Notesale: Turn your study into money

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Practice Exam 2
IE 2324
NOTE: Be aware there have been known to be mistakes in the author’s examples
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After looking around a bit, she finds that a wellequipped laptop with software can be purchased for $1,800 and that it should have a market value of at least
$300 if she wants to sell it when she graduates after 4 1/2 years
...
Use an interest rate of 12% with monthly compounding, and determine the present
cost of owning and operating the computer
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Salvage value = $300
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n = 4
...
t
nsa = 4
...

i = 12%
...
12/12)6 - 1 = 6
...

Present cost of ownership
= P - {S (P/F, i, n)} + {M&O (P/A, i, n)}
=1,800-{300(1 + 0
...
6152)-9} / 0
...
29 + 675
...
40
Problem 5-2
Given 2 alternatives:

First Cost
Annual Cost
Annual Benefit
Life
Salvage

A
4,000
1,000
2,000
5
3,000

B
6,000
500
2,200
10
1,000

If i = 10%, find the better alternative computing NPW of both alternatives Assume alternative A is replaced at
the end of its useful life
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NPW (Alt
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791) + 3,000{(0
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3855)]-4,000{1 + 0
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60
NPW (Alt
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145) + 1,000(0
...
B to maximize NPW

Problem 5-3
A person buys a $1,000 face value bond 2 years after its issue
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This bond pays 6% annually
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Interest / year = 0
...
372) + 1,000(0
...
52
Problem 5-4
The construction of a dam will cost $1,000 at time 0, $500 in year 1, $500 in year 2
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Beginning in year 3, maintenance costs will be $100 per year
through year 10
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Find the present worth of all the costs if i = 10%

Solution:
PW of costs = (150/0
...
10)(0
...
335)(0
...
736) +1,000 = $2,886
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The purchase price of the extruder is $350,000
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XYZ believes that the extruder can be sold for
$75,000 at the end of its 15-year service life
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Maintenance and operating costs are
expected to be $4,000 during the first year and to increase by $1,200 each year
...

What is the net present worth (NPW) of this investment to XYZ?
Solution:
P = $350,000
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M&O = $4,000
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Uniform Annual Benefit = $90,000
MARR = 12%
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25 × 350,000 = $87,500
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Interest rate for the loan = 10%
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2054) = $53,917
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NPW = Present Worth of Benefits - Present Worth of Costs
= {90,0000 (P/A, 12%, 15) + 75,000 (P/F, 12%, 15)}{(53,917
...
811) + 75,000 (0
...
50 + 4,000) (6
...
920)-87,500}
= 626, 692
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09 = $104, 012
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Problem 5-6
A taxicab company maintained accurate records of the expenses for one of its automobiles from January 1, 1996
through January 1, 2002
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The
maintenance and repair costs are $500 for 1986, $1,000 for year 1997, $1,500 for year 1998, and so on
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00 + 5,984
...
00
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The deluxe model costs $2,200 each and
will last for six years
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The deluxe model is
aluminum so it will have a scrap value of $150 at the end of its life
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The use of the deluxe model on the loading dock will also save your company $1,000 per year in
heating costs because of its ability to seal better
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MARR = 12%
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Standard Model:
NPC = 16 × [1,600 + {1,600 × (P/F, 12%, 4)} + {1,600 × (P/F, 12%, 8)}] = $52,209
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Problem 5-8
Find the capitalized cost for the cash flows shown below
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DISREGARD
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200 – 100 (0
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3021) = 296
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93/0
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30
WHY IS IT WRONG?

