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Solutions to end–of–chapter problems
Engineering Economy, 7th edition
Leland Blank and Anthony Tarquin
Chapter 11
Replacement and Retention Decisions
11.1 In taking a non-owner’s viewpoint, the analysis is done from the perspective of someone
who does not own any of the assets under consideration. This means that in order to
11.2 BV3 = 100,000 – 3(20,000) = $40,000
11.3 (a)x This type of thinking is improperly penalizing the challenger (Dodge Charger)
because he wants that deal to make up for the past bad investment he made in buying
(1) The services provided are needed for the indefinite future.
(2) The challenger is the best available challenger now and in the future. When this
11.5 P = market value = $39,000
11.6 (a) P = 90,000 – 8000(2) = $74,000
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11.7 P = 7000 + 17,000 = $24,000
11.8 AW1 = -10,000(A/P,10%,1) – 1000 + 7000(A/F,10%,1) = $-5000
AW2 = -10,000(A/P,10%,2) – 1000(P/F,10%,1)(A/P,10%,2)
11.9 (a) Find total AW for each year of ownership
AW1 = -345,000(A/P,10%,1) – 148,000 + 140,000(A/F,10%,1) = $-387,500
AW2 = -345,000(A/P,10%,2) – 148,000 + 140,000(A/F,10%,2) = $-280,119
11.10 For P: 18,899 = P(A/P,10%,3)
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11.11 Amortization of a $70,000,000 investment at 8% per year is constant at $-5,600,000.
Therefore, only consider maintenance cost:
AW1 = -83,000(A/F,8%,1) = $-83,000
11.12 AW1 = -65,000(A/P,10%,1) – 50,000 + 30,000(A/F,10%,1) = $-91,500
AW2 = -65,000(A/P,10%,2) – [50,000 + 10,000(A/G,10%,2)] + 30,000(A/F,10%,2)
11.13 (a) Use 1 year and AW of first cost P
-88,000 = -80,000(A/P,i,1)
11.14 AW1 = -70,000(A/P,12%,1) – 75,000 + 59,500(A/F,12%,1) = $-93,900
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11.15 (a) Solution by hand using regular AW computations
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40,000
100,000
5
20,000
110,000
6
120,000
7
130,000
AW6 = $-130,608
AW7 = $-130,552
(b) Spreadsheet screen shot utilizes the annual marginal costs to determine that ESL is 3
years with AW = $-127,489.
Year
Salvage
Value, $
AOC, $
per year
1
100,000
70,000
2
80,000
80,000
3
60,000
90,000
5
11.16 Set up AW equations for n = 1 through 7 and solve by hand.
AW1 = -100,000(A/P,14%,1) – 28,000 + 75,000(A/F,14%,1)
= $-67,000
11.18 (a) The three estimate changes are made in the spreadsheet: increase to $4
million for heating element exchange in year 5; market value retention of
only 50% starting with year 5; and, increases of 25% per year in maintenance
cost starting in year 5.
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(b) ESL has decreased from 12 to 8 or 9 years (about a 25 to 33% decrease); AW of costs
11.19 (a) If the year is nD, replace the defender, (b) if the year is not nD, retain the
11.20 (a) Purchase the challenger today because its AW of $-48,000 is lower
11.21 AWD = –(100,000 + 20,000)(A/P,20%,4) + 40,000(A/F,20%,4)
= -120,000(0.38629) + 40,000(0.18629)
11.22 AWD = -(9000 + 25,000)(A/P,10%,3) – 47,000 + 22,000(A/F,10%,3)
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11.23 AWC = -26,000(A/P,10%,5) –1200 + 8000(A/F,10%,5)
11.24 AWD = -25,000(A/P,15%,5) – 180,000
= $-187,458
11.25 AWD1 = -(8000 + 43,000)(A/P,10%,1) – 22,000 + 8000(A/F,10%,1)
= $-70,100
11.26 Defender estimates have changed; determine the ESL for the defender
AWD1 = -50,000(A/P,10%,1) – 37,000 + 10,000(A/F,10%,1)
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11.27 AWD = -25,000(A/P,10%,1) -15,000 + 14,000(A/F,10%,1)
11.28 Find AW of defender for keeping one or two more years and compare against
AW of challenger
AWD1 = -54,000(A/P,10%,1) -23,000 + 40,000(A/F,10%,1)
11.29 Determine cost of keeping defender one, two, or three more years and compare to
cost of challenger:
AWD1 = -30,000(A/P,10%,1) – 24,000 + 25,000(A/F,10%,1)
= -30,000(1.10) -24,000 + 25,000
11.30 AWD = -(50,000 + 200,000) (A/P,12%,3) + 40,000(A/F,12%,3)
11.31 Use Goal Seek to find the breakeven defender cost of $149,154. With the appraised
This is a maximum; any amount less than $99,154 will indicate selection of the upgraded
current system.
