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Engineering Economics

Presented by
Dr Sohail Malik

Lecture 1

Overview of Course CEM- 803:
ECONOMIC DECISION ANALYSIS IN
CONSTRUCTION
Instructor :
Text book :

:

Dr
...
Newnan, Ted Eschenbach
Jerome P
...

The objective of the project is to:
a) offer an opportunity to learn hands-on
b) study a subject of interest in great depth
c) become information literate
The final product of the project, as the basis for the grade, consists
of:
a) a comprehensive final written report
b) presentation of the project during class with
handouts/info distribution
c) Conference/Journal paper (Maximum Marks)

What is Engineering?
– Apply mathematics, natural sciences and judgment
gained by study, experience and practice
– Utilize the materials and forces of nature for the benefit
of mankind

– Has physical & economical aspects

5

What is Economics?
• The study of how individuals and societies choose to use
scarce resources that nature and previous generations have
provided

• Resources
–
–
–
–

Land
Labor
Capital
Entrepreneurial ability
6

The Foundation Of Economics

SOCIETY HAS VIRTUALLY
UNLIMITED WANTS
...
e
...

• Engineering Economy involves formulating,
estimating, and evaluating the economic
outcomes when alternatives to accomplish a
defined purpose are available
...


Also called No Free-Lunch Principle
13

The Cost-Benefit Principle
• An individual (or a firm or a society) should
take an action if and only if the extra benefits
are at least as great as the extra costs
• Costs and benefits are not just money

14

Problems
Simple Problems:
Simple
Intermediate
Complex

• Can generally be worked in one’s head
without extensive analysis
...

17

Role of Engineering Economic Analysis

• Assists in making decisions where:
– Decision is sufficiently important that serious
thought and effort is required
...


– ECONOMIC ISSUES are a significant component of
the analysis leading to a decision
...

2
...

4
...

6
...
Recognize the Problem

• A problem exists when:
– A standard or expectation is not being met
...
(An
opportunity)

2
...

– A goal is a general statement about what we
expect
...


– An objective is narrow and specific
...

22

3
...


– Deciding which data is relevant may be a
complex process
...

23

4
...

Occasionally this is to maintain the existing
situation
...

– Only feasible alternatives should be retained
for further analysis
...
Select the Criterion to Determine the
Best Alternative
– A criterion, or a set of criteria, is used to evaluate
the alternatives to determine which is best
...

– Selecting criteria to use is not easy because

different groups often support different criteria
...
”

25

Economic Decision-Making Problems
Fall Into Three Categories
1
...

2
...

3
...


26

6
...

• In economic decision making models are
usually mathematical
...

– Model must represent important parts of system
at hand
...

27

7
...
If
necessary, multiple criteria are combined
into a single criterion
...
e
...

• Costs and benefits may occur over a short
or long time period
...
Choosing the Best Alternative
– When choosing best alternative both economic &
non-economic criteria must be considered
...

– Elimination of feasible alternatives may

predetermine the outcome of decision making
process
...
Audit the Results
• Compare the results of changes to predictions
to assure that chosen alternative was
implement as planned and results are as
expected
...

– Make sure prediction errors are not repeated
...


• Audits promote realistic economic analysis &
implementation
...
Honda
1
...

3
...

5
...


Recognize a decision
problem
Define the goals or
objectives
Collect all the relevant
information
Identify a set of feasible
decision alternatives
Select the decision
criterion to use
Select the best
alternative

Need a car
Want mechanical security

Gather technical as well as
financial data
Choose between Saturn
and Honda
Want minimum total cash
outlay
Select Honda
31

Financial Data Required to Make an
Economic Decision

32

What Makes the Engineering Economic
Decision Difficult? - Predicting the Future
• Estimating a Required
investment
• Forecasting outcome
• Estimating a selling price
• Estimating a construction
cost
• Estimating a product life

33

Role of Engineers in Construction
Business
Create & Design
• Engineering Projects

Analyze
• Construction Methods
• Engineering Safety
• Environmental Impacts
• Market Assessment

Evaluate
• Expected
Profitability
• Timing of
Cash Flows
• Degree of
Financial Risk

Evaluate
• Impact on
Financial Statements
• Firm’s Market Value
• Stock Price

34

Accounting Vs
...

39

Equipment Replacement Problem
• Now is the time to
replace the old truck /
crane?
• If not, when is the right
time to replace the old
equipment?

40

Cost Reduction

• Should a company
buy equipment to
perform an operation
now done manually?
• Should spend money
now in order to save
more money later?

41

Wrecked Car” Example

“

Bad news – you just wrecked your car!
An automobile wholesaler offers you $3,000 for
the car “as is
...
The insurance company can fix the car
right away in a repair shop belonging to this
company
...
The odometer reading on your wrecked
car is 58,000 miles
...


You can buy a newer car for $10,000 with an odometer reading of 28,000 miles
...


A discount repair shop charges $1,100 and requires 1 month
...

42

“Wrecked Car” Example Alternatives
Buy newer car
...
Sell, and buy the newer car:
– Cash flow: = -$1,000 + $4,500 - $10,000 = -$6,500
• Repair at discount shop
...
Keep your car
...
Rent a car, then keep your car
...

43

?
Any Questions

Engineering Economics

Presented by
Sohail Malik

Lecture 2

Course topics and schedule
Week

Topic

Reading

1

Introduction, Study Skills , Engineering Economics, Decisi
Chapter 1
on Making Process

2

Cost Concept

Chapter 2

3

Accounting systems, Balance Sheet, Income sheet

Chapter 18,
Handouts

4

Time value of Money

Chapter 3

5

Interest and Equivalence

Chapter 2

7

Rate of Return

Chapter 7

8

Deprecation

Chapter 11

9

Profit Centre Analysis

Hand outs

10

Financial Analysis

Chapter 8,9

11

Cash Flow Analysis

Chapter 12

12

Benefit Cost Analysis

Chapter 9

13

Financing

Chapter 16

14

Financial Decision Making

Chapter 17

15

Project Presentations

16

Final Exam

Project

Phase 1 Due

Phase 2 Due

Phase 3 Due

Engineering Costs
• Evaluating a feasible alternative requires that many costs be
analyzed
•Initial investment
•New construction
•Facility modification
•General labor

•Parts and materials
•Inspection and quality
•Fixtures and tooling

•Data management
•Technical support

3
3

What is a Cost?

• Use of company resources (such as cash or cash-equivalent
value) to provide future company benefit
• Measured by resource given up
• Example

•Buy Equipment for $2,000, cost = $2,000,
• future benefit = use of equipment

4
4

What is an Expense?

Part of Cost which is “used up” for purpose of generating
revenue
• Example

Buy Equipment for $2,000,
cost = $2,000,
expense = $500 per year (assume 4 year life)

5
5

Costs
Costs:
Fixed and Variable
 Direct and Indirect
 Marginal and Average
 Sunk and Opportunity
 Recurring and NonRecurring
 Incremental
Cash and Book
 Life-Cycle

Cost Indices
Estimating Benefits
Cash Flow Diagrams

6

Direct Indirect Costs
Direct costs

Indirect costs

• Costs that can be
• Costs that cannot be easily
easily and conveniently
and conveniently traced to
traced to a unit of product or
a unit of product or other
other cost object
...

• Examples: direct material
• Example: manufacturing
and direct labor
overhead…
...