Problem 5-9
You have saved $2,500 toward a new car and you believe that you can afford monthly payments of $250
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What is the most you can pay for a
car on this basis?
Solution:
Part (a):
i = 12%/12 = 1% per month
...
75
...
75% per month
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75%, 36) + 2,500 = $10,361
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Problem 6-1
Lubbock County is planning to construct a bridge across the Rio de Lubbock to facilitate afternoon skiing in the
El Dusto Ski Basin
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Annual maintenance and repairs
will amount to $25,000 for each of the first five years, to $30,000 for each of the next ten years and to $35,000
for each of the next five years
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Use an interest rate of 5% and determine the equivalent uniform annual cost for a 20-year period
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n = 20 Years
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Annual Maintenance Cost for the first five years, A1 = $25,000
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Annual Maintenance Cost from year 16 thro' 20, A3 = $35,000
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EUAC = [6,500,000 + 500,000 (P/F, 5%, 10)] (A/P, 5%, 20) +
25,000 +[{5000 (F/A, 5%, 5) + 5000(F/A, 5%, 15)} (A/F, 5%, 20)]
= [6,500,000 + 500,000 (0
...
0802) +
25,000 +[{5000 (5
...
579)}(0
...
39 + 29,092
...
25
Hint: This way of solving the problem converts the difference between 30K and 35K (5K) in years 16 through
20 into a future value @ year 20 and the difference between 25K and 30K (5K) for years 6-20 (15 years) into a
future value at year 20
...

There are other ways of solving
...

Solution:
Need to look at one life cycle of each alternative
...
A)
= {(2,000 – 1,000) + 3,000 (A/F, 9%, 4)}- 4,000 (A/P, 9%, 4)
= {1,000 +3,000(0
...
3155) = $384
...
B)
= {(2,200 – 500) + 1,000 (A/F,9%,10)}- 6,000 (A/P,9%,10)
= {1,700 +1,000(0
...
1627) = $786
...

Problem 6-2
Two types of power converters are under consideration for a specific application
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(b)
Determine a Salvage Value for the Beta system such that the Beta system will have an Equivalent
Uniform Annual Cost equal to the Alpha system
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Alpha
EUACalpha = {P (A/P, i, n)} + A - {S (A/F, i, n)}
= {10,000-(A/P, 20%, 5)} + 2,500 - {0 (A/F, 20%, 5)} = 3,344 + 2,500 - 0
= $5,844
...
Beta
EUACbeta = {P (A/P, i, n)} + A - {S (A/F, i, n)}
= {20,000 (A/P, 20%, 9)} + 1,200 - {5,000 (A/F, 20%, 9)}
= 4,962 + 1,200 - 240
...
50
...

(b) EUACalpha = EUACbeta
...

S (A/F, 20%, 9) = $318
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0481 = $6,611
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Its initial cost is $35,000 and it has a salvage value of $6,000 at the
end of its useful life
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It requires $500 annual operating, maintenance and repair cost and has a zero
salvage value after its 10-year useful life
...
Present the economic equivalence
function required, showing the functional notation and then the numerical value
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System Service Life = 30 Years
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OMR Cost = $1,000 per year
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Salvage Value = $6,000
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80 = $4,604
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Component #2: Life = 10 Years
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Purchase Price = $18,000
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EUAC = {18,000 (A/P, 7%, 10)} + 500 = 2,563
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20
...

Problem 6-4

A farmer is considering the installation of a fuel storage system that will save $0
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The farmer uses about 20,000 gallons per year
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The annual operating and maintenance costs will be nothing in the first year but will increase by $25
each year thereafter
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The farmer's cost of funds is 12%
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065/gallon × 20,000 gallons/year = $1,300/Year
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10

25
50

...
63 = -$388
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The farmer should not purchase the fuel storage system
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The contract provides for the purchase of material at the beginning of the year for five
consecutive years for the amounts shown below
...
24) + (5,000 × 4
...

EUAC = NPW × (A/P, 9%, 4) = 462,755 × 0
...
47
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The engineer must prepare an estimate of an operating budget for a ten-year period,
if the signals were installed
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The estimated energy cost to operate the signals for the first year is $1,000
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5% of the first year's energy cost
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What is
the EUAC for the operating budget? Salvage value at the end of the ten-year period is negligible
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G = 1
...

EUAC = {P (A/P, i, n)} + M&O + {G (A/G, i, n)}
= {47,000 (A/P, 8%, 10)} + 1,000 +{15 (A/G, 8%, 10)}
= (47,000 × 0
...
871)
= 7,003 + 1,000 + 58
...
07
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wants to evaluate two methods of shipping their products
...

