $5000 salvage value
...
00%, 18
...
40%, respectively
...
46 A coal mine purchased for $5 million has enough
coal to operate for 10 years
...
The coal is
expected to sell for $150 per ton, with annual production expected to be 10,000 tons
...
The depletion
charge for year 6 according to the percentage depletion method would be closest to:
(a) $75,000
(b) $100,000
(c) $125,000
(d) $150,000
16
...
If the ﬁrst
cost was $20,000, the salvage value used in the depreciation calculation was closest to:
(a) $0
(b) $2500
(c) $5000
(d) $7500

16
...
If the ﬁrst cost of the asset was
$80,000, the salvage value that was used in the depreciation calculation was closest to:
(b) $2000
(a) $0
(c) $5000
(d) $8000
16
...
00%, 18
...
40%, 11
...
22%, respectively):
(a) $58,700
(b) $62,400
(c) $53,900
(d) $46,100
16
...
If the MACRS depreciation rates for
years 1, 2, and 3, were 0
...
32, and 0
...
51 A lumber company purchased a tract of land for
$70,000 that contained an estimated 25,000 usable
trees
...
In the ﬁrst year of operation, the lumber
company cut down 5000 trees
...
52 An asset with a ﬁrst cost of $50,000 and an estimated salvage value of $10,000 is depreciated by
the MACRS method
...
53 Under the General Depreciation System (GDS) of
asset classiﬁcation, any asset that is not in a stated
class is automatically assigned a recovery period of:
(a) 5 years
(b) 7 years
(c) 10 years
(d) 15 years
16
...

(b) Salvage value is neglected
...

(d) The straight line method is required
...

Sum-of-Years-Digits Depreciation
16A
...
2

16A
...
Use the SYD method
to tabulate annual depreciation and book value
...

The salvage value is expected to be 10% of the
ﬁrst cost
...

If B ϭ $12,000, n ϭ 6 years, and S is estimated
at 15% of B, use the SYD method to determine
(a) the book value after 3 years and (b) the rate
of depreciation and the depreciation amount in
year 4
...
4

16A
...
Volvo Motors decided to use the unit-of-production depreciation
method because the number of test crashes per
year in which the robot would be involved was
not estimable
...

A new hybrid car was purchased by Pedernales
Electric Cooperative as a courier vehicle to
transport items between its 12 city ofﬁces
...

Alternatively, it could have been retained for
100,000 miles
...
Five-year
DDB depreciation was applied
...
Use the actual annual miles driven,
listed below, to plot the book values for both
methods
...
Show hand or
spreadsheet solution, as instructed
...
6

An asset has a ﬁrst cost of $45,000, a recovery
period of 5 years, and a $3000 salvage value
...
7

If B ϭ $45,000, S ϭ $3000, and n ϭ 5-year
recovery period, use a spreadsheet and i ϭ
18% per year to maximize the present worth of
depreciation, using the following methods:
DDB-to-SL switching (this was determined in
Problem 16A
...
Given that
MACRS is the required depreciation system in
the United States, comment on the results
...
8

Hempstead Industries has a new milling machine
with B ϭ $110,000, n ϭ 10 years, and S ϭ $10,000
...
Use a spreadsheet to solve this problem
...
9

Reliant Electric Company has erected a large
portable building with a ﬁrst cost of $155,000
and an anticipated salvage of $50,000 after
25 years
...
10 Verify the 5-year recovery period rates for
MACRS given in Table 16–2
...

16A
...
A 5-year recovery
period and MACRS depreciation have been used
to write off the basis
...

Determine the MACRS depreciation, using the
switching rules to ﬁnd the difference between the
book value and the trade-in value after 3 years
...
12 Use the computations in Equations [16A
...
13] to determine the MACRS annual depreciation for the following asset data:
B ϭ $50,000 and a recovery period of 7 years
...
13 The 3-year MACRS recovery rates are 33
...
45%, 14
...
41%, respectively
...

CHAPTER 17

After-Tax
Economic
Analysis

L E A R N I N G

O U T C O M E S

Purpose: Perform an after-tax economic evaluation considering the impact of pertinent tax regulations, income taxes,
and depreciation
...
1

Terminology and rates

• Know the fundamental terms and relations of after-tax
analysis; use a marginal tax rate table
...
2

CFBT and CFAT

• Determine cash ﬂow series before taxes and after taxes
...
3

Taxes and depreciation

• Demonstrate the tax advantage of accelerated
depreciation and shortened recovery periods
...
4

Depreciation recapture

• Calculate the tax impact of DR; explain the use of
capital gains and capital losses
...
5

After-tax analysis

• Evaluate one project or multiple alternatives using
after-tax PW, AW, and ROR analysis
...
6

After-tax replacement

• Evaluate a defender and challenger in an after-tax
replacement study
...
7

EVA analysis

• Evaluate an alternative using after-tax economic valueadded analysis; compare to CFAT analysis
...
8

Taxes outside the United States

• Understand the fundamental practices for depreciation
and tax rates in international settings
...
9

VAT

• Demonstrate the use and computation of a valueadded tax on manufactured products
...
The change from estimating cash ﬂow before taxes (CFBT) to cash ﬂow after taxes (CFAT) involves a consideration of signiﬁcant tax effects that may alter the ﬁnal decision, and estimates the magnitude of the effect on cash ﬂow over the life of the alternative that taxes may have
...
Replacement studies are discussed
with tax effects that occur at the time that a defender is replaced
...
All
these methods use the procedures learned in earlier chapters, except now with tax effects
considered
...
Templates for tabulation of cash ﬂow after taxes by hand and by spreadsheet
are developed
...
S
...
irs
...
Publications 542, Corporations, and 544, Sales and Other
Dispositions of Assets, are especially applicable to this chapter
...

17
...
The perspective of a ﬁnancial study is that of the corporation and how the tax structure and laws affect
proﬁtability
...

There are many types of taxes levied upon corporations and individuals in all countries, including the United States
...
Federal governments rely on income taxes for a
signiﬁcant portion of their annual revenue
...
for the citizenry
...

Income tax is the amount of the payment (taxes) on income or proﬁt that must be delivered to a
federal (or lower-level) government unit
...
Corporate income taxes
are usually submitted quarterly, and the last payment of the year is submitted with the annual tax
return
...
S
...
The website www
...
gov provides information on tax laws, rates,
publications, etc
...

Though the formulas are much more complex when applied to a speciﬁc situation, two fundamental relations form the basis for income tax computations
...

Taxable income ϭ revenue Ϫ operating expenses Ϫ depreciation
These terms and relations for corporations are now described
...
) added; t is omitted here for simplicity
...
These incomes are listed in the income statement
...
) Other, nonoperating revenues such as sale of assets,
license fee income, and royalties are considered separately for tax purposes
...
These expenses are tax-deductible for corporations
...
Depreciation is not included here since it is not an operating expense
...

NOI ‫ ؍‬EBIT ‫ ؍‬GI ؊ OE

[17
...
A corporation is
allowed to remove depreciation, depletion and amortization, and some other deductibles
from net operating income in determining the taxable income for a year
...
2]

Though there may be subtleties and varying interpretations over time, in essence, the differences between NOI and TI are tax-law-allowed deductibles, such as depreciation
...
)
Tax rate T is a percentage, or decimal equivalent, of TI that is owed in taxes
...
The marginal tax rate is the percentage
paid on the last dollar of income
...
The general tax computation relation is
Income taxes ‫ ؍‬applicable tax rate ؋ taxable income
‫( ؍‬T)(TI)

[17
...

NOPAT ‫ ؍‬TI ؊ taxes ‫ ؍‬TI ؊ (T)(TI)
‫ ؍‬TI(1 ؊ T)

[17
...
It is also called net proﬁt after taxes (NPAT)
...
These are rates for the entire corporation, not for an individual
project, though they are often applied in the after-tax analysis of a single project
...
To illustrate the use of the graduated tax rate, assume a company is expected to generate a taxable income of $500,000 in 1 year
...
In this case, for TI ϭ $500,000,
Taxes ϭ 113,900 ϩ 0
...

Taxes ϭ 0
...
25(75,000 Ϫ 50,000) ϩ 0
...
39(335,000 Ϫ 100,000) ϩ 0
...
1

Income Tax Terminology and Basic Relations

TABLE 17–1

U
...
Corporate Income Tax Rate Schedule (2010)
If Taxable Income ($) Is:

Over

But Not
over

Tax Is

Of the
Amount over

0
50,000
75,000
100,000
335,000
10,000,000
15,000,000
18,333,333

50,000
75,000
100,000
335,000
10,000,000
15,000,000
18,333,333
—

15%
7,500 ϩ 25%
13,750 ϩ 34%
22,250 ϩ 39%
113,900 ϩ 34%
3,400,000 ϩ 35%
5,150,000 ϩ 38%
35%

0
50,000
75,000
100,000
335,000
10,000,000
15,000,000
0

Smaller businesses (with TI Ͻ $335,000) receive a slight tax advantage compared to large corporations
...
33 million, there is a ﬂat tax rate of 35%
...

The corporate tax rates apply to a corporation as a whole, not to a speciﬁc project, unless the
project is the company
...
Therefore, the
last dollar of TI is taxed at a marginal rate
...

Because the marginal tax rates change with TI, it is not possible to quote directly the percent of TI paid in income taxes
...
5]
Average tax rate ϭ ——————— ϭ ———
TI
taxable income
Referring to Table 17–1, for a small business with TI ϭ $100,000, the federal income tax
burden averages $22,250͞100,000 ϭ 22
...
If TI ϭ $15 million, the average tax rate is
$5
...
33%
...
For the sake of simplicity, the tax rate used in an economy study is often a single-ﬁgure effective tax rate Te, which
accounts for all taxes
...
One reason to use the
effective tax rate is that state taxes are deductible for federal tax computation
...
6]
[17
...
1
REI (Recreational Equipment Incorporated) sells outdoor equipment and sporting goods through
retail outlets, the Internet, and catalogs
...

Total revenue
Operating expenses
Depreciation and other allowed deductions

$19
...
6 million
$1
...

(b) Find the average federal tax rate paid for the year
...

(d) Estimate federal and state taxes using the single-value rate, and compare their total with the
total in part (a)
...
2] and use Table 17–1 rates for federal taxes due
...
9 million Ϫ 8
...
8 million
ϭ $9
...
06(TI) ϭ 0
...
34(8,930,000 Ϫ 335,000) ϭ $3,036,200
Total federal and state taxes ϭ 3,036,200 ϩ 570,000 ϭ $3,606,200
[17
...
5], the average tax rate paid is approximately 32% of TI
...
3196
(c) By Equation [17
...

Te ϭ 0
...
06)(0
...
3604

(36
...
5 million from part (a) in Equation [17
...

Taxes ϭ 0
...
8], this approximation is $182,400 low, a 5
...

It is interesting to understand how corporate tax and individual tax computations differ
...

However, for an individual’s taxable income, most of the expenses for living and working are not
tax deductible to the same degree as operating expenses are for corporations
...
Exemptions are yourself, your spouse, your children, and your other dependents
...

In the United States, the tax rates for individuals, like those for corporations, are graduated by
level of TI
...
These rates and TI levels are the subject of
ongoing political debates at the U
...
national level depending upon the balance of power in the
congressional bodies and the economic condition of the country
...
This process is
called indexing
...
Current information is available on the IRS website www
...
gov through
Publication 17, Your Federal Income Tax
...
2 Calculation of Cash Flow after Taxes
Early in the text, the term net cash ﬂow (NCF) was identiﬁed as the best estimate of actual cash
ﬂow each year
...
Since then, the annual NCF amounts have been used many times to perform alternative evaluations via the PW,
AW, ROR, and B/C methods
...
2

Calculation of Cash Flow after Taxes

will be considered, it is time to expand our terminology
...

CFBT and CFAT are actual cash ﬂows; that is, they represent the estimated actual ﬂow of
money into and out of the corporation that will result from the alternative
...
Once the
CFAT estimates are developed, the economic evaluation is performed using the same methods
and selection guidelines applied previously
...

We learned that net operating income (NOI) does not include the purchase or sale of capital
assets
...
Incorporating the deﬁnitions of gross income and
operating expenses from NOI, CFBT for any year is deﬁned as
CFBT ‫ ؍‬gross income ؊ operating expenses ؊ initial investment ؉ salvage value
‫ ؍‬GI ؊ OE ؊ P ؉ S
[17
...
Therefore, only in year 0 will the CFBT include P, and only in year n will an S value
be present
...
10]

where taxes are estimated using the relation (T)(TI) or (Te)(TI)
...
2] that depreciation D is considered when calculating TI
...

Depreciation is not an operating expense and is a noncash ﬂow
...

Therefore, the after-tax engineering economy study must be based on actual cash ﬂow estimates,
that is, annual CFAT estimates that do not include depreciation as an expense (negative cash ﬂow)
...
Equations [17
...
10] are now combined
...
11]

Suggested table column headings for CFBT and CFAT calculations by hand or by spreadsheet are
shown in Table 17–2
...
Expenses OE and initial investment P carry negative signs in all
tables and spreadsheets
...
It is possible to account for this in a detailed after-tax analysis using carry-forward
and carry-back rules for operating losses
...
Rather, the associated negative income tax is considered as a tax
savings for the year
...

TABLE 17–2

Year

Suggested Column Headings for Calculation of CFAT

Gross
Income
GI

Operating
Expenses
OE

Investment
and
Salvage
P and S

(1)

(2)

(3)

CFBT
(4) ϭ
(1) ϩ (2) ϩ (3)

Depreciation
D
(5)

Taxable
Income
TI

Taxes

CFAT

(6) ϭ
(7) ϭ (8) ϭ
(1) ϩ (2) Ϫ (5) Te(6) (4) Ϫ (7)

449

450

Chapter 17

After-Tax Economic Analysis

EXAMPLE 17
...
Wilson plans to purchase listening and detection
equipment for use in the 6-year contract
...
Based on the incentive clause in the contract, Wilson estimates that the equipment will increase contract revenue by $200,000 per year and require an additional M&O expense of $90,000 per year
...
Tabulate and plot the CFBT and CFAT series
...
The functions for year 6 are detailed in row 11
...

CFBT: The operating expenses OE and initial investment P are shown as negative cash ﬂows
...
CFBT is calculated by Equation
[17
...
In year 6, for example, when the equipment is sold, the function in row 11 indicates that
CFBT6 ϭ 200,000 Ϫ 90,000 ϩ 150,000 ϭ $260,000
CFAT: Column F for MACRS depreciation, which is determined using the VDB function
over the 6-year period, writes off the entire $550,000 investment
...

TI4 ϭ GI Ϫ OE Ϫ D ϭ 200,000 Ϫ 90,000 Ϫ 63,360 ϭ $46,640
Taxes4 ϭ (0
...
35) (46,640) ϭ $16,324
CFAT4 ϭ GI Ϫ OE Ϫ taxes ϭ 200,000 − 90,000 Ϫ 16,324 ϭ $93,676
In year 2, MACRS depreciation is large enough to cause TI to be negative ($−66,000)
...

Comment

MACRS depreciates to a salvage value of S ϭ 0
...

Figure 17–1
Computation of CFBT and CFAT using MACRS depreciation and Te ϭ 35%, Example 17
...

17
...
Larger rates in earlier
years of the recovery period require less taxes due to the larger reductions in taxable income
...
For the

17
...

t‫؍‬n

PWtax ‫؍‬

͚ (taxes in year t)(P͞F, i, t)

[17
...

Compare any two depreciation methods
...
On the basis of these
assumptions, the following statements are correct:
The total taxes paid are equal for all depreciation methods
...

As we learned in Chapter 16, MACRS is the prescribed tax depreciation method in the United
States, and the only alternative is MACRS straight line depreciation with an extended recovery
period
...
If the DDB method were still allowed directly, rather than embedded in
MACRS, DDB would not fare as well as MACRS
...
This is illustrated in Example 17
...

EXAMPLE 17
...
The CFBT for the machine is estimated at $20,000
...
Use a 6-year period for the comparison to accommodate the half-year
convention imposed by MACRS
...
For classical straight line depreciation with n ϭ 5, Dt ϭ $10,000 for 5 years and D6 ϭ
0 (column 3)
...

The classical DDB percentage of d ϭ 2͞n ϭ 0
...
The implied salvage
value is $50,000 − 46,112 ϭ $3888, so not all $50,000 is tax-deductible
...
35) ϭ $1361 larger than for the classical SL method
...
0 ؍‬
Taxes, $

(6)
D t, $

(7)
TI, $

(8) ‫)7(53
...

(11) ‫)01(53
...
3
...
Total taxes are $24,500,
the same as for classical SL depreciation
...
Note
the pattern of the curves, especially the lower total taxes relative to the SL model after year 1
for MACRS and in years 1 through 4 for DDB
...
The PWtax values at the bottom of Table 17–3 are calculated using
Equation [17
...
The MACRS PWtax value is the smallest at $18,162
...
It can be shown that a shorter
recovery period will offer a tax advantage over a longer period using the criterion to minimize
PWtax
...

The present worth of taxes is less for smaller n values
...
Example 17
...

EXAMPLE 17
...
This is
common for multinational corporations
...
The second set is for foreign government purposes, such as
depreciation and taxes
...
Demonstrate the
tax advantage for the smaller n if net operating income (NOI) is $30,000 per year, an effective
tax rate of 35% applies, invested money is returning 5% per year after taxes, and classical SL
depreciation is allowed
...

Solution
Determine the annual TI and taxes by Equations [17
...
3] and the present worth of
taxes using Equation [17
...

17
...
35)(20,000) ϭ $7000 per year
PWtax ϭ 7000(P͞A,5%,9) ϭ $49,755
Total taxes ϭ (7000)(9) ϭ $63,000
Recovery period n ϭ 5 years:
Use the same comparison period of 9 years, but depreciation occurs only during the ﬁrst 5 years
...
35)(30,000 Ϫ 18,000) ϭ $4200
(0
...
However, the more rapid write-off for
n ϭ 5 results in a present worth tax savings of nearly $2400 (49,755 Ϫ 47,356)
...
4 Depreciation Recapture and Capital
Gains (Losses)
All the tax implications discussed here are the result of disposing of a depreciable asset before, at,
or after its recovery period
...
The key is the size of the selling price (or salvage or market value)
relative to the current book value at disposal time and relative to the ﬁrst cost, which is called the
unadjusted basis B in depreciation terminology
...

Depreciation recapture DR, also called ordinary gain, occurs when a depreciable asset is
sold for more than the current book value BVt
...
13]

For an after-tax
study, the tax
effect is:

CG: Taxed at Te
(after CL offset)

plus
DR

DR: Taxed at Te

SP2
DR

Book
value BVt

CL
SP3

CL: Can only offset
CG

Zero, $0

Figure 17–3
Summary of calculations and tax treatment for depreciation recapture and capital gains (losses)
...
In the United States, an amount
equal to the estimated salvage value can always be anticipated as DR when the asset is disposed of after the MACRS recovery period
...
The amount of DR is treated as ordinary taxable income
in the year of asset disposal
...
See Figure 17–3
...
14]

Since future capital gains are difﬁcult to predict, they are usually not detailed in an after-tax economy study
...

When the selling price exceeds B, the TI due to the sale is the capital gain plus the depreciation recapture, as shown in Figure 17–3
...

Capital loss CL occurs when a depreciable asset is disposed of for less than its current book
value
...
15]
An economic analysis does not commonly account for capital loss, simply because it is not
estimable for a speciﬁc alternative
...
For the purposes of the
economic study, this provides a tax savings in the year of replacement
...
These savings are assumed to be offset elsewhere in the corporation by other income-producing assets that generate taxes
...

• U
...
tax law deﬁnes capital gains as long-term (items retained for more than 1 year) or short-term
...
The term
capital gain correctly applies at sale time to property such as investments (stocks and bonds),
art, jewelry, land, and the like
...
Corporate tax treatment at this
time is the same for both: taxed as ordinary income
...

• Capital gains are taxed as ordinary taxable income at the corporation’s regular tax rates
...
The terms used
then are net capital gains (losses)
...

• When an asset is disposed of, the tax treatment is referred to as a Section 1231 transaction,
which is the IRS rule section of the same number
...

• All these rules apply to corporations
...

If the three additional income and tax elements covered here are incorporated into Equation [17
...
16]

17
...
e
...
Only when a capital gain or loss must be included due to the nature of the problem will the calculations involve it
...
5
Biotech, a medical imaging and modeling company, must purchase a bone cell analysis system for
use by a team of bioengineers and mechanical engineers studying bone density in athletes
...
The effective tax rate is 35%
...

Analyzer 1

Analyzer 2

150,000
30,000
5

225,000
10,000
5

Basis B, $
Operating expenses, $ per year
MACRS recovery, years

Answer the following questions, solving by hand and spreadsheet
...
Which analyzer should be purchased?
(b) Assume that 3 years have now passed, and the company is about to sell the analyzer
...

Solution by Hand
(a) Table 17–4 details the tax computations
...
Equation [17
...
Taxes for the 3-year period are summed, with no consideration of
the time value of money
...

(b) When the analyzer is sold after 3 years of service, there is a depreciation recapture (DR)
that is taxed at the 35% rate
...
For each analyzer, account for the DR by Equation [17
...
16], TI ϭ GI − OE − D ϩ DR
...

TABLE 17–4

Year

Gross
Income
GI, $

Comparison of Total Taxes for Two Alternatives, Example 17
...
35TI, $

150,000
120,000
72,000
43,200

40,000
22,000
41,200

14,000
7,700
14,420

Analyzer 1
0
1
2
3

150,000
100,000
100,000
100,000

30,000
30,000
30,000

30,000
48,000
28,800

36,120
Analyzer 2
0
1
2
3

225,000
100,000
100,000
100,000

10,000
10,000
10,000

45,000
72,000
43,200

225,000
180,000
108,000
64,800

45,000
18,000
46,800

15,750
6,300
16,380
38,430

455

456

Chapter 17

After-Tax Economic Analysis

DR ϭ 130,000 Ϫ 43,200 ϭ $86,800
Year 3 TI ϭ 100,000 Ϫ 30,000 Ϫ 28,800 ϩ 86,800 ϭ $128,000
Year 3 taxes ϭ (0
...
35)(207,000) ϭ $72,450
Total taxes ϭ 15,750 ϩ 6300 ϩ 72,450 ϭ $94,500

Analyzer 2:

Now, analyzer 1 has a considerable advantage in total taxes ($94,500 versus $66,500)
...
Similar analysis in rows 14 to 18 results in total
taxes of $38,430 for analyzer 2, indicating that the company should select analyzer 1, based
on taxes only
...
The new TI in year 3 has the depreciation recapture incorporated as DR ϭ selling price Ϫ book value ϭ SP Ϫ BV3, which
is shown in the cell tag as the last term (D10 Ϫ F10)
...

Comment
Note that no time value of money is considered in these analyses, as we have used in previous
alternative evaluations
...
5 below we will rely upon PW, AW, and ROR analyses
at an established MARR to make an after-tax decision based upon CFAT values
...
5
...
5 After-Tax Evaluation
The required after-tax MARR is established using the market interest rate, the corporation’s effective tax rate, and its weighted average cost of capital
...
When positive and negative CFAT values are present, a
PW or AW Ͻ 0 indicates the MARR is not met
...
The guidelines are:

17
...
PW or AW Ն 0, the project is ﬁnancially viable because the after-tax MARR is
met or exceeded
...
Select the alternative with the best (numerically largest) PW or AW
value
...
Assign a plus sign to each saving and apply the guidelines above
...
This requirement must be met for every
analysis—before or after taxes
...

For AW analysis: Use the PMT function with an embedded NPV function over one life cycle
...

ϭ ϪPMT(MARR,n,NPV(MARR,year_1: year_n) ϩ year_0)

[17
...
(There is an LCM function in Excel
...
The general format is
ϭ ϪPV(MARR,LCM_years,PMT_result_cell )

[17
...
6
Paul is designing the interior walls of an industrial building
...
Two construction options—stucco on metal
lath (S) and bricks (B)—each have about the same transmission loss, approximately
33 decibels
...
Paul has estimated the ﬁrst costs and after-tax savings each year for both designs
...

(b) Use a spreadsheet to select the alternative and determine the required ﬁrst cost for the
plans to break even
...
Develop the AW relations using
the CFAT values over each plan’s life
...

AWS ϭ [−28,800 ϩ 5400(P͞A,7%,6) ϩ 2040(P͞A,7%,4)(P͞F,7%,6)
ϩ 2792(P͞F,7%,10)](A͞P,7%,10)
ϭ $422
AWB ϭ [−50,000 ϩ 14,200(P͞F,7%,1) ϩ · · · ϩ 10,600(P͞F,7%,5)](A͞P,7%,5)
ϭ $327
Both plans are ﬁnancially viable; select plan S because AWS is larger
...
Use the AW values and the P͞A factor for the
LCM of 10 years to select stucco on metal lath, plan S
...
0236) ϭ $2964
PWB ϭ AWB(P͞A,7%,10) ϭ 327(7
...
6
...
17], and row 15 shows the 10-year PW that results from the PV function in
Equation [17
...
Plan S is chosen by a relatively small margin
...
This is a small
reduction from the $−50,000 ﬁrst cost initially estimated
...
If the minus is omitted, the AW and PW values
have the wrong sign and it appears that the plans are not ﬁnancially viable in that they do not
return at least the after-tax MARR
...

To utilize the ROR method, apply exactly the same procedures as in Chapter 7 (single project) and Chapter 8 (two or more alternatives) to the CFAT series
...
Multiple roots may exist in the CFAT series, as they can for any cash ﬂow series
...

tϭn

Present worth:

0ϭ

͚ CFAT (P͞F,i*,t)
t

[17
...

[17
...
5

After-Tax Evaluation

Spreadsheet solution for i* is faster for most CFAT series
...
21]

If the after-tax ROR is important to the analysis, but the details of an after-tax study are not of
interest, the before-tax ROR (or MARR) can be adjusted with the effective tax rate Te by using
the approximating relation
after-tax ROR
Before-tax ROR ‫——————— ؍‬
1 ؊ Te

[17
...
To approximate the
effect of taxes without performing the details of an after-tax study, the before-tax MARR can be
estimated as
0
...
40
If the decision concerns the economic viability of a project and the resulting PW or AW value is
close to zero, the details of an after-tax analysis should be developed
...
7
A ﬁber optics manufacturing company operating in Hong Kong has spent $50,000 for a
5-year-life machine that has a projected $20,000 annual NOI and annual depreciation of
$10,000 for years 1 through 5
...
(a) Determine the after-tax rate
of return
...

Solution
(a) The CFAT in year 0 is $Ϫ50,000
...
(See Equations [17
...
9]
...

TI ϭ NOI Ϫ D ϭ 20,000 Ϫ 10,000 ϭ $10,000
Taxes ϭ Te (TI) ϭ 0
...
19]
...
125
Solution gives i* ϭ 18
...

(b) Use Equation [17
...

0
...
3005
1 Ϫ 0
...
05%)

The actual before-tax i* using CFBT ϭ $20,000 for 5 years is 28
...
05% is used in a before-tax
analysis
...
Solution by spreadsheet is accomplished using the incremental CFAT values
and the IRR function
...
4 through 8
...
You should review and understand these sections before proceeding with this section
...

Incremental ROR: Incremental analysis must be performed
...

Equal-service requirement: Incremental ROR analysis requires that the alternatives be evaluated over equal time periods
...
(The only exception, mentioned in Section 8
...
)
Revenue and cost alternatives: Revenue alternatives (positive and negative cash ﬂows) may
be treated differently from cost alternatives (cost-only cash ﬂow estimates)
...
Alternatives with
i* Ͻ MARR can be removed from further evaluation
...

Breakeven ROR

Once the CFAT series are developed, the breakeven ROR can be obtained using a plot of PW
versus i* by solving the PW relation for each alternative over the LCM at several interest rates
...

The next examples solve CFAT problems using incremental ROR analysis and the breakeven
ROR plot of PW versus i
...
8
In Example 17
...

Figure 17–5a presented both a PW analysis over 10 years and an AW analysis over the respective lives
...
After reviewing this earlier solution, (a) perform an ROR
evaluation at the after-tax MARR of 7% per year and (b) plot the PW versus ⌬i graph to determine the breakeven ROR
...
Apply the procedure in Section 8
...
Figure 17–6 shows the estimated CFAT for each alternative and the incremental
CFAT series
...
Row 14 indicates they do
...
35%
...
Plan S is
selected, the same as with the PW and AW methods
...

The graph indicates that the breakeven ⌬i* occurs at 6
...
Whenever the after-tax MARR is above 6
...

Comment
Note that the incremental CFAT series has three sign changes
...
Accordingly, there may be multiple ⌬i* values
...

17
...
35%

Figure 17–6
Incremental evaluation of CFAT and determination of breakeven ROR, Example 17
...

EXAMPLE 17
...
5 an after-tax analysis of two bone cell analyzers was initiated due to a new
3-year NBA contract
...

The complete solution is in Table 17–4 (hand) and Figure 17–4 (spreadsheet)
...
5: $130,000 for analyzer 1 and $225,000 for analyzer 2
...

Solution
A spreadsheet solution is presented here, but a hand solution is equivalent, just slower
...
The CFAT series (column I) are determined by the relation CFAT ϭ CFBT Ϫ
taxes, with the taxable income determined using Equation [17
...
For
example, in year 3 when analyzer 2 is sold for S ϭ $225,000, the CFAT calculation is
CFAT3 ϭ CFBT Ϫ (TI)(Te) ϭ GI Ϫ OE Ϫ P ϩ S Ϫ (GI Ϫ OE Ϫ D ϩ DR)(Te)
The depreciation recapture DR is the amount above the year 3 book value received at sale time
...

CFAT3 ϭ 100,000 Ϫ 10,000 ϩ 0 ϩ 225,000
Ϫ(100,000 Ϫ 10,000 Ϫ 43,200 ϩ 160,200)(0
...
35) ϭ $242,550
The cell tags in row 14 of Figure 17–7 follow this same progression
...

These are revenue alternatives, so the overall i* values indicate that both CFAT series are
acceptable
...
6% (cell J17) also exceeds MARR ϭ 10%, so analyzer 2 is
selected
...

461

462

Chapter 17

After-Tax Economic Analysis

Overall i*

ϭ B14ϩC14ϪE14ϩ(D14ϪF14)
This term
calculates
DR ϭ $160,200

CFAT calculation
ϭ B14ϩC14ϩD14ϪH14

Incremental i*
ϭ IRR(J11:J14)

Figure 17–7
Incremental ROR analysis of CFAT with depreciation recapture, Example 17
...

Comment
In Section 8
...
The incremental
ROR must be used
...
If the larger i* alternative
is chosen, analyzer 1 is incorrectly selected
...
For veriﬁcation, the PW at 10% is calculated
for each analyzer (column I)
...

17
...
The ﬁnal decision
may not be reversed by taxes, but the difference between before-tax AW values may be signiﬁcantly different from the after-tax difference
...
Additionally, the after-tax replacement study considers
tax-deductible depreciation and operating expenses not accounted for in a before-tax analysis
...

The same procedure as the before-tax replacement study in Chapter 11 is applied here, but for
CFAT estimates
...
Special attention to Sections 11
...
5 is recommended
...
10 presents a solution by hand of an after-tax replacement study using a simplifying assumption of classical SL (straight line) depreciation
...
11 solves the same problem by spreadsheet, but includes the detail of MACRS depreciation
...

EXAMPLE 17
...

Management has discovered that it is technologically and legally outdated now
...
If a market value of $400,000 is offered as the trade-in for the current
equipment, perform a replacement study using (a) a before-tax MARR of 10% per year and

17
...
Assume an effective tax rate of 34%
...

Defender
Market value, $
First cost, $
Annual cost, $͞year
Recovery period, years

Challenger

400,000
Ϫ100,000
8 (originally)

Ϫ1,000,000
Ϫ15,000
5

Solution
Assume that an ESL (economic service life) analysis has determined the best life values to be
5 more years for the defender and 5 years total for the challenger
...
The defender AW uses the market value as the ﬁrst cost, PD ϭ $Ϫ400,000
...
3), we select the better AW
value
...
The
defender has a $73,280 lower equivalent annual cost compared to the challenger
...

(b) For the after-tax replacement study, there are no tax effects other than income tax for the
defender
...

Dt ϭ 600,000͞8 ϭ $75,000

t ϭ 1 to 8 years

Table 17–5 shows the TI and taxes at 34%
...
(Remember that for tax savings in an economic
TA BLE 17–5

Before-Tax and After-Tax Replacement Analyses, Example 17
...
34TI, $

75,000
75,000
75,000
75,000
75,000
AW at 7%

Ϫ175,000
Ϫ175,000
Ϫ175,000
Ϫ175,000
Ϫ175,000

Ϫ59,500
Ϫ59,500
Ϫ59,500
Ϫ59,500
Ϫ59,500

ϩ25,000†
Ϫ215,000
Ϫ215,000
Ϫ215,000
−215,000
Ϫ215,000‡

8,500
Ϫ73,100
Ϫ73,100
Ϫ73,100
Ϫ73,100
Ϫ73,100

CFAT, $

Defender
3
4
5
6
7
8
AW at 10%

0
1
2
3
4
5

Ϫ400,000
Ϫ100,000
Ϫ100,000
Ϫ100,000
Ϫ100,000
Ϫ100,000

0

Ϫ400,000
Ϫ100,000
Ϫ100,000
Ϫ100,000
Ϫ100,000
Ϫ100,000
Ϫ205,520

Ϫ400,000
Ϫ40,500
Ϫ40,500
Ϫ40,500
Ϫ40,500
Ϫ40,500
Ϫ138,056

Challenger
0
1
2
3
4
5
AW at 10%
*

Ϫ1,000,000
Ϫ15,000
Ϫ15,000
Ϫ15,000
Ϫ15,000
Ϫ15,000

0

Ϫ1,000,000
Ϫ15,000
Ϫ15,000
Ϫ15,000
Ϫ15,000
Ϫ15,000
Ϫ278,800

Minus sign indicates a tax savings for the year
...

‡
Assumes challenger’s salvage actually realized is S ϭ 0; no tax
...
) Since only costs are estimated, the annual CFAT is negative, but the
$59,500 tax savings has reduced it
...
In year
0 for the challenger, Table 17–5 includes the following computations to arrive at a tax
of $8500
...
34(25,000) ϭ $8500

The SL depreciation is $1,000,000͞5 ϭ $200,000 per year
...
34) ϭ $Ϫ73,100
CFAT ϭ CFBT Ϫ taxes ϭ Ϫ15,000 Ϫ (Ϫ73,100) ϭ $ϩ58,100
In year 5, it is assumed the challenger is sold for $0; there is no depreciation recapture
...

Conclusion: By either analysis, retain the defender now and plan to keep it for 5 more
years
...
If and
when cash ﬂow estimates change signiﬁcantly, perform another replacement analysis
...
The resulting tax savings
would decrease the CFAT (which is to reduce costs if CFAT is negative) of the challenger
...
The CFAT is then $Ϫ1,000,000 Ϫ (Ϫ8500) ϭ
$Ϫ991,500
...
11
Repeat the after-tax replacement study of Example 17
...
Assume either
asset is sold after 5 years for exactly its book value
...

Solution
Figure 17–8 shows the complete analysis
...
Again the defender is selected for retention, but now by an advantage of $44,142 annually
...
Therefore, taxes and MACRS have reduced the defender’s economic advantage, but
not enough to reverse the decision to retain it
...
There is depreciation recapture in year 0 of the challenger due to trade-in of the defender at $400,000, a value larger than the book value of the 3-year-old defender
...
7

465

After-Tax Value-Added Analysis

Depreciation recapture
ϭϪC5 Ϫ F11

Challenger sales price
ϭ B15 Ϫ F24

Figure 17–8
After-tax replacement study with MACRS depreciation and depreciation recapture, Example 17
...

$137,620 (cell G18), is treated as ordinary taxable income
...
34)(137,620) ϭ $46,791
(cell H18)
See the cell tags and table notes that duplicate this logic
...
The entry $57,600 (C23) reﬂects this assumption, since the forgone
MACRS depreciation in year 6 would be 1,000,000(0
...
The spreadsheet
relation ϭ B15 Ϫ F24 determines this value using the accumulated depreciation in F24
...
This implies an additional tax saving of 57,600(0
...
Conversely, if the salvage value exceeds the book value, a depreciation recapture and associated
tax should be estimated
...
7 After-Tax Value-Added Analysis
When a person or company is willing to pay more for an item, it is likely that some processing
has been performed on an earlier version of the item to make it more valuable now to the purchaser
...

Value added is a term used to indicate that a product or service has added worth from the
perspective of a consumer, owner, investor, or purchaser
...

Value added

466

Chapter 17

After-Tax Economic Analysis

For an example of highly leveraged value-added activities, consider onions that are grown and
sold at the farm level for cents per pound
...
25 per pound
...
Thus, from the perspective of the consumer, there has been a large amount of value added by the processing from raw
onions in the ground into onion rings sold at a restaurant or fast-food shop
...
When value-added analysis is performed after taxes, the approach is somewhat different
from that of CFAT analysis developed previously in this chapter
...

Value-added analysis starts with Equation [17
...
Depreciation D is included in that TI ϭ GI Ϫ
OE − D
...

The term economic value added (EVA) indicates the monetary worth added by an alternative
to the corporation’s bottom line
...
) The technique discussed below was ﬁrst publicized in several articles1 in the mid-1990s, and it has since
become very popular as a means to evaluate the ability of a corporation to increase its economic
worth, especially from the shareholders’ viewpoint
...
That is, EVA indicates the project’s contribution to the net
proﬁt of the corporation after taxes
...
This is the interest incurred by the current
level of capital invested in the asset
...
) Computationally,
EVA ‫ ؍‬NOPAT ؊ cost of invested capital
‫ ؍‬NOPAT ؊ (after-tax interest rate)(book value in year t ؊ 1)
‫ ؍‬TI(1 ؊ Te) ؊ (i)(BVt؊1)
[17
...
This ﬁnancial worth is the amount used in public documents of the corporation
(balance sheet, income statement, stock reports, etc
...

The result of an EVA analysis is a series of annual EVA estimates
...
If only one project is evaluated, AW Ͼ 0 means the after-tax MARR is exceeded, thus making the project value-adding
...
Thus, either method can be used to make a decision
...
This comparison is made in Example 17
...

1

A
...
1997, pp
...
Freedman, “How Do You Add Up?”
Chemical Week, Oct
...
31–34
...
G
...
L
...
” The Engineering Economist, vol
...
2 (2000), pp
...

17
...
12
Biotechnics Engineering has developed two mutually exclusive plans for investing in new
capital equipment with the expectation of increased revenue from its medical diagnostic services to cancer patients
...
(a) Use classical straight line
depreciation, an after-tax MARR of 12%, and an effective tax rate of 40% to perform two annual worth after-tax analyses: EVA and CFAT
...