8

Examples of Manufacturing Overhead Costs
• Indirect Materials – materials that are part of a product, but
the costs are not easily or conveniently traced to the product
...
cost of various people such
as for supervisors, material handlers, design engineers
• Other Factory or Production Costs such as utilities,
depreciation, insurance, etc
...

– Total variable costs
change when activity
changes
...


10

Total Variable Cost

Total Long Distance
Telephone Bill

Your total mobile networks telephone bill is based on how
many minutes you talk
...
Cost per long distance
minute talked is constant
...


Minutes Talked

12

Total Fixed Cost

Monthly Basic
Telephone Bill

Total fixed cost is constant within the relevant range
...


Number of PTCL Calls
13

Fixed Cost Per Unit

Monthly Basic Telephone
Bill per Local Call

On a per unit basis, a fixed cost is inversely related to activity
...


14
Number of Local Calls

Engineering Costs

Marginal - variable cost for the next unit

• Depends on the next unit
Average - total cost/number of units

• Rent+ food+…+n/number of units

15

Engineering Costs and Cost Estimating
Key Question: Where do the numbers come from that we use in
engineering economic analysis?

• Cost estimating is necessary in an economic analysis

• When working in industry, you may need to consult with
professional accountants to obtain such information
16

Engineering Costs and Cost Estimating
Example 2-1
...
His wealthy uncle will reimburse him for his
personal time, so his time cost can be ignored
...
00

Item
Cost
Ticket
Refreshments $

Total Costs

$12
...
50

$20
...

Graph of Total Cost Equation:

Total cost

n

18

Albert’s Charter Bus Venture (example)
Marginal cost (marginal tax)
-The cost to take one more person
Average cost
- Average cost: the cost per person
Avg
...
Cost = ($225+$20n)/n

– For n = 30, TC = $885
Avg
...
50
Total cost cannot be calculated
from an average cost value
For n =35, TC  35*($29
...
50

19

Albert’s Charter Bus Venture (example)
Marginal and Average Costs
$300
...
00

Cost

$200
...
00

Marginal
Trip Ticket

$100
...
00
$0
...
00
$800
...
00
Cost

$400
...
00

Profit

$0
...
00)
($400
...
Where is the Profit Region?
...
Can you make this chart in Excel?
Albert's Charter Bus Venture

$1,000
...
00

Total Cost

$600
...
00

Revenue

$200
...
00
0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

($200
...
00)
Number of People

23

Sunk Costs
A sunk cost is money already spent due to a past decision
...

– How likely is it that you can sell it today for what it cost?
– Suppose you can sell the laptop today for $400
...
It has no influence on the present
24
opportunity to sell the laptop for $400
...

• Example: Suppose your wealthy uncle gives you $75,000 when
you graduate from high school
...
It is also enough for you to open
a business making web pages for small companies instead of
going to college
...

– If you decide to go to college you give up the
opportunity to make $20,000 per year
– Your opportunity cost is $20,000
– Your total cost per year is $35,000
25

Sunk and Opportunity Cost
Example 2-3
...
The
pumps are unused, but are three years old
...
Some pricing information is available as follows
...


Amount buyer offered for case
2 years ago

$5,000

A foregone opportunity

Case can currently be sold for

$3,000

Actual market value today
26

Recurring and Non-Recurring Costs
Recurring costs are those expenses that are known, anticipated,
and occur at regular intervals
...


Non-recurring costs are one-of-a-kind and occur at irregular
intervals
...

• Example
...
Which are recurring and which are non-recurring
costs you incur?
–
–
–
–
–
–
–

Remove existing trees, vegetation
Have land graded with bulldozer
Have yard planted with grass
Plant shrubs, trees
Mow grass
Fertilize grass, shrubs
Water grass, shrubs

27

Incremental Cost

• Incremental Cost is the additional cost that results from:
– Increasing the output of a system by one (or more) units
– Selecting one alternative over another
Example 2-4
...
The following
information is available
...
Book Costs
Cash costs
require the cash transaction of dollars from “one pocket to
another”
...

Example: You might use Edmond’s Used Car Guide to
conclude the book value of your car is $6,000
...
If you actually
sell the car to a friend for $5,500, then the cash cost to
29
your friend is $5,500
...
00%
100
...
00%
60
...
00%
20
...
00%

L
...
costs
committed
L
...
costs
spent

N
ee
C
on ds
ce De
f
pt

...

D
et
sg
ai
n
...

uc
O tion
pn
l
...
C
...

• Decisions made early in the life-cycle tend to “lock in”
costs incurred later in the life cycle:
Nearly 70 to 90% of all costs are set during the
design phases, while only 10 to 30% of the cumulative
life-cycle costs have been spent
...
When is the best time to consider all life-cycle
effects, and make design changes?
• Bottom Line
...

33

Cost Indices
• The Pakistan government publishes cost index data through the
State Bank of Pakistan
...

– The measure is scaled, so it is only the relative values of any two
measures that are meaningful
...
The conclusion is that one would have to spend
160/20, or 8 times as much in 1997 as in 1920 for the same
consumables
...


•

Cost indices are dimensionless
...
Develop the mathematical relationships for total cost and total
revenue for camp A
b
...
What is the profit or loss for the 12-week session if camp A
operates at 80% capacity?

d
...


36

Cash Flow Diagrams
•

•

•

Cash flow diagrams (CFD)
summarize the costs and
benefits of projects

Example:
Time Period

A CFD illustrates the size, sign,
and timing of individual cash
flows

Periods may be months,
quarters, years, etc
...
neg
...

pos
...

 Arrows signify cash flows and are placed at the end of
the period
 If a distinction needs to be made,



4

downward arrows represent expenses (negative cash flows or
cash outflows)
upward arrows represent receipts (positive cash flows or cash
inflows)

Notation and Cash Flow Diagram and Tables

5

EXAMPLE
 Before evaluating the economic merits of a proposed investment,

the XYZ Corporation insists that its engineers develops a cash
flow diagram of the proposal
...
Annual expenses will be $ 3000 at the end of each year
for operating and maintaining the project
...


6

SOLUTION EXAMPLE 3-1
2000

5310
1

3000

$10,000

7

5310
2

3000

5310

5310

5310

3

4

5

3000

3000

3000

Time Value of Money
Question: Would you prefer $100 today or $100 after 1 year?

There is a time value of money
...
The charge for its use
is called interest rate
...
Interest is a compensation for using
money
...

8

We know that receiving $1 today is worth more than $1
in the future
...

The opportunity cost of receiving $1 in the future is the
interest we could have earned if we had received the
$1 sooner
...

Today

Future

?

If we can MEASURE this
opportunity cost, we can:
 Translate $1 today into its equivalent in the future

(COMPOUNDING)
...

Today

?

Future

Simple Interest
Simple interest is interest that is computed on the original sum
...

Note: Interest is usually compound interest, not simple interest
...

At the end of each year your friend pays you 0
...
(this money is not paid until the end of the fifth year)
At the end of five years your friend also repays the $5000
...
08  5000
...

13

Simple Interest Example
 Example:
 P = $1,000
 i = 8%
 N = 3 years

End of
Year

Beginning Interest
Balance
earned

0

Ending
Balance
$1,000

1

$80

$1,080

2

$1,080

$80

$1,160

3

14

$1,000

$1,160

$80

$1,240

Compound Interest
Compounded interest is interest that is charged on the original sum and
un-paid interest
...