Using a MARR of 15%, calculate the equivalent uniform annual cash flow (EUAB -EUAC) for each
alternative
...

Solution:
Method A:
EUAB-EUAC= [154,000+142,000(A/F, 15%, 10)][{700,000(A/P, 15%, 10)}+18,000 + 900 (A/G, 15%, 10)]
= 154,000 + 7,000
...
70] = $445
...

Method B:
EUAB -EUAC = [303,000 +210,000 (A/F, 15%, 10)] [1,512,000 (A/P, 15%, 10)}+ 9,000 + {775 (A/G, 15%, 10)}
= [303,000 + 10,353] - [301,341
...
85] = $389
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The most desirable choice is Method A
...

Year

Cash flow
0
1
2
3
4
5

- $20,000
+ 15,000
+ 11,000
- 18,000
+ 9,000
+ 9,000

6
+ 9,000
Use an external rate of return (iext) of 10% and calculate the internal rate of return (iint)
...

Year
Cash Flow
Modified Cash Flow
0
- $20,000
1
2
3
4
5
6

15,000
0
- 5,900
9,000
9,000
9,000

= 11,000 - 11,000
= -18,000+{11,000(F/P, 10%, 1)}

Invest enough of the $15,000 positive cash flow, occurring at year 1, at iext to reduce the remaining $5,900
negative cash flow at year 3 to $0
...
24 = 15,000 -{5,900 (P/F, 10%, 2)}
2
0
3
0
= - 5,900 + 5,900
4
9,000
5
9,000
6
9,000
NPW =-20,000 + {10,124
...

i
NPW
20%
$591
...
99

18% < iint < 20%
...
Use an external rate of return
(iext) of 12%
...

Year Cash Flow
Modified Cash Flow
0
-$80,000
1
35,000
2
0
= 15,000 - 15,000
3
4

-13,200
75,000

= - 30,000 + {15,000 + (F/P, 12%, 1)}

Invest enough of the $35,000 positive cash flow, occurring at year 1, at iext to reduce the remaining $13,200
negative cash flow at year 3 to $0
...
96

2
3
4

0
0
75,000

Modified Cash Flow
= 35,000 - {13,200×(P/F, 12%,
2)}
= - 13,200 + 13,200

NPW = -80,000 + {24,476
...

i
10%
7%
8%
Problem 7-3

NPW
-$6,523
...
67
-2,211
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John had a little business on the side when he was a student and did not have a girl friend
when he was at school and hence he was able to save the $10,000 for his dream car, He expects to take good
care of the car and expects to have at least $8,000 for the vehicle after 4 years and the dealership agrees to take
the car back for $8,000 at the end of 4 years
...

If the monthly payment is $400, what is the nominal interest rate on this loan?
What is the effective interest rate?
Solution:
(30,000 -10,000) = 400 (P/ A, i, 48) + 8,000 (P/F, i, 48)
By trail and error, the unknown interest has to be computed
...
974) + 8,000 (0
...
25%
20,000 = 400 (35
...
5509)
20,000 = 18,780 ? NO
Interpolating between 1% and 1
...
0277 %
Nominal interest rate = 1
...
3324%
Effective interest rate = (1 + 0
...
1305 or 13
...

Year
1

Cash Flow
-$15,000

2
3
4

3,043
...
20 (P/F, i, 1)} + {11,000 (P/F, i, 3)} + (13,000 (P/F, i, 4)}
...
50
-1,608
...
, has developed a new Thing-A-May-Jig at a cost of $2,000,000
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At the end of the fifth year, the productive assets associated with the Thing-A-May-Jig project will
be disposed of for $750,000
...

Solution:
Year
0
1
2

Cash Flow
- $2,000,000
300,000
400,000

3
4
5

500,000
300,000
1,000,000

NPW(i) = - 2,000,000 + {300,000 (P/F, i, 1)} + {400,000 (P/F, i, 2)} +
{500,000 (P/F, i, 3)} + {300,000 (P/F, i, 4)} + {1,000,000 (P/F, i, 5)}
...