Plan A

Plan B

Initial investment, $
Gross income Ϫ expenses, $

Ϫ500,000
170,000 per year

Estimated life, years
Salvage value

None

Ϫ1,200,000
600,000 in year 1, decreasing by 100,000
per year thereafter
4
None

4

Solution by Spreadsheet
(a) Refer to the spreadsheet and function cells (row 22) in Figure 17–9
...
The net operating proﬁt after taxes (NOPAT) in column H is calculated by Equation [17
...
The book values (column E) are used to determine the
cost of invested capital in column I, using the second term in Equation [17
...
This represents the amount of interest at
12% per year, after taxes, for the currently invested capital as reﬂected by the book value
at the beginning of the year
...
Notice there is no EVA estimate for year 0, since NOPAT and the cost of invested
capital are estimated for years 1 through n
...

CFAT evaluation: As shown in function row 22 (plan B for year 3), CFAT estimates
(column K) are calculated as (GI Ϫ OE) Ϫ P Ϫ taxes
...

(b) What is the fundamental difference between the EVA and CFAT series in columns J and K?
They are clearly equivalent from the time value of money perspective since the AW values
are numerically the same
...
To obtain the AW of EVA estimate of $Ϫ12,617 for

Figure 17–9
Comparison of two plans using EVA and CFAT analyses, Example 17
...

467

468

Chapter 17

After-Tax Economic Analysis

years 1 through 4, the initial investment of $500,000 is distributed over the 4-year life using
the A͞P factor at 12%
...
In effect, the yearly
CFAT is reduced by this charge
...
However, the EVA method indicates an alternative’s yearly estimated contribution to the value of the corporation, whereas the CFAT method estimates the actual cash
ﬂows to the corporation
...

Comment

The calculation P(A͞P,i,n) ϭ $500,000(A͞P,12%,4) is exactly the same as the capital recovery
in Equation [6
...
Thus, the cost of invested
capital for EVA is the same as the capital recovery discussed in Chapter 6
...

17
...
7], taxes ϭ
(Te)(TI)
...
Expense deductions vary widely from country to
country
...

Canada
Depreciation: This is deductible and is normally based on DB calculations, although SL may be
used
...
The
annual tax-deductible allowance is termed capital cost allowance (CCA)
...
S
...

Class and CCA rate: Asset classes are deﬁned and annual depreciation rates are speciﬁed by
class
...
There are some 44 classes, and CCA rates vary from 4% per year (the equivalent of a
25-year-life asset) for buildings (class 1) to 100% (1-year life) for applications software, chinaware, dies, etc
...
Most rates are in the range of 10% to 30% per year
...
Expenses related to capital investments are not deductible, since they are accommodated through the CCA
...
cra
...
ca in the
Forms and Publications section
...
The selected industries and assets can change over time; currently favored industries serve areas such as technology and oil exploration, and equipment subjected to large vibrations during normal usage is
allowed accelerated depreciation
...
8

After-Tax Analysis for International Projects

Recovery period: Standardized recovery periods are published that vary from 3 years (electronic
equipment) to 10 years (aircraft, machinery, and other production equipment) to 20 years (buildings)
...

Expenses: Business expenses are deductible with some limitations and some special incentives
...
Incentives are generous in some cases; for example, 150% of actual expenses is deductible for new technology and new product R&D activities
...
worldwide-tax
...

Mexico
Depreciation: This is a fully deductible allowance for calculating TI
...
For some asset types, an immediate deduction of a percentage of the ﬁrst cost is allowed
...
)
Class and rates: Asset types are identiﬁed, though not as speciﬁcally deﬁned as in some countries
...
Most rates range from
10% to 30% per year
...
Most business expenses are deductible
...

Tax on Net Assets (TNA): Under some conditions, a tax of 1
...

Internet: The best information is via websites for companies that assist international corporations located in Mexico
...
pwcglobal

...

Japan
Depreciation: This is fully deductible and based on classical SL and DB methods
...
Switching to classical SL depreciation must take
place in the year that the accelerated rate amount falls below the corresponding SL amount
...

Expenses: Business expenses are deductible in calculating TI
...
mof
...
jp
...
Some countries levy taxes only
at the federal level, while others impose taxes at several levels of government (federal, state or
provincial, prefecture, county, and city)
...
These include income
taxes at all reported levels of government within each country; however, other types of taxes
may be imposed by a particular government
...
A close examination of international rates shows that they have decreased signiﬁcantly over the last decade
...
7% (1999) to 25
...
This has encouraged corporate investment and
business expansion within country borders and helped soften the massive economic downturn
experienced in recent years
...
kpmg
...
aspx) and from country websites on corporate taxation
...
These taxes are usually in the form of a value-added tax (VAT), goods and services tax (GST), and taxes on products imported from outside its borders
...
This has been especially true during the ﬁrst decade of the 21st
century
...
The VAT system is explained now
...
9 Value-Added Tax
A value-added tax (VAT) has facetiously been called a sales tax on steroids, because the VAT
rates on some items in some countries that impose a VAT can be as high as 90%
...

A value-added tax is an indirect tax; that is, it is a tax on goods and services rather than on
people or corporations
...
A speciﬁc percentage, say 10%, is a charge added to the price of the
item and paid by the buyer
...
This process of 10% VAT continues every time the item is resold—as purchased or in a modiﬁed form—thus the term value-added
...
In some countries, VAT is used in
lieu of business or individual income taxes
...
In
fact, the United States is the only major industrialized country in the world that does not have a
VAT system, though other forms of indirect taxation are used liberally
...

A sales tax is used by the U
...
government, by nearly all of the states, and by many local entities
...
That is, businesses do not pay a sales tax on raw material, unﬁnished goods,
or items they purchase that will ultimately be sold to an end user; only the end user pays the sales
tax
...
Total sales tax percentages imposed by multiple government levels can range from 5% to 11%, sometimes larger
on speciﬁc items
...
, Home Depot does not pay a sales tax on the microwave ovens because they will be sold
to HD customers who will pay the sales tax
...
Thus, a sales tax is paid only one time, and that

17
...
The sales tax is the responsibility of the merchant to collect and remand to the taxing entity
...
The seller sends the collected VAT to the taxing entity
...
Now, this second seller will send to the taxing entity an amount that is
equal to the total tax collected minus the amount of VAT already paid
...
S
...
Here is how the VAT
might work
...
As part of the price, Northshore collects $110,000, that is, $100,000 ϩ 0
...
Northshore remits the $10,000 VAT to the U
...
Treasury
...
Westfall collects $330,000 from GE and then sends $20,000 to the U
...
Treasury, that is, $30,000 it collected in VAT from GE minus $10,000 it paid in taxes to Northshore
Mining
...
GE collects $770,000 and then remits $40,000 to the U
...
Treasury, that is, $70,000 it collected in taxes from retailers minus $30,000 it paid in VAT to
Westfall Steel
...
S
...
For example, if GE
paid $5000 in taxes on motors it purchased for the refrigerators, the amount GE would remit
to the U
...
Treasury would be $35,000 (that is, $70,000 it collected from retailers minus
$30,000 it paid in taxes to Westfall Steel minus $5000 it paid in taxes on the motors)
...
The retailers remit $25,000 to the U
...
Treasury (that is, $95,000 they collected
minus $70,000 they paid previously)
...
S Treasury has received $10,000 from Northshore, $20,000
from Westfall Steel, $35,000 from GE, $5000 from the supplier of the motors, and $25,000
from the retailers, for a total of $95,000
...
The
VAT money was deposited into the Treasury at several different times from several different
companies
...

The taxes that a company collects are called output taxes, and these are forwarded to the taxing
entity, less the amount of input taxes the company paid
...

Several dimensions of a VAT distinguish it from a sales tax or corporate income taxes
...

• The end user pays all of the value-added taxes, but VATs are not as obvious as a sales tax that
is added to the price of the item at the time of purchase (and displayed on the receipt)
...

• Value-added taxes are generally considerably higher than sales taxes, with the average
European VAT rate at 20% and the worldwide average at 15
...

• The VAT is essentially a “sales tax,” but it is charged at each stage of the product development
process instead of when the product is sold
...

• VAT rates vary from country to country and from category to category
...

471

472

Chapter 17

After-Tax Economic Analysis

EXAMPLE 17
...
It has three different
manufacturing units that specialize in manufacturing different transportation-related products, such as trucks, engines and axles, commercial vehicles, utility vehicles, and passenger cars
...
In one
particular accounting period, Tata had invoices from four different suppliers (vendors A,
B, C, and D) in the respective amounts of $1
...
8 million, $1
...
The products Tata purchased were subject to VAT rates of 4%, 4%, 12
...

(a) How much total VAT did Tata pay to its vendors?
(b) Assume that Tata’s products have a VAT rate of 12
...
If Tata’s sales during the period
were $9
...
Solve for X and then subtract it
from the purchase amount to determine the VAT charged by each vendor
...
An example computation for vendor A is
as follows:
X ϩ 0
...
04X ϭ 1,500,000
X ϭ $1,442,308
VATA ϭ 1,500,000 Ϫ 1,442,308
ϭ $57,692
Total VAT paid ϭ 57,692 ϩ 146,154 ϩ 122,222 ϩ 162,295
ϭ $488,363
(b) Total from Tata ϭ total VAT Ϫ VAT paid by vendors
ϭ 9,200,000(0
...
13

Vendor

Purchases, $

VAT Rate, %

Price before VAT, X, $

VAT, $

A
B
C
D

1,500,000
3,800,000
1,100,000
900,000

4
...
0
12
...
0

1,442,308
3,653,846
977,778
737,705

57,692
146,154
122,222
162,295

Total

488,363

CHAPTER SUMMARY
After-tax analysis does not usually change the decision to select one alternative over another;
however, it does offer a much clearer estimate of the monetary impact of taxes
...

Income tax rates for U
...
corporations and individual taxpayers are graduated or progressive—
higher taxable incomes pay higher income taxes
...
Taxes are reduced because of tax deductible items,
such as depreciation and operating expenses
...
Accordingly, key general cash ﬂow after-tax relations for each year are as
follows:
NOI ϭ gross income Ϫ expenses
TI ϭ gross income Ϫ operating expenses Ϫ depreciation ϩ depreciation recapture
CFBT ϭ gross income Ϫ operating expenses Ϫ initial investment ϩ salvage value
CFAT ϭ CFBT Ϫ taxes ϭ CFBT Ϫ (Te)(TI)
If an alternative’s estimated contribution to corporate ﬁnancial worth is the economic measure,
the economic value added (EVA) should be determined
...
The equivalent annual worths of CFAT and EVA estimates are the same
numerically, because they interpret the annual cost of the capital investment in different, but
equivalent manners when the time value of money is taken into account
...
The replacement
study procedure of Chapter 11 is applied
...

International corporate tax rates have steadily decreased, but indirect taxes, such as valueadded tax (VAT), have increased
...
The United States currently has no VAT
...
1 (a)

Deﬁne the following tax terms: graduated tax
rates, marginal tax rate, and indexing
...

17
...

17
...

17
...

(a) A new machine had a ﬁrst-year write-off of
$10,500
...

(c) An asset with a book value of $8000 was retired and sold for $8450
...

(e) An asset with a MACRS recovery period of 7
years has been owned for 10 years
...

(f )
(g)

(h)

(i)

The cost of goods sold in the past year was
$3,680,200
...
Based on winners holding these tickets, a rebate of $350
was sent to the manager
...

The cost to maintain equipment during the
past year was $641,000
...
5 Two companies have the following values on their
annual tax returns
...

Determine the percent of sales revenue each
company will pay in federal income tax
...
Determine the percentage error made relative to the exact taxes in
part (a)
...
6 Last year, one division of Hagauer
...
This year, TI is
estimated to be $600,000
...

(a) What was the average federal tax rate paid
last year?
(b) What is the marginal federal tax rate on the
additional TI this year?
(c) What will be the average federal tax rate this
year?
(d) What will be the NOPAT on just the additional $350,000 in taxable income?
17
...
5 million for the year
...
3 million
...
2%
...

17
...

If the state income tax rate is 7%, determine the
(a) average federal tax rate, (b) overall effective
tax rate, (c) total taxes to be paid based on the effective tax rate, and (d) total taxes paid to the state
and paid to the federal government
...
9 The taxable income for a motorcycle sales and repair business is estimated to be $150,000 this year
...
A new engine diagnostics system will cost $40,000 and
have an average annual depreciation of $8000
...

Compute the expected change in income taxes for
the year if the new system is purchased
...
10 C
...
Jordon Construction Services has operated for
the last 26 years in a northern U
...
state where the
state income tax on corporate revenue is 6% per
year
...
F
...
Because of pressing labor cost increases, the president wants to move to another state to reduce the
total tax burden
...
You are an engineer with the company and are asked to do the following
...
F
...

(b) Estimate the state tax rate that would be necessary to reduce the overall effective tax rate
by 10% per year
...
F
...

After-Tax Economic Analysis

CFBT and CFAT
17
...

17
...
13 For a year in which there is no initial investment P
or salvage value S, derive an equation for CFBT
that contains only the following terms: CFAT,
CFBT, D, and Te
...
14 Determine the cash ﬂow before taxes for Anderson
Consultants when the cash ﬂow after taxes was
$600,000, asset depreciation was $350,000 and the
company’s effective tax rate was 36%
...
15 Four years ago Sierra Instruments of Monterey,
California spent $200,000 for equipment to manufacture standard gas ﬂow calibrators
...
The gross income for year 4 was
$100,000, with operating expenses of $50,000
...
The
MACRS depreciation rate for year 4 is 7
...

17
...
The total
revenue for year 2 was $48 million, depreciation
was $8
...
Use a federal tax rate of 35% and a
state tax rate of 6
...

17
...
, researchers in medical science, is contemplating a commercial venture concentrating on proteins based on the new
X-ray technology of free-electron lasers
...
5 million CFAT is needed
...
Over
a 3-year period, the deductible expenses and depreciation are estimated to total $1
...
Of this, 50% is expenses and 50% is depreciation
...
18
through 17
...
(Show hand and spreadsheet solutions, as
instructed)
After 4 years of use, Procter and Gamble has decided to
replace capital equipment used on its Zest bath soap line
...
After-tax MARR is 10% per year, and Te
is 35% in the United States
...

Year
0

1

2

3

4

Purchase, $
Ϫ1900
Gross income, $
800
950
600
300
Operating expenses, $
Ϫ100 Ϫ150 Ϫ200 Ϫ250
Salvage, $
700

17
...

year 2 if the depreciation method had been straight
line instead of MACRS
...
24 Cheryl, an electrical engineering student who is
working on a business minor, is studying depreciation and ﬁnance in her engineering management
course
...
Help her, using asset
estimates made for a 6-year study period: P ϭ
$65,000, S ϭ $5000, GI ϭ $32,000 per year, AOC
is $10,000 per year, SL depreciation, i ϭ 12% per
year, Te ϭ 31%
...

17
...

All cash ﬂows are in $1000 units
...
19 Calculate MACRS depreciation and estimate the
CFAT series over 4 years
...

Year

GI

17
...

0
1
2
3
4

—
8
15
12
10

17
...

17
...

Use a spreadsheet to tabulate CFBT, CFAT, NOPAT,
and i* before and after taxes for 6 years of ownership
...
Salvage is expected to be zero
...
23 An asset purchased by Stratasys, Inc
...
In year 2, the
revenue was $490,000 with operating expenses of
$140,000
...
26 Use an effective tax rate of 32% to determine the
CFAT and NOPAT associated with the asset shown
below under two different scenarios: (a) with depreciation at $6000 per year and (b) with depreciation at $6000, $9600, $5760, and $3456 in years 1
through 4, respectively
...

Estimates, $
Year GI OE
0
1
2
3
4

—
8
15
12
10

P

D

TI

— Ϫ30 — —
Ϫ2
Ϫ4
Ϫ3
Ϫ5

Taxes CFAT NOPAT
—

Ϫ30

—

17
...
B
...
, an overland freight company, has
purchased new trailers for $150,000 and expects to
realize a net $80,000 in gross income over operating expenses for each of the next 3 years
...
Assume an
effective tax rate of 35% and an interest rate of
15% per year
...
Since MACRS takes an additional year to fully depreciate the basis, assume no CFBT beyond year 3, but include
any negative tax as a tax savings
...

17
...
The machinery
costs $200,000, has no salvage value, and the
CFBT estimate is $75,000 per year for up to 10
years
...
The
two depreciation methods to consider are:
MACRS with n ϭ 5 years and SL with n ϭ
8 years (neglect the half-year convention effect)
...

17
...
The
CFBT is estimated at $10,000 for the ﬁrst 4 years
and $5000 thereafter as long as the asset is retained
...
In present worth dollars, how
much of the cash ﬂow generated by the asset over
its recovery period is lost to taxes?
Depreciation Recapture and Capital Gains (Losses)
17
...
According to tax law,
the simulation is MACRS-depreciated using a
3-year recovery period
...

(b) Determine by how much the sale will cause
TI and taxes to change in year 5 if Te ϭ 35%
...
31
through 17
...

Open Access, Inc
...
Different depreciation, recovery period, and tax law practices in the three
countries where depreciable assets are located are summarized in the table
...
After-tax MARR ϭ 9% per year and Te ϭ 30% can
be used for all countries
...
31 For country 1, SL depreciation is $20,000 per year
...

17
...
Determine the (a) CFAT series and (b) PW
of depreciation, taxes, and CFAT series
...
33 For country 3, DDB depreciation for 5 years is
$40,000, $24,000, $14,400, $1600, and 0, respectively
...

17
...
31 through 17
...
For each criterion, select the country that provides the best PW
value
...
(Hint: The PW should
be minimized for some criteria and maximized for
other criteria
...
)
17
...
The asset was depreciated by the MACRS method and has a book value
of $100,800 at the time of sale
...

17
...
In order to have fresh water at the site, the
company purchased a turnkey reverse osmosis (RO)
system for $355,000
...
Four years after the system
was purchased, water lines from a local water system were extended to the truck stop, so Sun-Tex
sold the RO system for $190,000
...
The
MACRS depreciation rates are 10%, 18%, 14
...
52% for years 1 through 4, respectively
...
37 Freeman Engineering paid $28,500 for specialized
equipment for use with its new GPS͞GIS system
...
The
company sold the equipment after 2 years for
$5000 when it purchased an upgraded system
...

(b) What tax effect will this amount have?
17
...
Use them to determine the amount of income tax effect, if the effective tax rate is 35%
...

(b) A hi-tech machine was sold internationally
for $10,000 more than its purchase price just
after it was in service 1 year
...

(c) Land purchased 4 years ago for $1
...

(d) A 21-year-old asset was removed from
service and sold for $500
...
Classical straight line depreciation
was used for the entire recovery period
...
It was sold in
the fourth year of use for $2000
...
39 Sunnen Products Co
...
Louis, Missouri, makes
actuator hones for gas meter tubes where light-duty
metal removal is needed
...
Use the information
shown to determine the presence and amount of any
depreciation recapture, capital gain, or capital loss
...
5
3
3

300,000
15,500
5,000

295,000
275,000
19,500
12,500

After-Tax Economic Analysis
17
...
2%
...

17
...
Assume that the company has an effective
tax rate of 35% and it uses MACRS depreciation
for an asset that has a $27,000 salvage value
...
42 Estimate the approximate after-tax rate of return
for a project that has a ﬁrst cost of $500,000, a
salvage value of 20% of the ﬁrst cost after 3 years,
and annual gross income less operating expenses
of (GIϪOE) ϭ $230,000
...

17
...
No taxes are paid on retirement earnings until they are withdrawn; however,
she was told by her brother, an accountant, that this
is the equivalent of an 8% per year after-tax return
...
44 Bart is an economic consultant to the textile industry
...
If the
stated MARR in both companies is 12% per year
after taxes, determine if management at both companies should accept the projects
...

Effective tax rates are 34% for the larger corporation and 28% for the small company
...
45 Elias wants to perform an after-tax evaluation of
equivalent methods A and B to electrostatically remove airborne particulate matter from clean rooms
used to package liquid pharmaceutical products
...
Any tax effects when the equipment
is salvaged were neglected
...
Now, use
classical SL depreciation with n ϭ 5 years to evaluate the alternatives
...
46 A corporation uses the following: before-tax
MARR of 14% per year, after-tax MARR of
7% per year, and Te of 50%
...

Machine A
First cost, $
Salvage value, $
AOC, $ per year
Life, years

Machine B

Ϫ15,000
3,000
Ϫ3,000
10

Ϫ22,000
5,000
Ϫ1,500
10

The machine is retained in use for a total of 10 years,
then sold for the estimated salvage value
...

(b) After-tax PW analysis, using classical SL depreciation over the 10-year life
...

17
...
The GIϪOE estimate
is made for only 3 years; it is zero when each asset
is sold in year 4
...
48 Offshore platform safety equipment, designed for
special jobs, will cost $2,500,000, will have no
salvage value, and will be kept in service for exactly 5 years, according to company policy
...
The effective tax rate for the multinational oil company is 30%
...

17
...
This represents a
return of 20%
...
If the
corporation wants to realize an after-tax return of
10% per year, for how many more years must the
equipment remain in service?

After-Tax Replacement
17
...
(a) How is the
gain or loss calculated, and (b) how does it affect
the AW values in the study?
17
...
52 An asset that was purchased 2 years ago was expected
to be kept in service for its projected life of 5 years, but
a new version (the challenger) of this asset promises
to be more efﬁcient and have lower operating costs
...
The
defender had a ﬁrst cost of $300,000, but its market
value now is only $150,000
...
To simplify calculations for this
problem only, assume that SL depreciation was
charged at $60,000 per year and that it will continue
at that rate for the next 3 years
...
Assume
the company’s effective tax rate is 35%, and its
after-tax MARR is 15% per year
...

(b) Determine the CFAT in years 1 through 3 for
the challenger and defender
...

17
...
The depreciation charges for the next 3 years will be $69,960,
$49,960, and $35,720
...
Assume the company’s aftertax MARR is 12%
...
54 Perform a PW replacement study (hand and
spreadsheet solutions, if instructed) from the information shown using an after-tax MARR ϭ 12%
per year, Te ϭ 35%, and a study period of 4 years
...
Since no revenues are
estimated, all taxes are negative and considered
“savings” to the alternative
...

The effective tax rate is 35%, and the after-tax
MARR is 10% per year
...

Defender Challenger

17
...
The devices
can be sold on the used equipment market for an
estimated $15,000
...
The upgrade
investment will be depreciated over 3 years with
no salvage value
...
The new units
will have operating expenses of $7000 per year
...
(b) If the
challenger is known to be salable after 5 years
for an amount between $2000 and $4000, will
the challenger AW value become more or less
costly? Why?

First cost, $1000
Estimated S at purchase, $1000
Market value now, $1000
AOC, $1000 per year
Depreciation method
Recovery period, years
Useful life, years
Years owned

Ϫ45
5
35
Ϫ7
SL
8
8
3

Ϫ24
0
—
Ϫ8
MACRS
3
5
—

17
...
Pete’s accountant used an
after-tax MARR of 8% per year, Te ϭ 30%, and a
current market value of $25,000 to determine
AW ϭ $2100
...
Estimated CFBT is $15,000 per year
...

From the accountant, Ramon learned the current
equipment cost $20,000 when purchased and
reached a zero book value several years ago
...

17
...
The contract is up for renewal
now for a period of 1 year or 2 years only
...

The ﬁnance vice president wants to renew the
contract for 2 years without further analysis, but
the vice president for engineering believes it is
more economical to perform the maintenance
in-house
...
The
estimates for the in-house (challenger) alternative are as follows:
First cost, $
AOC, $ per year
Life, years
Estimated salvage

MACRS depreciation

−800,000
−120,000
4
Loses 25% of P annually:
End year 1, S ϭ $600,000
End year 2, S ϭ $400,000
End year 3, S ϭ $200,000
End year 4, S ϭ $0
3-year recovery period

17
...
The
salvage value after 10 years at that time was estimated to be $50,000
...
The new president has recommended
early replacement of the system with one that
costs $400,000 and has a 12-year life, a $35,000
salvage value, and an estimated AOC of $50,000
per year
...
The president wishes to know the
replacement value that will make the recommendation to replace now economically advantageous
...
For solution purposes, use classical SL depreciation for both systems
...

Economic Value Added
17
...
Why are these preferences predictable?
17
...
S
...
One component is a 4-year
project for a special-purpose transport ship-crane
for use in building permanent storm surge protection against hurricanes on the New Orleans coastline
...
MACRS depreciation with a 3-year
recovery is indicated
...
The
CFAT is shown below
...
They should have
the same value
...
75%
and Te ϭ 35%
...
61 Triple Play Innovators Corporation (TPIC) plans
to offer IPTV (Internet Protocol TV) service to
North American customers starting soon
...
Let Te ϭ 30% and after-tax MARR ϭ
8%; use SL depreciation (neglect half-year convention and MACRS, for simplicity) and a study
period of 8 years
...
2 million
3
...
62 Sun Microsystems has developed partnerships
with several large manufacturing corporations to
use Java software in their consumer and industrial
products
...
One major project involves using Java in commercial and industrial
appliances that store and cook food
...
For t ϭ 1 to 6 years,
Annual gross income, GI ϭ 2,800,000 Ϫ 100,000t
Annual operating expenses, OE ϭ 950,000 ϩ 50,000t

The effective tax rate is 30%, the interest rate is
12% per year, and the depreciation method chosen
for the $3,000,000 capital investment is a 5-year
MACRS ADS alternative that allows straight line
write-off with the half-year convention in years 1
and 6
...

Value-Added Tax
17
...
64 In Denmark, VAT is applied at a rate of 25%, with
few exceptions
...
Determine
the following
...

(b) The amount of tax vendor B sends to
Denmark’s Treasury
...

The following information is used in Problems 17
...
70
Ajinkya Electronic Systems, a company in India that
manufactures many different electronic products, has to
purchase goods and services from a variety of suppliers
(wire, diodes, LED displays, plastic components, etc
...
It also shows the purchases in
$1000 units that Ajinkya made (before taxes) from each
supplier in the previous accounting period
...
2 million, and Ajinkya’s products carry a VAT of 12
...

Supplier

VAT Rate, %

Purchases by
Ajinkya, $1000

A
B
C
D
E

4
...
5
12
...
3
32
...
65 How much VAT did supplier C collect?
17
...
67 What was the total amount of taxes paid by Ajinkya
to the suppliers?

17
...
68 What was the average VAT rate paid by Ajinkya in
purchasing goods and services?

17
...
71 A graduated income tax system means:
(a) Only taxable incomes above a certain level
pay any taxes
...

(c) Higher tax rates go with higher taxable
incomes
...

17
...

(b) The end user pays value-added taxes
...

(d) Value-added taxes are charged only on the
raw materials for product development
...
73 A small company has a taxable income that places
it in the 35% tax bracket
...
74 A company that has a 50% effective tax rate had
income of $200 million in each of the last 2 years
...
In the other year, the company
had deductions of only $80 million
...
75 Taxable income (TI) is deﬁned as:
(a) TI ϭ revenue ϩ operating expenses
Ϫ depreciation
(b) TI ϭ revenue Ϫ operating expenses
ϩ depreciation

(c)
(d)

TI ϭ revenue Ϫ operating expenses
Ϫ depreciation ϩ amortization
TI ϭ revenue Ϫ operating expenses
Ϫ depreciation

17
...
77 A subcontractor with an effective tax rate of 25%
has gross income of $55,000, other income of
$4000, operating expenses of $13,000, and other
deductions and exemptions of $11,000
...
78 When a depreciable asset is disposed of for less than
its current book value, the transaction is known as:
(a) An after-tax expense
(b) Capital loss
(c) Capital gain
(d) Depreciation recapture
17
...

If the effective tax rate is 40%, the values for depreciation (D), taxable income (TI), and taxes for
year 1 are closest to:

Year
0
1
2

Investment, GI − OE,
$
$
D, $

TI, $

Taxes,
$

Ϫ60,000
30,000
35
...
80 If the after-tax rate of return for a cash ﬂow series
is 11
...
8%
(b) 5
...
0%
(d) 28
...
81 An asset purchased for $100,000 with S ϭ $20,000
after 5 years was depreciated using the 5-year
MACRS rates
...
The asset
was actually sold after 5 years of service for
$22,000
...
53%
and 5
...
The after-tax cash ﬂow
from the sale is closest to:

(a)
(b)
(c)
(d)

$27,760
$17,130
$26,870
$20,585

17
...
Of the following, the statements that are commonly incorrect are:
1
...

2
...

3
...

4
...

(a) 1, 2, and 3
(b) 1 and 4
(c) 2
(d) 4

CASE STUDY
AFTER-TAX ANALYSIS FOR BUSINESS EXPANSION
Background
Charles was always a hands-on type of person
...

After some 20 years, it has grown signiﬁcantly
...
in the Metroplex, specializing in custom-made metal and stone fencing for commercial and residential sites
...

Pro-Fence is privately owned by Charles; therefore, the question of how to ﬁnance such an expansion has been, and still is,
the major challenge
...
Taking capital from the retained earnings of Pro-Fence is a
second possibility, but taking too much will jeopardize the current business, especially if the expansion were not an economic
success and Pro-Fence were stuck with a large loan to repay
...
He knows you are quite economically oriented and
that you understand the rudiments of debt and equity ﬁnancing and economic analysis
...
You have agreed to help him, as much as you can
...
Between his accountant and a small market survey of the

business opportunities in Victoria, the following generalized
estimates seem reasonable
...
5 million
Annual gross income ϭ $700,000
Annual operating expenses ϭ $100,000
Effective income tax rate for Pro-Fence ϭ 35%
Five-year MACRS depreciation for all $1
...
Repayment would be in 5 equal payments of interest and principal
...
A range of D-E mixes should be analyzed
...

Debt
Loan
Percentage Amount, $
0
50
70
90

750,000
1,050,000
1,350,000

Equity
Percentage
100
50
30
10

Investment
Amount, $
1,500,000
750,000
450,000
150,000

Case Study

Case Study Exercises
1
...
An after-tax return
of 10% is expected
...
)
2
...
If the time value of money

483

is neglected, what is the constant amount by which
this sum changes for every 10% increase in equity
funding?
3
...

He wants to know why this phenomenon occurs
...
After deciding on the 50-50 split of debt and equity ﬁnancing, Charles wants to know what additional bottomline contributions to the economic worth of the company
may be added by the new Victoria site
...

SECTION

TOPIC

LEARNING OUTCOME

18
...

18
...

18
...

18
...

18
...

18
...

T

his chapter includes several related topics about alternative evaluation
...
Then the determination and use
of the expected value of a cash ﬂow series are treated
...

Economic decisions that involve staged funding are very common in professional and
everyday life
...

18
...
Example parameters are ﬁrst cost, salvage value, AOC, estimated life, production rate, and materials costs
...

Economic analysis uses estimates of a parameter’s future value to assist decision makers
...
The effect of variation may be determined by using sensitivity analysis
...
Usually
one parameter at a time is varied, and independence with other parameters is assumed
...

Sensitivity analysis

In reality, we have applied this approach (informally) throughout previous chapters to determine
the response to variation in a variety of parameters
...
However, variation in the n or AOC value may indicate that the alternative’s measure of worth is very sensitive
to the estimated life or annual operating costs
...
For example, if an operating level of 90%
of airline seating capacity for a domestic route is compared with 70% for a proposed international
route, the operating cost and revenue per passenger-mile will increase, but anticipated aircraft
life will probably decrease only slightly
...

Sensitivity analysis routinely concentrates on the variation expected in estimates of P, AOC,
S, n, unit costs, unit revenues, and similar parameters
...
Parameters that are interest rate–
based are not treated in the same manner
...
If performed, sensitivity analysis on them is for speciﬁc values or over a
narrow range of values
...

Plotting the sensitivity of PW, AW, or ROR versus the parameter(s) studied is very helpful
...

This is the value at which the two alternatives are economically equivalent
...
Thus, several charts are constructed, and independence of each parameter is assumed
...
However, if the results are sensitive to the parameter value, several
intermediate points should be used to better evaluate the sensitivity, especially if the relationships are not linear
...
It may
be performed one parameter at a time using a spreadsheet or computations by hand or calculator
...

Here is a general procedure to follow when conducting a thorough sensitivity analysis
...

2
...

4
...

Determine which parameter(s) of interest might vary from the most likely estimated value
...

Select the measure of worth
...

To better interpret the sensitivity, graphically display the parameter versus the measure of
worth
...
When there are two or more alternatives, it is better to use
the PW or AW measure of worth in step 3
...
Example 18
...

EXAMPLE 18
...
expects to purchase a new asset for automated rice handling
...
The MARR for the company varies over a
wide range from 10% to 25% per year for different types of investments
...
Evaluate the sensitivity of PW by varying
(a) MARR, while assuming a constant n value of 10 years, and (b) n, while MARR is constant
at 15% per year
...

Solution by Hand
(a) Follow the procedure above to understand the sensitivity of PW to MARR variation
...
MARR is the parameter of interest
...
Select 5% increments to evaluate sensitivity to MARR; the range is 10% to 25%
...
The measure of worth is PW
...
Set up the PW relation for 10 years
...
A plot of MARR versus PW is shown in Figure 18–1
...
If the MARR is established at the upper end of the range, the
investment is not attractive
...
Asset life n is the parameter
...
Select 2-year increments to evaluate PW sensitivity over the range 8 to 12 years
...
The measure of worth is PW
...
Set up the same PW relation as in part (a) at i ϭ 15%
...
1

487

Determining Sensitivity to Parameter Variation
Figure 18–1

30,000
MA

Plot of PW versus MARR
and n for sensitivity
analysis, Example 18
...

RR

20,000

Present worth, $

n

10,000

10

15

6

8

20

25

MARR %

12

Life n

0
10

– 10,000

– 20,000

5
...
Since the PW measure is positive for all
values of n, the decision to invest is not materially affected by the estimated life
...
This insensitivity to changes in cash ﬂow in the distant
future is a predictable observation, because the P͞F factor gets smaller as n increases
...
The NPV function calculates PW for i values from 10% to
25% and n values from 8 to 12 years
...

Figure 18–2
Sensitivity analysis of
PW to variation in
(a) MARR values and
(b) life estimates,
Example 18
...

PW computation
ϭ NPV(C6,B$4:B$13)
ϩB$3

(a)

PW computation
ϭ NPV(15%,B$4:B$15)
ϩB$3

(b)

488

Chapter 18

Figure 18–3

50%

Sensitivity analysis graph
of percent variation from
the most likely estimate
...
This is sometimes called a spider graph
...
The variation in each parameter is indicated as a percentage deviation from the most likely estimate on the horizontal axis
...
This is the conclusion for indirect
cost in Figure 18–3
...
A reduction of 30%
from the expected sales price reduces the ROR from approximately 20% to −10%, whereas a
10% increase in price raises the ROR to about 30%
...
Observe the general shape of the sample sensitivity graphs in
Figure 18–4
...
The
graph indicates that the PW of each plan is a nonlinear function of hours of operation
...

PW, $ ϫ 1000

Plan B

100
Pl

an

A

50

0

1000

2000

Hours of operation per year

3000

18
...
Plan B is more attractive due to its relative insensitivity
...
It may be necessary to plot the measure of worth at intermediate
points to better understand the nature of the sensitivity
...
2
Columbus, Ohio needs to resurface a 3-kilometer stretch of highway
...
The ﬁrst method is a concrete surface for a cost of
$1
...
The second method is an asphalt
covering with a ﬁrst cost of $1 million and a yearly maintenance of $50,000
...

The city uses the interest rate on bonds, 6% on its last bond issue, as the discount rate
...
If the city expects an interstate to replace this stretch of highway in 10 years, which method should be selected?
(b) If the touch-up cost increases by $5000 per kilometer every 3 years, is the decision sensitive to this increase?

Solution
(a) Use PW analysis to determine the breakeven n value
...
, n
...

Ϫ500,000 ϩ 40,000(P͞A,6%,n) ϩ 75,000

(P͞F,6%, j ) ϭ 0

j

[18
...
1] switches from negative to positive PW values
...
The NPV functions
in column C are the same each year, except that the cash ﬂows are extended 1 year for each
Figure 18–5
Sensitivity of the breakeven life between two alternatives, Example 18
...

PW for 11 years
ϭ NPV(6%,$B$5:$B15)ϩ$B$4

PW for 12 years
ϭ NPV(6%,$B$5:$B16)ϩ$B$4

Year counter advances by 1

490

Chapter 18

Sensitivity Analysis and Staged Decisions

present worth calculation
...
4 years, concrete and asphalt resurfacing break even economically
...

(b) The total touch-up cost will increase by $15,000 every 3 years
...
1] is now

[

(

jϪ3
Ϫ500,000 ϩ 40,000(P͞A,6%,n) ϩ 75,000 ϩ15,000 ———
3

) ] [ ͚(P͞F,6%, j ) ] ϭ 0
j

Now the breakeven n value is between 10 and 11 years—10
...
The decision has become marginal for asphalt, since the
interstate is planned for 10 years hence
...
One conclusion is that the asphalt decision becomes more questionable as the
asphalt alternative maintenance costs increase; that is, the PW value is sensitive to increasing touch-up costs
...
2 Sensitivity Analysis Using Three Estimates
We can thoroughly examine the economic advantages and disadvantages among two or more
alternatives by borrowing from the ﬁeld of project scheduling the concept of making three estimates for each parameter: a pessimistic, a most likely, and an optimistic estimate
...

This approach allows us to study measure of worth and alternative selection sensitivity within a
predicted range of variation for each parameter
...

EXAMPLE 18
...
She
has made three estimates for the salvage value, annual operating cost, and life
...
For example, alternative B
has pessimistic estimates of S ϭ $500, AOC ϭ $Ϫ4000, and n ϭ 2 years
...
Perform a sensitivity analysis and determine the most
economical alternative, using AW analysis at a MARR of 12% per year
...

18
...
3
Alternative AW Values, $
Estimates

A

B

C

P
ML
O

Ϫ19,327
Ϫ14,548
Ϫ9,026

Ϫ12,640
Ϫ8,229
Ϫ5,089

Ϫ19,601
Ϫ13,276
Ϫ8,927

20
18

AW of costs, Ϫ$1000

16
Alternative C

14
Alternative A

12
10
8

Alternative B

6
4
2
0
1

2

3

4

5

6

7

8

9

Life n, years

Figure 18–6
Plot of AW of costs for different-life estimates, Example 18
...

Solution
For each alternative in Table 18–1, calculate the AW value of costs
...
Figure 18–6 is a plot of AW versus the three estimates of
life for each alternative
...

Comment
While the alternative that should be selected here is quite obvious, this is not normally the case
...
In
this case, it would be necessary to select one set of estimates (P, ML, or O) upon which to base
the decision
...

18
...
This means that probability and
samples are used
...
The reason is not that the computations are difﬁcult to perform or understand, but that
realistic probabilities associated with cash ﬂow estimates are difﬁcult to assign
...

