Version A
Name_Key_____________________________
ME 353 ENGINEERING ECONOMICS
Final Exam – Sample
Scoring gives priority to the correct formulas. Numerical answers without the correct
formulas for justification receive no credit. Decisions without numerical justification
receive no credit. Interest tables and factor formulas are at the end of the exam.
1. (10 Points) A luxury apartment building project requires an investment of $1,250,000. The
building has 50 units. We expect the maintenance cost for the apartment building to be
$150,000 in the first year, $200,000 in the second year, and will continue to increase by
$50,000 per year in subsequent years. The cost to hire a manager for the building is estimated
to be $50,000 per year. After five years of operation the apartment building can be sold for
$750,000.
What is the annual rent per apartment unit that will provide a return on investment of 15%?
Assume the building will remain fully occupied during the five years.
The Annual Rent for the whole building is the amount that makes the NAW
= 0.
NAW = -1250(A/P, .15, 5) - 150 – 50(A/G, .15, 5) – 50
+ A + 750(A/F, .15, 5) = 0
A = 1250(A/P, .15, 5) + 200 + 50(A/G, .15, 5) – 750(A/F, .15, 5)
A = $547.8 thousand per year
A = $10,956 per apartment per year
Monthly rent is $913.
2. (10 Points) We use the cost of data of problem 1, but now we assume that each apartment unit
rents for $1000 per month. In this problem you are to include the effect of taxes. The tax rate
on net income per year is 30%. The building is depreciated with the straight-line method using
a tax life of 25 years and a tax salvage of zero. Tax on capital gains is 30%. The after tax
minimum acceptable rate of return is 12%. We repeat the other data from problem 1.
The building project requires an investment of $1,250,000. The building has 50 units. We
expect the maintenance cost for the apartment building to be $150,000 in the first year,
$200,000 in the second year, and will continue to increase by $50,000 per year in subsequent
years. The cost to hire a manager for the building is estimated to be $50,000 per year. After
five years of operation the apartment building can be sold for $750,000.
Show the after-tax cash flows in the table below. Also, write the formula for the after-tax net
present worth of this project. Use as few terms as possible and show time value of money
factors with the appropriate interest rates. You do not have to evaluate the formula.
Year
After Tax Cash
Flow
0
-1250
1
-295
2
-260
1
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Name_Key_____________________________
3
-225
4
-190
5
-980
Year
0
1
2
3
4
5
5 - salvage
BTCF
($1,250)
400
350
300
250
200
750
Depr
0
50
50
50
50
50
$1,000.00
TI
350
300
250
200
150
-250
Tax
0
105
90
75
60
45
-75
ATCF
($1,250)
$295.00
$260.00
$225.00
$190.00
$155.00
$825.00
NPW = -1250 + 295(P/A, .12, 5) - 35(P/G, .12, 5) + 980(P/F, .12, 5)
3. (10 Points) We use the cost of data of problem 1, but now we consider inflation. Do not
consider taxes. We first repeat the other data from problem 1.
These estimates of future values are estimated with today’s prices, however actual cash flows
escalate in the manner described in the next paragraph. The building project requires an
investment of $1,250,000. The building has 50 units. We expect the maintenance cost for the
apartment building to be $150,000 in the first year, $200,000 in the second year, and will
continue to increase by $50,000 per year in subsequent years. The cost to hire a manager for
the building is estimated to be $50,000 per year. After five years of operation the apartment
building can be sold for $750,000.
We assume that each apartment unit rents for $1000 per month. This rent amount is guaranteed
to the renters and is not affected by inflation during the five-year period. The maintenance and
management costs are expected to escalate in the future at the same rate as general inflation. The
resale value of the building will escalate at a 20% rate. The general inflation rate is expected to
be 6% per year. The after minimum acceptable rate of return is 18%. This is the market rate of
return.
Show the cash flows to be used for an analysis in the table below. Indicate whether the cash
flows are in real or actual dollars. Also, write the formula for the net present worth of this
project. Use as few terms as possible and show time value of money factors with the
appropriate interest rates. You do not have to evaluate the formula.