11.32 (a) By hand: Find ESL of the defender; compare with AWC over 5 years.
AWD1 = -8000(A/P,15%,1) – 50,000 + 6000(A/F,15%,1)
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AWC = -125,000(A/P,15%,5) – 31,000 + 10,000(A/F,15%,5)
(b) By spreadsheet: In order to obtain the defender ESL of 1 year, first enter market values
for each year in column B and AOC estimates in column C. Columns D determines
annual CR using the PMT function, and AW of AOC values are calculated in column
Select the defender now and replace it after one year.
11.33 The “opportunity” refers to the ability to receive money by selling the defender. In
11.34 The cash flow approach subtracts the market value of the defender from the first cost of
the challenger before amortizing the cost of the challenger.
It is not a good idea to do this approach because:
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(2) By subtracting the defender market value from the first cost of the challenger, the
11.35 There are four possibilities:
1. Keep the defender for 3 years
The PW cost for each scenario is as follows:
11.36 (a) PW C for 5 years = $-149,000
(b) The PW values are placed in the year cell prior to when the year starts for
11.37 (a) AWD = -17,000(A/P,10%,3) – 8000 + 9000(A/F,10%,3)
Keep the defender
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11.38 AWX = -82,000(A/P,15%,2) – 30,000 + 42,000(A/F,15%,2)
11.39 (a) AWD = -(70,000 + 40,000)(A/P,15%,3) – 85,000 + 30,000(A/F,15%,3)
(b) n = 3 years: CR = -220,000(A/P,15%,3) + 50,000(A/F,15%,3)
11.40 (a) For 2–year study period
AWK = -165,000(A/P,12%,2) – 69,000 + 40,000(A/F,12%,2)
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(b) For 3-year study period, must re-purchase K for only 1 year.
AWK = -165,000(A/P,12%,3) – 69,000
11.41 In $ million units, use the market value estimates in Example 11.3 (Figure 11–3) to
calculate CR for n = 6 and n = 12 years for the challenger GH.
11.42 (a) There are 6 options. Spreadsheet screen shot shows the AW of the current system
(defender D) for its retention period with close-down cost in last year, followed
11.43 -RV(A/P,12%,3) – 27,000 + 30,000(A/F,12%,3) = -400,000(A/P,12%,5)
11.44 -RV(A/P,12%,3) – 63,000 + 25,000(A/F,12%,3) = -130,000(A/P,12%,6) – 32,000
11.45 –RV(A/P,10%,2) – 75,000 = -220,000(A/P,10%,6) – 49,000 + 30,000(A/F,10%,6)
11.46 -RV(A/P,12%,3) – [140,000 + 2000(A/G,12%,3)] = -150,000(A/P,12%,8)
11.51 For a 3-year period, AWD = $-70,000 and AWC = $-75,000. Do not replace.
Four options are present, but they have the same conclusion.
Years kept
AW per year, $1000
11.52 For any time during the next 3, 4 or 5 years, the lowest AW of $-65,000 per year will
11.54 The company should never purchase the challenger, because its AW of $-86,000 is higher
than the defender’s 2–year ESL of $-81,000. The defender should be kept for 2 more years
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11.55 Defender: ESL is 2 years with AW = $-13,700
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Solution to Case Study, Chapter 11
Sometimes, there is not a definitive answer to a case study exercise. Here are example responses.
WILL THE CORRECT ESL PLEASE STAND?
2. Required MV = $1,420,983 found using Solver with F12 the target cell and B12 the
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3. Solver yields the base AOC = $-201,983 in year 1 with increases of 15% per year.
4. Compare the results in #2 and #3 with that in #1 and comment on them.