The expected value can be interpreted as a long-run average observable if the project is repeated
many times
...
However, even for a single occurrence, the expected value is a
meaningful number
...
2]

i‫1؍‬

Xi ϭ value of the variable X for i from 1 to m different values
P(Xi) ϭ probability that a speciﬁc value of X will occur

where

Probabilities are always correctly stated in decimal form, but they are routinely spoken of in
percentages and often referred to as chance, such as the chances are about 10%
...
2] or any other relation, use the decimal equivalent of 10%,
that is, 0
...
In all probability statements the P(Xi) values for a variable X must total to 1
...

iϭm

͚ P(X ) ϭ 1
...

If X represents the estimated cash ﬂows, some will be positive and others will be negative
...
If
the expected value is negative, the overall outcome is expected to be a cash outﬂow
...

EXAMPLE 18
...
The marketing director estimates that for a typical 24-hour period there is a 50% chance of having a net cash ﬂow of $5000 and a 35% chance of $10,000
...
Determine
the expected net cash ﬂow
...

Using Equation [18
...
5) ϩ 10,000(0
...
05) Ϫ 1000(0
...
0 and it makes the computation complete
...
4 Expected Value Computations for Alternatives
The expected value computation E(X) is utilized in a variety of ways
...

• Evaluate the expected viability of a fully formulated alternative
...
5 illustrates the ﬁrst situation, and Example 18
...

18
...
5
There are many government incentives to become more energy-efﬁcient
...
The
owner pays a portion of the total installation costs, and the government agency pays the
rest
...
She has exceeded the annual budgeted amount of $50 million per
year in each of the previous 2 years
...
Over the last 36 months, the amount of average monthly payout
and number of months are shown in Table 18–3
...
Provided the same pattern continues, what is the expected value of the dollar increase in annual budget that is needed to
meet the requests?

TA BLE 18–3

Solar Panel Incentive Payouts, Example 18
...
5
4
...
2
2
...
,very high) to estimate the probability P(POj)
for each level, and make sure the total is 1
...

Level, j
Very high
High
Moderate
Low

Probability of Payout Level, P(POj)
P(PO1) ϭ 15͞36 ϭ 0
...
278
P(PO3) ϭ 7͞36 ϭ 0
...
111
1
...
2]
...
5(0
...
7(0
...
2(0
...
9(0
...
711 ϩ 1
...
621 ϩ 0
...
961 ($4,961,000)
The annual expected budget need is 12 ϫ 4
...
532 million
...
532 million per year
...
6
Lite-Weight Wheelchair Company has a substantial investment in tubular steel bending equipment
...
Estimated cash ﬂows
(Table 18–4) depend on economic conditions classiﬁed as receding, stable, or expanding
...
Apply expected value and PW analysis to determine if the equipment should be purchased
...

493

494

Chapter 18

Sensitivity Analysis and Staged Decisions

TA BLE 18–4

Equipment Cash Flow and Probabilities, Example 18
...
ϭ 0
...
ϭ 0
...
ϭ 0
...
2]
...
The PW values for the three scenarios are
PWR ϭ Ϫ5000 ϩ 2500(P͞F,15%,1) ϩ 2000(P͞F,15%,2) ϩ 1000(P͞F,15%,3)
ϭ Ϫ5000 ϩ 4344 ϭ $Ϫ656
PWS ϭ Ϫ5000 ϩ 5708 ϭ $ϩ708
PWE ϭ Ϫ5000 ϩ 6309 ϭ $ϩ1309
Only in a receding economy will the cash ﬂows not return the 15% to justify the investment
...
4) ϩ 708(0
...
2)
ϭ $283
At 15%, E(PW) Ͼ 0; the equipment is justiﬁed, using an expected value analysis
...
Computing
E(cash ﬂow) ﬁrst may be easier in that it reduces the number of PW computations
...

E(CF0) ϭ $Ϫ5000
E(CF1) ϭ 2500(0
...
4) ϩ 2000(0
...
5 Staged Evaluation of Alternatives
Using a Decision Tree
Alternative evaluation may require a series of decisions in which the outcome from one stage is important to the next stage of decision making
...

A decision tree includes:
•
•
•
•
•
•

More than one stage of alternative selection
...

Expected results from a decision at each stage
...

Estimates of economic value (cost or revenue) for each outcome
...

18
...
5

Probability
node

0
...
3
(b) Probability node with outcomes

D

D
D
Final outcomes

(c) Tree structure

Figure 18–7
Decision and probability nodes used to construct a decision tree
...

• A square represents a decision node with the possible alternatives indicated on the branches
from the decision node (Figure 18–7a)
...

• The treelike structure in Figure 18–7c results, with outcomes following a decision
...
These cash ﬂows are expressed in terms of PW,
AW, or FW values and are shown to the right of each ﬁnal outcome branch
...
This process, called solving the tree or rollback, is explained
after Example 18
...

EXAMPLE 18
...
S
...
He was recently approached by an international supermarket chain that wants to
market in-country its own brand of frozen microwaveable dinners
...
The current decision involves two alternatives: (1) Lease a facility in the United Arab
Emirates (UAE) from the supermarket chain, which has agreed to convert a current processing
facility for immediate use by Jerry’s company; or (2) build and own a processing and packaging facility in the UAE
...

The decision choices 2 years hence are dependent upon the lease-or-own decision made
now
...
This will be a mutual decision between the supermarket chain and Jerry’s company
...
Outcomes for the
future decisions are, again, good and poor market responses
...
If market response is good, the decision alternatives are four or two times original levels
...

Construct the tree of decisions and outcomes for Hill Products and Services
...
Identify the decision nodes and branches, and then develop the tree using the branches and the outcomes of
good and poor market for each decision
...

Stage 1 (decision now):
Label it D1
...

Outcomes: good and poor markets
...
5ϫ

Poor

Good

Good
0
...

18
...

Outcomes: good market, poor market, and out of business
...
5ϫ); stop production (0ϫ)
The alternatives for future production levels (D2 through D5) are added to the tree and followed by the market responses of good and poor
...

To utilize the decision tree for alternative evaluation and selection, the following additional
information is necessary for each branch:
• The estimated probability that each outcome may occur
...
0
for each set of outcomes (branches) that result from a decision
...

Decisions are made using the probability estimate and economic value estimate for each outcome branch
...
2]
...
Start at the top right of the tree
...

2
...

E(decision) ‫(͚ ؍‬outcome estimate)P(outcome)

[18
...

3
...

4
...

5
...

EXAMPLE 18
...
If the product is marketed, the
next decision is to take it international or national
...
The probabilities for each outcome and PW of CFBT
(cash ﬂow before taxes) are indicated
...
Determine the
best decision at the decision node D1
...

1
...

2
...
3]
...
2 in ovals
are determined as
E(international decision) ϭ 12(0
...
5) ϭ 14
E(national decision) ϭ 4(0
...
4) Ϫ 1(0
...
2
The expected PW values of 4
...

497

498

Chapter 18

Sensitivity Analysis and Staged Decisions

14
14

International

D2

PW of
CFBT
($ million)
0
...
5

0
...
16

Low

Market

4
...
2
2

9
National

D1

0
...
2

0
...
4
0
...
2

4
...
4
0
...
2

12
16
4
–3
–1
6
–3
6
–2
2

9

Sell

1
...
8
...
Select the larger expected value at each decision node
...
2 (international) at D3
...
Calculate the expected PW for the two D1 branches
...
2) ϩ 4
...
8) ϭ 6
...
0) ϭ 9
The expected value for the sell decision is simple since the one outcome has a payoff of 9
...

5
...

18
...
When the decision to invest more or less can be delayed into the future, the
problem is called staged funding
...
Decision makers may opt to (1) build the capacity to
supply100,000 per month to market immediately or (2) build capacity to supply 25,000 per
month now and test the market’s receptivity
...
Of course, if an aggressive competitor
enters the scene, or the economy falters, the staged funding decision will change as warranted
...
Before we go further, some deﬁnitions are needed
...

18
...
The options usually involve physical (real) assets, buildings, equipment, materials, and
the like, thus, the word real
...
The investment alternatives present varying amounts of risk, which is estimated by probabilities of occurrence for predictable future events
...
The estimated cash ﬂows and other
consequences of these delays are analyzed with risk taken into account to the degree possible
...
g
...
A
decision may be to expand, continue as is, contract, abandon, or replicate the alternative at the
time the option must be exercised
...
After some illustrations of real options, we will discuss the probabilistic
dimensions
...

Industrial Setting
New markets—Purchase equipment and staff to enter an expanding international market over
the next 5 years
...

Removing car models—Ford Motor Company can decide to maintain production on an established car model with dwindling sales for the next 3 years or can opt to discontinue the model
in stages over a 1- or 2-year period
...
The drilling may not be justiﬁed at this time, but the contract offers
the option to drill were it to become economically advantageous based on events such as increased oil prices or improved recovery technology
...
The price of the option is
the cost of the extended warranty
...

House insurance—When a homeowner has no mortgage to pay, maintaining house insurance
is an option
...
g
...
Self-insurance, where
money is set aside for potential damages while accepting the risk that a major event will take
place, is an option for the homeowner
...

• Anticipated future options and cash ﬂow estimates (double production with annual net cash
ﬂows estimated)
...

• Market and risk-free interest rates (expected market MARR of 12% per year and inﬂation
rate estimate of 4% per year)
...

• Economic criterion used to make a decision (PW, ROR, or other measure of worth)
...
Example 18
...

EXAMPLE 18
...
, has
developed and ﬁeld-tested a modularized, scalar solar thermal electric (STE) generation
system that is relatively inexpensive to purchase and has an efﬁciency considerably better
than traditional photovoltaic (PV) panels
...
Additionally, a contract
with a consortium of sunbelt states has been offered, but not accepted thus far, for a total of
$1
...
By contract, the units will be marketed through
the state energy departments with all revenue going to the state treasuries
...

Condition

Option

CIF Funding

Consortium Contract

2 ϫ production
level

Additional $10 million in
year 2

Additional 8 years;
$4 million in years 3–10

Sales are excellent
1 ϫ production
(3000–5000 units͞year)
level

Nothing; no salvage after
10 years

Additional 8 years;
$1
...
5 million after 5 years

Additional 3 years;
$1
...

(b) The base case is the 1 ϫ production level with the 8-year follow-on contract from the consortium
...

(c) Determine the PW values for each possible ﬁnal outcome at 10% per year, and identify the
best economic option when the stage 2 funding decision must be made
...
There are 2 outcome branches initially (accept option; decline option) and four ﬁnal branches for the
accept decision at D1, based upon sales level
...

(SolarScale has other ways to pursue revenue that are not represented in this abbreviated example
...
The resulting PW1ϫ Ͻ 0 as shown
below indicates the contract is not justiﬁed economically
...
5(P͞A,10%,10)
ϭ $Ϫ0
...
The i* values are also shown using the IRR function
...
5 million (½ ϫ level) or after 2 years for $5 million
(stop) is included
...

18
...
9
...
9
...
If SolarScale and CIF, the ﬁnancial backers, are not convinced that the sales level will exceed
5000 units per year, the contract option should be declined
...

Of the characteristics listed above for real option situations, the primary one absent in
Example 18
...
Decision making under
risk is covered more extensively in the next chapter; however, we can use the following deﬁnition for this discussion of real options analysis
...

Risk

502

Chapter 18

Sensitivity Analysis and Staged Decisions

A coin has two sides
...
If the coin is intentionally biased in weight, such that 58% of
the time it lands heads up, then the long-run probability of heads is P(heads) ϭ 0
...
Since the sum
of probabilities across all possible values must add to 1, the biased coin has P(tails) ϭ 0
...

There are a couple of points worth mentioning about risk and the calculated PW values for real
options analysis
...
(We shall see this in Example 18
...
)
• Since one of the objectives of real options analysis is to evaluate the economic consequences
of delaying a decision, a PW value of the base case that is moderately positive means that the
project is justiﬁed and should be accepted immediately, without the decision delay
...
Thus, the project should be rejected now, not
delayed for a future decision
...

EXAMPLE 18
...
9 about SolarScale’s state-consortium contract offer, some expected sales information was collected
...
The results are as follows:
Probability of Outcome
Excellent
SolarScale
CIF
Consortium

Poor

0
...
8
0
...
5
0
...
4

Use these probability estimates to determine the expected PW value, provided equal weighting
is given to each representative’s input
...

Excellent: From 2 ϫ level and 1 ϫ level, select 2 ϫ with PW ϭ $1
...

Poor: From ½ ϫ level and stop now, select ½ ϫ with PW ϭ $Ϫ2
...

In $ million, E(PW) for each organization is
E(PW for SolarScale) ϭ 1
...
5) Ϫ 2
...
5) ϭ $Ϫ0
...
97(0
...
76(0
...
02
E(PW for consortium) ϭ 1
...
6) Ϫ 2
...
4) ϭ $0
...
33(Ϫ0
...
02 ϩ 0
...
23 ($230,000)
The base case of 1 ϫ level production in part (b) of Example 18
...
When compared with the positive E(PW) result here, we see that with consideration of the different options of production level and probabilities for sales level, the expected
PW has increased to a positive value
...

Problems

503

There are many other examples and dimensions of real options analysis in engineering economics and in the area of ﬁnancial analysis, where options analysis got its start some years ago
...

CHAPTER SUMMARY
In this chapter the emphasis is on sensitivity to variation in one or more parameters using a
speciﬁc measure of worth
...

When several parameters are expected to vary over a predictable range, the measure of worth
is plotted and calculated using three estimates for a parameter—most likely, pessimistic, and
optimistic
...
Independence between parameters is assumed in all these analyses
...

Decision trees are used to make a series of alternative selections
...
It is necessary to make several types of estimates for a decision tree: outcomes for each possible decision, cash ﬂows, and probabilities
...

Staged funding over time can be approached using the evolving area of real options
...

PROBLEMS
Sensitivity to Parameter Variation
18
...
The company will
make the investment only if it will result in a rate
of return of 15% per year or higher
...

18
...
For the past 5 years they have been
able to invest one of their salaries ($50,000 per
year, which includes employer contributions)
while living off the other one
...
If they
have gotten a rate of return of 10% per year on

their investments and expect to continue at this
ROR, is reaching their goal of $2
...
e
...

18
...
The company can invest $80,000 now, 1 year from now,
or 2 years from now
...
That is,
the savings will be $25,000, $26,000, or $29,000
per year if the investment is made now (year 0),
in 1 year, or in 2 years, respectively
...

18
...
Option 1 is a low-pressure seawater reverse osmosis (SWRO) system that will operate at
500 psi with a ﬁxed cost of $465 per day and an
operating cost of $0
...
A second
option is a higher-pressure SWRO system offered
by vendor X that operates at 800 psi and will have
a lower ﬁxed cost of $328 per day (because of
fewer membranes); however, its operating cost
will be $1
...
A third option is
also a high-pressure SWRO system from vendor Y,
who claims that its system will have a lower operating cost of $1
...
Determine if the
selection of a low- or high-pressure system is dependent on the lower operating cost offered by
vendor Y
...
5 A machine that is currently used in manufacturing
circuit board card locks has AW ϭ $–63,000 per
year
...
Three
different engineers have given their opinion, about
what the salvage value of the new machine will be
3 years from now: $10,000, $13,000, and $18,000
...
6 An equipment alternative is being economically
evaluated separately by three engineers at
Raytheon
...
The engineers disagree, however, on the
estimated revenue the equipment will generate
...
Jane
states that this is too low and estimates $14,000,
while Carlos estimates $18,000 per year
...

18
...
The ﬁrst
cost of the equipment is $250,000, and it will
likely have a salvage value of $90,000 in 3 years,
at which time he will not need the equipment
anymore
...
Alternatively, the owner can
subcontract the work for $175,000 per year
...
He thinks it
might be worth as little as $10,000 in 3 years
(a scrap value)
...

Sensitivity Analysis and Staged Decisions

18
...
5 million for a
plant expansion is not sure what the interest rate
will be when it applies for the loan
...
The company will only
move forward with the project if the annual worth
of the expansion is below $5
...
The M&O
cost is ﬁxed at $3
...
The salvage
could be $2 million if the interest rate is 10% or
$2
...
Is the decision to
move forward with the project sensitive to the interest rate and salvage value estimates?
18
...
Estimated cash ﬂows are as
follows:
Batch
First cost, $
Annual cost, $ per year
Salvage value for any year, $
Life, years

Continuous

Ϫ80,000
Ϫ55,000
10,000
3–10

Ϫ130,000
Ϫ30,000
40,000
5

The chief operating ofﬁcer (COO) has asked you
to determine if the batch option would ever have a
lower annual worth than the continuous ﬂow system, using interest rates over a range of 5% to 15%
for the batch option but only 15% for the continuous ﬂow system
...
)
18
...

Fixed cost ϭ $300,000 per year
Cost per unit ϭ $40
Revenue per unit ϭ $70
(a)

(b)

What is the range in breakeven quantity if
there is possible variation in the ﬁxed cost
from $200,000 to $400,000 per year? (Use
$50,000 increments
...
11
through 18
...

A new online patient diagnostics system for surgeons will
cost $200,000 to install, cost $5000 annually to maintain
and will have an expected life of 5 years
...
Examine the sensitivity of present worth
to variation in selected parameter estimates, while others
remain constant
...
11 Sensitivity to ﬁrst cost variation: $150,000 to
$250,000 (−25% to ϩ25%)
...
12 Sensitivity to revenue variation: $45,000 to $75,000
(−25% to ϩ25%)
...
13 Sensitivity to life variation: 4 years to 7 years (−20%
to ϩ40%)
...
18 (a)
18
...

18
...
She expects to retire and start to draw
a total of R ϭ $60,000 per year 1 year after the
20th deposit
...
Determine and comment on
the sensitivity of the size of the annual withdrawal R for variations in A and i
...

(a) Variation of Ϯ5% in the annual deposit A
...

18
...
Tests
on metal composites that rely upon scanning electron microscope results can be subcontracted, or
the labs can purchase new equipment
...
Use the
AW method and plot the results on a sensitivity
graph (like Figure 18–3)
...
17 Titan manufactures and sells gas-powered electricity generators
...
Cost and
savings estimates are made, but the savings estimate is unreliable at this time
...

(b)

Ϫ50,000
Ϫ7,500
15,000
5,000
5

Ϫ37,500
Ϫ8,000
13,000
3,700
5

Graph the sensitivity of what a person
should be willing to pay now for a 9%,
$10,000 bond due in 10 years if there is a
30% change in (1) face value, (2) dividend
rate, or (3) required nominal rate of return,
which is expected to be 8% per year, compounded semiannually
...

If the investor did purchase the $10,000 face
value bond at a premium of 5% (i
...
, 5%
above face value) and all your other estimates were correct, that is, 0% change, did
he pay too much or too little? How much?

Three Estimates
18
...

The company is considering making the parts
in-house through the purchase of equipment that
will have a ﬁrst cost of $240,000 with an estimated salvage value of $30,000 after 5 years
...
Determine if
the company should purchase the equipment
under any of the operating cost scenarios
...

18
...
The $30,000 per year
lease agreement will be a net, net, net lease,
which means that the lessee (Astor) will pay the
real estate taxes on the leased space, the building
insurance on the leased space, and the common
area maintenance
...
A new
building will cost $880,000 to purchase, but there
is considerable uncertainty about what it will be
worth in 20 years, which is the planning period
selected
...
Determine if Astor should
purchase the building under any of the estimated
resale values at i ϭ 10% per year
...
21 Holly Farms is considering two environmental
chambers to accomplish detailed laboratory conﬁrmations of online bacteria tests in chicken
meat for the presence of E
...
There is some uncertainty
about how long the D103 chamber will be useful
...
The estimated
salvage value will remain the same
...

Chamber D103 Chamber 490G
Installed cost, $
AOC, $ per year
Salvage value at 10% of P, $
Life, years

Ϫ400,000
Ϫ4,000
40,000
2, 3, or 6

Ϫ250,000
Ϫ3,000
25,000
2

18
...
However, in
a receding economy the expected return is 8%
...
An expanding
economy causes the estimates of asset life to go
down about 20%, and a receding economy makes
the n values increase about 10%
...
(c) Considering all the
analyses, under which scenario, if any, should plan
M or Q be rejected?
Plan M
Initial investment, $
Cash ﬂow, $ per year
Life, years

Plan Q

Ϫ100,000
ϩ15,000
20

Ϫ110,000
ϩ19,000
20

Expected Value
18
...

The probabilities are 20%, 50%, and 30% for revenues of $800,000, $1,000,000, and $1,100,000
per year, respectively, and operating expenses are
constant at $200,000 per year
...
24 A company that manufactures ampliﬁed pressure
transducers is trying to decide between a dualspeed and a variable-speed machine
...
The results can be summarized as follows: there is a 32% chance of getting $20,000, a
45% chance of getting $28,000, and a 13%
chance of getting $34,000
...
Calculate the
expected salvage value
...
25 The average success probability for a wildcat oil
well drilled in the Wind River basin 7 miles from
the nearest existing production well is estimated
to be 13%
...
5 million, $1
...
4 million, what is the expected income from
the well?
18
...
Determine the expected value of the
monthly income, if economic conditions remain
the same
...
27 Determine the expected maximum rainfall intensity in El Paso, Texas for the month of July using
the estimated probabilities shown
...
4

0
...
2

0
...
28 There are four estimates made for the anticipated
cycle time to produce a subcomponent
...
(a) If
equal weight is placed on each estimate, what is
the expected cycle time? (b) If the largest time is
disregarded, what is the percent reduction in the
expected time?
18
...

Your ofﬁce partner told you that the low bid is
$3200 per year
...
30 A total of 40 different proposals were evaluated by
the IRAD (Industrial Research and Development)
committee during the past year
...
Their rate of return estimates are summarized with the i* values rounded to the nearest

507

Problems

integer
...

Proposal
ROR, i*%

Number of
Proposals

Ϫ8
Ϫ5
0
5
8
10
15

1
1
5
5
2
3
3
20

18
...
The amount of rainfall experienced in a short time may cause damage in varying amounts, and the wall increases in cost in
order to protect against larger and faster rainfalls
...
31 Beckman Electronics has performed an economic
analysis of proposed service in a new region of the
country
...
The optimistic and pessimistic values each have an estimated 20% chance
of occurring
...

Optimistic
FW value, $

$35,000 per year
...

Most Likely
50,000

Ϫ25,000

Probability of
Greater Rainfall

First Cost
of Wall, $

2
...
25
2
...
0
3
...
3
0
...
05
0
...
005

200,000
225,000
300,000
400,000
450,000

Pessimistic

300,000

Rainfall, Inches
per 30 Minutes

18
...

Because of its location, there is a 30% chance of
a 120-day season of good outdoor weather, a
50% chance of a 150-day season, and a 20%
chance of a 165-day season
...
The
feature will cost $375,000 to construct and require a $25,000 rework each 4 years; and the annual maintenance and insurance costs will be
$56,000
...

If a life of 10 years is anticipated and a 12% per
year return is expected, determine if the addition
is economically justiﬁed
...
33 The owner of Ace Rooﬁng may invest $200,000 in
new equipment
...
The annual extra revenue will depend upon the state of
the housing and construction industry
...
Real
estate economists estimate a 50% chance of the
slump lasting 3 years and they give it a 20% chance
of continuing for 3 additional years
...
The principal and interest will be repaid over
a 10-year period
...
A discount rate of
6% per year is applicable
...

Decision Trees
18
...
(This decision branch is part of a
larger tree
...
4
0
...
3

Value, $
55
– 30
10

D3

0
...
4

– 17
0

18
...
If decisions D1, D2, and D3 are
all options in a 1-year period, ﬁnd the decision

508

Chapter 18

Sensitivity Analysis and Staged Decisions

path that maximizes the outcome value
...

annual equivalent cost for each alternative is dependent upon speciﬁc circumstances of the plant, producer, or contractor
...
Construct and solve a decision tree to determine the least-cost alternative to
provide the subassemblies
...
9

D3
0
...
6

0
...
Make

500

$80

90
2
...
3
0
...
4
0
...
5

$30

3
...
37 Decision D4, which has three possible alternatives—x, y, or z—must be made in year 3 of a 6-year
study period in order to maximize the expected
value of present worth
...

Investment Cash flow, $1000
Required, $
Outcome
Years
6 probability
3
4
5

50

50

50

0
...
3

30

40

50

0
...
55

– 350,000 190

170

150

0
...
3

– 200,000

Low
x
High
D4

y
Low
z
High
Low

– 75,000

Plant:
A
B
C
Quantity:
Ͻ5000, pay premium
5000 available
Ͼ5000, forced to buy
Delivery:
Timely delivery
Late delivery, then
buy some off shelf

Probability
0
...
5
0
...
2
0
...
1

Ϫ550,000
Ϫ250,000
Ϫ290,000

0
...
5

Ϫ175,000
Ϫ450,000

200
– 100

$20

High

Outcomes

Annual Cost for
5000 Units, $
per Year

18
...
The subassemblies can be obtained in one of three ways:
(1) Make them in one of three plants owned by the
company; (2) buy them off the shelf from the one
and only manufacturer; or (3) contract to have them
made to speciﬁcations by a vendor
...
39 The president of ChemTech is trying to decide
whether to start a new product line or purchase a
small company
...
To make the product for a 3-year period will
require an initial investment of $250,000
...
5), $90,000 (0
...
1)
...
Market surveys indicate a
55% chance of increased sales for the company and
a 45% chance of severe decreases with an annual
cash ﬂow of $25,000
...
Increased sales could be $100,000 the ﬁrst 2 years
...
This expansion could generate cash ﬂows
with indicated probabilities as follows: $120,000
(0
...
3), and $175,000 (0
...
If expansion is not chosen, the current size will be maintained with anticipated sales to continue
...
Use
the description given and a 15% per year return to
do the following
...

(b) Determine the expected PW values at the
“expansion/no expansion” decision node after
2 years, provided sales are up
...

509

Additional Problems and FE Exam Review Questions

(d)

Explain in words what would happen to the
expected values at each decision node if the
planning horizon were extended beyond
3 years and all cash ﬂow values continued as
forecasted in the description
...
40 A privately held company that makes chips that are
essential for high-volume data storage is valued at
$3 billion
...
1 billion
...
41 A company that is considering adding a new product line has determined that the ﬁrst cost would be
$80 million
...

Instead of expanding now, the company could implement a test program for 1 year in a limited area
that will cost $4 million
...
) This will provide the company
with the option to move forward or cancel the project
...
In this case, the pessimistic estimate will be eliminated, and equal probability will
be placed on the remaining revenue projections
...
42 Dow Chemical is considering licensing a low liquid discharge (LLD) water treatment system from a
small company that developed the process
...
8 million
plus 25% of sales
...
9 million plus 30% of sales
...

18
...
There is a 1-year sales warranty with the purchase; however, an extended warranty is available
for $2500 that will cover the same repairs and component failures as the 1-year warranty for 3 additional years
...
To help with
her decision, the salesman provided three typical
sets of historical data on estimated repair costs for
used cars
...

Year
1
Repair cost, $ per year:
A
B
C

2

3

4

0
0
0

Ϫ500
Ϫ1000
0

Ϫ1200
Ϫ1400
Ϫ500

Ϫ850
Ϫ400
Ϫ2000

The salesman said case C is the base case, since it
shows that the extended warranty is not needed because the cost of repairs equals the warranty cost
...

(a) If Abby assumes that each repair cost scenario
has equal probability of occurring with her
car, and money is worth 5% per year to her,
how much should she be willing to pay for the
extended warranty that is offered at $2500?
(b) If the base case actually occurs for her car
and she does not purchase the warranty, what
is the PW value of the expected future costs
at i ϭ 5% per year?

ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
18
...
45 When the measure of worth is plotted versus percent
change for several parameters, the parameter that is
the most sensitive in the economic analysis is the one:
(a) That has the steepest curve
(b) That has the ﬂattest curve
(c) With the largest present worth
(d) With the shortest life

510

Chapter 18

Sensitivity Analysis and Staged Decisions

18
...
47 For annual worth values of $30,000, $40,000, and
$50,000 with chances of 20%, 20%, and 60%, respectively, the expected AW is closest to:
(a) $34,000
(b) $40,000
(c) $44,000
(d) $48,000
18
...
A challenger will have a
ﬁrst cost of $90,000, an operating cost of $29,000,
and a salvage value after 5 years that may vary considerably
...
49 A decision tree includes all of the following except:
(a) Probability estimates for each outcome
(b) Measure of worth as the selection criterion
(c) Expected results from a decision at each
stage
(d) The MARR
18
...
51 A small manufacturing company needs to purchase a machine that will have a ﬁrst cost of
$70,000
...
If the
company’s MARR is 10% per year, the maximum
amount the company should pay for the option is
closest to:
(a) $5850
(b) $6365
(c) $6845
(d) $7295

CASE STUDY
SENSITIVITY TO THE ECONOMIC ENVIRONMENT
Background and Information

Case Study Questions

Berkshire Controllers usually ﬁnances its engineering projects with a combination of debt and equity capital
...
Normally, a 7% per year return is expected
...
The following estimates are the most likely values for two expansion
plans currently being evaluated
...
All monetary
estimates are in $1000 units
...

1
...
Are the PW values sensitive to varying life estimates?
3
...
As
cities grow and extend their boundaries to outlying areas,
they often inherit water systems that were not constructed
according to city codes
...
To avoid these problems, city ofﬁcials sometimes install water systems beyond the existing city limits
in anticipation of future growth
...

From about a dozen suggested plans, ﬁve methods were
developed by an executive committee as alternative ways
of providing water to the study area
...
Six attributes or factors were
used in the initial rating: ability to serve the area, relative
cost, engineering feasibility, institutional issues, environmental considerations, and lead time requirement
...
After the top three alternatives were identiﬁed, each was subjected to a detailed economic evaluation for selection of the best alternative
...
The annual cost (an AW value) was
then divided by the population served to arrive at a monthly
cost per household
...
Alternatives 1A, 3, and 4
were determined to be the three best and were chosen for further evaluation
...

Alternative 1A
Capital cost
Land with water rights: 1720 hectares
@ $5000 per hectare
Primary treatment plant
Booster station at plant
Reservoir at booster station
Site cost
Transmission line from river
Transmission line right-of-way
Percolation beds
Percolation bed piping
Production wells
Well ﬁeld gathering system
Distribution system
Additional distribution system
Reservoirs
Reservoir site, land, and development
Subtotal
Engineering and contingencies

$8,600,000
2,560,000
221,425
50,325
40,260
3,020,000
23,350
2,093,500
60,400
510,000
77,000
1,450,000
3,784,800
250,000
17,000
22,758,060
5,641,940

Total capital investment

$28,400,000

Results of Rating Six Factors for Each Alternative, Case Study
Factors

Alternative

Description

1A

Receive city
water and
recharge wells
Joint city and
county plant
County treatment
plant
Desalt
groundwater
Develop military
water

3
4
8
12

Ability
to
Supply Relative Engineering Institutional Environmental
Lead Time
Area
Cost
Feasibility
Issues
Considerations Requirement Total
5

4

3

4

5

3

24

5

4

4

3

4

3

23

4

4

3

3

4

3

21

1

2

1

1

3

4

12

5

5

4

1

3

1

19

512

Chapter 18

Sensitivity Analysis and Staged Decisions

Alternative 4

Maintenance and operation costs (annual)
Pumping 9,812,610 kWh per year
@ $0
...
76 per month

Household cost

ϭ 28,400,000(A͞P,8%,20)
ϩ 1,060,419

On the basis of the lowest monthly household cost, alternative 3 (joint city and county plant) is the most economically
attractive
...
95
ϭ $69
...
38 per month

TA BLE 18–6

1
...
If the ability to supply area and relative cost factors were
each weighted 20% and the other four factors 15% each,
which alternatives would be ranked in the top three?
3
...
If alternative 1A served 100% of the households instead
of 95%, by how much would the monthly household
cost decrease?
5
...
Three estimates are made for each
parameter in Table 18–6
...

The estimated number of households (4980) is determined to be the pessimistic estimate
...

(b) Consider the monthly cost per household for alternative 4, the optimistic estimate
...
What is
the number of households that would have to be
available in order for this option to have exactly
the same monthly household cost as that for alternative 3 at the optimistic estimate of 5230 households?

CHAPTER 19

More on
Variation and
Decision
Making under
Risk

L E A R N I N G

O U T C O M E S

Purpose: Incorporate decision making under risk into an engineering economy evaluation using probability, sampling,
and simulation
...
1

Risk versus certainty

• Understand the approaches to decision making
under risk and certainty
...
2

Probability and distributions

• Construct a probability distribution and
cumulative distribution for one variable
...
3

Random sample

• Obtain a random sample from a cumulative
distribution using a random number table
...
4

␮, ␴, and ␴2

• Estimate the population expected value, standard
deviation, and variance from a random sample
...
5

Simulation

• Use Monte Carlo sampling and spreadsheetbased simulation for alternative evaluation
...
Fundamentals discussed include variables; probability distributions, especially their graphs and properties
of expected value and dispersion; random sampling; and the use of simulation to
account for estimate variation in engineering economy studies
...
These techniques are more time-consuming than using estimates made with certainty, so they should be used primarily for critical parameters
...
1 Interpretation of Certainty,
Risk, and Uncertainty
All things in the world vary—one from another, over time, and with different environments
...
Except for the use of breakeven analysis, sensitivity analysis, and a very
brief introduction to expected values, virtually all our estimates have been certain; that is, no
variation in the amount has entered into the computations of PW, AW, ROR, or any relations
used
...
Decision making under certainty is, of course, not present in the real world now and surely not in the
future
...

To allow a parameter of an engineering economy study to vary implies that risk, and possibly
uncertainty, is introduced
...
Virtually all decision making is
performed under risk
...

Decision making under uncertainty means there are two or more values observable, but the
chances of their occurring cannot be estimated or no one is willing to assign the chances
...

For example, consider the states of nature to be the rate of national inﬂation in a particular
country during the next 2 to 4 years: remain low, increase 2% to 6% annually, or increase 6%
to 8% annually
...

Example 19
...

EXAMPLE 19
...
One of each piece of equipment must be purchased
...
The total for each piece of equipment is 100%
...

(a) Consider the ratings as the chance out of 100 that the bid will be chosen, and plot cost
versus chance for each vendor
...
Plot this
range with an equal chance of 1 in 14 that any amount in between these limits is possible
...

Risk

516

Chapter 19

Figure 19–1

Equipment A

Equipment B

Bid, $1000

Rating, %

Bid, $1000

Rating, %

3,000
5,000
10,000

65
25
10

8,000
10,000
15,000

33
...
3
33
...
3

33
...
3

7
8
9
Cost, $ million

10

11

12

25

27

29

20

0
1

2

3

4

5

6

13

14

15

16

(a) Specific values
Total cost

1
—
14
Chance

Plot of cost estimates
versus chance for (a)
each piece of equipment and (b) total cost
range, Example 19
...

More on Variation and Decision Making under Risk

0
7

9

11

13

15

17
19
21
Cost, $ million

23

(b) Continuous range

Solution
(a) Figure 19–1a plots the speciﬁc bids for equipment A and B
...
No values between the speciﬁc bids have any chance of occurring, according to the single-estimate bids from the three vendors
...

Tom decided to make his estimate of total cost continuous between these two extremes
...
Rather the entire range from $11 million to $25 million with a chance for
every total cost in between is included
...
Now, the sum is a continuous value
...
In the graph for the sum of the cost for equipment A and B (Figure 19–1b),
the y axis values are continuous over a speciﬁc range
...

19
...
A summary of the
meaning and use for each type of analysis follows
...

Deterministic estimates are made and entered into measure of worth relations—PW, AW, FW,
ROR, B͞C—and decision making is based on the results
...
A typical example is
an asset’s ﬁrst cost estimate made with certainty, say, P ϭ $50,000
...

The term deterministic, in lieu of certainty, is often used when single-value or single-point
estimates are used exclusively
...
The resulting measure of worth values are calculated and graphically
portrayed to determine the decision’s sensitivity to different estimates for one or more
parameters
...

However, it is more difﬁcult to make a clear decision because the analysis attempts to accommodate variation
...
The estimates will be expressed as in Example 19
...
Fundamentally,
there are two ways to consider risk in an analysis:
Expected value analysis
...
2]
...
To select the alternative, choose the most favorable
expected value of the measure of worth
...
The computations may become more elaborate, but the principle is fundamentally the same
...
Use the chance and parameter estimates to generate repeated computations of the measure of worth relation by randomly sampling from a plot for each varying
parameter similar to those in Figure 19–1
...
Usually, graphics are an
important part of decision making via simulation analysis
...

Decision Making under Uncertainty When chances are not known for the identiﬁed
states of nature (or values) of the uncertain parameters, the use of expected value–based decision making under risk as outlined above is not an option
...
If it is possible to agree that each state is
equally likely, then all states have the same chance, and the situation reduces to one of decision making under risk, because expected values can be determined
...

In an engineering economy study, observed parameter values will vary from the value estimated at the time of the study
...
Those that are estimable with a relatively high degree
of certainty should be ﬁxed for the study
...
Parameters such as P, AOC, material and unit costs, sales price, revenues, etc
...
Anticipated variation in interest rates is more commonly addressed by sensitivity analysis
...
Sections 19
...
4 provide foundation material necessary to design and correctly conduct a simulation analysis (Section 19
...

517

(a) Discrete and continuous variable scales and
(b) scales for a variable
versus its probability
...
0
0
...
6

Estimated life,
probability vs
...
4
0
...
0
0
...
6

Rate of return,
probability vs
...
4
0
...
2 Elements Important to Decision
Making under Risk
Some basics of probability and statistics are essential to correctly perform decision making under
risk via expected value or simulation analysis
...
(If you are already familiar with
them, this section will provide a review
...
Variables are classiﬁed as discrete or continuous
...

The estimated life of an asset is a discrete variable
...
The rate of return is an example of a
continuous variable; i can vary from −100% to ϱ, that is, −100% Յ i Ͻ ϱ
...
(In
probability texts, capital letters symbolize a variable, say X, and small letters x identify a
speciﬁc value of the variable
...
)
Probability is a number between 0 and 1
...
Probability is simply the amount of chance, divided by 100
...
(Actually, for a continuous variable, the probability at a
single value is zero, as shown in a later example
...
0,

19
...
The probability scale, like the percentage scale for chance in
Figure 19–1, is indicated on the ordinate (y axis) of a graph
...
0
range of probability for the variables n and i
...
Discrete variable distributions look signiﬁcantly different from continuous variable
distributions, as indicated by the inset at the right
...
1]

The distribution may be developed in one of two ways: by listing each probability value for each
possible variable value (see Example 19
...
3)
...

Identiﬁed by F(Xi), each cumulative value is calculated as
F(Xi) ‫ ؍‬sum of all probabilities through the value Xi
‫ ؍‬P(X Յ Xi)

[19
...
Examples 19
...
3 illustrate cumulative
distributions that correspond to speciﬁc probability distributions
...