 At the end of year 1 you have (1
...
06)  530 = $561
...
06)  $561
...
51
Note:
$595
...
06)  561
...
06) (1
...
06) (1
...
06) 500 = 500 (1
...
40

$1,166
...
40

$93
...
71

$1,259
...
08)3
$1,000

18

 $1, 259
...


The factor is used to compute F, given P, and given i and n
...

(F/P,i,n) = (1+i)n
F = P (1+i)n = P (F/P,i,n)
...
05, n = 4, F = $800
...
05)4 = 800 (1
...
8227) = $658
...

Single Payment PresentWorth Formula
P = F/(1+i)n = F(1+i)-n

21

Factors in the Book (page 677 in 8-th edition)

22

Economic Equivalence

23

Economic Equivalence
Which one would you prefer?
•$20,000 today
•$50,000 ten years from now
•$ 8,000 each year for the next ten years

We need to compare their economic worth!
Economic equivalence exists between cash flows if
they have the same economic effect
...
Requires conversion of multiple payment cash flows to

a single cash flow
26

Present Value
Example: You borrowed $5,000 from a bank at 8% interest rate and
you have to pay it back in 5 years
...


Plan A: At end of each year pay $1,000 principal
plus interest due
...

Plan C: Pay in five end-of-year payments
...


Example
You borrowed $5,000 from a bank at 8% interest rate and you have to pay it back
in 5 years
...

a
Year

b
Amnt
...
Owed

Total Owed
b+c

Princip
...

Plan B: Pay interest due at end of each year and principal at end of
five years
...

Owed

c

d

e

f

Int
...

Payment

Total
Payment

int*b

1
2
3

5,000
5,000
5,000

400
400
400

5,400
5,400
5,400

0
0
0

400
400
400

4

5,000

400

5,400

0

400

400
2,000

5,400
27,000

5,000
5,000

5,400
7,000

5
5,000
SUM 25,000
29

Example (cont'd)
You borrowed $5,000 from a bank at 8% interest rate and you have to pay it back
in 5 years
...

a
Year

b
Amnt
...
Owed

Total Owed
b+c

Princip
...

Plan D: Pay principal and interest in one payment at end of five years
...

Owed

c

d

e

f

Int
...

Payment

Total
Payment

int*b

1
2

5,000
5,400

400
432

5,400
5,832

0
0

0
0

3
4

5,832
6,299

467
504

6,299
6,802

0
0

0
0

544
2,347

7,347
31,680

5,000
5,000

7,347
7,347

5
6,802
SUM 29,333
31

?

COMPOUND INTEREST

Example (cont'd)
Plan 1: At end of each year pay $1,000 principal plus interest due

Year
1
2
3
4
5

EOY
Pay
...


Year
1
2
3
4
5

EOY
Owed
Pay
...

Owed
$1,252 $5,000
$1,252 $4,148
$1,252 $3,227
$1,252 $2,233
$1,252 $1,159
$6,261 $15,767

$ Owed
$6,000
$5,000
$4,000
$3,000

$ Owed

$2,000
$1,000
$0
1

2

3

4

Time (Years)

34

5

Example (cont'd)
Plan 4: Pay principal and interest in one payment at end of 5 years
...


Owed
$0 $5,000
$0 $5,400
$0 $5,832
$0 $6,299
$7,347 $6,802
$7,347 $29,333

$ Owed
$8,000
$7,000
$6,000
$5,000
$4,000
$3,000
$2,000
$1,000
$0

$ Owed

1

2

3

4

Time (Years)

35

5

Example (cont'd)
Summary of Payment Plans

Plan

(a)
Total
Paid

1
2
3
4

$6,200
$7,000
$6,261
$7,347

(b)

(c)

Tot
...
Area under
curve
$1,200
$15,000
$2,000
$25,000
$1,261
$15,767
$2,347
$29,333

(d)
Ratio:
(b)/(c)
0
...
08
0
...
08

Question
...
paid)/(area under curve)
or
total int
...

36

Example (cont'd)
 Ratio
...
You put $500 in a bank for 3 years at 6% compound
interest per year
...

The bank pays you i = 0
...
015 every 3 months;
1
...

At the end of three years you have:
F = P (1+i)n = 500 (1
...
19562)  $597
...
Usually the stated interest is for a 1-year period
...
If the interest is i per
year, each quarter the interest paid is i/4 since there are four 3-month
periods a year
...
In 3 years, you need $400 to pay a debt
...
How much should you put in the
bank today to meet these two needs if the bank pays 12% per year?

0

1

2

3

4

5

$400
$600

Interest is compounded yearly
P = 400(P/F,12%,3) + 600(P/F,12%,5)
= 400 (0
...
5674)
= 284
...
44 = $625
...
6989) + 600 (0
...
56 + 330
...
80

Example: Points of view
Borrower point of view:You borrow money from the bank to start a
business
...

40

Concluding Remarks
The yellow Pages in the text book tabulate:
Compound Amount Factor
(F/P,i,n) = (1+i)n

Present Worth Factor
(P/F,i,n) = (1+i)-n
These terms are in columns 2 and 3, identified as
Compound Amount Factor: “Find F Given P: F/P”
Present Worth Factor: “Find P Given F: P/F”

41

Interest Formulas
A uniform series of payments or receipts represents:
A collection of end-of-period cash payments or receipts arranged in a uniform
series and continuing for n periods
...

Consider a 4-yr period:
A
A
A
A
|
|
|
|
F = 0
...
2
...
4 + 0
...
2
...
4 + 0
...
2
...
4 + 0
...
2
...
4
|
|
|
|A
|
|
| A(1+i)1
|
| A(1+i)2
| A(1+i)3

Uniform series (contin
...
)
If we turn this around and solve for A, we obtain:



i
A  F

n
 1  i   1

\ uniform series
sinking-fund factor

Example: Set up a uniform-payment investment (college fund) with the goal
of having $80,000 after 20 years, invested at 6% compounded annually
...
06)/[1
...
06/2
...
77
A = F (A/F, 6%, 20) = $80,000(
...
)

Two Ways for Calculating the Time-Value
of Money
 Simple Interest
 Compound Interest

45

Ex 3-4
 How much interest is payable each year on a loan of $2,000 if

the interest rate is 10% per year when half of the loan
principal will be repaid as a lump sum at the end of 3 years
and the other half will be repaid in one lump-sum amount at
the end of 6 years? How much interest will be paid over the
6-year period?

46

Solution
EOY
1
2
3
4
5
6

Interest
200
200
200
100
100
100

Principal Payment

1,000

1,000

A total of $900 will be repaid over 6 years

47

Ex-3-12
 Suppose that $20,000 is to be repaid at a rate of $4000 per

year plus the interest 10 % that is owed and based on the
beginning of year unpaid principal
...


48

Solution
Year

EOY
Payment

1

$20,000

$6,000

2
3
4

16,000
12,000
8,000

1,600
1,200
800

4,000
4,000
4,000

5,600
5,200
4,800

5

4,000

400

4,000

4,400

$ 6,000

49

BOY Unpaid Interest Principal
Principal
Accrued Payment

$ 20,000

$ 2,000 $ 4,000

$ 26,000

50

Multiple Payments
 How much do you need to

$25,000

$5,000

$3,000
0
1

P

51

2

3

4

deposit today (P) to
withdraw $25,000 at n =1,
$3,000 at n = 2, and $5,000
at n =4, if your account
earns 10% annual interest?