Kind of dollars in the cash flow: ___Actual______________
Year
Cash Flow
0
-1250
1
388
2
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Name_Key_____________________________
2
319
3
243
4
158
5
1931
Year
0
1
2
3
4
5
5
Rent
0
600
600
600
600
600
Maint/Man
Capital
-1250
200
0
250
0
300
0
350
0
400
0
750
total 5
Actual
-1250
388
319
243
158
65
1866
1931
Real
-1250
366
284
204
125
48
1395
1443
ic = 0.18 Compute ir = (0.18 – 0.06)/(1 + 0.06) = 11.32 %
NPW = -1250 + 600(P/A, ic, 5) – 200(P/A, ir, 5) – 50((P/G, ir, 5)
+ 750(1 + .2)5(P/F, ic, 5)
or use the cash flow computed in the table. If actual cash flows are used,
compute NPW with ic. If real dollars are used, compute NPW with ir.
4. (12 Points) You are considering purchasing a bond with a face value of $1000. The company
issuing the bond will pay you an annual interest payment of 10% of the face value. The next
interest payment will occur one year from now. The bond matures in 9 years at which time you
will receive the face value of the bond.
a. What is the most that you should pay for the bond if your minimum acceptable rate of return
(MARR) is 15%?
The cash flow is -P, the cost of the bond, + 1000 per uear as the interest
payments, and +1000 at the end of nine years. The appropriate
purchase cost is the value of P that makes the present worth equal to
zero.
P = 100(P/A, .15, 9) + 1000(P/F, .15, 9)
P = 100(P/A, .15, 9) + 1000(P/F, .15, 9) = 100*4.772 + 1000*0.2843
= 477.2 + 284.3 = $761.5
b. Will the purchase price change up or down if the MARR decreases? Explain your answer.
The purchase price will increase as the MARR decreases. This occurs
because the present worth of the benefits received in the future
increase in value as the interest rate decreases.
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Name_Key_____________________________
c. In addition to the information above, you expect an inflation rate of 7%. The payments from the
bond are fixed and are not affected by inflation. You require a real rate of return of 15%. Write
a formula with which he can compute the purchase price of the bond in this case. Show
numerical interest rates in your formula.
The cash flow is in actual dollars. To evaluate P we must use u the interest
rate adjusted for inflation.
u = 0.15 + 0.07 + 0.15*0.07 = 0.15 + 0.07 + 0.011 = 0.231
P = 100(P/A, 0.231, 9) + 1000(P/F, 0.231, 9)
5. (12 Points) A company is considering producing a new product. The manufacturing process
can either be based on manual operations or robotic machines. The manual option costs
nothing to install, has operating cost of $300,000 per year and will last indefinitely. The robotic
option costs $1,000,000 to install, has no annual operating cost, but will last only 3 years. The
manual option has no salvage value, but the robotic option has a salvage value of $500,000
when you sell it after 3 years. Both methods produce the product equally well. Neglect
inflation for this problem.
Do an after tax analysis to help the company make the right choice. The robotic option is
depreciated with the ACRS method using the 3-year class. Depreciation percentages are shown
below. The tax rate is 40%. The company’s after tax MARR is 15% per year.
Year
1
2
3
4
Percent
33.33%
44.45%
14.81%
7.41%
Find the after tax cash flows for the two alternatives. Find the NPW and
then find the NAW. Compare.
For the Manual process:
Operating cost = 300, After tax operating cost is 300*.6 = 180/year.
Construct a table for the Robotic operation.
Year
0
1
2
3
3 - salvage
BTCF
-$1,000
0
0
0
500
Depr
0
$333.30
$444.50
$148.10
$74.10
TI
-333.3
-444.5
-148.1
425.9
Tax
0
-133.32
-177.8
-59.24
170.36
total 3
ATCF
-$1,000
$133.32
$177.80
$59.24
$329.64
$388.88
NPW = -1000 + 133(P/F, .15, 1) + 178(P/F,.15, 2) + 389(P/F, .15, 3) = 494
NAW = 493(A/P, .15, 3) = 216
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Name_Key_____________________________
Choose the manual process
6. (10 Points) You have three options to perform some function. The required investment and
annual costs are shown below. Use a rate of return analysis to select the best. All three options
have 0 salvage. Your MARR is 20%. Compute rates to nearest interest rates given in the tables
at the end of the test.