EXAMPLE 19
...
He is planning to start prescribing an antibiotic that may reduce infection in patients with ﬂesh wounds
...
If no drug is used, there is always a positive probability that the infection
will be reduced by a person’s own immune system
...
e
...
Use the probabilities listed below to construct a probability distribution and a cumulative distribution for
the total number of treatments per day
...
07
0
...
10
0
...
13
0
...
25

Solution
Deﬁne the random variable T as the number of added treatments per day
...
The probability of infection reduction is
listed for each value in column 2 of Table 19–1
...
2] by adding all P(Ti) values through Ti, as indicated in column 3
...
The summing of probabilities to obtain F(Ti) gives the cumulative distribution the
stair-stepped appearance, and in all cases the ﬁnal F(Ti) ϭ 1
...
0
...
2

(1)
Number per Day
Ti

(a) Probability distribution P(Ti) and
(b) cumulative distribution F(Ti) for
Example 19
...

Probability
P(Ti)

(3)
Cumulative
Probability
F(Ti)

0
1
2
3
4
5
6
Figure 19–3

(2)

0
...
08
0
...
12
0
...
25
0
...
07
0
...
25
0
...
50
0
...
00

F(Ti)

P(Ti)
0
...
00

1
...
25 0
...
8

0
...
2
0
...
50

0
...
13
0
...
37

0
...
07 0
...
3

0
...
15
0
...

{

0
...
08
0
...
12
0
...
25
0
...
07
0
...
25
F(Ti) ϭ 0
...
50
0
...
00

T1 ϭ 0
T2 ϭ 1
T3 ϭ 2
T4 ϭ 3
T5 ϭ 4
T6 ϭ 5
T7 ϭ 6

In basic engineering economy situations, the probability distribution for a continuous variable
is commonly expressed as a mathematical function, such as a uniform distribution, a triangular
distribution (both discussed in Example 19
...
For continuous variable distributions, the symbol f(X) is
routinely used instead of P(Xi), and F(X) is used instead of F(Xi), simply because the point probability for a continuous variable is zero
...

EXAMPLE 19
...
Sallie has concluded the following about the distribution of these monthly cash ﬂows:

19
...
Sallie assumes cash ﬂow to be
a continuous variable referred to as C
...

Solution
All cash ﬂow values are expressed in $1000 units
...
Probability and cumulative probability take the following general forms
...
3]

low value Յ C1 Յ high value
L Յ C1 Յ H

[19
...
Figure 19–4 is a plot of f(C1) and F(C1) from Equations [19
...
4]
...
2
5
C1 Ϫ 10
F(C1) ϭ ————
5

$10 Յ C1 Յ $15
$10 Յ C1 Յ $15

(b) The probability that client 1 has a monthly cash ﬂow of no more than $12 is easily determined from the F(C1) plot as 0
...
If the F(C1) relation is used directly, the
computation is
12 Ϫ 10
F($12) ϭ P(C1 Յ $12) ϭ ———— ϭ 0
...
3
...
0

0
...
6

0
...
2

0
...
This probability distribution has the shape of an upward-pointing triangle with the peak at the mode M, and downward-sloping lines joining the x axis on either
side at the low (L) and high (H) values
...

2
f(mode) ϭ f(M) ϭ ———
HϪL

[19
...
6]

For C2, the low value is L ϭ $20, the high is H ϭ $30, and the most likely cash ﬂow is the
mode M ϭ $28
...
5] is
2
2
f(28) ϭ ———— ϭ —— ϭ 0
...
Using Equation [19
...
8
30 Ϫ 20
Figure 19–5 presents the plots for f(C2) and F(C2)
...

(b) From the cumulative distribution in Figure 19–5, there is an estimated 31
...
Therefore,
F($30) Ϫ F($25) ϭ P(C2 Ն $25) ϭ 1 Ϫ 0
...
6875

Comment
The general relations f(C2) and F(C2) are not developed here
...
Therefore, it requires the use of an integral to ﬁnd cumulative probability values from the probability distribution f(C2)
...
0

0
...
6

0
...
3125

Mode

f (C2)

0
...
2

0

0
20

28

30

20

C2

Figure 19–5
Triangular distribution for client 2 monthly cash ﬂow, Example 19
...

25
C2

28

30

19
...
3 Random Samples
Estimating a parameter with a single value in previous chapters is the equivalent of taking a random sample of size 1 from an entire population of possible values
...
Each estimate is a sample of size 1
from an entire population of possible values for each parameter
...

If all values in the population were known, the probability distribution and cumulative distribution would be known
...

When we perform an engineering economy study and utilize decision making under certainty,
we use one estimate for each parameter to calculate a measure of worth (i
...
, a sample of size 1
for each parameter)
...
We know that all parameters will vary somewhat; yet some are important enough, or will
vary enough, that a probability distribution should be determined or assumed for it and the parameter treated as a random variable
...
This approach complicates the analysis somewhat; however, it also
provides a sense of conﬁdence (or possibly a lack of conﬁdence in some cases) about the decision
made concerning the economic viability of the alternative based on the varying parameter
...
)
A random sample of size n is the selection in a random fashion of n values from a population
with an assumed or known probability distribution, such that the values of the variable have the
same chance of occurring in the sample as they are expected to occur in the population
...
For a two-crew aircraft, there are three parachutes on board
...
”
Yvon is relatively sure that nationwide the probability distribution of N, the speciﬁc number of
chutes fully ready, may be described by the probability distribution

{

0
...
015
P(N ϭ Ni) ϭ
0
...
920

N ϭ 0 chutes ready
N ϭ 1 chute ready
N ϭ 2 chutes ready
N ϭ 3 chutes ready

This means that the safety standard is clearly not met nationwide
...
If the sample is truly random and Yvon’s
probability distribution is a correct representation of actual parachute readiness, the observed N values in the 200 aircraft will approximate the same proportions as the population
probabilities, that is, 1 aircraft with 0 chutes ready, etc
...
However, if the results are relatively close, the
study indicates that the sample results may be useful in predicting parachute safety across
the nation
...
1

for Xi ϭ 0, 1, 2, … , 9

In tabular form, the random digits so generated are commonly clustered in groups of two digits,
three digits, or more
...
This format is very useful because the numbers 00 to 99 conveniently relate to the cumulative distribution values 0
...
00
...
To apply this logic manually and develop a random sample of size n from a known

523

524

Chapter 19

TA BLE 19–2

More on Variation and Decision Making under Risk
Random Digits Clustered into Two-Digit Numbers

51 82 88 18 19 81 03 88 91 46 39 19 28 94 70 76 33 15 64 20 14 52
73 48 28 59 78 38 54 54 93 32 70 60 78 64 92 40 72 71 77 56 39 27
10 42 18 31 23 80 80 26 74 71 03 90 55 61 61 28 41 49 00 79 96 78
45 44 79 29 81 58 66 70 24 82 91 94 42 10 61 60 79 30 01 26 31 42
68 65 26 71 44 37 93 94 93 72 84 39 77 01 97 74 17 19 46 61 49 67
75 52 14 99 67 74 06 50 97 46 27 88 10 10 70 66 22 56 18 32 06 24

discrete probability distribution P(X ) or a continuous variable distribution f(X ), the following
procedure may be used
...
Develop the cumulative distribution F(X) from the probability distribution
...

2
...
For the parachute safety example, the probabilities from 0
...
15 are
represented by the random numbers 00 to 14
...

3
...
Any direction and pattern is acceptable, but the
scheme should be used consistently for one entire sample
...
Select the ﬁrst number from the RN table, enter the F(X ) scale, and observe and record the
corresponding variable value
...

5
...
These may include
•
•
•
•
•

Plotting the sample probability distribution
...

Comparing sample results with the assumed population distribution
...
4)
...
5)
...
4
Develop a random sample of size 10 for the variable N, number of months, as described by the
probability distribution

{

0
...
50
0
...
7]

Solution

Apply the procedure above, using the P(N ϭ Ni) values in Equation [19
...

1
...

2
...
2; 50 numbers to N2 ϭ 30; and 30 numbers to N3 ϭ 36
...
Initially select any position in Table 19–2, and go across the row to the right and onto the
row below toward the left
...
)
4
...

5
...

RN

45

44

79

29

81

58

66

70

24

82

N

30

30

36

30

36

30

30

36

30

36

19
...
0

Cumulative distribution
with random number values assigned in proportion to probabilities,
Example 19
...

70–99

F(Ni )

0
...
2
00–19
24

30

36

Ni, months

Now, using the 10 values, develop the sample probabilities
...
7]
Probability

24
30
36

0
6
4

0
...
60
0
...
2
0
...
3

With only 10 values, we can expect the sample probability estimates to be different from the
values in Equation [19
...
Only the value N ϭ 24 months is signiﬁcantly different, since no RN
of 19 or less occurred
...

To take a random sample of size n for a continuous variable, the procedure above is applied,
except the random number values are assigned to the cumulative distribution on a continuous
scale of 00 to 99 corresponding to the F(X ) values
...
3
...
2 for all values between L and H (all values are divided by $1000)
...
If the two-digit RN of 45 is chosen, the corresponding C1 is graphically estimated to be
$12
...
It can also be linearly interpolated as $12
...

1
...
6

59

0
...
2

Random numbers
assigned to the
continuous variable of
client 1 cash ﬂows in
Example 19
...

99

0
...
25

13

14

15

C1, $1000

526

Chapter 19

More on Variation and Decision Making under Risk

For greater accuracy when developing a random sample, especially for a continuous variable,
it is possible to use 3-, 4-, or 5-digit RNs
...
In computer-based sampling, most simulation software packages have an RN
generator built in that will generate values in the range of 0 to 1 from a continuous variable uniform distribution, usually identiﬁed by the symbol U(0, 1)
...
00000 and 0
...
The Excel functions RAND and RANDBETWEEN are described in Appendix A, Section A
...

An initial question in random sampling usually concerns the minimum size of n required to
ensure conﬁdence in the results
...
However, since reality does
not follow theory exactly, and since engineering economy often deals with sketchy estimates,
samples in the range of 100 to 200 are the common practice
...

19
...
If the entire population for a variable were known, these properties would be
calculated directly
...
The following is a brief introduction to the interpretation and calculation of these properties using a random
sample of size n from the population
...

True Population Measure
Symbol
Expected value

␮ or E(X)
______

Standard deviation ␴ or √Var(X)
___
or √␴2

Name

Sample Estimate
Symbol

Name

__

Mu or true mean
Sigma or true
standard deviation

Sample mean

X
__

s or √s

2

Sample standard
deviation

The expected value E(X) is the long-run expected average if the variable is sampled many times
...
Equation [18
...
8], is used to compute the E(X) of a probability distribution, and Equation [19
...

Population:
Probability distribution:
Sample:

␮
E(X) ‫ ؍‬⌺Xi P(Xi)
__
sum of sample values
X ‫—————————— ؍‬
sample size
⌺Xi ⌺fi Xi
‫——— ؍ —— ؍‬
n
n

[19
...
9]

The fi in the second form of Equation [19
...
The resulting X is not necessarily an observed value of the variable; it is the long-run average value and can take on any value within the range of the variable
...
)

19
...
5
Kayeu, an engineer with Paciﬁc NW Utilities, is planning to test several hypotheses about
residential electricity bills in North American and Asian countries
...
S
...
Two small samples have been collected from different countries of North America and Asia
...
Do the samples (from a nonstatistical viewpoint) appear to be
drawn from one population of electricity bills or from two different populations?
North American, Sample 1, $

40

66

75

92

107

Asian, Sample 2, $

84

90

104

187

190

159

275

Solution
Use Equation [19
...

nϭ7
nϭ5

Sample 1:
Sample 2:

⌺Xi ϭ 814
⌺Xi ϭ 655

__

X ϭ $116
...
00

Based solely on the small sample averages, the approximate $15 difference, which is
only11% of the larger average bill, does not seem sufﬁciently large to conclude that the two
populations are different
...
(Check a basic statistics text to learn
about them
...
The sample average
is the most popular, but the mode and the median are also good measures
...
3 for a triangular distribution
...
The
median is the middle value of the sample
...
The two medians in the samples are $92 and $104
...

The standard deviation __ or s(X ) is the dispersion or spread of values about the expected value
s
E(X ) or sample average X
...
A probability distribution for data with strong
central tendency is more closely clustered about the center of the data, and has a smaller s, than
a wider, more dispersed distribution
...

f (X)

P(X)
s2

s3

s4
s1
X1
s1 Ͼ s2

X2
s4 Ͼ s3

Figure 19–8
Sketches of distributions with different separate lines standard deviation values
...
The standard deviation
is simply the square root of the variance, so either measure can be used
...
Mathematically, the formulas
and symbols for variance and standard deviation of a discrete variable and a random sample of
size n are as follows:
Population:

___

␴2 ‫ ؍‬Var(X)

and

_______

␴ ‫ √ ؍ 2␴√ ؍‬Var(X)

Probability distribution:
Var(X) ‫ ؍‬⌺[Xi Ϫ E(X)]2P(Xi)
sum of (sample value Ϫ sample average)2
Sample:
s2 ‫—————————————————— ؍‬
sample size Ϫ 1

[19
...
11]

__

s ‫ √ ؍‬s2

Equation [19
...

⌺Xi2
⌺fiXi2
n __
n __
s2 ‫——— ؊ ——— ؍‬X 2 ‫——— ؊ ——— ؍‬X 2
n؊1 n؊1
n؊1 n؊1

[19
...
To accurately
__
measure the dispersion in both directions from the average, the quantity (X Ϫ X ) is squared
...
11] is extracted
...
The fi in the second form of Equation [19
...

One simple way to combine the average and standard deviation is to determine the percentage
or fraction of the sample that is within Ϯ1, Ϯ2, or Ϯ3 standard deviations of the average, that is,
__

X Ϯ ts

for t ϭ 1, 2, or 3

[19
...
14]
__

Virtually all the sample values will always be within the Ϯ3s range of X, but the percent within
__
Ϯ1s will vary depending on how the data points are distributed about X
...
6 illustrates
__
the calculation of s to estimate ␴ and incorporates s with the sample average using X Ϯ ts
...
6
(a) Use the two samples of Example 19
...
(b) Determine the percentages of each sample that are inside the ranges
of 1 and 2 standard deviations from the mean
...
For sample 1 (North American) with n ϭ 7, use X to identify the values
...
11], with X ϭ $116
...
The resulting s2 and s values are
37,743
...
57
6
s ϭ $79
...
4

Expected Value and Standard Deviation

TABLE 19–3 Computation of Standard Deviation
__
Using Equation [19
...
29, Example 19
...
29
Ϫ50
...
29
Ϫ24
...
29
ϩ42
...
71

5,820
...
08
1,704
...
00
86
...
14
25,188
...
40

TABLE 19–4 Computation of Standard Deviation
__

Using Equation [19
...
6

Y, $

Y2

84
90
104
187
190

7,056
8,100
10,816
34,969
36,100

655

97,041
__

For sample 2 (Asian), use Y to identify the values
...
12]
...
25 Ϫ 1
...
31)
...
13] determines the ranges of X Ϯ 1s and X Ϯ 2s
...
See Figure 19–9
for a plot of the data and the standard deviation ranges
...
29

– 80

– 40

0

40

80

120
–
X ± 1s
–
X ± 2s

160

200

240

280

X

240

Y

(a)
–
Y = 131

0

40

80

120
–
Y ± 1s
–
Y ± 2s

160

200

(b)

Figure 19–9
Values, averages, and standard deviation ranges for (a) North American and
(b) Asian samples, Example 19
...

529

530

Chapter 19

More on Variation and Decision Making under Risk

North American sample
__

X Ϯ 1s ϭ 116
...
31

for a range of $36
...
60

Six out of seven values are within this range, so the percentage is 85
...

__

X Ϯ 2s ϭ 116
...
62

for a range of $Ϫ42
...
91
__

There are still six of the seven values within the X Ϯ 2s range
...
33 is meaningful only from the probabilistic perspective; from the practical viewpoint, use zero, that
is, no amount billed
...

__

Y Ϯ 2s ϭ 131 Ϯ 106

for a range of $25 to $237

__

All ﬁve of the values are within the Y Ϯ 2s range
...
In the two samples here, the range estimates are $235 and $106
...
Calculators and
spreadsheets all have functions to determine these values by simply entering the data
...
8] through [19
...
The primary differences are that the
summation symbol is replaced by the integral over the deﬁned range of the variable, which we
identify as R, and that P(X) is replaced by the differential element f (X) dX
...
15]

Var(X) ϭ ͐R X2f(X) dX Ϫ [E(X)]2

Variance:

[19
...
3 (Figure 19–4) over
the range R from $10 to $15
...

1
$10 Յ X Յ $15
f(X) ϭ — ϭ 0
...
2) dX ϭ 0
...
1(225 Ϫ 100) ϭ $12
...
2 15
Var(X) ϭ ͐ X2(0
...
5)2 ϭ ——X3 10 Ϫ (12
...
06667(3375 Ϫ 1000) Ϫ 156
...
08

|

____

␴ ϭ √ 2
...
44
Therefore, the uniform distribution between L ϭ $10 and H ϭ $15 has an expected value of
$12
...
44
...
7
Christy is the regional safety engineer for a chain of franchise-based gasoline and food stores
...
) on concrete surfaces
...
4

Expected Value and Standard Deviation

Corporate management has authorized each regional engineer to contract locally to apply to all
exterior concrete surfaces a newly marketed product that absorbs up to 100 times its own
weight in liquid and to charge a home ofﬁce account for the installation
...

You have been asked to write a brief but thorough summary about the normal distribution,
explain the $8000 to $12,000 range statement, and explain the phrase “random samples that
assume a normal population
...

Normal distribution, probabilities, and random samples
The normal distribution is also referred to as the bell-shaped curve, the Gaussian distribution, or the error distribution
...
It places exactly one-half of the probability on either side of the mean or
expected value
...
The
normal distribution is found to accurately predict many types of outcomes, such as IQ values; manufacturing errors about a speciﬁed size, volume, weight, etc
...

The normal distribution, identiﬁed by the symbol N(␮, ␴2), where ␮ is the expected value or
mean and ␴2 is the variance, or measure of spread, can be described as follows:
• The mean ␮ locates the probability distribution (Figure 19–10a), and the spread of the distribution varies with variance (Figure 19–10b), growing wider and ﬂatter for larger variance
values
...

• The normal probability distribution f(X) for a variable X is quite complicated, because its
formula is

{[

(X Ϫ ␮)2
1___
f(X) ϭ ——— exp Ϫ ————
2␴2
␴√2␲

]}

where exp represents the number e ϭ 2
...

Since f(X) is so unwieldy, random samples and probability statements are developed using a
transformation, called the standard normal distribution (SND), which uses ␮ and ␴ (popula__
tion) or X and s (sample) to compute values of the variable Z
...
17]

__

Sample:

XϪX
Z ϭ ———
s

[19
...
Therefore, the probability
values under the SND curve can be stated exactly
...
17] for X:
X ϭ Z␴ ϩ ␮

[19
...

␴1

␴2

␮2

␮1

• Equal dispersion
(␴1 = ␴2 = ␴3)
• Increasing means

␴3

␮3

X

(a)
f (X)
• Increasing
dispersion
(␴1 < ␴2 < ␴3)
• Two different
means

␴1

␴2

␴3
␮1 = ␮3

␮2

X

(b)
f (X)
Normal distribution

␮ – 3␴

␮ + 3␴

␮

X

0
...
9546
f (Z)

0
...
3413 0
...
0013 0
...
1360

–2

0
...
0214 0
...

Variable X Range

Probability

Variable Z Range

␮ ϩ 1␴
␮ Ϯ 1␴
␮ ϩ 2␴
␮ Ϯ 2␴
␮ ϩ 3␴
␮ Ϯ 3␴

0
...
6826
0
...
9546
0
...
9974

0 to ϩ1
Ϫ1 to ϩ1
0 to ϩ2
Ϫ2 to ϩ2
0 to ϩ3
Ϫ3 to ϩ3

As an illustration, probability statements from this tabulation and Figure 19–10c for X and Z
are as follows:
The probability that X is within 2␴ of its mean is 0
...

The probability that Z is within 2␴ of its mean, which is the same as between the values
Ϫ2 and ϩ2, is also 0
...

19
...
(Tables of SND values are available in many statistics books
...
10,
ϩ1
...
Translation from the Z value back to the sample values for X is via Equation [19
...

Interpretation of the home ofﬁce memo
The statement that virtually all the local contract amounts should be between $8000 and
$12,000 may be interpreted as follows: A normal distribution is assumed with a mean of µ ϭ
$10,000 and a standard deviation for ␴ ϭ $667, or a variance of ␴2 ϭ ($667)2; that is, an
N[$10,000, ($667)2] distribution is assumed
...
74%) is within 3␴ of the mean, as stated above
...
19]
X ‫ ؍‬Z␴ ؉ ␮

Ϫ2
...
12
Ϫ0
...
24
Ϫ2
...
99

X ϭ (Ϫ2
...
12)(667) ϩ 10,000 ϭ $12,081
X ϭ (Ϫ0
...
24)(667) ϩ 10,000 ϭ $10,827
X ϭ (Ϫ2
...
99)(667) ϩ 10,000 ϭ $9340

In this sample of six typical concrete surfacing contract amounts for sites in our region, the
average is $9825 and ﬁve of six values are within the range of $8000 to $12,000, with the sixth
being only $81 above the upper limit
...
5 Monte Carlo Sampling and Simulation Analysis
Up to this point, all alternative selections have been made using estimates with certainty, possibly followed by some testing of the decision via sensitivity analysis or expected values
...

The random sampling technique discussed in Section 19
...

The general procedure outlined below uses Monte Carlo sampling to obtain samples of size n for
selected parameters of formulated alternatives
...
All other parameters in an alternative are considered certain; that is, they are known, or they can be estimated with enough precision to consider them certain
...

All parameters are independent; that is, one variable’s distribution does not affect the value of
any other variable of the alternative
...

The simulation approach to engineering economy analysis is summarized in the following
basic steps
...
Formulate alternative(s)
...
Determine the form of the relation(s) to calculate the measure
of worth
...
Parameters with variation
...
Estimate values for all other (certain) parameters for the analysis
...

Step 4
...

Step 6
...

More on Variation and Decision Making under Risk

Determine probability distributions
...
Use standard distributions, where possible, to simplify the sampling process and to prepare for computer-based simulation
...
Incorporate the random sampling procedure of Section 19
...
This results in the cumulative distribution,
assignment of RNs, selection of the RNs, and a sample of size n for each variable
...
Compute n values of the selected measure of
worth from the relation(s) determined in step 1
...
(This is when the property
of independent random variables is actually applied
...
Construct the probability distribution of the measure of worth, using between 10 and 20 cells of data, and calculate measures such as
__
__
X, s, X Ϯ ts, and relevant probabilities
...
Draw conclusions about each alternative, and decide which is to be
selected
...

Example 19
...
9 utilizes spreadsheet simulation for the same estimates
...
8
Yvonne Ramos is the CEO of a chain of 50 ﬁtness centers in the United States and Canada
...
As an enticement, the offer includes a guarantee of annual
revenue for one of the systems for the ﬁrst 5 years
...
Details for the two systems follow:
System 1
...

No guarantee for annual net revenue is offered
...
First cost is P ϭ $8000, there is no salvage value, and there is a guaranteed
annual net revenue of $1000 for each of the ﬁrst 5 years, but after this period, there is
no guarantee
...
Cancellation anytime after the initial 5 years is allowed, with no
penalty
...
If the
MARR is 15% per year, use PW analysis to determine if neither, one, or both of the systems
should be installed
...

Step 1
...
Using PW analysis, the relations for system 1 and system 2 are developed
...

PW1 ϭ ϪP1 ϩ NCF1(P͞A,15%,n1)

[19
...
21]

Step 2
...
Yvonne summarizes the parameters estimated with
certainty and makes distribution assumptions about three parameters treated as random variables
...
5

Monte Carlo Sampling and Simulation Analysis

System 1
Certainty
...

Variable
...

System 2
Certainty
...

Variable
...

Variable
...

Now, rewrite Equations [19
...
21] to reﬂect the estimates made with certainty
...
1604)

[19
...
4972)

[19
...
Figure 19–11 (left side) shows the assumed
probability distributions for NCF1, NCF2, and n2
...
Random sampling
...
3
...
The RNs
for NCF2 identify the x axis values so that all net cash ﬂows will be in even $1000
amounts
...
However, we round the number
to the next higher value of n2 because it is likely the contract may be canceled on an
anniversary date
...

Once the ﬁrst RN is selected randomly from Table 19–2, the sequence (step 3)
used will be to proceed down the RN table column and then up the column to the left
...

Step 5
...
With the ﬁve sample values in Table 19–5, calculate the PW values using Equations [19
...
23]
...

1
...

3
...

5
...

2
...

4
...

PW1 ϭ Ϫ12,000 ϩ (Ϫ2200)(4
...
1604)
PW1 ϭ Ϫ12,000 ϩ (Ϫ1100)(4
...
1604)
PW1 ϭ Ϫ12,000 ϩ 3100(4
...
4972)
PW2 ϭ Ϫ4648 ϩ 1000(P͞A,15%,5)(0
...
4972)
PW2 ϭ Ϫ4648 ϩ 3000(P͞A,15%,10)(0
...
4972)

ϭ $Ϫ21,153
ϭ $Ϫ3679
ϭ $Ϫ16,576
ϭ $Ϫ15,744
ϭ $ϩ897
ϭ $Ϫ2579
ϭ $Ϫ2981
ϭ $ϩ6507
ϭ $ϩ2838
ϭ $Ϫ107

Now, 25 more RNs are selected for each variable from Table 19–2, and the PW values are calculated
...
Measure of worth description
...

535

536

Chapter 19

More on Variation and Decision Making under Risk

Figure 19-11

F(NCF1 )

Distributions used for
random samples,
Example 19
...

RN

1
...
8
60–79
0
...
4
20–39

f (NCF1)
Continuous variable

1
10

0
...
00

83–99
0
...
67
50–66
0
...
33
P(NCF2)

17–32
Discrete variable

1
6

0
...
0

888–999
0
...
6

444–665

0
...
2
000–221
0

6

10
n2, years

15

6

8

10
12
n2, years

14 15

TA BLE 19–5 Random Numbers and Variable Values for NCF1, NCF2, and
n2, Example 19
...
3
9
...
7
14
...
3

12
10
13
15
8

*Randomly start with row 1, column 4 in Table 19–2
...

‡
Start with row 4, column 6
...

19
...
8
...
Sample values range from $−24,481 to $ϩ12,962
...
Sample values range from $−3031 to $ϩ10,324
...

Conclusions
...
Of course, many conclusions are possible once the PW distributions are
known, but the following seem clear at this point
...
Based on this small sample of 30 observations, do not accept this
alternative
...
27 (8 out of 30 values) that the PW will be
__
positive, and X1 is a large negative
...
A larger
sample may alter this analysis somewhat
...
If Yvonne is willing to accept the longer-term commitment that may
increase the NCF some years out, the sample of 30 observations indicates to accept this alternative
...

__
However, the probability of observing PW within the X Ϯ 1s limits ($Ϫ1612 and
$7060) is 0
...

Conclusion at this point
...

537

538

Chapter 19

More on Variation and Decision Making under Risk

Comment
The estimates in Example 13
...
The alternatives were
evaluated by the payback period method at MARR ϭ 15%, and the ﬁrst alternative was selected
...
5 selected alternative 2 based, in
part, upon the anticipated larger cash ﬂow in the later years
...
9
Help Yvonne Ramos set up a spreadsheet simulation for the three random variables and PW
analysis in Example 19
...
Does the PW distribution vary appreciably from that developed
using manual simulation? Do the decisions to reject the system 1 proposal and accept the system 2 proposal still seem reasonable?

Solution by Spreadsheet
Figures 19–13 and 19–14 are spreadsheet screen shots that accomplish the simulation portion
of the analysis described above in steps 3 (determine probability distribution) through 6 (measure of worth description)
...

Figure 19–13 shows the results of a small sample of 30 values from the three distributions
using the RAND and IF functions
...
3 in Appendix A
...
The spreadsheet relation in column B
translates RN1 values (column A) into NCF1 amounts
...
Column D cells display
NCF2 in the $1000 increments using the logical IF function to translate from the RN2 values
...
The results in column F are integer values obtained using the INT function operating on the RN3 values
...
9
...
5

Monte Carlo Sampling and Simulation Analysis

ϭSUM(G13:G42)

ϭAVERAGE(F13:F42)

ϭSTDEV(F13:F42)

Figure 19–14
Simulation results for 30 PW values, Example 19
...

Figure 19–14 presents the two alternatives’ estimates in the top section
...
22] and [19
...
The tabular approach used here tallies the number of PW values
below zero ($0) and equal to or exceeding zero using the IF operator
...
22]
...
Sample averages and standard deviations are also indicated
...

System 1 PW

System 2 PW

X, $

s, $

No
...
of
PW Ն 0

Ϫ7,729
Ϫ7,105

10,190
13,199

8
10

2,724
1,649

4,336
3,871

20
19

__

Hand
Spreadsheet

__

For the spreadsheet simulation, 10 (33%) of the PW1 values exceed zero, while the manual simulation included 8 (27%) positive values
...
(It is possible to deﬁne RAND to keep the same RN values
...
)
The conclusion to reject the system 1 proposal and accept system 2 is still appropriate for
the spreadsheet simulation as it was for the hand solution, since there are comparable chances
that PW Ն 0
...
Assumptions about the shape of the variable’s probability
distribution are used to explain how the estimates of parameter values may vary
...
In this chapter, we learned several of the simple, but useful, discrete and continuous population distributions used in engineering economy—uniform and triangular—as well
as specifying our own distribution or assuming the normal distribution
...

The results are used to make probability statements about the parameter, which help make the
ﬁnal decision with risk considered
...
The results of
such an analysis can then be compared with decisions when parameter estimates are made with
certainty
...
1 Identify the following variables as either discrete
or continuous
...
He wants to use E(output) in the decisionmaking process
...

Probability and Distributions
19
...
Data collected
from stripper wells in an established oil ﬁeld were
used to develop the probability-royalty relationship shown below
...
2 For each situation below, determine (1) if the variable is discrete or continuous and (2) if the information involves certainty, risk, and͞or uncertainty
...

(b) The raises for engineers and technical staff
employees will be 3%, or 5%, with one-half
getting 3% and one-half getting 5%
...

(d ) The salvage value for an old machine will be
$500 (i
...
, its asking price) or $0 (it will be
thrown away)
...

19
...
10 0
...
32

0
...
09

0
...

Determine the probability that the royalty
income will be at least $12,600 per year
...
5 Daily revenue from vending machines placed in
various buildings of a major university is as follows:
20, 75, 43, 62, 51, 52, 78, 33, 28, 39, 61, 56, 43, 49, 48, 49,
71, 53, 57, 46, 42, 41, 63, 36, 51, 59, 40, 32, 37, 29, 26

(a)

(b)
(c)
(d)

Construct a frequency distribution table with
a cell size of 12 starting with 19
...
e
...
5–31
...
5–43
...

Determine the probability distribution
...
6 A survey of households included a question about
the number of operating automobiles N currently
owned by people living at the residence and the
interest rate i on the lowest-rate loan for the cars
...

Number of Cars N

Households

0
1
2
3
Ն4

12
56
26
3
3

Purchase Alternative

13
14
19
38
12
4

Estimated Value

0
...
01Ϫ4
4
...
01Ϫ8
8
...
01Ϫ12

(b)
(c)

(d)

19
...

Use the parameter estimates and assumed distribution data shown to plot the probability distributions
on one graph for each corresponding parameter
...

Households

Loan Rate i, %

(a)

Parameter

Probability

(a)
(b)
(c)

0

2

5

10

100

0
...
045

0
...
013

0
...

Calculate the expected value of the distribution of dollars per ticket
...
8 Bob is working on two separate probability-related
projects
...
The
variable N is described by the formula (0
...
The second involves a battery
life L that varies between 2 and 5 months
...
7 An ofﬁcer of the state lottery commission sampled
lottery ticket purchasers over a 1-week period at
one location
...

Plot the probability distributions and cumulative distributions for N and i
...

Distribution, $

5 months, which is the design life
...
(a) Write out and plot the probability
distributions and cumulative distributions for Bob
...

Ϫ3,000

2,000

8

4

9,000

5,000

Life, years
AOC, $ per year

Assumed
Distribution
Uniform;
continuous
Triangular;
mode at $2500
Triangular;
mode at 6
Uniform;
continuous

Lease Alternative
Estimated Value
Parameter

High

Low

Lease ﬁrst cost, $

2000

1800

AOC, $ per year

9000

5000

Lease term, years

3

3

Assumed
Distribution
Uniform;
continuous
Triangular;
mode at $7000
Certainty

19
...
She has collected debt-to-equity mix data on mature (M) and
young (Y) companies
...
Carla has deﬁned
DM as a variable for the mature companies from
0 to 1, with DM ϭ 0 interpreted as the low of 20%
debt and DM ϭ 1
...
The
variable for young corporation debt percentages
DY is similarly deﬁned
...
(b) What can
you comment about the probability that a mature
company or a young company will have a low debt
percentage? A high debt percentage?

542

Chapter 19

More on Variation and Decision Making under Risk

19
...
A sample of size 50 results in the following
probability estimates:
Xi

1

2

3

6

9

10

P(Xi)

0
...
2

0
...
1

0
...
1

(a)
(b)

(c)

Write out and graph the cumulative
distribution
...

Use the cumulative distribution to show that
P(X ϭ 7 or 8) ϭ 0
...
Even though this probability is zero, the statement about X is that it
can take on integer values of 1 to 10
...
14 Develop a discrete probability distribution of your
own for the variable G, the expected grade in this
course, where G ϭ A, B, C, D, F, or I (incomplete)
...
Now plot the probability values from
the sample for each G value
...
15 Use the RAND function in Excel to generate
100 values from a U(0,1) distribution
...
5,
the expected value for a random sample between 0 and 1
...
1 width, that is 0
...
1,
0
...
2, etc
...
Determine the probability for each grouping from the results
...
12 A discrete variable X can take on integer values of
1 to 5
...

Xi

(b)

2

3

4

5

P(Xi)

(a)

1
0
...
3

0
...
16 An engineer was asked to determine whether the
average air quality in a vehicle assembly plant was
within OSHA guidelines
...
1

Use the following random numbers to estimate the probabilities for each value of X
...
Compare the sample results with
the probabilities in the problem statement
...
13 The percent price increase p on a variety of retail
food prices over a 1-year period varied from 5% to
10% in all cases
...
17 Carol sampled the monthly maintenance costs for automated soldering machines a total of 100 times during 1 year
...
She indicated the number of times
(frequency) each cell value was observed
...

Cell
Midpoint, $

when p ϭ 5%

1

when p ϭ 10%

For a continuous variable the cumulative distribution F(X) is the integral of f (X) over the same
range of the variable
...
Transform the X values into interest rates
...

(a)

(b)

Frequency

600
800
1000
1200
1400
1600
1800
2000

0ՅXՅ1

0

Determine the sample mean
...

Determine the number of values and percent
of values that fall within Ϯ1 standard deviation of the mean
...

What is the best estimate of the percentage of
costs that will fall within 2 standard deviations of the mean?

543

Additional Problems and FE Exam Review Questions

(c)

(d)

Develop a probability distribution of the
monthly maintenance costs from Carol’s
sample, and indicate the answers to the previous two questions on it
...

Simulation
19
...
The additional CFAT of $2800 in year 10 is
the salvage value of capital assets
...
18 (a) Determine the values of sample average and
standard deviation of the data in Problem 19
...

(b) Determine the values 1 and 2 standard deviations from the mean
...
19 (a) Use the relations in Section 19
...
10
...

19
...
4 for continuous variables to determine the expected value
and variance for the distribution of DM in
Problem 19
...

f (DM) ϭ 3(1 Ϫ DM)2
(b)

0 Յ DM Յ 1

Determine the probability that DM is within 2
standard deviations of the expected value
...
19
...
21 Calculate the expected value for the variable N in
Problem 19
...

19
...
Data collected over a 30-week
period are summarized by the following probability distribution
...

Y copies
P(Y)

3

7

10

1͞4

1͞3

CFAT, $

0
1Ϫ6
7Ϫ10
10

Ϫ28,800
5,400
2,040
2,800

The PW value at the current MARR of 7% per
year is
PW ϭ Ϫ28,800 ϩ 5400(P͞A,7%,6)
ϩ 2040(P͞A,7%,4)(P͞F,7%,6)
ϩ 2800(P͞F,7%,10)
ϭ $2966
Carl believes the MARR will vary over a relatively
narrow range, as will the CFAT, especially during
the out years of 7 through 10
...
Use the following probability distribution assumptions for MARR
and CFAT to perform a simulation—hand- or
spreadsheet-based
...
Uniform distribution over the range
6% to 10%
...
Uniform distribution over the range $1600 to $2400 for each
year
...
Should the plan
be accepted using decision making under certainty? Under risk?
19
...
23, except use the normal distribution for the CFAT in years 7 through 10 with
an expected value of $2000 and a standard deviation of $500
...
25 When there are at least two values for a parameter
and it is possible to estimate the chance that each
may occur, this situation is known as:
(a) Uncertainty
(b) Risk
(c) Standard deviation
(d) Cost estimating

19
...
27 Decision making under risk includes all of the following except:
(a) Expected value analysis
(b) Simulation
(c) Using only single-value estimates
(d) Probability
19
...
31 The revenue from an oil dispersant product has
averaged $15,000 per month for the past 12 months
...
32 A survey of the types of cars parked at an NFL
football stadium revealed that there were equal
probabilities of ﬁnding cars identiﬁed as type A, B,
C, and D
...

19
...
15
(b) 0
...
32
(d) 0
...
15

0
...
32

0
...
09

0
...
30 The symbol that represents the true population
mean is:
(a) ␴
(b) s
(c) ␮
__
(d) X

Car Type

Assigned RN

A
B
C
D

0 through 24
25 through 49
50 through 74
75 through 99

The sample probability of a type B car from the
12 random numbers shown is closest to:
RN: 75, 52, 14, 99, 67, 74, 06, 50, 97, 46, 27, 88

(a)
(b)
(c)
(d )

0
...
25
0
...
42

CASE STUDY
USING SIMULATION AND THREE-ESTIMATE SENSITIVITY ANALYSIS
Background
The Knox Brewing company makes specialty-named sodas
and ﬂavored drinks for retail grocery chains throughout
North and Central America
...
Dr
...
You just handed
him the ﬁrst cost bids from three vendors for the machine
...

You were quite surprised, as this was the ﬁrst time you had
been in his ofﬁce, and most other engineers at Knox have a
great fear of “the Old Man
...

After a few minutes, he told you to take these numbers
and use some of that “new engineering knowledge” you
acquired in college to determine which, if any, of these three
bids made the best economic sense
...
M
...
In addition, you
developed some possible distributions for the parameters that
Dr
...

These are summarized in Table 19–7
...