Uneven Payment Series

$25,000

$5,000

$3,000
0
1

2

3

4

P

$25,000

$5,000

$3,000
0
1

2

3

4

+

0
1

2

3

4

P2

 $22, 727

P2  $3, 000( P / F ,10%, 2)
 $2, 479

P  P  P2  P  $28,622
1
3
52

1

2

3

4

P4

P1
P  $25, 000( P / F ,10%,1)
1

+

0

P4  $5, 000( P / F ,10%, 4)
 $3, 415

ENGINEERING ECONOMICS

Presented by
Sohail Malik

Lecture 5a

Learning Objectives
 Arithmetic Gradient series
 Present value calculation for a
gradient series
 Examples

2

Arithmetic Gradient series
Definition: A collection of end-of-period, increasing cash
payments or receipts arranged in a uniformly increasing series
...

Consider a 4-yr period of uniformly increasing cash flows:
3G
Note: for n periods, there are (n-1) terms for G
2G
|
G
|
|
0
|
|
|
0…
...
…2…
...
…4
|
|
|
| F

Arithmetic Gradient
A uniform
increasing amount
...

G = the difference
between each cash
amount
...

•The “G” amount may be positive (receipts) or
negative (expense)!
•The present worth point is always one time
period to the left of the first cash flow in the
series or,
•Two periods to the left of the first gradient
cash (G) flow!
6
6

Arithmetic Gradient
• An arithmetic (linear) Gradient is a cash flow
series that either increases or decreases by a
constant amount over n time periods
...
Then add or subtract the
Present Value of the gradient to the Present
Value of the Uniform series

10

Example – Present value calculation for
a gradient series
$2,000

$1,250 $1,500

$1,750

$1,000
0
1

P =?

2

3

4

5

How much do you have to deposit
now in a savings account that
earns a 12% annual interest, if
you want to withdraw the annual
series as shown in the figure?
11

Method 1:
$2,000
$1,250 $1,500

$1,750

$1,000
0
1

P =?

2

3

4

5

$1,000(P/F, 12%, 1) = $892
...
49
$1,500(P/F, 12%, 3) = $1,067
...
16
$2,000(P/F, 12%, 5) = $1,134
...
03
12

Method 2:
P1  $1,000( P / A,12%,5)
 $3,604
...
20
P  $3,604
...
20
 $5,204

13

Arithmetic Gradient Uniform Series Factor

A pure gradient (uniformly increasing amount) can
also be converted into the equivalent present value
of uniform series:
AG = G(A/G, i, n)

14

Arithmetic Gradient series

 1  i n  ni  1
F  G

2
i



Example: For 6% interest rate over 4 yr
...
06)4 - 1 – 4(
...
06)2 = $200(6
...
72
15

Arithmetic Gradient series (contin
...
Such costs at the end of yr
...
Our plan is to make these payments by using
amounts based on an initial cash deposit in an investment account in the local credit
union paying 4% annual interest
...
1…
...
3…
...
5…
...
1…
...
3…
...
5…
...
242)+$50(12
...
20 + $625
...
50

Example: The Present Worth of $400 in year 1 and amounts increasing by
$30 per year thru year 5 at an interest rate of 12% per year is:
(a) $1532

(b) $1,634

(c) $1,744

(d) $1,829

Solution: The cash flow diagram is as follows:
P=?

i=12%

P=400(P/A,12%,5) + 30(P/G,12%,5)
1

2

3

4

5

Year

=400(3
...
3970)

0

= $1,633
...
7746)
=$453
...

 Economic equivalence exists between
individual cash flows and/or patterns of
cash flows that have the same value
...

 The purpose of developing various
interest formulas was to facilitate the
economic equivalence computation
...

 A GEOMETRIC gradient changes by a
fixed percentage each time period
...
of time periods – n

•The starting cash flow – A

24

Geometric Gradient Series
Geometric Gradients Change by the Same Percentage Each Period

Cash Flow Diagram for Present Worth of Geometric Gradients is as follows:
There are no Tables for Geometric Factors
Use Following Equation:

P=?
1
0

2

3

4

n

P=A{1-[(1+g)/(1+i)]n}/(i-g)

A
A(1+g)1
A(1+g)2

Where: A=Cash Flow in Period 1
g=Rate of increase
A(1+g)n-1

If g=i, P=An/(1+i)

Note: If g is negative, change signs in front of both g’s
25

Geometric Gradient Series
•Particularly relevant to construction costs
•Cash flows increase/decrease by a constant %(g); compound growth
•Example: price changes due to inflation
A1(1+g)N-1
A1

A1(1+g)

0

1
P

2 3
g>0

N-1 N

Find P, given A, g, i, N

26

Gradient Example:
Example: Find the present worth of $1,000 in year 1 and
amounts increasing by 7% per year thru year 10
...

(a) $5,670

(b) $7,335

(c) $12,670

(d)$13,550

Solution: P=1000[1-(1+0
...
12)10]/(0
...
07)
= $7,333
Answer is (b)

P=A{1-[(1+g)/(1+i)]n}/(i-g)
Where: A=Cash Flow in Period 1
g=Rate of increase

27

Example:
Airplane ticket price will increase 8%
in each of the next four years
...
How much should be put away
now to cover a students travel home
at the end of each year for the next
four years? Assume 5%
...
08) 4 (1
...
11928 
  180
 180
  $715
...
05 
...
03 



29

Check
As a check we can also solve this problem without
using the geometric gradient
Year
Ticket
1 A1 =
= 180
2 A2 = 180 + 8%(180)
= 194
...
40 + 8% (194
...
95
4 A4 = 209
...
95)
= 226
...
40(P/F,5%,2) +
209
...
75(P/F,5%,4)
=$715
...

30

Future worth Factor
Since F =P(1+i)
Multiplying (P/A,g,i,n) by (1+i) will give F

1  (1  g) (1  i) 
n
F  P(1  i)  A
(1  i)
ig


n

n

n

 (1  i) n  (1  g) n 
F  A

ig


31

Example
 A graduating CE is going to make $35,000/yr
with Granite Construction
...
The CE can count on a 3%
salary increase with the standard of living
increases for the next 30 years of employment
...
1 = 3,500
i = 12%
g = 3%
n = 30

 (1  i) n  (1  g ) n 
F  A

ig


 (1
...
03) 30 
 3500
  3500  305
...
09


33

ENGINEERING ECONOMICS

Presented by
Sohail Malik

Lecture 4

1

Annuity
2

Annuity
• A (Annuity) occurs at the end of each period for
N periods with interest rate at I % per period,
and such that
– P (present equivalent value) occurs one interest
period before the first A (uniform amount)
– F (future equivalent value) occurs at the same
time as the last A, and N periods after P, and
– A (annual equivalent value) occurs at the end of
periods 1 through N, inclusive
3

Annuity- Uniform Payment Series
F

0
A

A

1

2

N

A
P

0

1

2

N

0

N

4

Annuity- Interest Formulas (Uniform Payment)
A uniform series of payments or receipts represents:
A collection of end-of-period cash payments or receipts arranged in a uniform
series and continuing for n periods
...

