Tech ID#
TEST #3 (Practice) Answer Key
Name
ES 316 Engineering Economics
Spring 2020
To solve questions, refer to the compound interest tables. You need to show the problem solving process.
Otherwise, your score will be zero.
PART I: Short Questions (10x4=40 points)
1. Find the expected EUAW from the financial data provided in the table below for a new equipment. Because of
the uncertainty of technology being used in this equipment, it has not been possible to get the initial cost accurately.
The annual benefit, however, is estimated to be $25,000 with a possible equipment life of 5 years. The salvage value
is expected to be 10% of the initial cost. MARR = 5%
First Cost, $
Probability
$60,000
0.25
$80,000
0.35
$100,000
0.30
$120,000
0.10
Expected first cost = [60,000 (0.25) + 80,000 (0.35) + 100,000 (0.30) + 120,000 (0.10) = $85,000
Expected salvage value = 85,000 (0.10) = $8,500
Expected EUAW = 25,000 – [85,000 (A/P, 5%, 5) − 8,500 (A/F, 5%, 5)]
= 25,000 – [85,000 (0.2310) − 8,500 (0.1810)] = $6903.5
2. A project has the following costs and benefits. What is the payback period?
Year
0
1-2
3
4-10
Costs
$65,000
$15,000
$5,000
Benefits
$50,000
$10,000 in each year
Costs equate to the benefits by end of year 8. Payback of 8 years.
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ES 316 Engineering Economics
Spring 2020
3. Jain Mart is to depreciate an asset bought for $500,000 using the SOYD method over a life of 8 years. If the
depreciation charges in year 3 was $80,000, determine the salvage value used in computing the depreciation charges
in year 3.
SOYD = 8 (8+1) / 2 = 36
Depreciation in year 3: (8 – 3 + 1) (500,000 – X) / 36 = 80,000
X = 500,000 - 80,000 (36 / 6) = $20,000
4. An automated assembly line is purchased for $300,000. The company has decided to use units-of-production
depreciation. At the end of 5 years, the line will be scrapped for an estimated $100,000. Using the following
information, determine the depreciation schedule for the assembly line.
Year
Production Level (units)
Depreciation
1
5,000
20,000
2
10,000
40,000
3
15,000
60,000
4
15,000
60,000
5
5,000
20,000
Total production = 50,000 units
UOP depreciation in year “n” = (Production in year n /50,000)*($300,000 - $100,000)
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ES 316 Engineering Economics
Spring 2020
PART II: Long Questions (20x3=60 points)
1. Data for four mutually exclusive alternatives are given in the table below. Assume a life of 10 years and a
MARR of 10%. Use the present worth index in private sector.
Initial Cost
EUAB
Salvage Value
Alt. A
Alt. B
Alt. C
$7,000
$1,500
$2,000
$5,000
$1,000
$500
$1,500
$500
$0
Alt. D
Do Nothing
$0
$0
$0
1.1. What is the ∆B /∆C ratio for the first increment, (C-D)?
ΔB/ΔC = (500 − 0) / [1500 (A/P, 10%, 10)]
= 500 / [1,500 (0.1627)] = 2.05
1.2. What is the ∆B /∆C ratio for the second increment, (B-C)?
ΔB/ΔC = [(1,000 − 500) + 500 (A/F, 10%, 10)] / [(5,000 − 1,500) (A/P, 10%, 10)]
= [500 + 500 (0.0627)] / [3,500 (0.1627)] = 0.93
1.3. What is the ∆B /∆C ratio for the third increment, (A-C)?
ΔB/ΔC = [(1,500 − 1,000) + 2,000 (A/F, 10%, 10)] / [(7,000 − 1,500) (A/P, 10%, 10)]
= [500 + 2,000 (0.0627)] / [5,500 (0.1627)] = 0.70
1.4. What is the best alternative using B/C ratio analysis?
The better alternative between” C” and “D” was found to be “C”.
On the next increment, it was found that “C” was better.
On the final increment, it was found that “C” was better. Therefore, “C” is the ultimate winner.