Case Study Exercises
First, learn to use the RNG (random number generator) in
Excel, if you have not already done so
...
dev
...
dev
...
dev
...
RNG is part of the Analysis Tool-Pak
accessed through the Ofﬁce Button, Excel Options, AddIns path
...
Prepare the simulation using a spreadsheet; determine
which of the vendors offers the best machine from an

economic perspective, and take into account the estimates made by Dr
...
Use a sample size of at least
50, and base your conclusions on the AW measure of
worth
...
Prepare a short presentation for Dr
...

APPENDIX A
USING SPREADSHEETS
AND MICROSOFT EXCEL©
This appendix explains the layout of a spreadsheet and the use of Microsoft Excel (hereafter
called Excel) functions in engineering economy
...
Some speciﬁc commands and entries refer to Excel 2007 and
may differ slightly from your version
...
1 Introduction to Using Excel
Enter a Formula or Use an Excel Function
The ‫ ؍‬sign is required to perform any formula or function computation in a cell
...
The symbol ` is usually in the upper left of the keyboard with the ~ (tilde) symbol
...

1
...

2
...
(Move the pointer to C3 and left-click
...
Type ϭ PV(5%,12,10) and ϽEnterϾ
...

Another example: To calculate the future value of 12 payments of $10 at 6% per year interest, do
the following:
1
...

3
...

5
...

7
...

Move to cell B3, and type INTEREST
...

Move to cell B4, and type PAYMENT
...

Move to cell B5, and type NUMBER OF PAYMENTS
...

Move to cell B7, and type FUTURE VALUE
...
The answer will appear
in cell C7
...
Move to cell C3 and type ϭ5͞100 (the previous value will be replaced)
...
The value in cell C7 will update
...

This approach deﬁnes the referenced cell as a global variable for the worksheet
...

Relative References If a cell reference is entered, for example, A1, into a formula or function
that is copied or dragged into another cell, the reference is changed relative to the movement of
the original cell
...
This feature is used when dragging a function through several cells, and the source entries must change with the column or row
...
For example,

548

Appendix A

Using Spreadsheets and Microsoft Excel©

ϭ $A$1 will retain the formula when it is moved anywhere on the worksheet
...

Absolute references are used in engineering economy for sensitivity analysis of parameters
such as MARR, ﬁrst cost, and annual cash ﬂows
...

Print the Spreadsheet
First deﬁne the portion (or all) of the spreadsheet to be printed
...
Move the pointer to the top left cell of your spreadsheet
...
Hold down the left-click button
...
)
3
...

4
...
(It is ready to print
...
Left-click the Ofﬁce button (see Figure A–1)
...
Move the pointer down to select Print and left-click
...
In the dialog box, left-click the Print option (or similar command)
...

Save the Spreadsheet
You can save your spreadsheet at any time during or after completing your work
...

1
...

2
...
option
...
Type the ﬁle name, e
...
, Prob 7
...

To save the spreadsheet after it has been saved the ﬁrst time, i
...
, a ﬁle name has been assigned
to it, left-click the Ofﬁce button, move the pointer down, and left-click on Save
...
It plots pairs of data and can place multiple series of entries on the Y axis
...

1
...

2
...

Column A, cell A1 through A6: Rate i%, 4, 6, 8, 9, 10
Column B, cell B1 through B6: $ for A, 40, 55, 60, 45, 10
Column C, cell C1 through C6: $ for B, 100, 70, 65, 50, 30
...
Move the mouse to A1, left-click, and hold while dragging to cell C6
...

4
...
After dragging over one column of
data, momentarily release the left click, then move to the top of the next (nonadjacent) column of the chart
...

5
...

6
...
The graph appears with a
legend (Figure A–1)
...

Note that only the bottom row of the title can be highlighted
...
on the legend
...
2

Office button

Organization (Layout) of the Spreadsheet
Insert for
scatter chart

Figure A–1
Scatter chart for data entries and location of commonly used buttons
...
To get general help information, left-click on the “?” (upper right)
...
Enter the topic or phrase
...

3
...
You can browse through the options by left-clicking
on any item
...
2 Organization (Layout) of the Spreadsheet
A spreadsheet can be used in several ways to obtain answers to numerical questions
...

For example, to ﬁnd the future worth in a single-cell operation, move the pointer to any cell and
enter ϭ FV(8%,5,-2500)
...
50 is the 8% future worth at the end of year 5
of ﬁve equal payments of $2500 each
...
Some fundamental guidelines in spreadsheet
organization are presented here
...
As solutions become
more complex, organization of the spreadsheet becomes increasingly important, especially for
presentation to an audience via PowerPoint or similar software
...
It is advisable to organize the given or estimated data in the
top left of the spreadsheet
...
Then B1 can be the referenced cell for all
entries requiring the MARR
...
Often, the answers are best placed at the bottom or top of the column of entries
used in the formula or predeﬁned function
...
Each column or row should be labeled so its entries are
clear to the reader
...

Enter income and cost cash ﬂows separately
...

and ﬁrst cost, salvage value, and annual costs (usually negative, with salvage a positive number)
be entered into two adjacent columns
...
There are two immediate advantages to this practice: fewer errors are made
when performing the summation and subtraction mentally, and changes for sensitivity analysis
are more easily made
...
The use of absolute and relative cell references is a must when any changes
in entries are expected
...
The absolute cell reference
entry $B$1 in the three functions allows the MARR to be changed one time, not three
...
When the formulas and functions
are kept relatively simple, the ﬁnal answer can be obtained using the SUM function
...
This practice is especially useful when the cash
ﬂow series are complex
...
If a chart (graph) will be developed, plan ahead by leaving sufﬁcient
room on the right of the data and answers
...
Placement on the same worksheet is recommended, especially when the
results of sensitivity analysis are plotted
...
3 Excel Functions Important to Engineering
Economy (alphabetical order)
DB (Declining Balance)
Calculates the depreciation amount for an asset for a speciﬁed period n using the declining balance method
...
This is Equation [16
...
Threedecimal-place accuracy is used for d
...

Salvage value
...

The period, year, for which the depreciation is to be calculated
...

A
...
At the end of
10 years, the salvage value of the machine is $50,000
...
Simply use the optional factor entry for rates other than d ϭ 2͞n
...
A factor can also be entered for some other declining balance depreciation method by
specifying a factor in the function
...

Salvage value of the asset
...

The period, a year, for which the depreciation is to be calculated
...
If, for example,
the entry is 1
...

Example A new machine costs $200,000 and is expected to last 10 years
...
Calculate the depreciation of the machine for the ﬁrst and the eighth years
...

Depreciation for the ﬁrst year: ϭ DDB(200000,10000,10,1)
Depreciation for the eighth year: ϭ DDB(200000,10000,10,8)
Depreciation for the ﬁfth year using 175% DB: ϭ DDB(200000,10000,10,5,1
...
Excel uses Equation [4
...

‫ ؍‬EFFECT(nominal, npery)
nominal
npery

Nominal interest rate for the year
...

Example Claude has applied for a $10,000 loan
...
What effective annual rate will Claude pay?
Effective annual rate: ϭ EFFECT(8%,12)
EFFECT can also be used to ﬁnd effective rates other than annually
...

Example Interest is stated as 3
...
Find the effective
semiannual rate
...

Effective semiannual rate: ϭ EFFECT(7%,2)

551

552

Using Spreadsheets and Microsoft Excel©

Appendix A

FV (Future Value)
Calculates the future value (worth) based on periodic payments at a speciﬁc interest rate
...

Number of compounding periods
...

The present value amount
...

(optional entry) Either 0 or 1
...
If omitted, 0 is assumed
...
He will deposit
$12,000 to start the account and plans to add $500 to the account at the beginning of each month
for the next 24 months
...
25% per month
...
25%,24,500,12000,1)

IF (IF Logical Function)
Determines which of two entries is entered into a cell based on the outcome of a logical check on
the outcome of another cell
...
If the response is a text string, place it between quote marks
(“ ”)
...
Up to seven IF functions can be nested for
very complex logical tests
...

Result if the logical_test argument is true
...

Example The entry in cell B4 should be “selected” if the PW value in cell B3 is greater than
or equal to zero and “rejected” if PW Ͻ 0
...

Entry in cell C5:

ϭ IF(C4Ͻ0,“rejected”,IF(C4Ͼϭ200,“fantastic”,“selected”))

IPMT (Interest Payment)
Calculates the interest accrued for a given period n based on constant periodic payments and
interest rate
...

Period for which interest is to be calculated
...

Present value
...

Future value
...
The fv
can also be considered a cash balance after the last payment is made
...
A 0 represents payments made at the
end of the period, and 1 represents payments made at the beginning of
the period
...

A
...
The interest rate is 0
...

Interest due: ϭ IPMT(0
...

‫ ؍‬IRR(values, guess)
values

guess

A set of numbers in a spreadsheet column (or row) for which the rate of
return will be calculated
...
Negative numbers denote a payment
made or cash outﬂow, and positive numbers denote income or cash
inﬂow
...
In most cases, a guess is not required, and a
10% rate of return is initially assumed
...
Inputting different guess values makes
it possible to determine the multiple roots for the rate of return equation
of a nonconventional cash ﬂow series
...
He will need $25,000 in capital and anticipates that the business will generate the following incomes during the ﬁrst 5 years
...

Year 1
Year 2
Year 3
Year 4
Year 5

$5,000
$7,500
$8,000
$10,000
$15,000

Set up an array in the spreadsheet
...

In cell A2, type 5000 (positive for income)
...

In cell A4, type 8000
...

In cell A6, type 15000
...
Note that any years with a zero cash ﬂow must have a zero entered to ensure that
the year value is correctly maintained for computation purposes
...

To calculate the internal rate of return after 5 years and specify a guess value of 5%, move to
cell A8, and type ϭ IRR(A1:A6,5%)
...

‫ ؍‬MIRR(values, ﬁnance_rate, reinvest_rate)
values

Refers to an array of cells in the spreadsheet
...
The series of payments and income must occur at
regular periods and must contain at least one positive number
and one negative number
...
9])
...
9])
...

See Section 7
...
)

Example Jane opened a hobby store 4 years ago
...
Since then, the business has yielded $10,000 the
ﬁrst year, $15,000 the second year, $18,000 the third year, and $21,000 the fourth year
...
What is the modiﬁed rate of return after 3 years and
after 4 years?
In cell A1, type Ϫ50000
...

In cell A3, type 15000
...

In cell A5, type 21000
...

To calculate the modiﬁed rate of return after 4 years, move to cell A7, and type
ϭ MIRR(A1:A5,12%,8%)
...
This function is designed to display only nominal annual rates
...

Number of times that interest is compounded per year
...
55% per year
...

Nominal annual rate, quarterly compounding: ϭ NOMINAL(12
...
55%,100000)

NPER (Number of Periods)
Calculates the number of periods for the present worth of an investment to equal the future value
speciﬁed, based on uniform regular payments and a stated interest rate
...

Amount paid during each compounding period
...

(optional entry) Future value or cash balance after the last payment
...

(optional entry) Enter 0 if payments are due at the end of the
compounding period and 1 if payments are due at the beginning of the
period
...

A
...
25% per month
...
How many payments
does she have to make to accumulate $25,000 to buy a new car?
Number of payments: ϭ NPER(0
...

‫ ؍‬NPV(rate, series)
rate
series

Interest rate per compounding period
...

Example Mark is considering buying a sports store for $100,000 and expects to receive the
following income during the next 6 years of business: $25,000, $40,000, $42,000, $44,000,
$48,000, $50,000
...

In cells A1 through A7, enter Ϫ100,000, followed by the six annual incomes
...
Any year with a zero cash ﬂow must have a 0 entered
to ensure a correct result
...

‫ ؍‬PMT(rate, nper, pv, fv, type)
rate
nper
pv
fv
type

Interest rate per compounding period
...

Present value
...

(optional entry) Enter 0 for payments due at the end of the
compounding period and 1 if payment is due at the start of the
compounding period
...

Example Jim plans to take a $15,000 loan to buy a new car
...

He wants to pay the loan off in 5 years (60 months)
...

‫ ؍‬PPMT(rate, per, nper, pv, fv, type)
rate
per
nper
pv
fv
type

Interest rate per compounding period
...

Total number of periods
...

Future value
...
If omitted, 0 is assumed
...
The interest rate is 5%
...

‫ ؍‬PV(rate, nper, pmt, fv, type)
rate
nper
pmt
fv
type

Interest rate per compounding period
...

Cash ﬂow at regular intervals
...

Future value or cash balance at the end of the last period
...
If omitted, 0 is assumed
...

Example Jose is considering leasing a car for $300 a month for 3 years (36 months)
...
Using an interest rate of 8% per year, ﬁnd
the present value of this option
...

RAND (Random Number)
Returns an evenly distributed number that is (1) Ն 0 and Ͻ 1; (2) Ն 0 and Ͻ 100; or (3) between
two speciﬁed numbers
...

Example Grace needs random numbers between 5 and 10 with 3 digits after the decimal
...

Random number:

ϭ RAND()*5 ϩ 5

Example Randi wants to generate random numbers between the limits of Ϫ10 and 25
...

Random number:

ϭ RAND()*35 Ϫ 10

A
...

‫ ؍‬RATE(nper, pmt, pv, fv, type, guess)
nper
pmt
pv
fv
type

guess

Total number of periods
...

Present value
...

(optional entry) Enter 0 for payments due at the end of the
compounding period and 1 if payments are due at the start of each
compounding period
...

(optional entry) To minimize computing time, include a guessed
interest rate
...
This function usually converges to a solution if
the rate is between 0% and 100%
...
She will make an initial deposit of
$1000 to open the account and plans to deposit $100 at the beginning of each month
...
At the end of 3 years, she wants to have at least $5000
...

‫ ؍‬SLN(cost, salvage, life)
cost
salvage
life

First cost or basis of the asset
...

Depreciation life
...
The machine has an allowed depreciation life of 8 years and an estimated salvage value of $15,000
...

‫ ؍‬SYD(cost, salvage, life, period)
cost
salvage
life
period

First cost or basis of the asset
...

Depreciation life
...

Example Jack bought equipment for $100,000 that has a depreciation life of 10 years
...
What is the depreciation for year 1 and year 9?
Depreciation for year 1: ϭ SYD(100000,10000,10,1)
Depreciation for year 9: ϭ SYD(100000,10000,10,9)

VDB (Variable Declining Balance)
Calculates the depreciation using the declining balance method with a switch to straight line
depreciation in the year in which straight line has a larger depreciation amount
...

‫ ؍‬VDB (cost, salvage, life, start_period, end_period, factor, no_switch)
cost
salvage
life
start_period
end_period
factor

no_switch

First cost of the asset
...

Depreciation life
...

Last period for depreciation to be calculated
...
Other entries
deﬁne the declining balance method, for example, 1
...

(optional entry) If omitted or entered as FALSE, the function will
switch from declining balance to straight line depreciation when
the latter is greater than DB depreciation
...

Example Newly purchased equipment with a ﬁrst cost of $300,000 has a depreciable life of
10 years with no salvage value
...

Depreciation for ﬁrst year, with switching: ϭ VDB(300000,0,10,0,1,1
...
75)
Depreciation for ﬁrst year, no switching: ϭ VDB(300000,0,10,0,1,1
...
75,TRUE)

VDB (for MACRS Depreciation)
The VDB function can be adapted to generate the MACRS annual depreciation amount, when the
start_period and end_period are replaced with the MAX and MIN functions, respectively
...

The VDB format is
‫ ؍‬VDB(cost,0,life,MAX(0,t؊1
...
5),factor)
Example Determine the MACRS depreciation for year 4 for a $350,000 asset that has a 20%
salvage value and a MACRS recovery period of 3 years
...

Depreciation for year 4: ϭ VDB(350000,0,3,MAX(0,4Ϫ1
...
5),2)
Example Find the MACRS depreciation in year 16 for a $350,000-asset with a recovery
period of n ϭ 15 years
...
5 is required here, since MACRS starts with
150% DB for n ϭ 15-year and 20-year recovery periods
...

Depreciation for year 16: ϭ VDB(350000,0,15,MAX(0,16Ϫ1
...
5),1
...
These
can be viewed by clicking the Formulas tab on the Excel toolbar
...
4 Goal Seek—A Tool for Breakeven and
Sensitivity Analysis
Goal Seek is found on the Excel toolbar labeled Data, followed by What-if Analysis
...
It is a good tool for sensitivity analysis, breakeven analysis, and “what if?” questions

A
...

ϭ B$5

ϭ B$5ϩ500

Figure A–4
Use of Goal Seek to determine an annual cash ﬂow to increase the rate of return
...
The initial Goal Seek template is pictured
in Figure A–3
...
Only a single cell can be identiﬁed
as the changing cell; however, this limitation can be avoided by using equations rather than speciﬁc numerical inputs in any additional cells also to be changed
...

Example A new asset will cost $25,000, generate an annual cash ﬂow of $6000 over its 5-year
life, and have an estimated $500 salvage value
...
94%
...

Figure A–4 (top left) shows the cash ﬂows and return displayed using the function
ϭ IRR(B4:B9) prior to the use of Goal Seek
...
The $500 salvage is added for the
last year
...
The tool ﬁnds the required cash ﬂow of $6506 to approximate the 10% per
year return
...
Clicking OK saves all
changed cells; clicking Cancel returns to the original values
...
5 Solver—An Optimizing Tool for Capital Budgeting,
Breakeven, and Sensitivity Analysis
Solver is a powerful spreadsheet tool to change the value in multiple (one or more) cells based on
the value in a speciﬁc (target) cell
...
(Section 12
...
) The initial Solver template is shown in Figure A–5
...

Set Target Cell box
...
The target cell itself must contain a formula or function
...

By Changing Cells box
...
Each cell must be directly or indirectly related to the target cell
...
The
Guess button will list all possible changing cells related to the target cell
...
Enter any constraints that may apply, for example,
$C$1 Ͻ $50,000
...

Options box
...
Also, linear and
nonlinear model assumptions can be set here
...
0001
...
If tolerance remains at
the default value of 5%, a project may be incorrectly included in the solution set at a very low
level
...
This appears after Solve is clicked and a solution appears
...
It is possible to update the
spreadsheet by clicking Keep Solver Solution, or return to the original entries using Restore
Original Values
...
6 Error Messages
If Excel is unable to complete a formula or function computation, an error message is displayed
...

#N͞A
Refers to a value that is not available
...

#NULL!
Speciﬁes an invalid intersection of two areas
...

#REF!
Refers to a cell that is not valid
...

#####
Produces a result, or includes a constant numeric value,
that is too long to ﬁt in the cell
...
)

APPENDIX B
BASICS OF ACCOUNTING
REPORTS AND BUSINESS RATIOS
This appendix provides a fundamental description of ﬁnancial statements
...

B
...
The ﬁscal year (FY) is commonly not the calendar year (CY) for a corporation
...
S
...
For example, October
2011 through September 2012 is FY2012
...

At the end of each ﬁscal year, a company publishes a balance sheet
...
This is a yearly presentation of the state of the
ﬁrm at a particular time, for example, May 31, 2012; however, a balance sheet is also usually
prepared quarterly and monthly
...

Assets
...
There
are two main classes of assets
...
), which is more easily converted to cash, usually within 1 year
...
Conversion of these holdings
to cash in a short time would require a major corporate reorientation
...
This section is a summary of all ﬁnancial obligations (debts, mortgages, loans,
etc
...
Bond indebtedness is included here
...
Also called owner’s equity, this section provides a summary of the ﬁnancial
value of ownership, including stocks issued and earnings retained by the corporation
...
For example,
current assets is comprised of cash, accounts receivable, etc
...

B
...
The income statement summarizes the proﬁts or losses of the corporation for a stated period of time
...
The major categories of an income statement are
Revenues
...

Expenses
...
Some expense amounts are itemized in other statements, for example, cost of
goods sold
...
1 and 17
...
The income statement, published at the same time as the balance sheet, uses the basic equation
Revenues ؊ expenses ‫ ؍‬proﬁt (or loss)
The cost of goods sold is an important accounting term
...
Cost of goods sold may also be called factory cost
...
The total of
the cost of goods sold statement is entered as an expense item on the income statement
...
1]

B
...

Indirect cost allocation methods are discussed in Chapter 15
...
3 Business Ratios
Accountants, ﬁnancial analysts, and engineering economists frequently utilize business ratio
analysis to evaluate the ﬁnancial health (status) of a company over time and in relation to industry norms
...
For comparison purposes, it is necessary
to compute the ratios for several companies in the same industry
...
The ratios are classiﬁed according to their role in measuring the corporation
...
Assess ability to meet short-term and long-term ﬁnancial obligations
...
Measure management’s ability to use and control assets
...
Evaluate the ability to earn a return for the owners of the corporation
...

Current Ratio This ratio is utilized to analyze the company’s working capital condition
...
Note that
only balance sheet data are utilized in the current ratio; that is, no association with revenues or
expenses is made
...
06
26,700
Since current liabilities are those debts payable in the next year, the current ratio value of 3
...
Current ratio
values of 2 to 3 are common
...
Often, however, a better idea of a company’s immediate ﬁnancial position can
be obtained by using the acid test ratio
...
For JAGBA Corporation,
81,700 Ϫ 52,000
Acid test ratio ϭ ——————— ϭ 1
...
However, an acid test ratio of approximately 1
...

Debt Ratio This ratio is a measure of ﬁnancial strength since it is deﬁned as
total liabilities
Debt ratio ϭ ——————
total assets
For JAGBA Corporation,
62,700
Debt ratio ϭ ———— ϭ 0
...
6% creditor-owned and 86
...
A debt ratio in the range of 20%
or less usually indicates a sound ﬁnancial condition, with little fear of forced reorganization because
of unpaid liabilities
...
The debt-equity (D-E) mix is another measure of ﬁnancial strength
...
It
is deﬁned as
net proﬁt
Return on sales ϭ ———— (100%)
net sales
Net proﬁt is the after-tax value from the income statement
...
For JAGBA Corporation,
78,925
Return on sales ϭ ———— (100%) ϭ 15
...
5% to 4
...
In truth, for a relatively large-volume, high-turnover business, an income ratio of 3% is quite healthy
...

Return on Assets Ratio This is the key indicator of proﬁtability since it evaluates the ability of the corporation to transfer assets into operating proﬁt
...
1%
462,700
Efﬁcient use of assets indicates that the company should earn a high return, while low returns
usually accompany lower values of this ratio compared to the industry group ratios
...
They both indicate the number
of times the average inventory value passes through the operations of the company
...
3

Business Ratios

where average inventory is the ﬁgure recorded in the balance sheet
...
71
52,000
This means that the average value of the inventory has been sold 9
...

Values of this ratio vary greatly from one industry to another
...
This ratio is commonly used as a measure of the inventory
turnover rate in manufacturing companies
...
For JAGBA, using the values in Table B–3,
290,000
Cost of goods sold to inventory ϭ ————————— ϭ 5
...

EXAMPLE B
...

Compare the corresponding JAGBA Corporation values with these norms, and comment on
differences and similarities
...
4
1
...
3%
40
...
4
0
...
8%
8
...
2
1
...
1%
8
...
6
1
...
4%
5
...

SOURCE: L
...

Solution
It is not correct to compare ratios for one company with indexes in different industries, that is,
with indexes for different NAICS codes
...
The corresponding values for JAGBA are
Current ratio ϭ 3
...
11
Debt ratio ϭ 13
...
1%

JAGBA has a current ratio larger than all four of these industries, since 3
...
6 and much less in the case of the “average”
air transportation corporation
...
Return on assets, which is a measure of ability to turn assets into proﬁtability, is not as high at JAGBA as motor vehicles, but
JAGBA competes well with the other industry sectors
...
Corporate assets are classiﬁed in categories by
$100,000 units, such as 100 to 250, 1001 to 5000, over 250,000, etc
...
nspe
...

Appendix C

Code of Ethics for Engineers

567

Code of Ethics for Engineers
Preamble
Engineering is an important and learned profession
...
Engineering has a direct and vital impact on the quality of life for
all people
...
Engineers must perform under a standard of
professional behavior that requires adherence to the highest principles of ethical
conduct
...
Fundamental Canons
Engineers, in the fulfillment of their professional duties, shall:
1
...

2
...

3
...

4
...

5
...

6
...

II
...
Engineers shall hold paramount the safety, health, and welfare
of the public
...
If engineers’ judgment is overruled under circumstances that
endanger life or property, they shall notify their employer or client
and such other authority as may be appropriate
...
Engineers shall approve only those engineering documents that are
in conformity with applicable standards
...
Engineers shall not reveal facts, data, or information without the
prior consent of the client or employer except as authorized or
required by law or this Code
...
Engineers shall not permit the use of their name or associate in
business ventures with any person or firm that they believe is
engaged in fraudulent or dishonest enterprise
...
Engineers shall not aid or abet the unlawful practice of engineering
by a person or firm
...
Engineers having knowledge of any alleged violation of this Code
shall report thereon to appropriate professional bodies and, when
relevant, also to public authorities, and cooperate with the proper
authorities in furnishing such information or assistance as may be
required
...
Engineers shall perform services only in the areas of their
competence
...
Engineers shall undertake assignments only when qualified by
education or experience in the specific technical fields involved
...
Engineers shall not affix their signatures to any plans or documents
dealing with subject matter in which they lack competence, nor to
any plan or document not prepared under their direction and
control
...
Engineers may accept assignments and assume responsibility for
coordination of an entire project and sign and seal the engineering
documents for the entire project, provided that each technical
segment is signed and sealed only by the qualified engineers who
prepared the segment
...
Engineers shall issue public statements only in an objective and
truthful manner
...
Engineers shall be objective and truthful in professional reports,
statements, or testimony
...

b
...

c
...

4
...

a
...

b
...

c
...

d
...

e
...

5
...

a
...
They
shall not misrepresent or exaggerate their responsibility in or for the
subject matter of prior assignments
...

b
...

They shall not offer any gift or other valuable
consideration in order to secure work
...

III
...
Engineers shall be guided in all their relations by the highest standards
of honesty and integrity
...
Engineers shall acknowledge their errors and shall not distort or
alter the facts
...
Engineers shall advise their clients or employers when they believe
a project will not be successful
...
Engineers shall not accept outside employment to the detriment of
their regular work or interest
...

d
...

e
...

2
...

a
...

b
...
If the client or employer insists on such unprofessional
conduct, they shall notify the proper authorities and withdraw from
further service on the project
...
Engineers are encouraged to extend public knowledge and
appreciation of engineering and its achievements
...
Engineers are encouraged to adhere to the principles of sustainable
development1 in order to protect the environment for future
generations
...
Engineers shall avoid all conduct or practice that deceives the public
...
Engineers shall avoid the use of statements containing a material
misrepresentation of fact or omitting a material fact
...
Consistent with the foregoing, engineers may advertise for
recruitment of personnel
...
Consistent with the foregoing, engineers may prepare articles for
the lay or technical press, but such articles shall not imply credit to
the author for work performed by others
...
Engineers shall not disclose, without consent, confidential information
concerning the business affairs or technical processes of any present or
former client or employer, or public body on which they serve
...
Engineers shall not, without the consent of all interested parties,
promote or arrange for new employment or practice in connection
with a specific project for which the engineer has gained particular
and specialized knowledge
...
Engineers shall not, without the consent of all interested parties,
participate in or represent an adversary interest in connection with a
specific project or proceeding in which the engineer has gained
particular specialized knowledge on behalf of a former client or
employer
...
Engineers shall not be influenced in their professional duties by
conflicting interests
...
Engineers shall not accept financial or other considerations,
including free engineering designs, from material or equipment
suppliers for specifying their product
...
Engineers shall not accept commissions or allowances, directly or
indirectly, from contractors or other parties dealing with clients or
employers of the engineer in connection with work for which the
engineer is responsible
...
Engineers shall not attempt to obtain employment or advancement or
professional engagements by untruthfully criticizing other engineers,
or by other improper or questionable methods
...
Engineers shall not request, propose, or accept a commission on a
contingent basis under circumstances in which their judgment may
be compromised
...
Engineers in salaried positions shall accept part-time engineering
work only to the extent consistent with policies of the employer and
in accordance with ethical considerations
...
Engineers shall not, without consent, use equipment, supplies,
laboratory, or office facilities of an employer to carry on outside
private practice
...

Engineers shall not attempt to injure, maliciously or falsely, directly
or indirectly, the professional reputation, prospects, practice, or
employment of other engineers
...

a
...

b
...

c
...

8
...

a
...

b
...

9
...

a
...

b
...

c
...

d
...
The employer
should indemnify the engineer for use of the information for any
purpose other than the original purpose
...
Engineers shall continue their professional development throughout
their careers and should keep current in their specialty fields by
engaging in professional practice, participating in continuing
education courses, reading in the technical literature, and attending
professional meetings and seminars
...

As Revised July 2007
“By order of the United States District Court for the District of Columbia,
former Section 11(c) of the NSPE Code of Ethics prohibiting competitive
bidding, and all policy statements, opinions, rulings or other guidelines
interpreting its scope, have been rescinded as unlawfully interfering with the
legal right of engineers, protected under the antitrust laws, to provide price
information to prospective clients; accordingly, nothing contained in the NSPE
Code of Ethics, policy statements, opinions, rulings or other guidelines prohibits
the submission of price quotations or competitive bids for engineering services
at any time or in any amount
...
”
It is further noted that as made clear in the Supreme Court decision:
1
...

2
...

3
...

4
...

5
...
State registration boards with authority to
adopt rules of professional conduct may adopt rules governing procedures to
obtain engineering services
...
As noted by the Supreme Court, “nothing in the judgment prevents NSPE and
its members from attempting to influence governmental action
...
The Code deals with professional services, which services must be performed by real
persons
...
The Code is clearly written to apply to the Engineer,
and it is incumbent on members of NSPE to endeavor to live up to its provisions
...

1420 King Street
Alexandria, Virginia 22314-2794
703/684-2800 • Fax:703/836-4875
www
...
org
Publication date as revised: July 2007 • Publication #1102

APPENDIX D
ALTERNATE METHODS FOR
EQUIVALENCE CALCULATIONS
Throughout the text, engineering economy factor formulas, tabulated factor values, or built-in
spreadsheet functions have been used to obtain a value of P, F, A, i, or n
...
An overview of the possibilities is presented here
...
From the summation of
the series, it is possible to perform calculator-based computations to obtain P, F, or A values
...
2
...
1 Using Programmable Calculators
A basic way to calculate one parameter, given the other four, is to use a calculator that allows a
present worth relation to be encoded
...
For example, consider a PW relation in which all ﬁve parameters are included
...
The interest rate i can be coded for entry as a percent
or decimal
...

This is the approach taken by relatively simple scientiﬁc calculators, such as the HewlettPackard (HP) scientiﬁc series, for example, HP 33s
...
The initial investment is called B rather than P, and the equal uniform amount is termed P rather than A
...

Another example that offers freedom from the spreadsheet and tables is an engineering calculator that has the same functions built in as those on a spreadsheet to determine P, F, A, i, or n
...
The functions are basically
the same as those on a spreadsheet
...
For example, the tvmPV function format is
tvmPV(n,i,Pmt,FV,PpY,CpY,PmtAt)
where

n ϭ number of periods
i ϭ annual interest rate as a percent
Pmt ϭ equal uniform periodic amount A
FV ϭ future amount F
PpY ϭ payments per year (optional; default is 1)
CpY ϭ compounding periods per year (optional; default is 1)
PmtAt ϭ beginning- or end-of-period payments (optional; default is 0 ϭ end)

570

Appendix D

Alternate Methods for Equivalence Calculations

It is easy to understand why it is possible to do a lot on a calculator with relatively wellbehaved cash ﬂow series
...
However, the use of tables or factor formulas is not necessary
...
2 Using the Summation of a Geometric Series
A geometric progression is a series of n terms with a common ratio or base r
...
1]

Ristroph1 and others have explained how the recognition that equivalence computations are
simple applications of geometric series can be used to determine F, P, and A using Equation
[D
...

Before explaining how to apply this approach, we deﬁne the base r as follows when a future
worth F or present worth P is sought
...
This is the same as using a geometric
series of only one term
...
1)10
ϭ 100(2
...
37
This is identical to using the tabulated value (or formula) for the F/P factor
...
5937)
ϭ $259
...
To calculate F, we can move
each A value forward to year 10
...

F ϭ A1(1 ϩ i)9 ϩ A2(1 ϩ i)8 ϩ ⋅ ⋅ ⋅ ϩ A9(1 ϩ i)1 ϩ A10(1 ϩ i)0
Figure D–1

F=?

Future worth of a single
amount in year 0
...
H
...

D
...

P= ?

i = 10%
0 1

2

3

4

5

6

7

8

9

10

Year

A ϭ $100

5

6

7

8

9

10

Times
discounted

Figure D–3
Present worth of a shifted uniform series
...
With r ϭ 1 ϩ i, the geometric series
and its sum S (from Equation [D
...
Removing the subscript on A,
jϭ9

F = A[r9 ϩ r8 ϩ ⋅ ⋅ ⋅ ϩ r1 ϩ r0] ϭ A

⌺r

j

jϭ0

10
r10 Ϫ r0 (1
...
9374
rϪ1
0
...

F ϭ 100(S) ϭ 100(15
...
74
F ϭ 100(F/A,10%,10) ϭ 100(15
...
74
As a ﬁnal demonstration of the geometric series approach to equivalence computations, consider the shifted cash ﬂow series in Figure D–3
...
If the tables are used, the solution is
P ϭ A(P/A,10%,6)(P/F,10%,4) = 100(4
...
6830)
ϭ $297
...
The last A term is discounted 10 years, making n ϭ 10
...
Again using Equation [D
...
1)11 Ϫ (1/1
...
1) Ϫ 1

]

]

(0
...
9091)5
ϭ 100 —————————— ϭ 100(2
...
9091 Ϫ 1
ϭ $297
...
Proponents of this approach point out the use of standard
mathematical notation; the removal of a need to derive any factors; no need to remember the
placement of P, F, and A values based on factor formula development; and the easy use of a calculator to determine the equivalence relations
...
When the series become
quite involved, or when sensitivity analysis is required to reach an economic decision, it may
be beneﬁcial to use a spreadsheet
...
However, once again, the use of tables and factors is not necessary
...
1 Important Concepts and Guidelines
The following elements of engineering economy are identiﬁed throughout the text in the margin
by this checkmark and a title below it
...

Time Value of Money It is a fact that money makes money
...
(1)
Economic Equivalence A combination of time value of money and interest rate that makes
different sums of money at different times have equal economic value
...
Revenues are
cash inﬂows and carry a positive (+) sign; expenses are outﬂows and carry a negative (−) sign
...
g
...
(1, 9)
End-of-Period Convention To simplify calculations, cash ﬂows (revenues and costs) are
assumed to occur at the end of a time period
...
A half-year convention is often used in depreciation calculations
...
COC is usually
a weighted average that involves the cost of debt capital (loans, bonds, and mortgages) and equity capital (stocks and retained earnings)
...
Also called the hurdle rate, MARR is based on cost of
capital, market trend, risk, etc
...
(1, 10)
Opportunity Cost A forgone opportunity caused by the inability to pursue a project
...
Stated differently, it is the ROR of the ﬁrst project rejected because of unavailability of
funds
...
Effective interest
rate is the actual rate over a period of time because compounding is imputed; for example, 1% per
month, compounded monthly, is an effective 12
...
Inﬂation or deﬂation is not considered
...
The A or AW is a series of equal,
end-of-period cash ﬂows for n consecutive periods, expressed as money per time (say,
$/year; /year)
...
(2, 3)

Title

574

Appendix E

Glossary of Concepts and Terms

Placement of Gradient Present Worth (PG ; Pg) The (P/G,i%,n) factor for an arithmetic
gradient ﬁnds the PG of only the gradient series 2 years prior to the ﬁrst appearance of the constant gradient G
...

The (P/A,g,i,n) factor for a geometric gradient determines Pg for the gradient and initial amount
A1 two years prior to the appearance of the ﬁrst gradient amount
...
(2, 3)
Equal-Service Requirement Identical capacity of all alternatives operating over the
same amount of time is mandated by the equal-service requirement
...
PW analysis requires evaluation over the same
number of years (periods) using the LCM (least common multiple) of lives; AW analysis is
performed over one life cycle
...
(5, 6, 8)
LCM or Study Period To select from mutually exclusive alternatives under the equal-service
requirement for PW computations, use the LCM of lives with repurchase(s) as necessary
...
(5, 6, 11)
Salvage/Market Value Expected trade-in, market, or scrap value at the end of the estimated
life or the study period
...
MACRS depreciation always reduces the book value to a salvage of zero
...
DN is status quo; it generates no new costs, revenues, or savings
...
Incremental evaluation requires comparison with DN for revenue alternatives
...
(5)
Rate of Return An interest rate that equates a PW or AW relation to zero
...
(7, 8)
Project Evaluation For a speciﬁed MARR, determine a measure of worth for net cash ﬂow
series over the life or study period
...
(5, 6, 7, 9, 17)
Present worth: If PW ≥ 0
Future worth: If FW ≥ 0
Beneﬁt/cost: If B/C ≥ 1
...
0

ME Alternative Selection For mutually exclusive (select only one) alternatives, compare
two alternatives at a time by determining a measure of worth for the incremental (∆) cash ﬂow
series over the life or study period, adhering to the equal-service requirement
...

Rate of return: Order by initial cost, perform pairwise ∆i* comparison; if ∆i* ≥ MARR,
select larger cost alternative; continue until one remains
...
0, select larger cost alternative; continue until one remains
...
1

Important Concepts and Guidelines

Cost-effectiveness ratio: For service sector alternatives; order by effectiveness measure;
perform pairwise ∆C/E comparison using dominance; select from nondominated alternatives
without exceeding budget
...
Calculate a measure of worth and select using the guidelines below
...

Rate of return: No incremental comparison; select all projects with overall i * ≥ MARR
...
0
...

When a capital budget limit is deﬁned, independent projects are selected using the capital budgeting process based on PW values
...

Capital Recovery CR is the equivalent annual amount an asset or system must earn to recover
the initial investment plus a stated rate of return
...
The salvage value is considered in CR calculations
...
(11)
Sunk Cost Capital (money) that is lost and cannot be recovered
...
They should be handled using tax laws and write-off
allowances, not the economic study
...
Inﬂation
occurs when the value of a currency decreases
...
(1, 14)
Breakeven For a single project, the value of a parameter that makes two elements equal,
e
...
, sales necessary to equate revenues and costs
...
Breakeven analysis is fundamental to make-buy decisions, replacement studies, payback analysis, sensitivity analysis,
breakeven ROR analysis, and many others
...
(8, 13)
Payback Period Amount of time n before recovery of the initial capital investment is expected
...
(13)
Direct / Indirect Costs Direct costs are primarily human labor, machines, and materials associated with a product, process, system, or service
...
(15)
Value Added Activities have added worth to a product or service from the perspective of a
consumer, owner, or investor who is willing to pay more for an enhanced value
...
Parameters may be any cost factor, revenue,
life, salvage value, inﬂation rate, etc
...
Risk represents an absence of or deviation from certainty
...
(10, 18, 19, 20)

E
...

The numbers in parentheses indicate sections where the term is introduced and used in various
applications
...
5, 6
...