Consider a 4-yr period:
A
A
A
A
|
|
|
|
F = 0
...
2
...
4 + 0
...
2
...
4 + 0
...
2
...
4 + 0
...
2
...
4
|
|
|
|A
|
|
| A(1+i)1
|
| A(1+i)2
| A(1+i)3
5

Uniform series (contin
...
)
If we turn this around and solve for A, we obtain:



i
A  F

n
 1  i   1

\ uniform series
sinking-fund factor

7

Sinking Funds
This is an annuity that is invested for a specific
purpose and is continued for a predefined
period
...


8

Sinking Funds
• The constant periodic amount, at a constant
interest rate that must be deposited to
accumulate a future value
...
50
11

Finding an Annuity Value (Sinking Fund)
Example: Set up a uniform-payment investment (college
fund) with the goal of having $80,000 after 20 years, invested
at 6% compounded annually
...
06)/[1
...
06/2
...
77
OR

A = F (A/F, 6%, 20) = $80,000(
...
This
expression is used to calculate the present worth, given
the regular annuity payment
...
Which option do you think the testing lab would prefer,
assuming it has to replace the sold machine?
Ploan = $1200 [P/A, 0
...
871) = $39,445
The lab would prefer the $40k payment now, because it is greater
than the present worth of the proposed loan terms
...
5%, 36] = $1200(39
...


18

Finding A when Given P

19

Sinking Fund Factor pg 85
Find A, given F, I, N


i
A  F

N  1
 (1  i)



Capital Recovery (Annuity) Factor
Find A, given P, I, N
 i(1  i) N 
A  P
N

 (1  i)  1

Present Worth Factor
Find P, given A, I, N
 (1  i) N  1
P  A
N 
 i(1  i) 
20

Convert Present to Annual
• How is it possible to calculate a constant amount to
repay, and have the total be exactly equivalent to P?
– It is sort of like magic!
• The calculations would be easier if you paid an equal
fraction of the principal P every year:
– Plus whatever interest is owed on the unpaid principal
• But in that case nobody could afford a mortgage:
– The payments would be too high in the first few years!

21

Example (cont'd)
You borrowed $5,000 from a bank at 8% interest rate and you
have to pay it back in 5 years
...

a
Year

b
Amnt
...
Owed

Total Owed
b+c

Princip
...

Plan C: Pay in five end-of-year payments
...
05)5 + $1200/(1
...
05)15 + $1200/(1
...
06)64 = $41,647
– Things get big over time!
• Invest $1000 each year for 64 years at 6%:
– F = A [(1+i)n - 1]/i
• = $1000 [(1
...
06 = $677,450
– This is really big!
27

Types of Annuities


An Annuity represents a series of equal payments (or
receipts) occurring over a specified number of
equidistant periods
...

• Annuity Due: Payments or receipts occur at the beginning of
each period
...
46
31

Validation
$5,000(1  0
...
38
4

F =?

$5,000(1  0
...
08
3

$5,000(1  0
...
00

i = 6%
0

1

2

3

4

5

$5,000(1  0
...
00
$5,000(1  0
...
00
$28
...
46
0

$5,000 $5,000 $5,000 $5,000 $5,000

32

Deferred Annuities
• A deferred annuity is the same as any other
annuity, except that its payments do not
begin until some later period
• The timeline shows a five-period deferred
annuity
100
0

1

2

100

100

100

100

3

4

5

6

7

33

Deferred Annuities
• Find P given A for an ordinary annuity if the
annuity is deferred j periods, where j < N
• P = A ( P/A, i%, N-j )
• P = A ( P/A, i%, N-j ) ( P/F, i%, j )

34

PV of a Deferred Annuity (cont
...
08

PV0 = 313
...
50
37

Find F When A is Given (Future Value of an annuity)
F
0

1

2

3
N

A

(1  i )  1
FA
i
 A( F / A, i , N )
N

Example :
• Given: A = $5,000, N = 5 years, and i = 6%
• Find: F
• Solution: F = $5,000(F/A,6%,5) = $28,185
...
144

•

Use the look-up tables in the back of the book to find
interest rates that give values near
...
FACTOR

WORTH

FUND

AMOUNT

RECOVERY

F/P

N

PRESENT
P/F

A/F

F/A

A/P

1
...
5083

0
...
8164

0
...
14238

45

Interpolation example
•

For i = 8%, we observe:
COMPOUND

10

SINKING

COMPOUND

CAPITAL

AMT
...
1589

0
...
0690

14
...
14903

(A/P, 8%, 10) = 0
...
14238 (<
...
14903 (>
...
144 lets us “eyeball” how close
the IRR is to 7%
Interpolation just lets us do that more exactly!

Interest rate, i

•

8
...
5
7
6
...
142

0
...
146

0
...
15

(A/P, i, 10)

48

Interest rate, i

Interpolation example
8
...
5
7
6
...
142

0
...
146

0
...
15

(A/P, i, 10)

•

As the factor A/P increases, the interest rate i also
increases

•

Use the slope of this line to figure out how much to add to
7% to get A/P =
...
5

0
...
5
7
6
...
142

0
...
144

0
...
148

0
...
14903-
...
144-
...
24%
50

Interpolation example
• The formula for interpolation tells us:
• i  7%
+ [(8%-7%))/(
...
14238)] (
...
14238)
• i  7
...
2459%

51

Interpolation Review
• Interpolation is approximate, not exact!
• It requires:
– One interest rate that gives a smaller ratio
– One interest rate that gives a larger ratio
• The answer must lie between the two:
– Results will be more accurate when the two starting points
are close together

52

ENGINEERING ECONOMICS
Presented by
Sohail Malik

Lecture 6

PRESENT WORTH ANALYSIS

2

MARR
 A policy issue resolved by top management in view of
the following considerations








The amount of money available for investment, and the
source and cost of these funds (i
...
debt or equity)
The number of good projects available for investment and
their purposes (i
...
whether they sustain present operations
and are essential, or whether they expand on present
operations and are elective)
The amount of perceived risk associated with investment
opportunities available and the estimated cost of
administering the projects over short planning horizons
versus long planning horizons
The type of organization involved (i
...
government, public
utility, or private industry)
3

MARR
 MARR (sometimes called the hurdle rate) should be
chosen to maximize the well-being of an organization,
subject to the considerations above

 Select only those projects which provide annual rate of
return in excess of MARR

 As amount of investment capital and opportunities
available change over time, a firm’s MARR will also change

4

PRESENT WORTH ANALYSIS

5

ECONOMIC CRITERIA FOR DECISION MAKING

6

Present Worth Method
 Based on concept of equivalent worth of all cash flows
relative to the present as a base

 All cash in/out-flows are discounted with MARR
 Present worth (PW) is a measure of how much money an
individual can afford for the investment in excess of its cost

 It is assumed that cash generated by the alternative is
placed in reserve and earns interest at a rate equal to the

MARR
7

DIFFERENT ANALYSIS-PERIOD SITUATIONS
1
...


2
...


3
...


8

Present Worth Method

9

Example 1
An investment of $10,000 can be made in a project that will
produce a uniform revenue of $5,310 for 5 years and then
have a salvage value of $2,000
...
The
company is willing to accept any project that will earn an
annual return of 10% or more, before income taxes, on all
invested capital
...