Option
Investment
Annual Cost
Life
A
250
50
5
B
300
55
10
C
215
60
5
Use C as the defender. Challenger is A.
Compute ROR A - C
NAWA - NAWC = -35(A/P, i, 5) + 10 = 0. (A/P, i, 5) = 10/35 = 0.286
Looking in the 5-year table we find A/P to between 12% and 15%.
Reject A - C:
Use B as the challenger.
NAWB – NAWC = -300(A/P, i, 10) -55 – [215(A/P, i. 5) – 60]
-300(A/P, i, 10) + 215(A/P, i, 5) + 5 = 0. Trial and error yields 26%
(between 25% and 30%%).
Accept B.
7. (12 Points) A chemical plant stores raw materials, finished goods and intermediate products in
tanks. The materials are pumped to and from the various processing operations through pipes.
There are several pumps in the system. The original investment in the pumps was $500,000
when they were purchased 3 years ago. Unfortunately, there must be a design problem because
the pumps are leaking. Last year $180,000 was spent to replace seals and to clean up leaked
materials at the pumps.
The plant management is tired of the expense of pump repair and is considering two
alternatives. The first alternative is to extensively rebuild the pumps. Management likes this idea
because the original cost of the pumps will not have been entirely wasted. The renovation cost
for the pumps will be $200,000. If this is done the maintenance cost will be reduced to $60,000
next year. If the renovated pumps are kept for a second year, the maintenance cost will increase
to $120,000. For the third year, the maintenance cost will be back to $180,000. The renovated
pumps will last no more than 3 years. The scrap value of the pumps is $100,000 if they are
removed now or at any time in the future.
The second alternative is to replace the pumps with an advanced design with gas seals. The cost
to purchase and install the gas seal pumps and the gas distribution system that they require is
$600,000. The annual cost of maintaining the gas system is $40,000. Its salvage value is zero.
The economic life of the new pump system is five years.
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Name_Key_____________________________
Using a 15% MARR make the most economic decision for the plant. Should the pumps be
renovated or replaced? Justify your conclusion.
Find the NAC of the new system, the challenger.
NAC = 600(A/P, .15, 5) + 40 = 218
Find the NAC of the renovated system, the defender. The investment in the
defender is 100 (the scrap value) + 200 (the renovation cost) = 300
NAC(1) = 300(A/P, .15, 1) +60 - 100 = 305
NPC(2) = 300 +60(P/F, .15, 1) + 120(P/F, .15, 2) - 100(P/F, .15, 2) = 367
NAC(2) = 367(A/P, .15, 2) = 225
NPC(3) = 300 +60(P/F, .15, 3) + 60(P/G, .15, 3) - 100(P/F, .15, 3) = 495.5
NAC(3) = 495.5 (A/P, .15, 3) = 217
Defender wins. Keep it for another year.
8.
(12 Points) You own a site at which there is the possibility of discovering oil. Your options
are:
q a1: drill for oil yourself,
q a2: lease the site to someone else to drill,
q a3: lease the site, but maintain an interest in the results.
There are four possible outcomes regarding the success of the well.
q q1: 600,000 barrel well,
q q2: 400,000 barrel well,
q q3: 100,000 barrel well,
q q4: dry hole.
Your net present worth of the profit depends on which development option you choose and the
results of the drilling.
NPW Profit
Well success
q1
q2
q3
q4
a1
800
400
-50 -100
Option
a2
60
60
60
60
a3
300
100
0
0
Based on information about the site we make estimates of probabilities of the four results
P(q1) = 0.1, P(q2) = 0.15, P(q3) = 0.25, P(q4) = 0.5.
a.
Compute the expected value of the three drilling options. In order to maximize your expected
value, which option should you choose?
E(Profit|a1) = 77.5, E(Profit|a2) = 60, E(Profit|a3) = 45. The best decision
is to drill the well yourself.
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b.
Name_Key_____________________________
Compute the standard deviations of the option that you chose in part a and the option with the
second highest expected value? Comment on what role the standard deviation might play in
your decision process.
Option
a1
a2
a3
Var.