Description

Annual operating cost

AOC

Estimated annual costs to maintain and support an
alternative (1
...

Beneﬁt/cost ratio

B/C

Ratio of a project’s beneﬁts to costs expressed in PW, AW,
or FW terms (9
...

Book value

BV

Remaining capital investment in an asset after depreciation
is accounted for (16
...

Breakeven point

QBE

Quantity at which revenues and costs are equal, or two
alternatives are equivalent (13
...

Capital budget

b

Amount of money available for capital investment projects
(12
...

Capital recovery

CR or A

Equivalent annual cost of owning an asset plus the required
return on the initial investment (6
...

Capitalized cost

CC or P

Present worth of an alternative that will last forever (or a
long time) (5
...

Cash ﬂow

CF

Actual cash amounts that are receipts (inﬂow) and disbursements (outﬂow) (1
...

Cash ﬂow before or
after taxes

CFBT or
CFAT

Cash ﬂow amount before relevant taxes or after taxes are
applied (17
...

Compounding
frequency

m

Number of times interest is compounded per period (year)
(4
...

Cost-effectiveness ratio

CER

Ratio of equivalent cost to effectiveness measure to evaluate service sector projects (9
...

Cost estimating relationships

C2 or CT

Relations that use design variables and changing costs over
time to estimate current and future costs (15
...

Cost of capital

WACC

Interest rate paid for the use of capital funds; includes both
debt and equity funds
...
9, 10
...

Debt-equity mix

D-E

Percentages of debt and equity investment capital used by a
corporation (10
...

Depreciation

D

Reduction in the value of assets using speciﬁc models and
rules; there are book and tax depreciation methods (16
...

Depreciation rate

dt

Annual rate for reducing the value of assets using different
depreciation methods (16
...

Economic service life

ESL or n

Number of years at which the AW of costs is a minimum
(11
...

Effectiveness measure

E

A nonmonetary measure used in the cost-effectiveness ratio
for service sector projects (9
...

E
...
3, 19
...

Expenses, operating

OE

All corporate costs incurred in transacting business (17
...

First cost

P

Total initial cost—purchase, construction, setup, etc
...
3, 16
...

Future amount or worth

F or FW

Amount at some future date considering time value of
money (1
...
4)
...
5)
...
6)
...
1)
...
1)
...
4)
...
4, 4
...

Interest rate, inﬂationadjusted

if

Interest rate adjusted to take inﬂation into account (14
...

Life (estimated)

n

Number of years or periods over which an alternative or
asset will be used; the evaluation time (1
...

Life-cycle cost

LCC

Evaluation of costs for a system over all stages: feasibility
to design to phaseout (6
...

Measure of worth

Varies

Value, such as PW, AW, i*, used to judge economic
viability (1
...

Minimum attractive
rate of return

MARR

Minimum value of the rate of return for an alternative to be
ﬁnancially viable (1
...
1)
...
5)
...
6)
...
1)
...
1)
...

Payback period

np

Number of years to recover the initial investment and a
stated rate of return (13
...

Present amount
or worth

P or PW

Amount of money at the current time or a time denoted as
present (1
...
2)
...
2)
...
2, 12
...

Random variable

X

Parameter or characteristic that can take on any one of
several values; discrete and continuous (19
...

Rate of return

i* or ROR

Compound interest rate on unpaid or unrecovered balances
such that the ﬁnal amount results in a zero balance (7
...

577

578

Appendix E

Glossary of Concepts and Terms

Term

Symbol

Description

Recovery period

n

Number of years to completely depreciate an asset (16
...

Return on invested
capital

i'' or ROIC

Unique ROR when a reinvestment rate ii is applied to
multiple-rate cash ﬂows (7
...

Salvage/market value

S or MV

Expected trade-in or market value when an asset is traded
or disposed of (6
...
1, 16
...

Standard deviation

s or σ

Measure of dispersion or spread about the expected value
or average (19
...

Study period

n

Speciﬁed number of years over which an evaluation takes
place (5
...
5)
...
1)
...
1)
...
1)
...
7)
...
1)
...
7)
...
9)
...
T
...
Tarquin: Basics of Engineering Economy, McGraw-Hill, New York, 2008
...
S
...
, Pearson
Prentice Hall, Upper Saddle River, NJ, 2003
...
E
...
G
...
, Pearson
Prentice Hall, Upper Saddle River, NJ, 1992
...
R
...
G
...
J
...
A
...
, Pearson Prentice Hall, Upper Saddle River, NJ, 2005
...
A
...
R
...
, Pearson Prentice Hall,
Upper Saddle River, NJ, 1999
...
F
...
Loulakis: Design-Build Contracting Handbook, 2d ed
...

Eschenbach, T
...
: Engineering Economy: Applying Theory to Practice, 3d ed
...

Fabrycky, W
...
, G
...
Thuesen, and D
...
, Pearson Prentice Hall,
Upper Saddle River, NJ, 1998
...
M
...
M
...
Bernhardt, and M
...

Hartman, J
...
: Engineering Economy and the Decision Making Process, Pearson Prentice Hall, Upper
Saddle River, NJ, 2007
...
M
...

Newnan, D
...
, J
...
Lavelle, and T
...
Eschenbach: Engineering Economic Analysis, 10th ed
...

Ostwald, P
...
: Construction Cost Analysis and Estimating, Pearson Prentice Hall, Upper Saddle River,
NJ, 2001
...
F
...
S
...

Park, C
...
: Contemporary Engineering Economics, 5th ed
...

Park, C
...
: Fundamentals of Engineering Economics, 2d ed
...

Peurifoy, R
...
, and G
...
Oberlender: Estimating Construction Costs, 5th ed
...

Riggs, J
...
, D
...
Bedworth, and S
...
Randhawa: Engineering Economics, 4th ed
...

Stewart, R
...
, R
...
Wyskida, and J
...
Johannes: Cost Estimator’s Reference Manual, 2d ed
...

Sullivan, W
...
, E
...
Wicks, and C
...
Koelling: Engineering Economy, 15th ed
...

Thuesen, G
...
, and W
...
Fabrycky: Engineering Economy, 9th ed
...

White, J
...
, K
...
Case, D
...
Pratt, and M
...
Agee: Principles of Engineering Economic Analysis,
5th ed
...

Using Excel 2007
Gottfried, B
...
: Spreadsheet Tools for Engineers Using Excel® 2007, McGraw-Hill, New York, 2010
...
E
...
S
...
J
...
,
Wadsworth Cengage Learning, Belmont, CA, 2009
...
W
...
Schinzinger: Introduction to Engineering Ethics, 2d ed
...

580

Reference Materials

Websites
Construction cost estimation index: www
...
com
The Economist: www
...
com
For this textbook: www
...
com/blank
Plant cost estimation index: www
...
com/pci
Revenue Canada: www
...
gc
...
S
...
irs
...
online
...
com

U
...
Government Publications (available at
www
...
gov)
Corporations, Publication 544, Internal Revenue Service, GPO, Washington, DC, annually
...

Your Federal Income Tax, Publication 17, Internal Revenue Service, GPO, Washington, DC, annually
...

Harvard Business Review, Harvard University Press, Boston, bimonthly
...

581

Compound Interest Factor Tables

0
...
25%

Discrete Cash Flow: Compound Interest Factors

Single Payments

Uniform Series Payments

Arithmetic Gradients

n

Compound
Amount
F͞P

Present
Worth
P͞F

Sinking
Fund
A͞F

Compound
Amount
F͞A

Capital
Recovery
A͞P

Present
Worth
P͞A

Gradient
Present Worth
P͞G

Gradient
Uniform Series
A͞G

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
84
90
96
100
108
120
132
144
240
360
480

1
...
0050
1
...
0100
1
...
0151
1
...
0202
1
...
0253
1
...
0304
1
...
0356
1
...
0408
1
...
0460
1
...
0512
1
...
0565
1
...
0618
1
...
0671
1
...
0724
1
...
0778
1
...
1050
1
...
1330
1
...
1472
1
...
1969
1
...
2334
1
...
2709
1
...
3095
1
...
3904
1
...
8208
2
...
3151

0
...
9950
0
...
9901
0
...
9851
0
...
9802
0
...
9753
0
...
9705
0
...
9656
0
...
9608
0
...
9561
0
...
9513
0
...
9466
0
...
9418
0
...
9371
0
...
9325
0
...
9278
0
...
9050
0
...
8826
0
...
8717
0
...
8355
0
...
8108
0
...
7869
0
...
7636
0
...
7192
0
...
5492
0
...
3016

1
...
49938
0
...
24906
0
...
16563
0
...
12391
0
...
09888
0
...
08219
0
...
07028
0
...
06134
0
...
05438
0
...
04882
0
...
04427
0
...
04048
0
...
03727
0
...
03452
0
...
03214
0
...
02380
0
...
01880
0
...
01698
0
...
01269
0
...
01071
0
...
00923
0
...
00808
0
...
00640
0
...
00305
0
...
00108

1
...
0025
3
...
0150
5
...
0376
7
...
0704
9
...
1133
11
...
1664
13
...
2298
15
...
3035
17
...
3876
19
...
4822
21
...
5872
23
...
7028
25
...
8290
27
...
9658
30
...
1133
37
...
0132
50
...
1887
55
...
8819
64
...
7794
82
...
3419
100
...
3474
113
...
8093
139
...
1582
173
...
3020
582
...
0595

1
...
50188
0
...
25156
0
...
16813
0
...
12641
0
...
10138
0
...
08469
0
...
07278
0
...
06384
0
...
05688
0
...
05132
0
...
04677
0
...
04298
0
...
03977
0
...
03702
0
...
03464
0
...
02630
0
...
02130
0
...
01948
0
...
01519
0
...
01321
0
...
01173
0
...
01058
0
...
00890
0
...
00555
0
...
00358

0
...
9925
2
...
9751
4
...
9478
6
...
9107
8
...
8639
10
...
8073
12
...
7410
14
...
6650
16
...
5795
18
...
4845
20
...
3800
22
...
2660
24
...
1426
26
...
0099
27
...
8679
34
...
0199
45
...
9462
48
...
3264
55
...
8169
68
...
6813
80
...
2546
88
...
5453
103
...
3121
120
...
3109
237
...
3418

0
...
9801
5
...
9007
14
...
7223
27
...
4061
44
...
9133
64
...
2053
88
...
2441
116
...
9917
148
...
4106
183
...
4634
222
...
1131
264
...
3230
310
...
0566
360
...
2776
413
...
4988
728
...
06
1125
...
59
1353
...
08
2265
...
61
3029
...
87
3886
...
24
4829
...
11
6950
...
41
19399
36264
53821

0
...
9983
1
...
9950
2
...
9900
3
...
9834
4
...
9750
5
...
9650
6
...
9534
7
...
9401
8
...
9251
9
...
9085
10
...
8901
11
...
8702
12
...
8485
13
...
8252
14
...
2306
19
...
0209
23
...
9377
26
...
7514
34
...
8305
40
...
8162
45
...
4216
51
...
5084
61
...
1949
107
...
8902
192
...
5%

TABLE 2

0
...
0050
1
...
0151
1
...
0253
1
...
0355
1
...
0459
1
...
0564
1
...
0670
1
...
0777
1
...
0885
1
...
0994
1
...
1104
1
...
1216
1
...
1328
1
...
1442
1
...
1556
1
...
1967
1
...
2705
1
...
2961
1
...
3489
1
...
4536
1
...
5666
1
...
6467
1
...
8194
1
...
0508
3
...
0226
10
...
9950
0
...
9851
0
...
9754
0
...
9657
0
...
9561
0
...
9466
0
...
9372
0
...
9279
0
...
9187
0
...
9096
0
...
9006
0
...
8916
0
...
8828
0
...
8740
0
...
8653
0
...
8356
0
...
7871
0
...
7716
0
...
7414
0
...
6879
0
...
6383
0
...
6073
0
...
5496
0
...
4876
0
...
1660
0
...
00000
0
...
33167
0
...
19801
0
...
14073
0
...
10891
0
...
08866
0
...
07464
0
...
06436
0
...
05651
0
...
05030
0
...
04528
0
...
04113
0
...
03765
0
...
03469
0
...
03213
0
...
02542
0
...
01849
0
...
01689
0
...
01433
0
...
01102
0
...
00883
0
...
00773
0
...
00610
0
...
00476
0
...
00100
0
...
0000
2
...
0150
4
...
0503
6
...
1059
8
...
1821
10
...
2792
12
...
3972
14
...
5365
16
...
6973
18
...
8797
20
...
0840
23
...
3104
25
...
5591
27
...
8304
29
...
1244
32
...
3361
44
...
0978
56
...
2180
63
...
7700
86
...
7265
104
...
3109
122
...
3337
142
...
8793
186
...
1502
462
...
52
1991
...
00500
0
...
33667
0
...
20301
0
...
14573
0
...
11391
0
...
09366
0
...
07964
0
...
06936
0
...
06151
0
...
05530
0
...
05028
0
...
04613
0
...
04265
0
...
03969
0
...
03713
0
...
03042
0
...
02349
0
...
02189
0
...
01933
0
...
01602
0
...
01383
0
...
01273
0
...
01110
0
...
00976
0
...
00600
0
...
9950
1
...
9702
3
...
9259
5
...
8621
7
...
7791
9
...
6770
11
...
5562
13
...
4166
15
...
2586
17
...
0824
18
...
8880
20
...
6757
22
...
4456
24
...
1980
26
...
9330
27
...
8710
36
...
5803
44
...
6897
47
...
7256
60
...
4136
68
...
3313
76
...
5426
83
...
0735
96
...
4747
139
...
7916
181
...
9901
2
...
9011
9
...
6552
20
...
1755
34
...
3865
52
...
2136
74
...
5835
99
...
4238
128
...
6634
160
...
2322
195
...
0611
233
...
0820
275
...
2281
319
...
4332
367
...
6324
557
...
3347
959
...
70
1113
...
27
1448
...
35
2163
...
66
2976
...
18
3562
...
37
4823
...
59
6451
...
4988
0
...
4938
1
...
4855
2
...
4738
3
...
4589
4
...
4406
5
...
4190
6
...
3940
7
...
3658
8
...
3342
9
...
2993
10
...
2611
11
...
2195
12
...
1747
13
...
1265
16
...
8359
22
...
4624
24
...
7447
28
...
3504
34
...
5763
41
...
6845
45
...
6758
53
...
3103
62
...
1131
128
...
7949

583

Compound Interest Factor Tables

0
...
75%

Discrete Cash Flow: Compound Interest Factors

Single Payments

Uniform Series Payments

Arithmetic Gradients

n

Compound
Amount
F͞P

Present
Worth
P͞F

Sinking
Fund
A͞F

Compound
Amount
F͞A

Capital
Recovery
A͞P

Present
Worth
P͞A

Gradient
Present Worth
P͞G

Gradient
Uniform Series
A͞G

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
84
90
96
100
108
120
132
144
240
360
480

1
...
0151
1
...
0303
1
...
0459
1
...
0616
1
...
0776
1
...
0938
1
...
1103
1
...
1270
1
...
1440
1
...
1612
1
...
1787
1
...
1964
1
...
2144
1
...
2327
1
...
2513
1
...
3483
1
...
4530
1
...
5083
1
...
7126
1
...
8732
1
...
0489
2
...
2411
2
...
6813
2
...
0092
14
...
1099

0
...
9852
0
...
9706
0
...
9562
0
...
9420
0
...
9280
0
...
9142
0
...
9007
0
...
8873
0
...
8742
0
...
8612
0
...
8484
0
...
8358
0
...
8234
0
...
8112
0
...
7992
0
...
7416
0
...
6883
0
...
6630
0
...
5839
0
...
5338
0
...
4881
0
...
4462
0
...
3730
0
...
1664
0
...
0277

1
...
49813
0
...
24721
0
...
16357
0
...
12176
0
...
09667
0
...
07995
0
...
06801
0
...
05906
0
...
05210
0
...
04653
0
...
04198
0
...
03818
0
...
03498
0
...
03223
0
...
02985
0
...
02153
0
...
01656
0
...
01476
0
...
01053
0
...
00859
0
...
00715
0
...
00604
0
...
00446
0
...
00150
0
...
00021

1
...
0075
3
...
0452
5
...
1136
7
...
2132
9
...
3443
11
...
5076
13
...
7034
15
...
9323
18
...
1947
20
...
4912
22
...
8223
25
...
1885
27
...
5903
29
...
0282
32
...
5029
41
...
4465
57
...
3943
63
...
7688
75
...
0070
100
...
4269
127
...
8562
148
...
4832
193
...
1748
257
...
8869
1830
...
32

1
...
50563
0
...
25471
0
...
17107
0
...
12926
0
...
10417
0
...
08745
0
...
07551
0
...
06656
0
...
05960
0
...
05403
0
...
04948
0
...
04568
0
...
04248
0
...
03973
0
...
03735
0
...
02903
0
...
02406
0
...
02226
0
...
01803
0
...
01609
0
...
01465
0
...
01354
0
...
01196
0
...
00900
0
...
00771

0
...
9777
2
...
9261
4
...
8456
6
...
7366
8
...
5996
10
...
4349
12
...
2430
14
...
0243
15
...
7792
17
...
5080
19
...
2112
21
...
8891
22
...
5422
24
...
1707
25
...
7751
31
...
4469
40
...
5664
42
...
9316
48
...
4768
57
...
1540
65
...
2584
70
...
8394
78
...
6064
87
...
1450
124
...
6409

0
...
9408
5
...
7058
14
...
1808
26
...
2544
42
...
8174
61
...
7632
84
...
9876
110
...
3887
139
...
8671
171
...
3253
206
...
6682
243
...
8029
284
...
6387
327
...
0867
373
...
9924
637
...
8404
953
...
59
1128
...
52
1791
...
22
2308
...
00
2853
...
75
3419
...
56
4583
...
58
9494
...
4981
0
...
4907
1
...
4782
2
...
4608
3
...
4384
4
...
4110
5
...
3786
6
...
3413
7
...
2989
8
...
2516
9
...
1994
10
...
1422
11
...
0800
12
...
0128
13
...
9407
16
...
5058
22
...
9476
23
...
1223
27
...
2882
33
...
1357
39
...
8107
43
...
3154
50
...
8232
58
...
4210
107
...
6620

584

Compound Interest Factor Tables

1%

TABLE 4

1%

Discrete Cash Flow: Compound Interest Factors

Single Payments

Uniform Series Payments

Arithmetic Gradients

n

Compound
Amount
F͞P

Present
Worth
P͞F

Sinking
Fund
A͞F

Compound
Amount
F͞A

Capital
Recovery
A͞P

Present
Worth
P͞A

Gradient
Present Worth
P͞G

Gradient
Uniform Series
A͞G

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
84
90
96
100
108
120
132
144
240
360
480

1
...
0201
1
...
0406
1
...
0615
1
...
0829
1
...
1046
1
...
1268
1
...
1495
1
...
1726
1
...
1961
1
...
2202
1
...
2447
1
...
2697
1
...
2953
1
...
3213
1
...
3478
1
...
4889
1
...
6446
1
...
7285
1
...
0471
2
...
3067
2
...
5993
2
...
9289
3
...
7190
4
...
8926
35
...
6477

0
...
9803
0
...
9610
0
...
9420
0
...
9235
0
...
9053
0
...
8874
0
...
8700
0
...
8528
0
...
8360
0
...
8195
0
...
8034
0
...
7876
0
...
7720
0
...
7568
0
...
7419
0
...
6717
0
...
6080
0
...
5785
0
...
4885
0
...
4335
0
...
3847
0
...
3414
0
...
2689
0
...
0918
0
...
0084

1
...
49751
0
...
24628
0
...
16255
0
...
12069
0
...
09558
0
...
07885
0
...
06690
0
...
05794
0
...
05098
0
...
04542
0
...
04086
0
...
03707
0
...
03387
0
...
03112
0
...
02875
0
...
02046
0
...
01551
0
...
01373
0
...
00955
0
...
00765
0
...
00625
0
...
00518
0
...
00368
0
...
00101
0
...
00008

1
...
0100
3
...
0604
5
...
1520
7
...
2857
9
...
4622
11
...
6825
13
...
9474
16
...
2579
18
...
6147
20
...
0190
23
...
4716
25
...
9735
28
...
5256
30
...
1291
33
...
7849
43
...
8864
61
...
4632
67
...
8525
81
...
7099
110
...
6723
144
...
9273
170
...
8926
230
...
8959
319
...
2554
3494
...
01000
0
...
34002
0
...
20604
0
...
14863
0
...
11674
0
...
09645
0
...
08241
0
...
07212
0
...
06426
0
...
05805
0
...
05303
0
...
04889
0
...
04541
0
...
04245
0
...
03990
0
...
03321
0
...
02633
0
...
02476
0
...
02224
0
...
01902
0
...
01690
0
...
01587
0
...
01435
0
...
01313
0
...
01029
0
...
9901
1
...
9410
3
...
8534
5
...
7282
7
...
5660
9
...
3676
11
...
1337
13
...
8651
14
...
5623
16
...
2260
18
...
8570
19
...
4558
21
...
0232
22
...
5596
24
...
0658
25
...
1075
32
...
9740
39
...
3942
42
...
9550
51
...
5871
56
...
1609
61
...
0289
65
...
7005
73
...
1372
90
...
2183
99
...
9803
2
...
8044
9
...
3205
19
...
3812
33
...
8435
50
...
5687
71
...
4221
94
...
2734
120
...
9957
149
...
4664
181
...
5663
216
...
1800
252
...
1957
292
...
5047
333
...
0021
494
...
8561
820
...
4176
939
...
81
1192
...
87
1702
...
32
2240
...
43
2605
...
42
3334
...
69
4177
...
60
8720
...
16

0
...
9934
1
...
9801
2
...
9602
3
...
9337
4
...
9005
5
...
8607
6
...
8143
7
...
7613
8
...
7017
9
...
6354
10
...
5626
11
...
4831
11
...
3971
12
...
3044
13
...
4285
18
...
5976
22
...
2686
24
...
5333
31
...
3793
35
...
8724
39
...
3426
44
...
8349
51
...
8676
75
...
6995
95
...
25%

TABLE 5

1
...
0125
1
...
0380
1
...
0641
1
...
0909
1
...
1183
1
...
1464
1
...
1753
1
...
2048
1
...
2351
1
...
2662
1
...
2981
1
...
3307
1
...
3642
1
...
3985
1
...
4337
1
...
5639
1
...
8154
1
...
9078
1
...
1072
2
...
5388
2
...
0588
3
...
4634
3
...
4402
5
...
9825
19
...
5410
388
...
9877
0
...
9634
0
...
9398
0
...
9167
0
...
8942
0
...
8723
0
...
8509
0
...
8300
0
...
8096
0
...
7898
0
...
7704
0
...
7515
0
...
7330
0
...
7150
0
...
6975
0
...
6394
0
...
5509
0
...
5242
0
...
4746
0
...
3939
0
...
3269
0
...
2887
0
...
2252
0
...
1672
0
...
0114
0
...
00000
0
...
32920
0
...
19506
0
...
13759
0
...
10567
0
...
08537
0
...
07132
0
...
06103
0
...
05316
0
...
04696
0
...
04194
0
...
03780
0
...
03432
0
...
03137
0
...
02882
0
...
02217
0
...
01533
0
...
01377
0
...
01129
0
...
00812
0
...
00607
0
...
00507
0
...
00363
0
...
00251
0
...
00014
0
...
0000
2
...
0377
4
...
1266
6
...
2680
8
...
4634
10
...
7139
12
...
0211
15
...
3863
17
...
8111
20
...
2968
22
...
8450
25
...
4574
27
...
1354
30
...
8809
33
...
6954
36
...
1155
51
...
2284
68
...
6271
78
...
5745
115
...
1035
147
...
7050
183
...
0723
226
...
2171
332
...
6021
1497
...
28
31016

1
...
50939
0
...
25786
0
...
17403
0
...
13213
0
...
10700
0
...
09026
0
...
07831
0
...
06935
0
...
06238
0
...
05682
0
...
05227
0
...
04849
0
...
04529
0
...
04255
0
...
04018
0
...
03192
0
...
02702
0
...
02525
0
...
02115
0
...
01930
0
...
01795
0
...
01692
0
...
01551
0
...
01317
0
...
01253

0
...
9631
2
...
8781
4
...
7460
6
...
5681
8
...
3455
10
...
0793
11
...
7706
13
...
4203
15
...
0295
16
...
5993
18
...
1306
19
...
6242
21
...
0813
22
...
5025
24
...
8889
28
...
3269
35
...
0129
38
...
6017
42
...
2925
48
...
8222
53
...
7246
56
...
0865
61
...
4781
66
...
9423
79
...
7942

0
...
9023
5
...
5160
14
...
6571
25
...
1487
41
...
8201
59
...
5072
80
...
0519
104
...
3021
130
...
1115
159
...
3392
191
...
8499
224
...
5132
260
...
2040
298
...
8019
337
...
2830
559
...
2296
811
...
9409
946
...
84
1428
...
79
1778
...
83
2127
...
24
2468
...
57
3109
...
61
5101
...
90
6284
...
4969
0
...
4845
1
...
4638
2
...
4348
3
...
3975
4
...
3520
5
...
2982
6
...
2362
7
...
1659
8
...
0874
9
...
0006
10
...
9056
11
...
8024
12
...
6911
13
...
5715
16
...
8515
21
...
9295
22
...
8936
25
...
2047
31
...
3258
36
...
1793
39
...
7737
45
...
2234
51
...
1764
75
...
7619

586

Compound Interest Factor Tables

1
...
5%

Discrete Cash Flow: Compound Interest Factors

Single Payments

Uniform Series Payments

Arithmetic Gradients

n

Compound
Amount
F͞P

Present
Worth
P͞F

Sinking
Fund
A͞F

Compound
Amount
F͞A

Capital
Recovery
A͞P

Present
Worth
P͞A

Gradient
Present Worth
P͞G

Gradient
Uniform Series
A͞G

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
84
90
96
100
108
120
132
144
240
360
480

1
...
0302
1
...
0614
1
...
0934
1
...
1265
1
...
1605
1
...
1956
1
...
2318
1
...
2690
1
...
3073
1
...
3469
1
...
3876
1
...
4295
1
...
4727
1
...
5172
1
...
5631
1
...
8140
2
...
1052
2
...
2679
2
...
9212
3
...
4926
3
...
1758
4
...
9927
5
...
1370
8
...
6328
212
...
70

0
...
9707
0
...
9422
0
...
9145
0
...
8877
0
...
8617
0
...
8364
0
...
8118
0
...
7880
0
...
7649
0
...
7425
0
...
7207
0
...
6995
0
...
6790
0
...
6591
0
...
6398
0
...
5513
0
...
4750
0
...
4409
0
...
3423
0
...
2863
0
...
2395
0
...
2003
0
...
1401
0
...
0281
0
...
0008

1
...
49628
0
...
24444
0
...
16053
0
...
11858
0
...
09343
0
...
07668
0
...
06472
0
...
05577
0
...
04881
0
...
04325
0
...
03870
0
...
03492
0
...
03173
0
...
02900
0
...
02664
0
...
01843
0
...
01357
0
...
01183
0
...
00781
0
...
00602
0
...
00472
0
...
00376
0
...
00244
0
...
00043
0
...
00001

1
...
0150
3
...
0909
5
...
2296
7
...
4328
9
...
7027
11
...
0412
14
...
4504
16
...
9324
19
...
4894
21
...
1237
24
...
8376
27
...
6335
30
...
5140
32
...
4815
35
...
5387
47
...
2679
69
...
6828
77
...
5296
96
...
0772
136
...
1726
187
...
7202
228
...
1778
331
...
1354
502
...
85
14114
84580

1
...
51128
0
...
25944
0
...
17553
0
...
13358
0
...
10843
0
...
09168
0
...
07972
0
...
07077
0
...
06381
0
...
05825
0
...
05370
0
...
04992
0
...
04673
0
...
04400
0
...
04164
0
...
03343
0
...
02857
0
...
02683
0
...
02281
0
...
02102
0
...
01972
0
...
01876
0
...
01744
0
...
01543
0
...
01501

0
...
9559
2
...
8544
4
...
6972
6
...
4859
8
...
2222
10
...
9075
11
...
5434
13
...
1313
14
...
6726
16
...
1686
17
...
6208
19
...
0304
20
...
3986
22
...
7267
23
...
0158
27
...
9158
34
...
9997
35
...
2715
39
...
8447
44
...
5786
49
...
7017
51
...
3137
55
...
3257
58
...
7957
66
...
6142

0
...
8833
5
...
4229
13
...
4018
25
...
6125
40
...
8568
58
...
9454
78
...
6974
101
...
9400
126
...
5084
154
...
2453
184
...
0006
216
...
6310
249
...
0002
284
...
9779
321
...
8303
524
...
5462
749
...
8774
868
...
1674
1279
...
56
1568
...
54
1847
...
45
2112
...
71
2588
...
58
3870
...
72
4415
...
4963
0
...
4814
1
...
4566
2
...
4219
3
...
3772
4
...
3227
5
...
2582
6
...
1839
7
...
0997
8
...
0057
9
...
9018
10
...
7881
11
...
6646
12
...
5313
12
...
3883
15
...
5277
20
...
4277
22
...
2894
25
...
1893
30
...
9668
34
...
4381
37
...
6171
42
...
1579
47
...
7368
64
...
2883

587

Compound Interest Factor Tables

2%

TABLE 7

2%

Discrete Cash Flow: Compound Interest Factors

Single Payments

Uniform Series Payments

Arithmetic Gradients

n

Compound
Amount
F͞P

Present
Worth
P͞F

Sinking
Fund
A͞F

Compound
Amount
F͞A

Capital
Recovery
A͞P

Present
Worth
P͞A

Gradient
Present Worth
P͞G

Gradient
Uniform Series
A͞G

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
84
90
96
100
108
120
132
144
240
360
480

1
...
0404
1
...
0824
1
...
1262
1
...
1717
1
...
2190
1
...
2682
1
...
3195
1
...
3728
1
...
4282
1
...
4859
1
...
5460
1
...
6084
1
...
6734
1
...
7410
1
...
8114
2
...
2080
2
...
6916
2
...
9717
3
...
1611
4
...
2773
5
...
6929
7
...
4883
10
...
6528
17
...
8887
1247
...
9804
0
...
9423
0
...
9057
0
...
8706
0
...
8368
0
...
8043
0
...
7730
0
...
7430
0
...
7142
0
...
6864
0
...
6598
0
...
6342
0
...
6095
0
...
5859
0
...
5631
0
...
4902
0
...
3865
0
...
3571
0
...
3048
0
...
2265
0
...
1683
0
...
1380
0
...
0929
0
...
0578
0
...
0008
0
...
00000
0
...
32675
0
...
19216
0
...
13451
0
...
10252
0
...
08218
0
...
06812
0
...
05783
0
...
04997
0
...
04378
0
...
03878
0
...
03467
0
...
03122
0
...
02829
0
...
02578
0
...
01923
0
...
01260
0
...
01111
0
...
00877
0
...
00586
0
...
00405
0
...
00320
0
...
00205
0
...
00123
0
...
00002

1
...
0200
3
...
1216
5
...
3081
7
...
5830
9
...
9497
12
...
4121
14
...
9739
17
...
6393
20
...
4123
22
...
2974
25
...
2990
28
...
4219
32
...
6709
35
...
0512
38
...
5681
51
...
4020
79
...
5794
90
...
5865
114
...
0570
170
...
8666
247
...
6467
312
...
4129
488
...
6415
815
...
44
62328

1
...
51505
0
...
26262
0
...
17853
0
...
13651
0
...
11133
0
...
09456
0
...
08260
0
...
07365
0
...
06670
0
...
06116
0
...
05663
0
...
05287
0
...
04970
0
...
04699
0
...
04465
0
...
03656
0
...
03182
0
...
03014
0
...
02633
0
...
02468
0
...
02351
0
...
02267
0
...
02158
0
...
02017
0
...
02000

0
...
9416
2
...
8077
4
...
6014
6
...
3255
8
...
9826
9
...
5753
11
...
1062
12
...
5777
14
...
9920
15
...
3514
17
...
6580
18
...
9139
19
...
1210
20
...
2813
21
...
3965
25
...
3555
30
...
4236
32
...
1748
34
...
9841
38
...
5255
41
...
5294
43
...
1095
45
...
3378
47
...
5686
49
...
9963

0
...
8458
5
...
2403
13
...
9035
24
...
5720
38
...
9977
55
...
9475
74
...
2021
96
...
5554
119
...
8139
144
...
7959
171
...
3309
199
...
2592
229
...
4311
259
...
7064
291
...
0405
461
...
9657
642
...
7849
733
...
6975
1034
...
64
1230
...
17
1409
...
75
1569
...
42
1833
...
79
2374
...
57
2498
...
4950
0
...
4752
1
...
4423
2
...
3961
3
...
3367
4
...
2642
5
...
1786
6
...
0799
7
...
9681
8
...
8433
9
...
7055
10
...
5547
10
...
3910
11
...
2145
12
...
0251
15
...
8885
19
...
4420
21
...
1057
23
...
2234
28
...
3616
31
...
1370
33
...
5774
37
...
5676
41
...
9110
49
...
9643

588

Compound Interest Factor Tables

3%

TABLE 8

3%

Discrete Cash Flow: Compound Interest Factors

Single Payments

Uniform Series Payments

Arithmetic Gradients

n

Compound
Amount
F͞P

Present
Worth
P͞F

Sinking
Fund
A͞F

Compound
Amount
F͞A

Capital
Recovery
A͞P

Present
Worth
P͞A

Gradient
Present Worth
P͞G

Gradient
Uniform Series
A͞G

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
40
45
50
55
60
65
70
75
80
84
85
90
96
108
120

1
...
0609
1
...
1255
1
...
1941
1
...
2668
1
...
3439
1
...
4258
1
...
5126
1
...
6047
1
...
7024
1
...
8061
1
...
9161
1
...
0328
2
...
1566
2
...
2879
2
...
4273
2
...
5751
2
...
7319
2
...
2620
3
...
3839
5
...
8916
6
...
9178
9
...
6409
11
...
3357
14
...
0755
24
...
7110

0
...
9426
0
...
8885
0
...
8375
0
...
7894
0
...
7441
0
...
7014
0
...
6611
0
...
6232
0
...
5874
0
...
5537
0
...
5219
0
...
4919
0
...
4637
0
...
4371
0
...
4120
0
...
3883
0
...
3660
0
...
3066
0
...
2281
0
...
1697
0
...
1263
0
...
0940
0
...
0811
0
...
0586
0
...
0288

1
...
49261
0
...
23903
0
...
15460
0
...
11246
0
...
08723
0
...
07046
0
...
05853
0
...
04961
0
...
04271
0
...
03722
0
...
03275
0
...
02905
0
...
02594
0
...
02329
0
...
02102
0
...
01905
0
...
01732
0
...
01326
0
...
00887
0
...
00613
0
...
00434
0
...
00311
0
...
00265
0
...
00187
0
...
00089

1
...
0300
3
...
1836
5
...
4684
7
...
8923
10
...
4639
12
...
1920
15
...
0863
18
...
1569
21
...
4144
25
...
8704
28
...
5368
32
...
4265
36
...
5530
40
...
9309
45
...
5754
50
...
5028
55
...
7302
60
...
4013
92
...
7969
136
...
0534
194
...
5941
272
...
3630
365
...
8570
443
...
8502
778
...
70

1
...
52261
0
...
26903
0
...
18460
0
...
14246
0
...
11723
0
...
10046
0
...
08853
0
...
07961
0
...
07271
0
...
06722
0
...
06275
0
...
05905
0
...
05594
0
...
05329
0
...
05102
0
...
04905
0
...
04732
0
...
04326
0
...
03887
0
...
03613
0
...
03434
0
...
03311
0
...
03265
0
...
03187
0
...
03089

0
...
9135
2
...
7171
4
...
4172
6
...
0197
7
...
5302
9
...
9540
10
...
2961
11
...
5611
13
...
7535
14
...
8775
15
...
9369
16
...
9355
17
...
8768
18
...
7641
19
...
6004
20
...
3888
20
...
1318
21
...
1148
24
...
7298
26
...
6756
28
...
1234
29
...
2008
30
...
6312
31
...
3812
31
...
3730

0
...
7729
5
...
8888
13
...
9547
23
...
6119
36
...
5330
51
...
4196
68
...
0002
86
...
0280
106
...
2788
126
...
5496
148
...
6566
170
...
4336
194
...
7309
217
...
4137
241
...
3609
265
...
4642
289
...
6267
361
...
6325
477
...
7411
583
...
2010
676
...
6978
756
...
5434
791
...
6302
858
...
6013
963
...
4926
0
...
4631
1
...
4138
2
...
3450
3
...
2565
4
...
1485
5
...
0210
6
...
8742
7
...
7081
8
...
5229
8
...
3186
9
...
0954
10
...
8535
11
...
5930
11
...
3141
12
...
0169
13
...
7018
14
...
6502
17
...
5575
19
...
0674
22
...
2145
24
...
0353
25
...
8349
26
...
3615
28
...
7737

589

Compound Interest Factor Tables

4%

TABLE 9

4%

Discrete Cash Flow: Compound Interest Factors

Single Payments

Uniform Series Payments

Arithmetic Gradients

n

Compound
Amount
F͞P

Present
Worth
P͞F

Sinking
Fund
A͞F

Compound
Amount
F͞A

Capital
Recovery
A͞P

Present
Worth
P͞A

Gradient
Present Worth
P͞G

Gradient
Uniform Series
A͞G

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
40
45
50
55
60
65
70
75
80
85
90
96
108
120
144

1
...
0816
1
...
1699
1
...
2653
1
...
3686
1
...
4802
1
...
6010
1
...
7317
1
...
8730
1
...
0258
2
...
1911
2
...
3699
2
...
5633
2
...
7725
2
...
9987
3
...
2434
3
...
5081
3
...
7943
3
...
8010
5
...
1067
8
...
5196
12
...
5716
18
...
0498
28
...
1193
43
...
1195
110
...
6618

0
...
9246
0
...
8548
0
...
7903
0
...
7307
0
...
6756
0
...
6246
0
...
5775
0
...
5339
0
...
4936
0
...
4564
0
...
4220
0
...
3901
0
...
3607
0
...
3335
0
...
3083
0
...
2851
0
...
2636
0
...
2083
0
...
1407
0
...
0951
0
...
0642
0
...
0434
0
...
0293
0
...
0145
0
...
0035

1
...
49020
0
...
23549
0
...
15076
0
...
10853
0
...
08329
0
...
06655
0
...
05467
0
...
04582
0
...
03899
0
...
03358
0
...
02920
0
...
02559
0
...
02257
0
...
02001
0
...
01783
0
...
01595
0
...
01431
0
...
01052
0
...
00655
0
...
00420
0
...
00275
0
...
00181
0
...
00121
0
...
00059
0
...
00014

