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Chapter 11

Replacement and Retention Decisions

Solutions to Problems

1. Specific assumptions about the challenger are:
1. Challenger is best alternative to defender now and it will be the best for all
succeeding life cycles.
2. Cost of challenger will be same in all future life cycles.

2. The defender’s value of P is its fair market value. If the assetmust be updated or augmented, this cost is added to the first cost. Obtain market value estimates from expert appraisers, resellers or others familiar with the asset being evaluated.

11.3 The consultant’s (external or outsider’s) viewpoint is important to provide an unbiased analysis for both the defender and challenger, without owning or using either one.

11.4 (a) Defender first cost = bluebook value = 10,000 – 3,000 = \$7000

(b) Since the trade-in is inflated by \$3000 over market value (blue book value)

Challenger first cost = sales price – (trade-in value – market value)
= P – (TIV – MV)
= 28,000 – (10,000 – 7000)
= \$25,000

11.5 P = market value = \$350,000
AOC = \$125,000 per year
n = 2 years
S =\$5,000

11.6 (a) Now, k = 2, n = 3 years more. Let MVk = market value k years after purchase

P = MV2 = 400,000 – 50,000(2)1.4 = \$268,050
S = MV5 = 400,000 – 50,000(5)1.4 = \$-75,913
AOC = 10,000 + 100(k)3 for k = year 3, 4, and 5

k 3 4 5

años 1 2 3COA \$12,700 16,400 22,500

(b) In 2 years, k = 4, n = 1 since it had an expected life of 5 years. more.

P = MV4 = 400,000 –50,000(4)1.4 = \$51,780
S = MV5 = 400,000 – 50,000(5)1.4 = \$-75,913
AOC = 10,000 + 100(5)3 = \$22,500 for year 5 only
11.7 P = MV = 85,000 – 10,000(1) = \$75,000
AOC = \$36,500 +1,500k (k = 1 to 5)
n = 5 years
S = 85,000 – 10,000(6) = \$25,000

11.8 Set up AW equations for 1 through 5 years and solve by hand.

For n=1: Total AW1 = -70,000(A/P,10%,1) – 20,000 + 10,000(A/F,10%,1)
= -70,000(1.10) – 20,000 + 10,000(1.0)
= \$-87,000

For n=2: Total AW2 = -70,000(A/P,10%,2) – 20,000 +10,000(A/F,10%,2)
= \$-55,571

For n=3: Total AW3 = \$-45,127

For n=4: Total AW4 = \$-39,928

For n=5: Total AW5 = \$-36,828

Economic service life is 5 years with Total AW5 = \$-36,828

11.9 (a) Set up the spreadsheet using Figure 11-2 as a template and develop the cell formulas indicated in Figure 11-2 (a). The ESL is 5 years, as in Problem 11.8.[pic]

(b) On the same spreadsheet, decrease salvage by \$1000 each year,
and increase AOC by 15% per year. Extend the years to 10. The ESL is
relatively insensitive between years 5 and 7, but the conclusion is:

ESL = 6 years with Total AW6 = \$43,497
[pic]

11.10 (a) Set up AW relations for each year.

For n = 1: AW1 = -250,000(A/P,4%,1) – 25,000 + 225,000(A/F,4%,1)= \$- 60,000

For n = 2: AW2 = -250,000(A/P,4%,2) – 25,000 + 200,000(A/F,4%,2)
= \$- 59,510

For n = 3: AW3 = -250,000(A/P,4%,3) – 25,000 + 175,000(A/F,4%,3)
= \$- 59,029

For n = 4: AW4 = -250,000(A/P,4%,4) – 25,000 + 150,000(A/F,4%,4)
= \$- 58,549

For n = 5: AW5 = -250,000(A/P,4%,5) –25,000 + 125,000(A/F,4%,5)
= \$- 58,079

For n = 6: AW6 = -250,000(A/P,4%,6) – [25,000(P/A,4%,5) +
25,000(1.25)(P/F,4%,6)](A/P,4%,6) + 100,000(A/F,4%,6)
= \$- 58,556

AW values will increase, so ESL = 5 years with AW5 = \$-58,079.

No, the ESL is not sensitive since AW values are within a percent or two of each other for values of n...