87619
0
8475
Std. Dev.
296
0
92
Only two of these need be computed. The standard deviation is much
higher for the drill-it-yourself option. This might cause you to choose
the leasing option which has 0 standard deviation.
9. (12 Points) The questions in this
problem use the cash flow shown at
the right. When formulas are
requested in the question, use as few
factors as possible.
300
0
1
600
2
900
3
1200
4
5
-1200
a. What is the NAW of the cash flow
for a 4-year study period when the
MARR is 0%?
Sum up the cash flows and divide by 4.
sum = 1800. A = 1800/4 = 450.
b. Write the formula for the NAW. Use
i as the interest rate in the formula.
NAW = -1200(A/P, i, 4) + 300 +
300(A/G, i, 4)
c. We compute the NAW of the cash
flow using i = 30% to be $99.53. Now
we want to analyze the cash flow with
inflation.
Say the general inflation rate is 10% and
the cash flow above is expressed in real
dollars. The value of ir = 30%. Write the
formula for the NPW expressed in
actual dollars. Use a numerical interest
rate in your formula.
NPW = 99.53(P/A, .3, 4)
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rate in your formula.
d. The $1200 value at time 0 is an
investment. We depreciate the investment
using the SYD method with a tax life of
4 years and a tax salvage of 0. Write the
formula for the NAW of the after tax
cash flows. The tax rate is 50%. Use i as
the after-tax MARR.
Name_Key_____________________________
D1 = (4/10)1200 = 480
AT1 = 300 – (300-480).5 = 390
D2 = (3/10)1200 = 360
AT2 = 600 – (600 – 360).5 = 480
The gradient is 90.
NAW = -1200(A/P, i, 4) + 390
+ 90(A/G, i, 4)
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Compound Interest Factors
Single Payment Compound
Amount Factor
Single Payment Present
Worth Factor
Uniform Series Compound
Amount Factor
Uniform Series Sinking
Fund Factor
Uniform Series Present
Worth Factor
(F/P, i, n) = (1 + i)n
1
1
= (F/P,!i,!n)!
n
(1!+!i)
(1!+!i)n!–!1
(F/A, i, n) =
i
i
1
(A/F, i, n) =
=
(1!+!i)n!–!1 (F/A,!i,!n)!
(P/F, i, n) =
(1!+!i)n!–!1
i(1!+!i)n
i(1!+!i)n
1
(A/P, i, n) =
=
(P/A,!i,!n)!
(1!+!i)n!–!1
(P/A, i, n) =
Uniform Series Capital
Recovery Factor
Arithmetic Gradient Present
Worth Factor
(1!+!i)n!–!in !!–!1
i2(1!+!i)n
(1!+!i)n!–!in !!–!1
(A/G, i, n) =
i(1!+!i)n!–!i
(P/G, i, n) =
Arithmetic Gradient Uniform
Series Factor
Factor Table for 10 and 5 periods at various interest rates
Interest
Periods
F/P
10
P/F
A/F
A/P
F/A
P/A
A/G
P/G
10
Interest
0.00%
1.000
1.0000
0.1000
0.1000
10.000
10.000
4.500
45.000 0 . 0 0 %
0.25%
1.025
0.9753
0.0989
0.1014
10.113
9.864
4.479
44.184 0 . 2 5 %
0.50%
1.051
0.9513
0.0978
0.1028
10.228
9.730
4.459
43.386 0 . 5 0 %
0.75%
1.078
0.9280
0.0967
0.1042
10.344
9.600
4.438
42.606 0 . 7 5 %
1%
1.105
0.9053
0.0956
0.1056
10.462
9.471
4.418
41.843
1%
2%
1.219
0.8203
0.0913
0.1113