10

Solution Example 1
Present Worth
Outflows
Annual Revenue : $5,310(P/A,10%,5)

Inflows
$20,125

Salvage Value : $2,000(P/F,10%,5)

1,245

Investment

-10,000

Annual expenses :
$3,000(P/A,10%,5)

-11,370
Total -$21,370

$21370

Total Present Worth= $0
Therefore, the project is marginally acceptable
11

Example 2
A piece of new eqpt has been proposed by engineers to increase the
productivity of a certain manual welding operation
...
Increased productivity
attributable to the eqpt will amount to $8,000 per year after extra operating
costs have been subtracted from the value of the additional production
...
A cash flow diagram for this equipment is
given below
...


12

Solution Example 2
Total PW = PW of cash receipts – PW of cash outlays
Total PW (20%)

= $8,000(P/A,20%,5)
+$5,000(P/F,20%,5) - $25,000
= $934
...
It has two options
...


 Let’s see what the consequences are for each option
...

 Applications for the FW approach:
Projects that do not come on line until the end
of the investment (construction) period:
• Power Generation Facilities
• Toll Roads
• Large building projects

17

Example 3
Evaluate the FW of equipment described in Example 2
...

FW(20%) = - $25,000(F/P,20%,5) + $8,000(F/P,20%,4)
+ $8,000(F/P,20%,3) + $8,000(F/P,20%,2)
+ $8,000(F/P,20%,1) + $13,000
= - $25,000(F/P,20%,5) + $8,000(F/A,20%,5)
+ $5,000
= $2,324
...
80(P/F, 20%, 5) = $934
...
10, 5) – 200 (P|F,
...
7908) - 200 (
...
72 - 124
...
10, 5) – 350 (P|F,
...
56 - 217
...


24

Present Worth: Example Unequal Lives
Machine A
0

$11,000

F6=$1,000

1

2

3

4

5

6

A 1-6 =$3,500
F6=$2,000

0

1

2

Machine B

$18,000

3

4

5

6

7

8

9

A 1-9

=$3,100

i = 15% per year

LCM(6,9) = 18 year study period will apply for present worth
25

Present Worth: Example Unequal Lives
Unequal Lives: 2 Alternatives
Machine A
6 years

Cycle 1 for A

6 years

Cycle 2 for A

6 years

Cycle 3 for A

Machine B
9 years

9 years

Cycle 1 for B

Cycle 2 for B

18 years

i = 15% per year

LCM(6,9) = 18 year study period will apply for present worth

Present Worth: Example Unequal Lives
Unequal Lives: 2 Alternatives
 Since the leases have different terms (lives), compare
them over the LCM of 18 years
...

 These are years 6 and 12 for Machine A and year 9 for
B
...
15, 6) – 1,000 (P|F,
...
7845) – 1,000 (
...
15, 6)+23,813 (P|F,
...
15, 9) – 2,000(P|F,
...
7716) - 2,000(
...
GatorCo is considering buying device A or B
...
Each device has a useful life of five years, and no salvage
value
...
Interest is 7%
...
1000) = $1230
Device B:
NPW = 400 (P/A,7%,5) - 50 (P/G,7%,5) = 400(4
...
647) =
$1257
...

Device B gives more of its benefits in the earlier years
...
RCB can:
a) spend $300 million now, and enlarge the aqueduct in 25 years for
$350 million more,
b) construct a full-size aqueduct now for $400 million
...
We ignore maintenance costs
...
There is no salvage value
...
6 million
b) NPW = $400 million
This is an example of stage construction
...

34

Same-Length Analysis Periods
Example
MCE construction company needs new Bulldozer
...

Make

Cost

Useful life

EOL salvage value

Speedy

$1500

5 years

$200

Allied

$1600

5 years

$325

Speedy: NPW = 1500 – 200 (P/F,7%,5) = 1500 – 200 (0
...

Allied: NPW = 1600 – 325 (P/F,7%,5) = 1600 – 325 (0
...

Remark
...

Assuming both pieces of equipment have the same annual
maintenance costs, why is this omission justifiable?
35

Same-Length Analysis Periods
Suppose each has a maintenance cost of C per year
...

The revised PW of the costs would be:

Speedy: $1358 + C (P/A,7%,5)
...


The difference between the PW’s remains the same
...

36

PV Formula
Example
We must choose a weighing scale to install in a package
filling operation in a plant
...

Each scale has a life of 6 years
...

Alternative

Cost

Uniform annual
benefit

EOL salvage
value

Atlas

$2000

$450

$100

Tom Thumb

$3000

$600

$700

We use the formula:
NPW = PW of benefits – PW of costs
37

PV Formula
Atlas:
NPW = 450 (P/A,8%,6) + 100 (P/F,8%,6) – 2000
= 450 (4
...
6302) - 2000
= 2080 + 63 – 2000 = $143
Tom Thumb:
NPW = 600 (P/A,8%,6) + 700 (P/F,8%,6) – 3000
= 600 (4
...
6302) – 3000
= 2774 + 441 – 3000 = $215
Tom Thumb looks preferable
...
The NPV formula is of fundamental importance
...

Example MCE construction company needs new equipment
...

Speedy equipment for five years is not equivalent to Allied
equipment for ten years
...
5083)
= 1600 – 165 = $1435
...
7130) = $1368
...

Two Speedy’s:
PW = 1500 + (1500 – 200) (P/F,7%,5) – 200 (P/F,7%,10)
= 1500 + 1300 (0
...
508)
= 1500 + 927 – 102 = $2325
...
5083)
= 1600 – 165 = $1435
...

In the above example, it made some sense to use 10 years as
the analysis period
...
But an analysis period of 91 years is not too
realistic
...

• In governmental analyses, a service or condition
sometimes must be maintained for an infinite period
...

• In these situations a present worth of cost analysis would
have an infinite analysis period
...

44

Infinite Analysis Period
 Capitalized cost is the present sum of money that would need

to be set aside now, at some interest rate, to yield the funds
required to provide the service (or whatever)indefinitely
...
The interest received on the
money set aside can be spent, but not the principal
...

45

Infinite Analysis Period
In Chapter 4 we saw that
principal sum + interest for the period = amount at end of period,
or
P + iP = P + iP

Capitalized cost is therefore the P in the equation A = i P
...
xchng
4

Investment Appraisal
• A means of assessing whether an investment project is worthwhile or not
• Investment project could be the purchase of a new PC for a small firm, a new
piece of equipment in a manufacturing plant, a whole new factory, etc
• Used in both public and private sector

5

Investment Appraisal

• Types of investment appraisal:
• Future worth analysis
• Benefit-cost ratio analysis

• Sensitivity and breakeven analysis
• Payback Period
• Accounting Rate of Return (ARR)
• Internal Rate of Return (IRR)
• Profitability Index
What factors need to be considered before
investing in equipment such as this?

Copyright: Gergely Erno, stock
...
g
...
g
...
23 months
• (2 yrs, 6¾ months)

11

Accounting Rate of Return

12

Accounting Rate of Return
• A comparison of the profit generated by the investment with the cost of the
investment
Average annual return or annual profit
•

ARR = -------------------------------------------Initial cost of investment

13

Accounting Rate of Return
• e
...

• An investment is expected to yield cash flows of £10,000 annually for the
next 5 years

• The initial cost of the investment is £20,000
• Total profit therefore is: £30,000
• Annual profit = £30,000 / 5 = £6,000

ARR = 6,000/20,000 x 100 = 30%
A worthwhile return?

14

Investment Appraisal
• To make a more informed decision,
more sophisticated techniques need to

be used
...