Problem 7-11
A $5,000 face value industrial bond can be purchased for $4,920
...
What are the rate of return and the effective rate of return that the
purchaser can expect to receive if the bond is purchased?
Solution:
n= 8 Years
...

mA = 2
...


Face Value = $5,000
Purchase Price = $4,920
...
08/2 = $200
...

i

NPW

0%

$3,280
...
05

4%

79
...
5%

-200
...
5%
...

The effective interest rate (ie) = (1+i)m - 1
...
04)2 - 1 = 8
...
045)2 - 1 = 9
...

Problem 7-12
Barry, a Texas Crude Engineer who did not take Engineering Economy while studying at Tech, recommended
that Texas Crude purchase a special tool to reduce the cost of pumping oil out of the bayous of St
...
As a result of Barry's recommendation, Texas Crude purchased the tool for $30,000 on January 1, 1998
...
After going on line full
time, the tool saved Texas Crude $9,000 each year for the next three years and Barry was happy
...
It was scrapped as being unusable at the
end of year five and had a zero salvage value
...

Use a MARR of 10% and evaluate the effectiveness of the tool and the correctness of Barry's recommendation
...

i

NPW

0%

$6,000
...
50

7%

-401
...


Barry should have taken Engineering Economics so that he would have known what he was talking about
...
The first costs represent the expenses of rearranging the current layout to the alternative
new layout and the annual savings represent the reduction in the production costs of the new layout compared to
the current layout
...

ROR (A1) = 12,500 / 110,000 = 11
...
04%
Since both the alternatives have RoRs greater than MARR
...

Incremental cost = 115,000–110,000 = $5,000
Incremental benefit = 15,000–12,500 = $2,500
ROR = 2,500/5,000 = 0
...
A2
...

Year
0
1 thro’ 9
10

Cash Flow
-1,000
100
600

Solution:
1,000 = 100 (P/A, i%, 10) + 500 (P/F, i%, 10)
Try i = 10%
1,000=100 (6
...
3855) ?
=807
...
910) + 500 (0
...
6?No
Try i = 6%
1,000=100 (7
...
5584) ?
=1015
...
024) + 500 (0
...
55?No
i = 6+{(1015
...
2 - 956
...
35%
Problem 7-22
For the cash flow diagram shown below, compute the unknown interest rate to the second decimal place
...


$400
$350

$300

$250

$200

$150

0
1

2

3

4

5

6

$1,225

Solution:
1,225=400 (P/A, i, 6) - 50 (P/G, i, 6)
The ROR has to be computed by trial and error
...
Since there
is only one sign change, there is one unique interest rate to satisfy the relationship above
...
355) - 50 (9
...
8 ? No
i = 12%
1,225 = 400 (4
...
930) ?
= 1,197
...
4 ? No
i is between 10% and 12%
...

i = 10 + {(2) (1257
...
8 - 1197
...
10%
Problem 7-23

For the cash flow diagram given below, compute the rate of return
...

1,000= 100 (P/A, i%, 5) + 50 (P/G, i%, 5) + [250 (P/A, i%, 5)] [P/E, i%, s] [100 (P/A, i, 1) + 150 (P/i, i%, 2)]
Try i = 10%
1000 = 100 (3
...
862) + [250 (3
...
862)] [0
...
9091) + 150 (0
...
76 ? No
Try i = 9%
1000 = 100 (3
...
111) + [250 (3
...
111)] [0
...
9170) + 150 (0
...
5? No
Try i = 8%
1000 =100 (3
...
372) + [250 (3
...
372)] [0
...
9259) + 150 (0
...
25? No
Try i = 7%
1000 = 100 (4
...
647) + [250 (4
...
647)] [0
...
9346) + 150 (0
...
09? No
" i" is between 7% and 8%
...
09 -1000)/ (1026
...
25)} = 7
...
51 x 4 = 30

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