10.950
8.983
4.337
38.955
2%
3%
1.344
0.7441
0.0872
0.1172
11.464
8.530
4.256
36.309
3%
4%
1.480
0.6756
0.0833
0.1233
12.006
8.111
4.177
33.881
4%
5%
1.629
0.6139
0.0795
0.1295
12.578
7.722
4.099
31.652
5%
6%
1.791
0.5584
0.0759
0.1359
13.181
7.360
4.022
29.602
6%
7%
1.967
0.5083
0.0724
0.1424
13.816
7.024
3.946
27.716
7%
8%
2.159
0.4632
0.0690
0.1490
14.487
6.710
3.871
25.977
8%
9%
2.367
0.4224
0.0658
0.1558
15.193
6.418
3.798
24.373
9%
10%
2.594
0.3855
0.0627
0.1627
15.937
6.145
3.725
22.891
10%
12%
3.106
0.3220
0.0570
0.1770
17.549
5.650
3.585
20.254
12%
15%
4.046
0.2472
0.0493
0.1993
20.304
5.019
3.383
16.979
15%
18%
5.234
0.1911
0.0425
0.2225
23.521
4.494
3.194
14.352
18%
20%
6.192
0.1615
0.0385
0.2385
25.959
4.192
3.074
12.887
20%
25%
9.313
0.1074
0.0301
0.2801
33.253
3.571
2.797
9.987
25%
30%
13.786
0.0725
0.0235
0.3235
42.619
3.092
2.551
7.887
30%
35%
20.107
0.0497
0.0183
0.3683
54.590
2.715
2.334
6.336
35%
40%
28.925
0.0346
0.0143
0.4143
69.814
2.414
2.142
5.170
40%
45%
41.085
0.0243
0.0112
0.4612
89.077
2.168
1.973
4.277
45%
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Version A
50%
Interest
Name_Key_____________________________
57.665
Periods
F/P
0.0173
0.0088
0.5088 113.330
P/F
A/F
A/P
1.965
1.824
3.584
5
F/A
P/A
A/G
50%
5
Interest
P/G
0.00%
1.000
1.0000
0.2000
0.2000
5.000
5.000
2.000
10.000 0 . 0 0 %
0.25%
1.013
0.9876
0.1990
0.2015
5.025
4.963
1.995
9.901 0 . 2 5 %
0.50%
1.025
0.9754
0.1980
0.2030
5.050
4.926
1.990
9.803 0 . 5 0 %
0.75%
1.038
0.9633
0.1970
0.2045
5.076
4.889
1.985
9.706 0 . 7 5 %
1%
1.051
0.9515
0.1960
0.2060
5.101
4.853
1.980
9.610
1%
2%
1.104
0.9057
0.1922
0.2122
5.204
4.713
1.960
9.240
2%
3%
1.159
0.8626
0.1884
0.2184
5.309
4.580
1.941
8.889
3%
4%
1.217
0.8219
0.1846
0.2246
5.416
4.452
1.922
8.555
4%
5%
1.276
0.7835
0.1810
0.2310
5.526
4.329
1.903
8.237
5%
6%
1.338
0.7473
0.1774
0.2374
5.637
4.212
1.884
7.935
6%
7%
1.403
0.7130
0.1739
0.2439
5.751
4.100
1.865
7.647
7%
8%
1.469
0.6806
0.1705
0.2505
5.867
3.993
1.846
7.372
8%
9%
1.539
0.6499
0.1671
0.2571
5.985
3.890
1.828
7.111
9%
10%
1.611
0.6209
0.1638
0.2638
6.105
3.791
1.810
6.862
10%
12%
1.762
0.5674
0.1574
0.2774
6.353
3.605
1.775
6.397
12%
15%
2.011
0.4972
0.1483
0.2983
6.742
3.352
1.723
5.775
15%
18%
2.288
0.4371
0.1398
0.3198
7.154
3.127
1.673
5.231
18%
20%
2.488
0.4019
0.1344
0.3344
7.442
2.991
1.641
4.906
20%
25%
3.052
0.3277
0.1218
0.3718
8.207
2.689
1.563
4.204
25%
30%
3.713
0.2693
0.1106
0.4106
9.043
2.436
1.490
3.630
30%
35%
4.484
0.2230
0.1005
0.4505
9.954
2.220
1.422
3.157
35%
40%
5.378
0.1859
0.0914
0.4914
10.946
2.035
1.358
2.764
40%
45%
6.410
0.1560
0.0832
0.5332
12.022
1.876
1.298
2.434
45%
50%
7.594
0.1317
0.0758
0.5758
13.188
1.737
1.242
2.156
50%
10
Version A
Name_Key_____________________________
Factor Table for an interest rate of 15%.