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ES 316 Engineering Economics
Spring 2020
2. A new equipment is being considered at a local company at a cost of $200,000. The operation and maintenance
costs of this equipment are estimated to be $20,000 per year. Salvage value is expected to be $40,000 at the end of
its useful life. The life of this equipment is estimated to vary anywhere from 5 to 7 years with the associated
probabilities as shown in the table below. Assume an interest rate is 10%.
Life, Years
Probability
5
0.3
6
0.4
7
0.3
2.1. what is the expected EUAC for this equipment?
EUAC for a 5-year life = 200,000 (A/P, 10%, 5) + 20,000 – 40,000 (A/F, 10%, 5)
= 200,000 (0.2638) + 20,000 – 40,000 (0.1638) = $66,208
EUAC for a 6-year life = 200,000 (0.2296) + 20,000 – 40,000 (0.1296) = $60,734
EUAC for a 7-year life = 200,000 (0.2054) + 20,000 – 40,000 (0.1054) = $56,864
Expected EUAC = 0.3 (66,208) + 0.4 (60,734) + 0.3 (56,864)
= $61,215.2
2.2. Determine the associated risk measure in this equipment investment in terms of standard deviation.
Standard deviation, σ = [EV (X2) – {EV(X)}2]1/2
= [0.3 (66,208)2 + 0.4 (60,734)2 + 0.30 (56,864)2 – {61,215.2}2]1/2 = $3,640.2
3. New Mexico Tech purchased a computer (5-year MACRS property) at a cost of $2,000.
3.1. Develop the MACRS percentage rates (rt) for the asset with a cost basis of $100. Note the switch from
DDB to SL can be done in Year 4.
Year
1
2
3
4
5
6
Calculation
(1/2) (2/5) (100 – 0) = 20.00
(2/5) (100 – 20.00) = 32.00
(2/5) (100 – 52.00) = 19.20
(2/5) (100 – 71.20) = 11.52
11.52
(1/2) (11.52) = 5.76
MACRS percentage rate (rt)
20.00
32.00
19.20
11.52
11.52
5.76
3.2. The computer was sold after 3 years for $700. Determine the depreciation recapture on this equipment.
The accumulated depreciation through year 3 = (0.20 + 0.32 + 0.192) (2,000) = $1,424
Book value at the end of year 3 = 2,000 – 1,424 = $576
Depreciation recapture = 700 – 576 = $124
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ES 316 Engineering Economics
Spring 2020
PART III: Extra Credit Questions (10x2=20 points)
1. A robot has just been installed at a cost of $81,000. It will have no salvage value at the end of its useful life.
Savings per year
$18,000
$20,000
$22,000
Probability
0.2
0.7
0.1
Useful life (years)
12
5
4
Probability
1/6
2/3
1/6
1.1 What is the joint probability distribution for savings per year and useful life?
1.2 What is the rate of return for most likely scenario? Use interpolation with the interest rates of 7% and 8%.
Most likely: NPW = 0 = -81,000 + 20,000 (P/A, i, 5)
i) Try i = 7%: NPW = -81,000 + 20,000 (4.100) = 1,000
ii) Try i = 8%: NPW = -81,000 + 20,000 (3.993) = -1,140
RoR = 7% + [1,000/(1,000 + 1,140)] (8% - 7%) = 7.47%
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ES 316 Engineering Economics
Spring 2020
2. The tree in Fig. P10-41 has probabilities after each chance node and PW values for each terminal node. What
decision should be made? What is the expected value?
At decision node D2 we must decide between Pick Z1 and Pick Z2 based on PW:
PW(C4) = (0.2) ($12,000) + (0.8) ($9,000) = $9,600
PW(C5) = (0.3) ($14,000) + (0.7) ($4,000) = $7,000
Pick Z1, greatest PW of $9,600
At decision node D1 we must decide between Pick X, Y, or Z based on PW:
PW(C1) = (0.2) ($15,000) + (0.5) ($12,000) + (0.3) ($9,000) = $11,700
PW(C2) = (0.4) ($16,000) + (0.6) ($8,000) = $11,200
PW(C3) = (0.3) (-$10,000) + (0.7) ($9,600) = $3,720
So, PW is greatest if we Pick X
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