• At an interest rate usually equal to or greater than the Organization’
established MARR

18

Present Worth –
A Function of the assumed interest rate
• If the cash flow contains a mixture of positive and negative cash flows –
• We calculate:
– PW(+ Cash Flows) at i%;
– PW( “-” Cash Flows) at i%;
– Add the result!
• We can add the two results since the equivalent cash flows occur at the
same point in time!

19

Present Worth –
A Function of the assumed interest rate
• If P(i%) > 0 then the project is deemed acceptable
...


20

Net Present Value
• e
...

• Project A costs £1,000,000
• After 5 years the cash returns = £100,000 (10%)
• If you had invested the £1 million into a bank offering interest at

12% the returns would be greater
• You might be better off re-considering your investment!
21

Net Present Value
• The principle:
• How much would you have to invest now to earn £100 in one year’s time if
the interest rate was 5%?
• The amount invested would need to be: £95
...
e the present
...
9090
• If you invested 0
...
g
...
25%

= 500 x
...
77
• £329
...
25% to earn £500 in 10 years time
• PVs can be found through valuation tables (e
...
Parry’s Valuation Tables)

24

Cash Flows

25

Discounted Cash Flow
• An example:

• A firm is deciding on investing in an energy efficiency system
...
75%
(1 / 1+r) n

Present Value (£)
(CF x DF)

0

- 600,000

1
...
9546539

71,599
...
9113641

91,136
...
8700374

130,505
...
8305846

166,116
...
7929209

166,513
...
7569650

113,544
...
75%
(1 / 1+r) n

Present Value (£)
(CF x DF)

0

- 600,000

1
...
9546539

23,866
...
9113641

68,352
...
8700374

73,953
...
8305846

83,058
...
7929209

118,938
...
7569650

340,634
...
70
29

Discounted Cash Flow
• System A represents the better investment
• System B yields the same return after six years but the returns of

System A occur faster and are worth more to the firm than returns
occurring in future years even though those returns are greater

30

Internal Rate of Return (IRR)

31

Internal Rate of Return
• Allows the risk associated with an investment project to be assessed
• The IRR is the rate of interest (or discount rate) that makes the net present
value = to zero
• Helps measure the worth of an investment
• Allows the firm to assess whether an investment in the machine, etc
...
What was
the rate of return on the investment?
PW of benefits = 1
=
2000(P / A, i, 5) = 1
PW of costs
8200

(PIA, i, 5) = 8200/ 2000 = 4
...
1; if
no tabulated value of i gives this value, we will then find values on either side of the
desired value (4
...

34

CALCULATING RATE OF RETURN
From interest tables we find:

i

(P / A, i, 5)

6%

4
...
100

8%

3
...


35

Profitability Index

36

Profitability Index
• Allows a comparison of the costs and benefits of different projects to
be assessed and thus allow decision making to be carried out
Net Present Value
Profitability Index = --------------------Initial Capital Cost

37

ENGINEERING ECONOMICS

Presented by
Sohail Malik

Lecture 5

NOMINAL AND EFFECTIVE
INTEREST RATE

Nominal and Effective Interest Rate
Effective Interest Rate
Nominal Interest Rate
Ten thousand dollars is borrowed for two years at an interest rate of
24% per year compounded quarterly
...
48

Compounding is not less
important than interest

interest charges for annually compounding:
$10,000(1+24%)2 - $10,000 = $5376
...
48 - $5376 = $562
...
Sometimes one interest rate is
quoted, sometimes another is quoted
...

A bank pays 5% compounded semi-annually
...
5% each six months
...
5% interest per period for two periods
...
025) = 1,025  1025(1
...
60
With i = 0
...
05,
P  (1 + i) P  (1+r/2)2 P = (1+ 0
...
050625) P

Nominal and Effective Interest Rate
Terms the example illustrates:
r = 5% is called the nominal interest rate per interest period
(usually one year)
i = 2
...
0625% is called the effective interest rate per year

In the example: m = 2 is the number of compounding subperiods
per time period
...
050625) – 1 = (1
...
05/2)2 – 1
The term i we have used up to now is more precisely defined as the effective interest rate per interest period If the interest period is
one year
...


r = nominal interest rate per year
m = number of compounding sub-periods per year
i = r/m = effective interest rate per compounding sub-period
...
5% interest every three months
...
5% = 6% a year
Effective interest rate per year:
ia = (1 + r/m)m – 1 = (1
...
06136  6
...


Nominal and Effective Interest Rate
Table Nominal & Effective Interest expressed in percent

Nominal rate Yearly
r
1
2
3
4
5
6
8
10
15
25

m=1
1
...
0000
3
...
0000
5
...
0000
8
...
0000
15
...
0000

Effective rates, ia = (1 + r/m)m - 1
SemiMonthly
Daily
Continuously
ann
...
0025
1
...
0050
1
...
0100
2
...
0201
2
...
0225
3
...
0453
3
...
0400
4
...
0808
4
...
0625
5
...
1267
5
...
0900
6
...
1831
6
...
1600
8
...
3278
8
...
2500
10
...
5156
10
...
5625
16
...
1798
16
...
5625
28
...
3916
28
...
The formula used for the continuous rate is er - 1, expressed
in percent
...


Nominal and Effective Interest Rate
Example Joe Loan Shark lends money on the following terms
...
”
1
...
2
...

We know m = 52, so r = 52  i = 10
...

2
...
4/52)52 – 1  13,104
...


Nominal and Effective Interest Rate
Suppose Joe can keep the $50, as well as all the money he
receives in payments, out in loans at all times? How much
would Joe have at the end of the year?
We use F = P(1+i)n to get F = 50(1
...
We put $5000 in an account paying 8%
interest, compounded annually
...

We used A = P(A/P,8%,5) = 5000  (0
...


Nominal and Effective Interest Rate
Example

Sally deposits $5000 in a saving account paying 8%

nominal interest, compounded quarterly
...

31 of the first year
...

W

W

W

i = 2%, n = 20

$5000

W

W

Nominal and Effective Interest Rate
The withdrawal periods and the compounding periods are not
the same
...

Solution 1
...

(We don’t, but suppose we do
...
0612) =
$306
...

W = A(F/A,i,n) = 306 (F/A,2%,4) = 306 (4
...

Sally should withdraw $1260 at the end of each year
...
24%, n = 5
$5000

ia = (1 + r/m)m – 1 = (1 + i)m – 1 = (1
...
0824  8
...
24%,5) = P {[i (1+i)n]/[(1+i)n – 1]} = 5000(0
...


Summary Notation
i: effective interest rate per interest period (stated as a
decimal)

n: number of interest periods
P: present sum of money
F: future sum of money: an amount, n interest periods from the
present, that is equivalent to P with interest rate i
A: end-of-period cash receipt or disbursement amount in a
uniform series, continuing for n periods, the entire series
equivalent to P or F at interest rate i
...



...



...
05)n
We can solve for n

n

Example (continued)
• (1
...
05) = ln(2)
• n = ln(2)/ln(1
...
693147/0
...
2067 years
• With compounding every year:
• It will take 15 years to amass $2,000

• (Actually, a little more than $2,000!)