Interest
n
Rate
F/P
15%
P/F
A/F
A/P
F/A
P/A
A/G
P/G
15%
n
1
1.150
0.8696
1.0000
1.1500
1.000
0.870
0.000
0.000
1
2
1.323
0.7561
0.4651
0.6151
2.150
1.626
0.465
0.756
2
3
1.521
0.6575
0.2880
0.4380
3.473
2.283
0.907
2.071
3
4
1.749
0.5718
0.2003
0.3503
4.993
2.855
1.326
3.786
4
5
2.011
0.4972
0.1483
0.2983
6.742
3.352
1.723
5.775
5
6
2.313
0.4323
0.1142
0.2642
8.754
3.784
2.097
7.937
6
7
2.660
0.3759
0.0904
0.2404
11.067
4.160
2.450
10.192
7
8
3.059
0.3269
0.0729
0.2229
13.727
4.487
2.781
12.481
8
9
3.518
0.2843
0.0596
0.2096
16.786
4.772
3.092
14.755
9
10
4.046
0.2472
0.0493
0.1993
20.304
5.019
3.383
16.979
10
11
4.652
0.2149
0.0411
0.1911
24.349
5.234
3.655
19.129
11
12
5.350
0.1869
0.0345
0.1845
29.002
5.421
3.908
21.185
12
13
6.153
0.1625
0.0291
0.1791
34.352
5.583
4.144
23.135
13
14
7.076
0.1413
0.0247
0.1747
40.505
5.724
4.362
24.972
14
15
8.137
0.1229
0.0210
0.1710
47.580
5.847
4.565
26.693
15
16
9.358
0.1069
0.0179
0.1679
55.717
5.954
4.752
28.296
16
17
10.761
0.0929
0.0154
0.1654
65.075
6.047
4.925
29.783
17
18
12.375
0.0808
0.0132
0.1632
75.836
6.128
5.084
31.156
18
19
14.232
0.0703
0.0113
0.1613
88.212
6.198
5.231
32.421
19
20
16.367
0.0611
0.0098
0.1598 102.444
6.259
5.365
33.582
20
21
18.822
0.0531
0.0084
0.1584 118.810
6.312
5.488
34.645
21
22
21.645
0.0462
0.0073
0.1573 137.632
6.359
5.601
35.615
22
23
24.891
0.0402
0.0063
0.1563 159.276
6.399
5.704
36.499
23
24
28.625
0.0349
0.0054
0.1554 184.168
6.434
5.798
37.302
24
25
32.919
0.0304
0.0047
0.1547 212.793
6.464
5.883
38.031
25
26
37.857
0.0264
0.0041
0.1541 245.712
6.491
5.961
38.692
26
27
43.535
0.0230
0.0035
0.1535 283.569
6.514
6.032
39.289
27
28
50.066
0.0200
0.0031
0.1531 327.104
6.534
6.096
39.828
28
29
57.575
0.0174
0.0027
0.1527 377.170
6.551
6.154
40.315
29
30
66.212
0.0151
0.0023
0.1523 434.745
6.566
6.207
40.753
30
31
76.144
0.0131
0.0020
0.1520 500.957
6.579
6.254
41.147
31
32
87.565
0.0114
0.0017
0.1517 577.100
6.591
6.297
41.501
32
33
100.700
0.0099
0.0015
0.1515 664.666
6.600
6.336
41.818
33
34
115.805
0.0086
0.0013
0.1513 765.365
6.609
6.371
42.103
34
35
133.176
0.0075
0.0011
0.1511 881.170
6.617
6.402
42.359
35
36
153.152
0.0065
0.0010
0.1510 1014.35
6.623
6.430
42.587
36
40
267.864
0.0037
0.0006
0.1506 1779.09
6.642
6.517
43.283
40
48
819.401
0.0012
0.0002
0.1502 5456.00
6.659
6.608
44.00
48
50
1083.7
0.0009
0.0001
0.1501 7217.72
6.661
6.620
44.10
50
inf.
inf.
0.0000
0.0000
0.1500
6.6667
6.667
44.44
inf.
inf.
11