1
...
0400
3
...
2465
5
...
6330
7
...
2142
10
...
0061
13
...
0258
16
...
2919
20
...
8245
23
...
6454
27
...
7781
31
...
2480
36
...
0826
41
...
3117
47
...
9676
52
...
0849
59
...
7015
66
...
8579
73
...
0255
121
...
6671
191
...
9907
294
...
2905
448
...
2450
676
...
9833
1054
...
99
2741
...
55

1
...
53020
0
...
27549
0
...
19076
0
...
14853
0
...
12329
0
...
10655
0
...
09467
0
...
08582
0
...
07899
0
...
07358
0
...
06920
0
...
06559
0
...
06257
0
...
06001
0
...
05783
0
...
05595
0
...
05431
0
...
05052
0
...
04655
0
...
04420
0
...
04275
0
...
04181
0
...
04121
0
...
04059
0
...
04014

0
...
8861
2
...
6299
4
...
2421
6
...
7327
7
...
1109
8
...
3851
9
...
5631
11
...
6523
12
...
6593
13
...
5903
14
...
4511
14
...
2470
15
...
9828
16
...
6631
16
...
2920
17
...
8736
18
...
4112
18
...
7928
20
...
4822
22
...
6235
23
...
3945
23
...
9154
24
...
2673
24
...
6383
24
...
9119

0
...
7025
5
...
5547
12
...
0657
22
...
8013
33
...
3772
47
...
4546
61
...
7355
77
...
9581
94
...
8933
111
...
3414
129
...
1284
147
...
1040
165
...
1385
183
...
1206
201
...
9556
218
...
5634
236
...
8768
286
...
4028
361
...
6890
422
...
2014
472
...
0408
511
...
9384
540
...
9312
576
...
2428
610
...
4902
0
...
4510
1
...
3857
2
...
2944
3
...
1773
4
...
0343
5
...
8659
6
...
6720
7
...
4530
7
...
2091
8
...
9407
9
...
6479
9
...
3312
10
...
9909
11
...
6274
11
...
2411
12
...
8324
13
...
4765
15
...
8122
17
...
6972
19
...
1961
20
...
3718
21
...
2826
22
...
4146
23
...
4906

590

Compound Interest Factor Tables

5%

TABLE 10

5%

Discrete Cash Flow: Compound Interest Factors

Single Payments

Uniform Series Payments

Arithmetic Gradients

n

Compound
Amount
F͞P

Present
Worth
P͞F

Sinking
Fund
A͞F

Compound
Amount
F͞A

Capital
Recovery
A͞P

Present
Worth
P͞A

Gradient
Present Worth
P͞G

Gradient
Uniform Series
A͞G

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
40
45
50
55
60
65
70
75
80
85
90
95
96
98
100

1
...
1025
1
...
2155
1
...
3401
1
...
4775
1
...
6289
1
...
7959
1
...
9799
2
...
1829
2
...
4066
2
...
6533
2
...
9253
3
...
2251
3
...
5557
3
...
9201
4
...
3219
4
...
7649
5
...
2533
5
...
0400
8
...
4674
14
...
6792
23
...
4264
38
...
5614
63
...
7304
103
...
1864
119
...
5013

0
...
9070
0
...
8227
0
...
7462
0
...
6768
0
...
6139
0
...
5568
0
...
5051
0
...
4581
0
...
4155
0
...
3769
0
...
3418
0
...
3101
0
...
2812
0
...
2551
0
...
2314
0
...
2099
0
...
1904
0
...
1420
0
...
0872
0
...
0535
0
...
0329
0
...
0202
0
...
0124
0
...
0092
0
...
0076

1
...
48780
0
...
23201
0
...
14702
0
...
10472
0
...
07950
0
...
06283
0
...
05102
0
...
04227
0
...
03555
0
...
03024
0
...
02597
0
...
02247
0
...
01956
0
...
01712
0
...
01505
0
...
01328
0
...
01176
0
...
00828
0
...
00478
0
...
00283
0
...
00170
0
...
00103
0
...
00063
0
...
00047
0
...
00038

1
...
0500
3
...
3101
5
...
8019
8
...
5491
11
...
5779
14
...
9171
17
...
5986
21
...
6575
25
...
1324
30
...
0660
35
...
5052
41
...
5020
47
...
1135
54
...
4026
62
...
4388
70
...
2988
80
...
0670
90
...
7998
159
...
3480
272
...
5837
456
...
5285
756
...
2288
1245
...
61
2040
...
73
2365
...
03

1
...
53780
0
...
28201
0
...
19702
0
...
15472
0
...
12950
0
...
11283
0
...
10102
0
...
09227
0
...
08555
0
...
08024
0
...
07597
0
...
07247
0
...
06956
0
...
06712
0
...
06505
0
...
06328
0
...
06176
0
...
05828
0
...
05478
0
...
05283
0
...
05170
0
...
05103
0
...
05063
0
...
05047
0
...
05038

0
...
8594
2
...
5460
4
...
0757
5
...
4632
7
...
7217
8
...
8633
9
...
8986
10
...
8378
11
...
6896
12
...
4622
12
...
1630
13
...
7986
14
...
3752
14
...
8981
15
...
3725
15
...
8027
16
...
1929
16
...
1591
17
...
2559
18
...
9293
19
...
3427
19
...
5965
19
...
7523
19
...
8151
19
...
8479

0
...
6347
5
...
2369
11
...
2321
20
...
1268
31
...
4988
43
...
9879
56
...
2880
70
...
1405
84
...
3275
98
...
6673
112
...
0087
127
...
2275
141
...
2226
155
...
9126
168
...
2333
181
...
1351
194
...
5807
229
...
3145
277
...
5104
314
...
6910
340
...
0721
359
...
8007
372
...
6774
378
...
2139
381
...
4878
0
...
4391
1
...
3579
2
...
2445
3
...
0991
4
...
9219
5
...
7133
6
...
4736
6
...
2034
7
...
9030
8
...
5730
8
...
2140
9
...
8266
10
...
4114
10
...
9691
11
...
5005
11
...
0063
12
...
3775
14
...
2233
15
...
6062
17
...
6212
18
...
3526
18
...
8712
19
...
1044
19
...
2337

591

Compound Interest Factor Tables

6%

TABLE 11

6%

Discrete Cash Flow: Compound Interest Factors

Single Payments

Uniform Series Payments

Arithmetic Gradients

n

Compound
Amount
F͞P

Present
Worth
P͞F

Sinking
Fund
A͞F

Compound
Amount
F͞A

Capital
Recovery
A͞P

Present
Worth
P͞A

Gradient
Present Worth
P͞G

Gradient
Uniform Series
A͞G

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
40
45
50
55
60
65
70
75
80
85
90
95
96
98
100

1
...
1236
1
...
2625
1
...
4185
1
...
5938
1
...
7908
1
...
0122
2
...
2609
2
...
5404
2
...
8543
3
...
2071
3
...
6035
3
...
0489
4
...
5494
4
...
1117
5
...
7435
6
...
4534
6
...
2510
7
...
2857
13
...
4202
24
...
9877
44
...
0759
79
...
7960
141
...
4645
253
...
7590
301
...
3021

0
...
8900
0
...
7921
0
...
7050
0
...
6274
0
...
5584
0
...
4970
0
...
4423
0
...
3936
0
...
3503
0
...
3118
0
...
2775
0
...
2470
0
...
2198
0
...
1956
0
...
1741
0
...
1550
0
...
1379
0
...
0972
0
...
0543
0
...
0303
0
...
0169
0
...
0095
0
...
0053
0
...
0037
0
...
0029

1
...
48544
0
...
22859
0
...
14336
0
...
10104
0
...
07587
0
...
05928
0
...
04758
0
...
03895
0
...
03236
0
...
02718
0
...
02305
0
...
01968
0
...
01690
0
...
01459
0
...
01265
0
...
01100
0
...
00960
0
...
00646
0
...
00344
0
...
00188
0
...
00103
0
...
00057
0
...
00032
0
...
00022
0
...
00018

1
...
0600
3
...
3746
5
...
9753
8
...
8975
11
...
1808
14
...
8699
18
...
0151
23
...
6725
28
...
9057
33
...
7856
39
...
3923
46
...
8156
54
...
1564
63
...
5281
73
...
0582
84
...
8898
97
...
1838
111
...
7620
212
...
3359
394
...
1282
719
...
9322
1300
...
60
2342
...
08
4209
...
65
5016
...
37

1
...
54544
0
...
28859
0
...
20336
0
...
16104
0
...
13587
0
...
11928
0
...
10758
0
...
09895
0
...
09236
0
...
08718
0
...
08305
0
...
07968
0
...
07690
0
...
07459
0
...
07265
0
...
07100
0
...
06960
0
...
06646
0
...
06344
0
...
06188
0
...
06103
0
...
06057
0
...
06032
0
...
06022
0
...
06018

0
...
8334
2
...
4651
4
...
9173
5
...
2098
6
...
3601
7
...
3838
8
...
2950
9
...
1059
10
...
8276
11
...
4699
11
...
0416
12
...
5504
12
...
0032
13
...
4062
13
...
7648
13
...
0840
14
...
3681
14
...
0463
15
...
7619
15
...
1614
16
...
3845
16
...
5091
16
...
5787
16
...
6047
16
...
6175

0
...
5692
4
...
9345
11
...
4497
19
...
5768
29
...
8702
40
...
9629
51
...
5546
63
...
4011
75
...
3062
87
...
1136
98
...
7007
110
...
9732
121
...
8600
132
...
3096
142
...
2864
152
...
7681
161
...
7427
185
...
1096
217
...
3222
239
...
9450
253
...
4527
262
...
8096
268
...
4375
270
...
4491
272
...
4854
0
...
4272
1
...
3304
2
...
1952
3
...
0220
4
...
8113
5
...
5635
5
...
2794
6
...
9597
7
...
6051
7
...
2166
8
...
7951
9
...
3414
9
...
8568
10
...
3422
10
...
7988
11
...
2276
11
...
3590
13
...
7964
14
...
7909
15
...
4613
15
...
9033
16
...
1891
16
...
3081
16
...
3711

592

Compound Interest Factor Tables

7%

TABLE 12

7%

Discrete Cash Flow: Compound Interest Factors

Single Payments

Uniform Series Payments

Arithmetic Gradients

n

Compound
Amount
F͞P

Present
Worth
P͞F

Sinking
Fund
A͞F

Compound
Amount
F͞A

Capital
Recovery
A͞P

Present
Worth
P͞A

Gradient
Present Worth
P͞G

Gradient
Uniform Series
A͞G

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
40
45
50
55
60
65
70
75
80
85
90
95
96
98
100

1
...
1449
1
...
3108
1
...
5007
1
...
7182
1
...
9672
2
...
2522
2
...
5785
2
...
9522
3
...
3799
3
...
8697
4
...
4304
4
...
0724
5
...
8074
6
...
6488
7
...
6123
8
...
7153
9
...
9781
10
...
9745
21
...
4570
41
...
9464
81
...
9894
159
...
2344
314
...
1030
618
...
9766
757
...
7163

0
...
8734
0
...
7629
0
...
6663
0
...
5820
0
...
5083
0
...
4440
0
...
3878
0
...
3387
0
...
2959
0
...
2584
0
...
2257
0
...
1971
0
...
1722
0
...
1504
0
...
1314
0
...
1147
0
...
1002
0
...
0668
0
...
0339
0
...
0173
0
...
0088
0
...
0045
0
...
0023
0
...
0015
0
...
0012

1
...
48309
0
...
22523
0
...
13980
0
...
09747
0
...
07238
0
...
05590
0
...
04434
0
...
03586
0
...
02941
0
...
02439
0
...
02041
0
...
01719
0
...
01456
0
...
01239
0
...
01059
0
...
00907
0
...
00780
0
...
00501
0
...
00246
0
...
00123
0
...
00062
0
...
00031
0
...
00016
0
...
00011
0
...
00008

1
...
0700
3
...
4399
5
...
1533
8
...
2598
11
...
8164
15
...
8885
20
...
5505
25
...
8881
30
...
9990
37
...
9955
44
...
0057
53
...
1767
63
...
6765
74
...
6977
87
...
4608
102
...
2182
118
...
2588
138
...
6351
285
...
5289
575
...
5204
1146
...
13
2269
...
06
4478
...
19
8823
...
52
10813
12382

1
...
55309
0
...
29523
0
...
20980
0
...
16747
0
...
14238
0
...
12590
0
...
11434
0
...
10586
0
...
09941
0
...
09439
0
...
09041
0
...
08719
0
...
08456
0
...
08239
0
...
08059
0
...
07907
0
...
07780
0
...
07501
0
...
07246
0
...
07123
0
...
07062
0
...
07031
0
...
07016
0
...
07011
0
...
07008

0
...
8080
2
...
3872
4
...
7665
5
...
9713
6
...
0236
7
...
9427
8
...
7455
9
...
4466
9
...
0591
10
...
5940
10
...
0612
11
...
4693
11
...
8258
11
...
1371
12
...
4090
12
...
6466
12
...
8540
12
...
3317
13
...
8007
13
...
0392
14
...
1604
14
...
2220
14
...
2533
14
...
2641
14
...
2693

0
...
5060
4
...
6467
10
...
7149
18
...
1404
27
...
4665
37
...
3302
47
...
4461
57
...
5923
67
...
5991
77
...
3393
87
...
7201
96
...
6765
104
...
1656
113
...
1622
120
...
6550
128
...
6435
134
...
1353
152
...
7559
172
...
1243
185
...
1452
193
...
1035
198
...
5717
200
...
5581
201
...
9651
202
...
4831
0
...
4155
1
...
3032
2
...
1465
3
...
9461
4
...
7025
5
...
4167
5
...
0897
6
...
7225
7
...
3163
7
...
8725
8
...
3923
8
...
8773
9
...
3289
9
...
7487
9
...
1381
10
...
4987
10
...
4233
12
...
5287
12
...
2321
13
...
6662
13
...
9273
14
...
0812
14
...
1405
14
...
1703

593

Compound Interest Factor Tables

8%

TABLE 13

8%

Discrete Cash Flow: Compound Interest Factors

Single Payments

Uniform Series Payments

Arithmetic Gradients

n

Compound
Amount
F͞P

Present
Worth
P͞F

Sinking
Fund
A͞F

Compound
Amount
F͞A

Capital
Recovery
A͞P

Present
Worth
P͞A

Gradient
Present Worth
P͞G

Gradient
Uniform Series
A͞G

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
40
45
50
55
60
65
70
75
80
85
90
95
96
98
100

1
...
1664
1
...
3605
1
...
5869
1
...
8509
1
...
1589
2
...
5182
2
...
9372
3
...
4259
3
...
9960
4
...
6610
5
...
4365
5
...
3412
6
...
3964
7
...
6271
9
...
0627
10
...
7371
12
...
6901
14
...
7245
31
...
9016
68
...
2571
148
...
6064
321
...
9548
693
...
92
1497
...
89
1885
...
76

0
...
8573
0
...
7350
0
...
6302
0
...
5403
0
...
4632
0
...
3971
0
...
3405
0
...
2919
0
...
2502
0
...
2145
0
...
1839
0
...
1577
0
...
1352
0
...
1159
0
...
0994
0
...
0852
0
...
0730
0
...
0460
0
...
0213
0
...
0099
0
...
0046
0
...
0021
0
...
0010
0
...
0006
0
...
0005

1
...
48077
0
...
22192
0
...
13632
0
...
09401
0
...
06903
0
...
05270
0
...
04130
0
...
03298
0
...
02670
0
...
02185
0
...
01803
0
...
01498
0
...
01251
0
...
01049
0
...
00883
0
...
00745
0
...
00630
0
...
00386
0
...
00174
0
...
00080
0
...
00037
0
...
00017
0
...
00008
0
...
00005
0
...
00004

1
...
0800
3
...
5061
5
...
3359
8
...
6366
12
...
4866
16
...
9771
21
...
2149
27
...
3243
33
...
4502
41
...
7620
50
...
4568
60
...
7648
73
...
9544
87
...
3388
103
...
2832
123
...
2135
145
...
6267
172
...
0565
386
...
7702
848
...
21
1847
...
08
4002
...
94
8655
...
08000
0
...
38803
0
...
25046
0
...
19207
0
...
16008
0
...
14008
0
...
12652
0
...
11683
0
...
10963
0
...
10413
0
...
09983
0
...
09642
0
...
09368
0
...
09145
0
...
08962
0
...
08811
0
...
08685
0
...
08580
0
...
08259
0
...
08118
0
...
08054
0
...
08025
0
...
08012
0
...
08005
0
...
08004
0
...
9259
1
...
5771
3
...
9927
4
...
2064
5
...
2469
6
...
1390
7
...
9038
8
...
5595
8
...
1216
9
...
6036
9
...
0168
10
...
3711
10
...
6748
10
...
9352
11
...
1584
11
...
3498
11
...
5139
11
...
6546
11
...
1084
12
...
3186
12
...
4160
12
...
4611
12
...
4820
12
...
4917
12
...
4934
12
...
8573
2
...
6501
7
...
5233
14
...
8061
21
...
9768
30
...
6339
39
...
4723
47
...
2640
56
...
8426
65
...
0898
73
...
9257
80
...
2997
87
...
1842
94
...
5687
100
...
4558
106
...
8575
111
...
7924
116
...
0422
133
...
5928
144
...
3000
149
...
5326
152
...
8001
154
...
9925
155
...
4112
155
...
6107

0
...
9487
1
...
8465
2
...
6937
3
...
4910
3
...
2395
4
...
9402
5
...
5945
5
...
2037
6
...
7697
7
...
2940
7
...
7786
8
...
2254
8
...
6363
8
...
0133
9
...
3584
9
...
6737
9
...
9611
10
...
0447
11
...
6902
11
...
0602
12
...
2658
12
...
3772
12
...
4365
12
...
4480
12
...
0900
1
...
2950
1
...
5386
1
...
8280
1
...
1719
2
...
5804
2
...
0658
3
...
6425
3
...
3276
4
...
1417
5
...
1088
6
...
2579
7
...
6231
9
...
2451
11
...
1722
13
...
4618
15
...
1820
18
...
4140
31
...
3273
74
...
4083
176
...
8460
416
...
1909
986
...
93
2335
...
50
3916
...
68
5529
...
9174
0
...
7722
0
...
6499
0
...
5470
0
...
4604
0
...
3875
0
...
3262
0
...
2745
0
...
2311
0
...
1945
0
...
1637
0
...
1378
0
...
1160
0
...
0976
0
...
0822
0
...
0691
0
...
0582
0
...
0490
0
...
0207
0
...
0087
0
...
0037
0
...
0016
0
...
0007
0
...
0003
0
...
0002
0
...
00000
0
...
30505
0
...
16709
0
...
10869
0
...
07680
0
...
05695
0
...
04357
0
...
03406
0
...
02705
0
...
02173
0
...
01762
0
...
01438
0
...
01181
0
...
00973
0
...
00806
0
...
00669
0
...
00556
0
...
00464
0
...
00190
0
...
00079
0
...
00033
0
...
00014
0
...
00006
0
...
00003
0
...
00002
0
...
0000
2
...
2781
4
...
9847
7
...
2004
11
...
0210
15
...
5603
20
...
9534
26
...
3609
33
...
9737
41
...
0185
51
...
7645
62
...
5319
76
...
7009
93
...
7231
112
...
1354
136
...
5752
164
...
8003
196
...
7108
337
...
8587
815
...
09
1944
...
29
4619
...
23
10951
16855
25939
39917
43510
51696
61423

1
...
56847
0
...
30867
0
...
22292
0
...
18067
0
...
15582
0
...
13965
0
...
12843
0
...
12030
0
...
11421
0
...
10955
0
...
10590
0
...
10302
0
...
10072
0
...
09885
0
...
09734
0
...
09610
0
...
09508
0
...
09296
0
...
09123
0
...
09051
0
...
09022
0
...
09009
0
...
09004
0
...
09002
0
...
09002

0
...
7591
2
...
2397
3
...
4859
5
...
5348
5
...
4177
6
...
1607
7
...
7862
8
...
3126
8
...
7556
8
...
1285
9
...
4424
9
...
7066
9
...
9290
10
...
1161
10
...
2737
10
...
4062
10
...
5178
10
...
7574
10
...
9617
11
...
0480
11
...
0844
11
...
0998
11
...
1064
11
...
1083
11
...
1091

0
...
3860
4
...
1110
10
...
3746
16
...
5711
24
...
2481
32
...
0731
39
...
8069
47
...
2821
54
...
3868
61
...
0509
68
...
2359
74
...
9265
79
...
1241
84
...
8422
89
...
1024
93
...
9314
96
...
3590
105
...
5561
114
...
0362
118
...
3344
121
...
9646
122
...
7533
122
...
1287
123
...
1963
123
...
4785
0
...
3925
1
...
2498
2
...
0512
3
...
7978
4
...
4910
4
...
1326
5
...
7245
6
...
2687
6
...
7674
7
...
2232
7
...
6384
7
...
0156
8
...
3571
8
...
6657
8
...
9436
9
...
1933
9
...
7957
10
...
4295
10
...
7683
10
...
9427
10
...
0299
11
...
0726
11
...
0866
11
...
0930

595

Compound Interest Factor Tables

10%

TABLE 15

10%

Discrete Cash Flow: Compound Interest Factors

Single Payments

Uniform Series Payments

Arithmetic Gradients

n

Compound
Amount
F͞P

Present
Worth
P͞F

Sinking
Fund
A͞F

Compound
Amount
F͞A

Capital
Recovery
A͞P

Present
Worth
P͞A

Gradient
Present Worth
P͞G

Gradient
Uniform Series
A͞G

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
40
45
50
55
60
65
70
75
80
85
90
95
96
98
100

1
...
2100
1
...
4641
1
...
7716
1
...
1436
2
...
5937
2
...
1384
3
...
7975
4
...
5950
5
...
5599
6
...
7275
7
...
1403
8
...
8497
10
...
9182
13
...
4210
15
...
4494
19
...
1138
23
...
5477
28
...
2593
72
...
3909
189
...
4816
490
...
7470
1271
...
40
3298
...
02
8556
...
34
11389
13781

0
...
8264
0
...
6830
0
...
5645
0
...
4665
0
...
3855
0
...
3186
0
...
2633
0
...
2176
0
...
1799
0
...
1486
0
...
1228
0
...
1015
0
...
0839
0
...
0693
0
...
0573
0
...
0474
0
...
0391
0
...
0221
0
...
0085
0
...
0033
0
...
0013
0
...
0005
0
...
0002
0
...
0001
0
...
0001

1
...
47619
0
...
21547
0
...
12961
0
...
08744
0
...
06275
0
...
04676
0
...
03575
0
...
02782
0
...
02193
0
...
01746
0
...
01401
0
...
01130
0
...
00916
0
...
00745
0
...
00608
0
...
00497
0
...
00407
0
...
00226
0
...
00086
0
...
00033
0
...
00013
0
...
00005
0
...
00002
0
...
00001
0
...
00001

1
...
1000
3
...
6410
6
...
7156
9
...
4359
13
...
9374
18
...
3843
24
...
9750
31
...
9497
40
...
5992
51
...
2750
64
...
4027
79
...
4973
98
...
1818
121
...
2099
148
...
4940
181
...
1378
222
...
4767
271
...
5926
718
...
91
1880
...
82
4893
...
47
12709
20474
32980
53120
85557
94113

1
...
57619
0
...
31547
0
...
22961
0
...
18744
0
...
16275
0
...
14676
0
...
13575
0
...
12782
0
...
12193
0
...
11746
0
...
11401
0
...
11130
0
...
10916
0
...
10745
0
...
10608
0
...
10497
0
...
10407
0
...
10226
0
...
10086
0
...
10033
0
...
10013
0
...
10005
0
...
10002
0
...
10001
0
...
10001

0
...
7355
2
...
1699
3
...
3553
4
...
3349
5
...
1446
6
...
8137
7
...
3667
7
...
8237
8
...
2014
8
...
5136
8
...
7715
8
...
9847
9
...
1609
9
...
3066
9
...
4269
9
...
5264
9
...
6086
9
...
7791
9
...
9148
9
...
9672
9
...
9873
9
...
9951
9
...
9981
9
...
9989
9
...
9993

0
...
3291
4
...
8618
9
...
7631
16
...
4215
22
...
3963
29
...
3772
36
...
1520
43
...
5819
49
...
5827
55
...
1095
60
...
1462
65
...
6964
69
...
7773
73
...
4146
77
...
6395
80
...
4856
82
...
9872
88
...
4544
94
...
5619
97
...
4705
98
...
3317
99
...
7120
99
...
8773
99
...
9052
99
...
4762
0
...
3812
1
...
2236
2
...
0045
3
...
7255
4
...
3884
4
...
9955
5
...
5493
5
...
0526
6
...
5081
6
...
9189
7
...
2881
7
...
6186
7
...
9137
8
...
1762
8
...
4091
8
...
6149
8
...
0962
9
...
5704
9
...
8023
9
...
9113
9
...
9609
9
...
9831
9
...
9898
9
...
9927

596

Compound Interest Factor Tables

11%

TABLE 16

11%

Discrete Cash Flow: Compound Interest Factors

Single Payments

Uniform Series Payments

Arithmetic Gradients

n

Compound
Amount
F͞P

Present
Worth
P͞F

Sinking
Fund
A͞F

Compound
Amount
F͞A

Capital
Recovery
A͞P

Present
Worth
P͞A

Gradient
Present Worth
P͞G

Gradient
Uniform Series
A͞G

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
40
45
50
55
60
65
70
75
80
85

1
...
2321
1
...
5181
1
...
8704
2
...
3045
2
...
8394
3
...
4985
3
...
3104
4
...
3109
5
...
5436
7
...
0623
8
...
9336
11
...
2392
13
...
0799
16
...
5799
20
...
8923
25
...
2056
31
...
7521
38
...
0009
109
...
5648
311
...
0572
883
...
02
2507
...
11
7119
...
9009
0
...
7312
0
...
5935
0
...
4817
0
...
3909
0
...
3173
0
...
2575
0
...
2090
0
...
1696
0
...
1377
0
...
1117
0
...
0907
0
...
0736
0
...
0597
0
...
0485
0
...
0394
0
...
0319
0
...
0259
0
...
0091
0
...
0032
0
...
0011
0
...
0004
0
...
0001

1
...
47393
0
...
21233
0
...
12638
0
...
08432
0
...
05980
0
...
04403
0
...
03323
0
...
02552
0
...
01984
0
...
01558
0
...
01231
0
...
00979
0
...
00781
0
...
00626
0
...
00502
0
...
00404
0
...
00326
0
...
00172
0
...
00060
0
...
00021
0
...
00007
0
...
00003
0
...
0000
2
...
3421
4
...
2278
7
...
7833
11
...
1640
16
...
5614
22
...
2116
30
...
4054
39
...
5008
50
...
9395
64
...
2651
81
...
1479
102
...
4133
127
...
0786
159
...
3972
199
...
9132
247
...
5292
306
...
5896
581
...
6386
1668
...
20
4755
...
79
13518
22785
38401
64714

1
...
58393
0
...
32233
0
...
23638
0
...
19432
0
...
16980
0
...
15403
0
...
14323
0
...
13552
0
...
12984
0
...
12558
0
...
12231
0
...
11979
0
...
11781
0
...
11626
0
...
11502
0
...
11404
0
...
11326
0
...
11172
0
...
11060
0
...
11021
0
...
11007
0
...
11003
0
...
9009
1
...
4437
3
...
6959
4
...
7122
5
...
5370
5
...
2065
6
...
7499
6
...
1909
7
...
5488
7
...
8393
7
...
0751
8
...
2664
8
...
4217
8
...
5478
8
...
6501
8
...
7331
8
...
8005
8
...
8552
8
...
0079
9
...
0617
9
...
0806
9
...
0873
9
...
0896

0
...
2740
4
...
6240
9
...
1872
15
...
3520
21
...
6945
27
...
9290
33
...
8709
39
...
4095
45
...
4856
49
...
0771
54
...
1864
58
...
8322
61
...
0433
64
...
8542
67
...
3016
69
...
4228
71
...
2538
75
...
1551
79
...
7712
81
...
8819
82
...
3397
82
...
5245

0
...
9306
1
...
7923
2
...
5863
2
...
3144
3
...
9788
4
...
5822
4
...
1275
5
...
6180
5
...
0574
6
...
4491
6
...
7969
6
...
1045
7
...
3754
7
...
6131
7
...
8210
7
...
0021
8
...
1594
8
...
6763
8
...
9135
8
...
0172
9
...
0610
9
...
0790

597

Compound Interest Factor Tables

12%

TABLE 17

12%

Discrete Cash Flow: Compound Interest Factors

Single Payments

Uniform Series Payments

Arithmetic Gradients

n

Compound
Amount
F͞P

Present
Worth
P͞F

Sinking
Fund
A͞F

Compound
Amount
F͞A

Capital
Recovery
A͞P

Present
Worth
P͞A

Gradient
Present Worth
P͞G

Gradient
Uniform Series
A͞G

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
40
45
50
55
60
65
70
75
80
85

1
...
2544
1
...
5735
1
...
9738
2
...
4760
2
...
1058
3
...
8960
4
...
8871
5
...
1304
6
...
6900
8
...
6463
10
...
1003
13
...
1786
17
...
0401
21
...
8839
26
...
9599
33
...
5817
42
...
1425
52
...
0510
163
...
0022
509
...
5969
1581
...
80
4913
...
48
15259

0
...
7972
0
...
6355
0
...
5066
0
...
4039
0
...
3220
0
...
2567
0
...
2046
0
...
1631
0
...
1300
0
...
1037
0
...
0826
0
...
0659
0
...
0525
0
...
0419
0
...
0334
0
...
0266
0
...
0212
0
...
0107
0
...
0035
0
...
0011
0
...
0004
0
...
0001
0
...
00000
0
...
29635
0
...
15741
0
...
09912
0
...
06768
0
...
04842
0
...
03568
0
...
02682
0
...
02046
0
...
01576
0
...
01224
0
...
00956
0
...
00750
0
...
00590
0
...
00466
0
...
00369
0
...
00292
0
...
00232
0
...
0074
0
...
00024
0
...
00008
0
...
00002
0
...
00001

1
...
1200
3
...
7793
6
...
1152
10
...
2997
14
...
5487
20
...
1331
28
...
3926
37
...
7533
48
...
7497
63
...
0524
81
...
5026
104
...
1552
133
...
3339
169
...
6989
214
...
3327
271
...
8477
342
...
5210
431
...
0914
1358
...
02
4236
...
64
13174
23223
40934
72146

1
...
59170
0
...
32923
0
...
24323
0
...
20130
0
...
17698
0
...
16144
0
...
15087
0
...
14339
0
...
13794
0
...
13388
0
...
13081
0
...
12846
0
...
12665
0
...
12524
0
...
12414
0
...
12328
0
...
12260
0
...
12130
0
...
12042
0
...
12013
0
...
12004
0
...
12001
0
...
8929
1
...
4018
3
...
6048
4
...
5638
4
...
3282
5
...
9377
6
...
4235
6
...
8109
6
...
1196
7
...
3658
7
...
5620
7
...
7184
7
...
8431
7
...
9426
7
...
0218
8
...
0850
8
...
1354
8
...
1755
8
...
2825
8
...
3170
8
...
3281
8
...
3316
8
...
3328

0
...
2208
4
...
3970
8
...
6443
14
...
3563
20
...
1288
25
...
7024
31
...
9202
36
...
6973
40
...
9979
44
...
8188
48
...
1776
51
...
1046
54
...
6369
56
...
8141
58
...
6761
60
...
2612
61
...
6052
65
...
7342
67
...
4082
68
...
0581
69
...
3031
69
...
3935

0
...
9246
1
...
7746
2
...
5512
2
...
2574
3
...
8953
4
...
4683
4
...
9803
5
...
4353
5
...
8375
6
...
1913
6
...
5010
6
...
7708
6
...
0049
7
...
2071
7
...
3811
7
...
5302
7
...
6577
7
...
0572
8
...
2251
8
...
2922
8
...
3181
8
...
3278

598

Compound Interest Factor Tables

14%

TABLE 18

14%

Discrete Cash Flow: Compound Interest Factors

Single Payments

Uniform Series Payments

Arithmetic Gradients

n

Compound
Amount
F͞P

Present
Worth
P͞F

Sinking
Fund
A͞F

Compound
Amount
F͞A

Capital
Recovery
A͞P

Present
Worth
P͞A

Gradient
Present Worth
P͞G

Gradient
Uniform Series
A͞G

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
40
45
50
55
60
65
70
75
80
85

1
...
2996
1
...
6890
1
...
1950
2
...
8526
3
...
7072
4
...
8179
5
...
2613
7
...
1372
9
...
5752
12
...
7435
15
...
8610
20
...
2122
26
...
1666
34
...
2045
44
...
9502
58
...
2148
75
...
0528
98
...
8835
363
...
2330
1348
...
92
4998
...
64
18530
35677
68693

0
...
7695
0
...
5921
0
...
4556
0
...
3506
0
...
2697
0
...
2076
0
...
1597
0
...
1229
0
...
0946
0
...
0728
0
...
0560
0
...
0431
0
...
0331
0
...
0255
0
...
0196
0
...
0151
0
...
0116
0
...
0053
0
...
0014
0
...
0004
0
...
0001
0
...
00000
0
...
29073
0
...
15128
0
...
09319
0
...
06217
0
...
04339
0
...
03116
0
...
02281
0
...
01692
0
...
01266
0
...
00954
0
...
00723
0
...
00550
0
...
00419
0
...
00320
0
...
00245
0
...
00188
0
...
00144
0
...
00039
0
...
00010
0
...
00003
0
...
00001

1
...
1400
3
...
9211
6
...
5355
10
...
2328
16
...
3373
23
...
2707
32
...
5811
43
...
9804
59
...
3941
78
...
0249
104
...
4360
138
...
6586
181
...
3327
238
...
8892
312
...
7868
407
...
8202
532
...
5199
693
...
03
2590
...
52
9623
...
14000
0
...
43073
0
...
29128
0
...
23319
0
...
20217
0
...
18339
0
...
17116
0
...
16281
0
...
15692
0
...
15266
0
...
14954
0
...
14723
0
...
14550
0
...
14419
0
...
14320
0
...
14245
0
...
14188
0
...
14144
0
...
14039
0
...
14010
0
...
14003
0
...
14001
0
...
14000

0
...
6467
2
...
9137
3
...
8887
4
...
6389
4
...
2161
5
...
6603
5
...
0021
6
...
2651
6
...
4674
6
...
6231
6
...
7429
6
...
8351
6
...
9061
6
...
9607
6
...
0027
7
...
0350
7
...
0599
7
...
1050
7
...
1327
7
...
1401
7
...
1421
7
...
1427
7
...
7695
2
...
8957
5
...
2511
10
...
1028
15
...
9906
20
...
6399
24
...
9009
28
...
7057
32
...
0380
35
...
9135
38
...
3658
40
...
4371
42
...
1728
43
...
6176
45
...
8132
46
...
7979
47
...
6053
47
...
2376
49
...
4375
50
...
8357
50
...
9632
50
...
0030
51
...
4673
0
...
3370
1
...
1218
2
...
8246
3
...
4490
3
...
9998
4
...
4819
4
...
9011
5
...
2630
5
...
5734
5
...
8381
5
...
0624
6
...
2514
6
...
4100
6
...
5423
6
...
6522
6
...
7431
6
...
9300
7
...
0714
7
...
1197
7
...
1356
7
...
1406
7
...
1500
1
...
5209
1
...
0114
2
...
6600
3
...
5179
4
...
6524
5
...
1528
7
...
1371
9
...
7613
12
...
2318
16
...
8215
21
...
8915
28
...
9190
37
...
5353
50
...
5755
66
...
1435
87
...
6998
115
...
1755
267
...
7693
1083
...
62
4384
...
79
17736
35673
71751

0
...
7561
0
...
5718
0
...
4323
0
...
3269
0
...
2472
0
...
1869
0
...
1413
0
...
1069
0
...
0808
0
...
0611
0
...
0462
0
...
0349
0
...
0264
0
...
0200
0
...
0151
0
...
0114
0
...
0086
0
...
0037
0
...
0009
0
...
0002
0
...
0001

1
...
46512
0
...
20027
0
...
11424
0
...
07285
0
...
04925
0
...
03448
0
...
02469
0
...
01795
0
...
01319
0
...
00976
0
...
00727
0
...
00543
0
...
00407
0
...
00306
0
...
00230
0
...
00173
0
...
00131
0
...
00056
0
...
00014
0
...
00003
0
...
00001

1
...
1500
3
...
9934
6
...
7537
11
...
7268
16
...
3037
24
...
0017
34
...
5047
47
...
7175
65
...
8364
88
...
4436
118
...
6316
159
...
1678
212
...
7120
283
...
1041
377
...
7451
500
...
1005
664
...
3654
881
...
09
3585
...
72
14524
29220
58779

1
...
61512
0
...
35027
0
...
26424
0
...
22285
0
...
19925
0
...
18448
0
...
17469
0
...
16795
0
...
16319
0
...
15976
0
...
15727
0
...
15543
0
...
15407
0
...
15306
0
...
15230
0
...
15173
0
...
15131
0
...
15056
0
...
15014
0
...
15003
0
...
15001
0
...
15000
0
...
8696
1
...
2832
2
...
3522
3
...
1604
4
...
7716
5
...
2337
5
...
5831
5
...
8474
5
...
0472
6
...
1982
6
...
3125
6
...
3988
6
...
4641
6
...
5135
6
...
5509
6
...
5791
6
...
6005
6
...
6166
6
...
6543
6
...
6636
6
...
6659
6
...
6665
6
...
6666

0
...
0712
3
...
7751
7
...
1924
12
...
7548
16
...
1289
21
...
1352
24
...
6930
28
...
7828
31
...
4213
33
...
6448
35
...
4988
37
...
0314
38
...
2890
39
...
3146
40
...
1466
41
...
8184
42
...
3586
43
...
8051
44
...
2558
44
...
3903
44
...
4292
44
...
4402

0
...
9071
1
...
7228
2
...
4498
2
...
0922
3
...
6549
3
...
1438
4
...
5650
4
...
9251
5
...
2307
5
...
4883
5
...
7040
5
...
8834
5
...
0319
6
...
1541
6
...
2541
6
...
3357
6
...
4019
6
...
5830
6
...
6414
6
...
6593
6
...
6646
6
...
6661

600

Compound Interest Factor Tables

16%

TABLE 20

16%

Discrete Cash Flow: Compound Interest Factors

Single Payments

Uniform Series Payments

Arithmetic Gradients

n

Compound
Amount
F͞P

Present
Worth
P͞F

Sinking
Fund
A͞F

Compound
Amount
F͞A

Capital
Recovery
A͞P

Present
Worth
P͞A

Gradient
Present Worth
P͞G

Gradient
Uniform Series
A͞G

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
22
24
26
28
30
32
34
35
36
38
40
45
50
55
60