Example of the NPER function
• In Excel, you can do this using NPER:

NPER(i, A, P, F)

=NPER(C23,C22,C20,C21)

SOLVING FOR AN UNKNOWN
INTEREST RATE

Uncertain interest rates
• It may be of interest to determine the value of i that makes

two streams of payments equivalent
• Why?
– Compare a series of payments to a fixed interest rate
– Determine attractiveness of the payments
– Provide a simple summary measure

Example: Solving for i
• Assume you can invest $3,000 now in a friend’s business,
and will get paid back $5,000 in 5 years:
• For what interest rate or “internal rate of return” (IRR) are
these amounts equivalent?

$5,000

0

$3,000

1

2

3

4

5

Example (continued)
$5,000
0

1

2

3

4

5

$3,000

• F = P (1+i)n
• 5,000 = 3,000 (1+i)5
• (1+i)5 = 5,000/3000 = 1
...
66670
...
1076
• i = 1
...
1076 = 10
...
76% per year

Example of the IRR Function

=IRR($D6:$D11)

Uncertain interest rates
• If there is an annual series of payments, this gets more complicated:

• Can find n (using logarithms), but not i!
– Because i appears twice:
• Once with an exponent
• Once without

Interest rate i is unknown
• Usually, you will need some other approach:

• Trial and error (try different values of i until you converge
on the unknown interest rate)
• Use look-up tables at back of textbook, and interpolate
(needed for exams!)
• Use IRR function in Excel

Uncertain interest rates
• How to solve?
• Two options
– Trial and error:
• Try different values of i until you converge
• I
...
, until P = A [1 - 1/(1+i)n]/i
– Use look-up tables
• (e
...
, tables at back of textbook)
and interpolate

Interest rate i is unknown
• When annual payments are involved:

• Interpolation or trial and error must be used

 i(1  i)n 
A P

n
 (1  i)  1 



i
AF
(1  i ) n  1 



Review of Interpolation

ENGINEERING ECONOMICS

Presented by
Sohail Malik

1

Lecture 8

2

COMPARING ALTERNATIVES
WITH RATE OF RETURN

Use Incremental Analysis

Rate of Return Analysis
3







Recall that the ROR of an investment is the interest
rate that makes
In rate of return analysis no interest rate is
introduced in calculations, rather we compute a rate
of return
...
The project is accepted if IRR > MARR
...

Since we want to look at increments of investment, the
cash flow for the difference between the alternatives
is computed by taking the higher initial-cost
alternative minus the lower initial cost alternative
...
If IRR is less than the
MARR, choose the lower-cost alternative
...
Rank the alternatives in increasing order of
investment
...
Select as the defender the alternative with the
smallest investment
...
Let the challenger be the alternative with the next
higher investment
...
Accept or reject the challenger on the basis of the
return on the extra investment
...

5
...
Otherwise return to step 3
...

Alternative

Investment

Operating Cost

D

$2500

$1000

B

$3000

$875

C

$4000

$500

A

$5000

$240

Example 2: Incremental Analysis
9





Defender: Project D
...

Is the extra investment in B over D justified?
 Incremental

Investment (D-B):
-$500
 Incremental Benefit (D-B):
$125
 NAW= -500 (A/P, i, 5) + 125=
...


10

Example 2: Incremental Analysis
(cont’d)




Defender: Project D
...

Is the extra investment in C over D justified?
 Incremental

Investment (D-C):
-$1500
 Incremental Benefit (D-C):
$500
 NAW= -1500 (A/P, i, 5) + 500 =
...


11

Example 2: Incremental Analysis
(cont’d)




Defender: Project C
...

Is the extra investment in A over C justified?
 Incremental

Investment (C-A):
-$1000
 Incremental Benefit (C-A):
$260
 NAW= -1000 (A/P, i, 5) + 260 =
...
)

Usually rate of return, or net
present value
Mainly economic

Estimating costs
18

• Costs:
• Construction, operations, maintenance
• Initial costs fairly well known:
• Future operations and maintenance costs are usually less
well known
• Must be estimated!

Estimating benefits
19

• BENEFITS to the public (users) must be estimated in
terms of dollar values:

• Difficult to do!

Estimating “disbenefits”
20

• Disbenefits:
• Expected undesirable or negative consequences to the
public
• For example, there may
disadvantages to the public

be

indirect

economic

• Difficult to estimate and convert to dollar values

General observation
21

It is difficult to estimate and reach
agreement on the economic benefits and
disbenefits of public-sector projects

Funding of public projects
22

• Generally low interest charges
• Investments backed by public agencies
• Cost-sharing arrangements often exist
• Less perceived risk

Determining an interest rate
23

• Different than in the private sector:

• Called the “social discount rate”
• For federal projects:
• A typical rate is 10% per year
• Often less (as low as 3% per year)

Example
24

An $8200 investment returned $2000 per year over a 5-year
useful life
...
1

2000

Then look at the compound interest tables for the value of i

where (PIA, i, 5) = 4
...
Both devices cost $1000
and have useful lives of 5 years and no salvage value
...
Device B
will provide cost savings of $400 the first year, but savings will
decline by $50 annually, making the second year savings
$350, the third-year savings $300, and so forth
...
100) =$1230
B = PW of benefit = 1230 = 1
...
100)- 50(7
...
26

1000

To maimize the benefit-cost ratio, select Device B
...

If the MARR is 10%, which alternative should be selected?
Solve the problem by benefit-cost ratio analysis
...
0

Uniform annual
benefit

$158
...
0

C
$200
...
7

$58
...
3/[$600 (A/P, 10%, 5)]
B/COF B = $138
...
3/[$200 (A/P, 10%, 5)]

= 1
...
05
= 1
...
00
...

Cost
Uniform Annual Benefit

B- C
$300
$80
...
6

B/COF B-C
= $80
...
Reject C
...
02

B/COF A-B
= $19
...
Reject A
...
74

Conclusion: Select B
...
The benefits at the end of
the first year are estimated to be $10,000, increasing
at a 10% uniform rate in subsequent years
...


Solution
31

A1 = $10,000

i = 10%

g = 1% n = 8 yrs

A1

P = $50,000

Geometric gradient at a 10% uniform rate
...
10)-1]/$50,000 = 1
...
’ This is called sensitivity analysis
...
To
illustrate the sensitivity of a decision between
alternatives to particular estimates, breakeven analysis
is often presented as a breakeven chart
...

Should a facility be constructed now to meet its future
full- scale requirement, Or should it be constructed in
stages as the need for the increased capacity arises?
Three examples of this situation are:

Sensitivity and Breakeven Analysis
36

 Should

we install a cable with 400 circuits now or a 200circuit cable now and another 200-circuit cable later?
 A 10 cm water main is needed to serve a new area of
homes
...
The
firm could have a warehouse built now and later enlarged,
or have the warehouse with capacity for expanded
operations built right away
...

 Construction Costs
 Two-stage construction
Construct first stage now
 Construct second stage


$100,000
120,000

n years from now
 Full-capacity construction


Construct full capacity now

140,000

Sensitivity and Breakeven Analysis
38

Example
 Other Factors
 1
...
zero salvage value
...
The annual cost of operation and maintenance is the same for
both two-stage construction and full-capacity construction
...
Assume an 8% interest rate
...
" Mark the break even point on your graph
...
Present
worth calculations appear simpler and are used here
...
Compute the PW of cost for several
values of n (years)
...
6806) = $181,700
PW = 100,000+120,000 (0
...
2145) = 125,700
PW = 100,000+120,000(0

---