1
...
3456
1
...
8106
2
...
4364
2
...
2784
3
...
4114
5
...
9360
6
...
9875
9
...
7480
12
...
4625
16
...
4608
26
...
2364
47
...
8004
85
...
5196
155
...
3141
209
...
4515
378
...
4438
1670
...
05
7370
...
8621
0
...
6407
0
...
4761
0
...
3538
0
...
2630
0
...
1954
0
...
1452
0
...
1079
0
...
0802
0
...
0596
0
...
0382
0
...
0211
0
...
0116
0
...
0064
0
...
0048
0
...
0026
0
...
0006
0
...
0001

1
...
46296
0
...
19738
0
...
11139
0
...
07022
0
...
04690
0
...
03241
0
...
02290
0
...
01641
0
...
01188
0
...
00867
0
...
00467
0
...
00255
0
...
00140
0
...
00089
0
...
00057
0
...
00020
0
...
00005
0
...
0000
2
...
5056
5
...
8771
8
...
4139
14
...
5185
21
...
7329
30
...
7862
43
...
6595
60
...
6730
84
...
6032
115
...
4150
213
...
0883
392
...
3117
715
...
2698
1120
...
03
1752
...
76
4965
...
16000
0
...
44526
0
...
30541
0
...
24761
0
...
21708
0
...
19886
0
...
18718
0
...
17936
0
...
17395
0
...
17014
0
...
16635
0
...
16345
0
...
16189
0
...
16104
0
...
16077
0
...
16042
0
...
16010
0
...
16002

0
...
6052
2
...
7982
3
...
6847
4
...
3436
4
...
8332
5
...
1971
5
...
4675
5
...
6685
5
...
8178
5
...
9288
6
...
0726
6
...
1520
6
...
1959
6
...
2153
6
...
2278
6
...
2421
6
...
2482
6
...
7432
2
...
6814
5
...
6380
9
...
8962
13
...
0399
17
...
8472
21
...
2175
24
...
1241
27
...
5828
29
...
6321
32
...
6970
34
...
7073
36
...
9930
37
...
6327
37
...
0799
38
...
6598
38
...
9534
39
...
4630
0
...
3156
1
...
0729
2
...
7388
3
...
3187
3
...
8189
4
...
2464
4
...
6086
4
...
9130
5
...
1666
5
...
5490
5
...
8041
5
...
9706
6
...
0548
6
...
1145
6
...
1934
6
...
2343
6
...
1800
1
...
6430
1
...
2878
2
...
1855
3
...
4355
5
...
1759
7
...
5994
10
...
9737
14
...
6722
19
...
2144
27
...
1421
53
...
9490
102
...
3706
199
...
9638
327
...
0368
538
...
3783
1716
...
36
8984
...
8475
0
...
6086
0
...
4371
0
...
3139
0
...
2255
0
...
1619
0
...
1163
0
...
0835
0
...
0600
0
...
0431
0
...
0262
0
...
0135
0
...
0070
0
...
0036
0
...
0026
0
...
0013
0
...
0003
0
...
00000
0
...
27992
0
...
13978
0
...
08236
0
...
05239
0
...
03478
0
...
02369
0
...
01640
0
...
01149
0
...
00810
0
...
00485
0
...
00247
0
...
00126
0
...
00065
0
...
00047
0
...
00024
0
...
00005
0
...
0000
2
...
5724
5
...
1542
9
...
1415
15
...
0859
23
...
7551
34
...
2187
50
...
9653
72
...
0680
103
...
4135
146
...
3448
289
...
2721
566
...
9480
1103
...
69
1816
...
65
2988
...
21
9531
...
18000
0
...
45992
0
...
31978
0
...
26236
0
...
23239
0
...
21478
0
...
20369
0
...
19640
0
...
19149
0
...
18810
0
...
18485
0
...
18247
0
...
18126
0
...
18065
0
...
18047
0
...
18024
0
...
18005
0
...
18001

0
...
5656
2
...
6901
3
...
4976
3
...
0776
4
...
4941
4
...
7932
4
...
0081
5
...
1624
5
...
2732
5
...
3527
5
...
4509
5
...
5016
5
...
5277
5
...
5386
5
...
5452
5
...
5523
5
...
5549
5
...
7182
1
...
4828
5
...
0834
8
...
8292
12
...
3525
15
...
4811
18
...
1576
21
...
3885
23
...
2123
24
...
6813
26
...
7725
28
...
0537
29
...
8191
30
...
1773
30
...
4152
30
...
7006
30
...
8268
30
...
4587
0
...
2947
1
...
0252
2
...
6558
2
...
1936
3
...
6470
3
...
0250
4
...
3369
4
...
5916
4
...
7978
4
...
0950
5
...
2810
5
...
3945
5
...
4485
5
...
4849
5
...
5293
5
...
5494
5
...
2000
1
...
7280
2
...
4883
2
...
5832
4
...
1598
6
...
4301
8
...
6993
12
...
4070
18
...
1861
26
...
9480
38
...
2061
79
...
4755
164
...
3763
341
...
2235
590
...
8019
1020
...
77
3657
...
44
22645

0
...
6944
0
...
4823
0
...
3349
0
...
2326
0
...
1615
0
...
1122
0
...
0779
0
...
0541
0
...
0376
0
...
0261
0
...
0126
0
...
0061
0
...
0029
0
...
0017
0
...
0010
0
...
0003
0
...
00000
0
...
27473
0
...
13438
0
...
07742
0
...
04808
0
...
03110
0
...
02062
0
...
01388
0
...
00944
0
...
00646
0
...
00369
0
...
00176
0
...
00085
0
...
00041
0
...
00028
0
...
00014
0
...
00002
0
...
0000
2
...
6400
5
...
4416
9
...
9159
16
...
7989
25
...
1504
39
...
4966
59
...
0351
87
...
9306
128
...
7400
186
...
0307
392
...
3773
819
...
88
1704
...
12
2948
...
01
5098
...
86
18281
45497

1
...
65455
0
...
38629
0
...
30071
0
...
26061
0
...
23852
0
...
22526
0
...
21689
0
...
21144
0
...
20781
0
...
20536
0
...
20255
0
...
20122
0
...
20059
0
...
20034
0
...
20020
0
...
20005
0
...
20001

0
...
5278
2
...
5887
2
...
3255
3
...
8372
4
...
1925
4
...
4392
4
...
6106
4
...
7296
4
...
8122
4
...
8696
4
...
9371
4
...
9697
4
...
9854
4
...
9915
4
...
9951
4
...
9986
4
...
9998

0
...
8519
3
...
9061
6
...
2551
9
...
4335
12
...
2330
15
...
5883
17
...
5095
19
...
0419
20
...
2439
21
...
5546
23
...
6460
23
...
2628
24
...
6038
24
...
7108
24
...
8469
24
...
9698
24
...
4545
0
...
2742
1
...
9788
2
...
5756
2
...
0739
3
...
4841
3
...
8175
3
...
0851
4
...
2975
4
...
4643
4
...
6943
4
...
8291
4
...
9061
4
...
9406
4
...
9627
4
...
9877
4
...
9976

603

Compound Interest Factor Tables

22%

TABLE 23

22%

Discrete Cash Flow: Compound Interest Factors

Single Payments

Uniform Series Payments

Arithmetic Gradients

n

Compound
Amount
F͞P

Present
Worth
P͞F

Sinking
Fund
A͞F

Compound
Amount
F͞A

Capital
Recovery
A͞P

Present
Worth
P͞A

Gradient
Present Worth
P͞G

Gradient
Uniform Series
A͞G

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
22
24
26
28
30
32
34
35
36
38
40
45
50
55

1
...
4884
1
...
2153
2
...
2973
4
...
9077
5
...
3046
8
...
8722
13
...
1822
19
...
0856
29
...
8490
43
...
3576
79
...
2050
175
...
8637
389
...
1156
863
...
40
1285
...
82
2847
...
71
20797
56207

0
...
6719
0
...
4514
0
...
3033
0
...
2038
0
...
1369
0
...
0920
0
...
0618
0
...
0415
0
...
0279
0
...
0187
0
...
0085
0
...
0038
0
...
0017
0
...
0009
0
...
0005
0
...
0001

1
...
45045
0
...
18102
0
...
09576
0
...
05630
0
...
03489
0
...
02228
0
...
01449
0
...
00953
0
...
00631
0
...
00420
0
...
00188
0
...
00084
0
...
00038
0
...
00021
0
...
00012
0
...
00003
0
...
0000
2
...
7084
5
...
7396
10
...
7396
17
...
6700
28
...
9620
44
...
7459
69
...
1922
104
...
0201
158
...
2535
237
...
4432
532
...
1653
1185
...
08
2632
...
20
4783
...
05
8690
...
22000
0
...
48966
0
...
34921
0
...
29278
0
...
26411
0
...
24781
0
...
23794
0
...
23174
0
...
22775
0
...
22515
0
...
22281
0
...
22126
0
...
22057
0
...
22026
0
...
22017
0
...
22008
0
...
22001
0
...
8197
1
...
0422
2
...
8636
3
...
4155
3
...
7863
3
...
0354
4
...
2028
4
...
3152
4
...
3908
4
...
4415
4
...
4882
4
...
5196
4
...
5338
4
...
5402
4
...
5419
4
...
5439
4
...
5452
4
...
6719
1
...
1275
4
...
1239
7
...
0417
10
...
6100
12
...
7438
14
...
4519
16
...
7838
17
...
8025
18
...
5702
19
...
5635
19
...
0962
20
...
3748
20
...
4905
20
...
5601
20
...
6319
20
...
6563

0
...
8683
1
...
6090
1
...
2297
2
...
7409
2
...
1551
3
...
4855
3
...
7451
3
...
9465
4
...
1009
4
...
2649
4
...
3968
4
...
4683
4
...
5060
4
...
5174
4
...
5314
4
...
5431
4
...
2400
1
...
9066
2
...
9316
3
...
5077
5
...
9310
8
...
6571
13
...
3863
20
...
1956
31
...
7408
48
...
5679
73
...
5735
174
...
5121
412
...
8199
976
...
85
1861
...
71
3548
...
91
15995
46890

0
...
6504
0
...
4230
0
...
2751
0
...
1789
0
...
1164
0
...
0757
0
...
0492
0
...
0320
0
...
0208
0
...
0135
0
...
0057
0
...
0024
0
...
0010
0
...
0005
0
...
0003
0
...
0001

1
...
44643
0
...
17593
0
...
09107
0
...
05229
0
...
03160
0
...
01965
0
...
01242
0
...
00794
0
...
00510
0
...
00329
0
...
00138
0
...
00058
0
...
00025
0
...
00013
0
...
00007
0
...
00002
0
...
0000
2
...
7776
5
...
0484
10
...
6153
19
...
7125
31
...
2379
50
...
1097
80
...
8151
126
...
2534
195
...
0328
303
...
0563
723
...
63
1716
...
92
4062
...
38
7750
...
28
14781
22729
66640

1
...
68643
0
...
41593
0
...
33107
0
...
29229
0
...
27160
0
...
25965
0
...
25242
0
...
24794
0
...
24510
0
...
24329
0
...
24138
0
...
24058
0
...
24025
0
...
24013
0
...
24007
0
...
24002
0
...
24000

0
...
4568
1
...
4043
2
...
0205
3
...
4212
3
...
6819
3
...
8514
3
...
9616
4
...
0333
4
...
0799
4
...
1103
4
...
1428
4
...
1566
4
...
1624
4
...
1664
4
...
1655
4
...
1664
4
...
1666

0
...
6993
2
...
3327
5
...
0392
8
...
4458
10
...
4313
12
...
9960
13
...
1915
14
...
0846
15
...
7406
15
...
4011
16
...
8930
17
...
1369
17
...
2552
17
...
2886
17
...
3274
17
...
3563
17
...
4464
0
...
2346
1
...
8898
2
...
4236
2
...
8499
3
...
1843
3
...
4420
3
...
6376
3
...
7840
3
...
8922
3
...
0284
4
...
0987
4
...
1338
4
...
1479
4
...
1560
4
...
1639
4
...
1663

605

Compound Interest Factor Tables

25%

TABLE 25
Single Payments

n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
22
24
26
28
30
32
34
35
36
38
40
45
50
55

25%

Discrete Cash Flow: Compound Interest Factors
Uniform Series Payments

Arithmetic Gradients

Compound
Amount
F͞P

Present
Worth
P͞F

Sinking
Fund
A͞F

Compound
Amount
F͞A

Capital
Recovery
A͞P

Present
Worth
P͞A

Gradient
Present Worth
P͞G

Gradient
Uniform Series
A͞G

1
...
5625
1
...
4414
3
...
8147
4
...
9605
7
...
3132
11
...
5519
18
...
7374
28
...
5271
44
...
5112
69
...
7362
135
...
7582
330
...
9879
807
...
18
1972
...
19
3081
...
82
7523
...
8000
0
...
5120
0
...
3277
0
...
2097
0
...
1342
0
...
0859
0
...
0550
0
...
0352
0
...
0225
0
...
0144
0
...
0074
0
...
0030
0
...
0012
0
...
0005
0
...
0003
0
...
0001

1
...
44444
0
...
17344
0
...
08882
0
...
05040
0
...
03007
0
...
01845
0
...
01150
0
...
00724
0
...
00459
0
...
00292
0
...
00119
0
...
00048
0
...
00020
0
...
00010
0
...
00005
0
...
00001

1
...
2500
3
...
7656
8
...
2588
15
...
8419
25
...
2529
42
...
2077
68
...
9495
109
...
1085
173
...
0446
273
...
9447
538
...
0329
1319
...
95
3227
...
71
7884
...
76
12322
19255
30089
91831

1
...
69444
0
...
42344
0
...
33882
0
...
30040
0
...
28007
0
...
26845
0
...
26150
0
...
25724
0
...
25459
0
...
25292
0
...
25119
0
...
25048
0
...
25020
0
...
025010
0
...
25005
0
...
25001
0
...
25000

0
...
4400
1
...
3616
2
...
9514
3
...
3289
3
...
5705
3
...
7251
3
...
8241
3
...
8874
3
...
9279
3
...
9539
3
...
9811
3
...
9923
3
...
9968
3
...
9984
3
...
9992
3
...
9998
3
...
0000

0
...
6640
2
...
2035
5
...
7725
7
...
0207
9
...
8460
11
...
2617
12
...
3260
13
...
1085
14
...
6741
14
...
2326
15
...
6373
15
...
8316
15
...
9229
15
...
9481
15
...
9766
15
...
9969
15
...
4444
0
...
2249
1
...
8683
2
...
3872
2
...
7971
2
...
1145
3
...
3559
3
...
5366
3
...
6698
3
...
7667
3
...
8861
3
...
9457
3
...
9746
3
...
9858
3
...
9921
3
...
9980
3
...
9997

606

Compound Interest Factor Tables

30%

TABLE 26

30%

Discrete Cash Flow: Compound Interest Factors

Single Payments

Uniform Series Payments

Arithmetic Gradients

n

Compound
Amount
F͞P

Present
Worth
P͞F

Sinking
Fund
A͞F

Compound
Amount
F͞A

Capital
Recovery
A͞P

Present
Worth
P͞A

Gradient
Present Worth
P͞G

Gradient
Uniform Series
A͞G

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
22
24
25
26
28
30
32
34
35

1
...
6900
2
...
8561
3
...
8268
6
...
1573
10
...
7858
17
...
2981
30
...
3738
51
...
5417
86
...
4554
146
...
0496
321
...
8008
705
...
3333
1550
...
00
4427
...
97
9727
...
7692
0
...
4552
0
...
2693
0
...
1594
0
...
0943
0
...
0558
0
...
0330
0
...
0195
0
...
0116
0
...
0068
0
...
0031
0
...
0014
0
...
0006
0
...
0002
0
...
0001

1
...
43478
0
...
16163
0
...
07839
0
...
04192
0
...
02346
0
...
01345
0
...
00782
0
...
00458
0
...
00269
0
...
00159
0
...
00055
0
...
00033
0
...
00011
0
...
00004
0
...
0000
2
...
9900
6
...
0431
12
...
5828
23
...
0150
42
...
4053
74
...
6250
127
...
2863
218
...
0139
371
...
9734
630
...
28
1806
...
80
3054
...
31
8729
...
30000
0
...
55063
0
...
41058
0
...
35687
0
...
33124
0
...
31773
0
...
31024
0
...
30598
0
...
30351
0
...
30207
0
...
30094
0
...
30043
0
...
30019
0
...
30007
0
...
30003

0
...
3609
1
...
1662
2
...
6427
2
...
9247
3
...
0915
3
...
1903
3
...
2487
3
...
2832
3
...
3037
3
...
3158
3
...
3272
3
...
3297
3
...
3321
3
...
3329
3
...
5917
1
...
5524
3
...
6656
5
...
4800
7
...
8872
8
...
9173
9
...
6437
9
...
1426
10
...
4788
10
...
7019
10
...
9433
10
...
0045
11
...
0687
11
...
0945
11
...
4348
0
...
1783
1
...
7654
2
...
2156
2
...
5512
2
...
7952
2
...
9685
3
...
0892
3
...
1718
3
...
2275
3
...
2890
3
...
3050
3
...
3219
3
...
3288
3
...
3500
1
...
4604
3
...
4840
6
...
1722
11
...
8937
20
...
1439
36
...
4697
66
...
1585
121
...
3138
221
...
4619
404
...
7886
1342
...
78
2447
...
11
8128
...
7407
0
...
4064
0
...
2230
0
...
1224
0
...
0671
0
...
0368
0
...
0202
0
...
0111
0
...
0061
0
...
0033
0
...
0014
0
...
0006
0
...
0002
0
...
0001

1
...
42553
0
...
15076
0
...
06926
0
...
03489
0
...
01832
0
...
00982
0
...
00532
0
...
00290
0
...
00158
0
...
00087
0
...
00026
0
...
00014
0
...
00004
0
...
00001
0
...
0000
2
...
1725
6
...
9544
14
...
4919
28
...
6964
54
...
6967
101
...
4848
187
...
7385
344
...
6109
630
...
7483
1152
...
25
3833
...
50
6989
...
35000
0
...
58966
0
...
45046
0
...
39880
0
...
37519
0
...
36339
0
...
35722
0
...
35393
0
...
35214
0
...
35117
0
...
35048
0
...
35019
0
...
35008
0
...
35002
0
...
35001

0
...
2894
1
...
9969
2
...
3852
2
...
5982
2
...
7150
2
...
7792
2
...
8144
2
...
8337
2
...
8443
2
...
8501
2
...
8550
2
...
8560
2
...
8568
2
...
8570
2
...
5487
1
...
2648
3
...
9828
4
...
3515
5
...
3363
6
...
0049
7
...
4421
7
...
7206
7
...
8946
7
...
0017
8
...
1061
8
...
1296
8
...
1517
8
...
1594
8
...
4255
0
...
1341
1
...
6698
1
...
0597
2
...
3338
2
...
5205
2
...
6443
2
...
7246
2
...
7756
2
...
8075
2
...
8393
2
...
8465
2
...
8535
2
...
8559
2
...
4000
1
...
7440
3
...
3782
7
...
5414
14
...
6610
28
...
4957
56
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3715
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90
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...
6195
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...
8939
656
...
2613
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...
84
3325
...
83
16834
25251
37877
85223

0
...
4444
0
...
1975
0
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0
...
0390
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...
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...
0077
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...
0034
0
...
0015
0
...
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0
...
0003
0
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40000
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...
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00882
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...
00172
0
...
00076
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...
00034
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00015
0
...
00003
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00001
0
...
0000
2
...
7500
8
...
1875
20
...
1719
49
...
8867
113
...
9951
257
...
2390
581
...
7878
1311
...
52
2953
...
68
6648
...
50000
0
...
71053
0
...
57583
0
...
53108
0
...
51335
0
...
50585
0
...
50258
0
...
50114
0
...
50051
0
...
50023
0
...
50007
0
...
50002
0
...
50001
0
...
50000
0
...
50000

0
...
1111
1
...
6049
1
...
8244
1
...
9220
1
...
9653
1
...
9846
1
...
9931
1
...
9970
1
...
9986
1
...
9994
1
...
9999
1
...
9999
2
...
0000
2
...
0000
2
...
4444
1
...
6296
2
...
5953
2
...
2196
3
...
5838
3
...
7842
3
...
8904
3
...
9452
3
...
9729
3
...
9868
3
...
9969
3
...
9985
3
...
9997
3
...
9999
3
...
4000
0
...
0154
1
...
4226
1
...
6752
1
...
8235
1
...
9068
1
...
9519
1
...
9756
1
...
9878
1
...
9940
1
...
9986
1
...
9993
1
...
9998
1
...
0000
2
...

Chapter 2
Opener: Royalty-Free/CORBIS
...

Chapter 4
Opener: Chad Baker/Getty Images
...

Chapter 6
Opener: Stockdisc
...

Chapter 8
Opener: PhotoLink/Getty Images
...

Chapter 10
Opener: © Digital Vision/PunchStock
...

Chapter 12
Opener: Ariel Skelley/Blend Images/Getty Images
...

Chapter 14
Opener: Photodisc/Getty Images
...

Chapter 16
Opener: Arthur S
...

Chapter 17
Opener: Photodisc/Getty Images
...

Chapter 19
Opener: © Brand X/JupiterImages
...
See also Gradient, arithmetic
Alternative depreciation system (ADS), 426–27
Allocation variance, 399
Alternatives
cost, 131
deﬁned, 6
do-nothing, 130–31
independent, 130–31, 132, 157, 203, 207, 244, 247 (see also
Capital budgeting)
inﬁnite life, 138, 157–60
mutually exclusive, 130–31, 132, 155, 208, 238, 248
revenue, 131
selection, 6
service, 131
in simulation, 533–40
Amortization, 415
Annual interest rate
effective, 99–105
nominal, 99–105
Annual operating costs (AOC), 6, 153, 297–98
and estimation, 388
Annual Percentage Rate (APR), 97
Annual Percentage Yield (APY), 97
Annual worth
advantages, 151
after-tax analysis, 456–58
of annual operating costs, 297–98
and B/C analysis, 235, 238–42
and breakeven analysis, 345–48
and capital-recovery-plus-interest, 153–54
equivalent uniform, 151
evaluation by, 155–60
and EVA, 153, 465–68
and future worth, 151
and incremental rate of return, 213–14
of inﬁnite-life projects, 157–60

and inﬂation, 151, 377–78
and present worth, 151
and rate of return, 175, 213–14
and replacement analysis, 302–06, 307–10, 462–65
spreadsheet solutions, 156, 159, 458
when to use, 262
AOC
...
See Gradient, arithmetic
Assets
...
See Expected value
Average cost per unit, 345
Average tax rate, 447

B
Balance sheet, 267, 561–63
Base amount
deﬁned, 50
and shifted gradients, 80–82
Basis, unadjusted, 416
B/C
...
See Normal distribution
Beneﬁt and cost difference, 236
Beneﬁt/cost ratio
calculation, 235–36
conventional, 235
incremental analysis, 238–39
modiﬁed, 235–36
for three or more alternatives, 242–46
for two alternatives, 238–42
when to use, 262
Beneﬁts
direct versus implied, 242
in public projects, 231, 235, 242
␤ (beta), 274
Bonds
and debt ﬁnancing, 271–72
and inﬂation, 372, 385
interest computation, 190
payment periods, 190
present worth, 191–92
for public sector projects, 232
rate of return, 190–92
types, 191
Book depreciation, 415, 417, 466
Book value
declining balance method, 419–22
deﬁned, 416
double declining balance method, 419–22
and EVA, 466

612

Index

Book value (continued)
MACRS method, 423
versus market value, 416
straight line method, 418
sum-of-years-digits method, 430, 557
unit-of-production method, 431
Borrowed money, 267
Borrowing rate, 185–87
Bottom-up approach, 388–89
Breakeven analysis
...
i graph
average cost per unit, 345
ﬁxed costs, 341
and Goal Seek, 353–54
and make-buy decisions, 341, 347
and payback, 351–54
and rate of return, 210–12, 460
versus sensitivity analysis, 341, 485
single project, 341–45
spreadsheet application, 212, 352–54
three or more alternatives, 348
two alternatives, 345–47
variable costs, 341
Breakeven point, 341, 489
Budgeting
...
See also Cost of capital
debt, 27, 267, 271–73
equity, 27, 267, 273–75
mixed (debt and equity), 275–77
Capital gains
deﬁned, 454
short-term and long-term, 454
taxes for, 454
Capital losses
deﬁned, 454
taxes for, 454
Capital rationing
...
See also A/P factor; Depreciation
deﬁned, 153
and economic service life, 297–99
and EVA, 468
and inﬂation, 377–78
and replacement analysis, 306
Capital recovery factor, 43–45
and equivalent annual worth, 153

Capitalized cost
in alternative evaluation, 138–42
and annual worth, 138
and public projects, 244–46
CAPM
...
See Cash ﬂow after taxes
CFBT
...
See also Compounding
Compounding
annual, 99–105
continuous, 114–16
and effective interest rate, 96
frequency, 97–98
interperiod, 113–14
period, 97–98
and simple interest, 21–23
Compounding period
continuous, 114–16
deﬁned, 97
and effective annual rate, 102
and payment period, 106–14
Concepts, fundamental, summary, 573–76
Contingent projects, 323
Continuous compounding, 114–16
Contracts, types, 234
Conventional beneﬁt/cost ratio, 235
Conventional cash ﬂow series, 180
Conventional gradient, 51–57
Corporations
and capital, 4
ﬁnancial worth, 466–68
leveraged, 275–77
Cost alternative, 131, 204, 216, 457, 460
Cost-effectiveness
analysis, 246–50
ratio, 246
Cost, life-cycle, 160–63
Cost-capacity equations, 394–95
Cost centers, 397, 401
Cost components, 387–88
Cost depletion, 427–29
Cost drivers, 401
Cost-estimating relationships, 394–97
Cost estimation
accuracy, 389
approaches, 388–89
cost-capacity method, 394–95
and cost indexes, 391–94
factor method, 395–97
and inﬂation, 12, 377
unit method, 390
Cost of capital
and debt-equity mix, 269–71
for debt ﬁnancing, 271–73
deﬁned, 26, 267
for equity ﬁnancing, 273–75
versus MARR, 26, 267
weighted average, 27, 270, 275
Cost of goods sold, 397, 562–63, 565
Cost of invested capital, 466–67
Costs
...
See also Depreciation recapture; Rate of
depreciation; Replacement analysis
accelerated, 416, 419
ACRS, 422
alternative system, 426–27
and amortization, 415
basis, 416
book, 415, 417, 466
declining balance, 419–22, 550–51
deﬁned, 415
double declining balance, 419–22, 551
and EVA, 466
general depreciation system (GDS), 426
half-year convention, 416, 424, 427
and income taxes, 415, 417, 445–68
MACRS, 417, 422–27
present worth of, 432–35
property class, 426
recovery period for, 416, 418, 426
rate of, 416
recovery rate, 416

613

614

Index

Depreciation (continued)
straight line, 418–19, 557
straight line alternative, 426–27
sum-of-years digits, 430, 557
switching methods, 432–38, 557–58
tax, 415, 417
unit-of-production, 431
Depreciation recapture
deﬁnition, 453
in replacement studies, 462–65
and taxes, 453, 461
Descartes’ rule, 181–84, 219
Design-build contracts, 234
Design stages, preliminary and detailed, 161, 163, 389
Design-to-cost approach, 388–89
Direct beneﬁts, 242, 244
Direct costs, 387, 390–97
Disbeneﬁts, 231, 235
Disbursements, 15, 177
Discount rate, 129, 232
Discounted cash ﬂow, 129
Discounted payback analysis, 349
Discrete cash ﬂows
compound interest factors (tables), 581–609
discrete versus continuous compounding, 114–16
and end-of-period convention, 15–16
Disposal phase, 161
Distribution
...
See Equivalence
Economic service life (ESL), 294, 296–302
Economic value added, 153, 465–68
EFFECT function, 103, 551
Effective interest rate
annual, 99–100
for any time period, 105–06
of bonds, 191–92
and compounding periods, 100, 105
for continuous compounding, 114–16
deﬁned, 96
and nominal rate, 96–97
Effective tax rate, 447, 462
Efﬁciency ratios, 563

End-of-period convention, 15–16
Engineering economy
Concepts, summary, 573–76
deﬁned, 3
study approach, 4–7
terminology and symbols, 13
Equal service requirement, 131, 151, 213, 217, 240, 457
Equity ﬁnancing, 26, 267
cost of, 273–75
Equivalence
calculations without tables, 569–72
compounding period greater than payment period, 112–14
compounding period less than payment period, 107–12
deﬁned, 19–21
Equivalent uniform annual cost
...
See Annual worth
Error distribution
...
See Annual worth
EUAW
...
See also Spreadsheet, usage in examples
absolute cell reference, 9, 547, 550
basics, 547–50
charts, 548–49
displaying functions, 30
embedding functions, 75, 156–57
error messages, 560
functions, in engineering economy, 28, 550–58
Goal Seek tool, 558–59
introduction, 27–30, 547–50
and linear programming, 330–31
random number generation, 556
Solver tool, 559–60
spreadsheet layout, 549–50
Expected value
computation, 492–94, 526–27, 530
and decisions under risk, 517
and decision trees, 497–98
deﬁned, 492, 526
and real options, 501–02
in simulation, 538–39
Expenses, operating, 6, 153, 297–98
External rate of return, 185–90
...
See also Bonds
Fixed percentage method
...
See also Single payment factors
Future worth
and annual worth, 151
and effective interest rate, 100
evaluation by, 137
and inﬂation, 374–77
and multiple rates of return, 186–87
from present worth, 137
of shifted series, 73, 77–78
and spreadsheet solutions, 42, 48
when to use, 262
FV function, 28, 552
and shifted uniform series, 78–79
and single payments, 41
and uniform series, 46

G
Gains and losses, 454
Gaussian distribution
...
See Public sector projects
Gradient, arithmetic
base amount, 50
conventional, 51
decreasing, 51, 56
deﬁned, 50
factors, 52–54
increasing, 51–55
shifted, 80–82, 83–86
spreadsheet use, 57
Graduated tax rates, 446–47, 448
Gross income, 445, 448

615

H
Half-year convention, 416, 424, 427
Highly leveraged corporations, 275–77
Hurdle rate
...
See First cost
Installment ﬁnancing, 175
Intangible factors, 6
...
See also Interest rate(s)
compound, 22, 24, 29
continuous compounding, 114–16
deﬁned, 10
interperiod, 112–14
rate, 10, 12
simple, 21, 24
Interest period, 10, 12
Interest rate(s)
...
See also Rate of return
and annual worth, 175
deﬁnition, 173
versus external ROR, 185
and present worth, 175
spreadsheet solution, 177
International aspects
contracts, 234
corporate taxes, 468–70
cost estimation, 388–89
deﬂation, 368–69
depreciation, 416–17, 470–72
dumping, 368
inﬂation aspects, 368, 377
value-added tax, 470–72

Interperiod interest, 112–14
Interpolation, in interest rate tables, 48–50
Inventory turnover ratio, 564–65
Invested capital, cost of, 466–67
Investment(s)
...
S
...
See also Annual
operating cost
Make-or-buy decisions, 341, 347
...
See Minimum attractive rate of return
Mean
...
See Economic service life
MIRR function, 186, 553–54
M&O costs
...
See Depletion
Net cash ﬂow, 15
Net operating income (NOI), 446, 459
Net investment procedure, 187–90
Net operating proﬁt after taxes (NOPAT), 446, 466
Net present value
...
i graphs, 178–79
sensitivity analysis, 486–87, 489
and shifted series, 75, 86
NSPE (National Society of Professional Engineers), 7,
404, 566–68

O
Obsolescence, 294
One-additional-year replacement study, 302–05
Operating costs
...
See Indirect costs
Owner’s equity, 267, 561

P
P, 13, 40
P/A factor,43, 53
...
See Payback analysis
Percentage depletion
...
See also Gradient, arithmetic
Phaseout phase, 161
Planning horizon
...
See Present worth
Present worth
after-tax analysis, 456–58
and annual worth, 151
assumptions, 134
and B/C analysis, 235
of bonds, 191–92
and breakeven analysis, 345–46
and capital budgeting, 325–29
of depreciation, 432
for equal lives, 132–33
evaluation method, 261–62
geometric gradient series, 82–86
income taxes, 451–53
and independent projects, 325–29
index, 332
and inﬂation, 369–74
and multiple interest rates, 181–84
and proﬁtability index, 237
and rate of return, 175–76, 182–84, 458–62
and sensitivity analysis, 486–90
in shifted series, 73, 76, 80–86
in simulation, 534–39
single-payment factor, 40–41
for unequal lives, 133–37
Present worth factors
gradient, 50–53, 58–60
single payment factor, 40–41
uniform series, 43–45
Probability
in decision trees, 495–98
deﬁned, 492, 518
and expected value, 492–94, 526–27
and standard deviation, 528
Probability distribution
of continuous variables, 519–22
deﬁned, 519
of discrete variables, 519–20
properties, 526–28, 530
and samples, 523–26
in simulation, 533–39

Probability node, 495
Productive hour rate, 399
Proﬁtability index, 237, 332
Proﬁtability ratios, 563
Proﬁt-and-loss statement, 562
Project net-investment, 187–90
Property class, 426
Property of independent random variables, 533
Public-private partnerships, 234–35
Public sector projects, 230–35
and annual worth, 157–58
B/C analysis, 235–38
capitalized cost, 138–42
characteristics, 231–32
design-build contracts, 234–35
proﬁtability index, 237
public-private partnerships, 234–35
Purchasing power, 367, 374, 376
PV function, 28, 44, 556
versus NPV function, 556
and present worth, 41, 44
and single payment, 41
and uniform series present worth, 44
PW vs
...
See also Incremental rate of return
after-tax, 458–62
and annual worth, 175, 213–14, 218
of bonds, 190–92, 272
breakeven, 210–12, 460
in capital budgeting, 332–34
cautions, 179–80
on debt capital, 272–73, 276
deﬁned, 12, 173, 175
on equity capital, 273–74, 276
evaluation method, 261–62
external, 185–90
on extra investment, 206
incremental, 206, 207–10
and independent projects, 207
and inﬂation, 12–13, 368, 374–75

Index
installment ﬁnancing, 175
internal, 173, 175, 185
minimum attractive (see Minimum attractive rate of return)
modiﬁed ROR approach, 185–87
multiple, 180–90
and mutually exclusive alternatives, 206, 207–18
and present worth, 175, 177–79, 207–09, 218
ranking inconsistency, 213, 216, 462
return on invested capital (ROIC) approach, 185, 187–90
spreadsheet solution, 177, 179, 211–13, 216–19
Ratios, accounting, 563–65
Real interest rate, 368, 370, 374–75
Real options, 498–503
and decision trees, 500
deﬁnition, 499
Real property, 416, 423–24, 426
Recovery period
deﬁned, 416
effect on taxes, 452–53
MACRS, 423–24, 426–27
straight line option, 426
Recovery rate
...
See Investment rate
Repayment of loans, 24–25
Replacement analysis, 292–313, 462–65
after-tax, 462–65
annual worth, 295, 302–06
and capital losses, 454, 462
cash ﬂow approach, 306
depreciation recapture, 462–64
and economic service life, 296–99, 305
ﬁrst costs, 294–96
and marginal costs, 300–01
market value, 294, 301
need for, 294
one-additional year, 302–05
opportunity cost approach, 306
overview, 302
and study periods, 307–12
sunk costs, 295
terminology, 294
viewpoint, 295
Replacement life
...
See Economic service life
Return on assets ratio, 564
Return on invested capital, 187–90
Return on investment (ROI), 12, 173
...
See Return on investment

Root mean square deviation, 528
ROR
...
See also Market value
and capital recovery, 153, 297
deﬁned, 6, 153
and depreciation, 416, 418, 420, 423, 430
and market value, 294, 297–98
and public projects, 235
in PW analysis, 132, 134–37
in replacement analysis, 294, 297, 303–06, 464
and trade-in value, 294, 464
Sampling, 523–26
Savings, tax, 449, 463–65
Scatter charts
...
See also Breakeven analysis
description, 485
and Excel cell referencing, 29, 547
of one parameter, 485–87
spider graph, 488
with three estimates, 490–91
two alternatives, 488–90
Service alternative
...
See also Excel
annual worth, 155–56, 159, 218, 299, 301, 304,
305, 353, 458, 467
B/C analysis, 245
breakeven analysis, 352, 353–54
cash ﬂow after tax (CFAT), 450, 456, 458, 461, 462
compound interest, 29–30
depreciation, 422, 425, 434
EVA, 467
and factor values, 49
independent projects, 328, 331
inﬂation, 372
layout, 549–50
multiple attributes, 283
nominal and effective interest, 103–04

619

620

Index

Spreadsheet (continued)
present worth, 136, 209, 212, 218, 311, 354, 458,
461, 489, 501, 539
rate of return, 63, 182, 184, 189, 192, 209, 212,
217, 218, 461, 462, 501
replacement analysis, 299, 301, 304, 305, 311, 465
replacement value, 312
sensitivity analysis, 487, 489
simulation, 538–39
Staged funding, 494–503
Standard deviation
for continuous variable, 530
deﬁnition, 527–28
for discrete variable, 528–29
Standard normal distribution, 531–33
Stocks
CAPM model, 274
common, 267, 274
in equity ﬁnancing, 273–75, 276
preferred, 267, 273
Straight line alternative, in MACRS, 426–27
Straight line depreciation, 418–19
Straight line rate, 418
Study period
and AW evaluation, 155
and equal service, 133
and FW analysis, 137
and PW evaluation, 133–34
and replacement analysis, 302, 307–11
and salvage value, 153
spreadsheet example, 136, 310–11
Sum-of-years digits depreciation, 430
Sunk costs, 295
SYD function, 430, 557
System, phases of, 160–61

U

T

WACC
...
See After-tax; Income tax; Taxable income
Time, 13
Time value of money
deﬁned, 4
and equivalence, 19
factors to account for, 39–61
and no-return payback, 349–50
Total cost relation, 342–45
...
See also Market value; Salvage value
Treasury securities, 26, 190
Triangular distribution, 520, 522

Unadjusted basis, 416
Uncertainty, 515, 517
Uniform distribution, 520–21, 535, 538
Uniform gradient
...
See Declining balance depreciation
Uniform series
compound amount (F/A) factor, 46
compounding period greater than payment period, 112–14
compounding period less than payment period, 109–12
description, 13
present worth (P/A) factor, 43
shifted, 73–80
Unit method, 390–91
Unit-of-production depreciation, 431
Unknown interest rate, 61–63
Unknown years (life), 61, 63–64
Unrecovered balance, 173–74

V
Value, resale, 6
...
See Economic value added
Value-added tax, 470–72
Variable
...
Place a minus in front of the
function to retain the same sign
...
5),
MIN(n, t؊0
...
2
...
2
...
2
...
2
...
2
...
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Title: Solution manual of Engineering Economics 7th Ed
Description: Solution manual of Engineering Economics 7th Ed by Leland Blank and Anthony

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