11.36 (a) Two options; (1) upgrade and retain defender for 3 years, or (2) buy challenger now
Select option 2; replace the defender now
(b) AW2 = $-126,957 for 3 years
11.37 All evaluations are performed at an effective i = 1% per month
(a) For 1–year study period
(b) For 2–year study period
AWK,2 = -160,000(A/P,1%,24) – 7000 + 40,000(A/F,1%,24)
(c) For 3–year study period, repurchase process K machine for 1 year after 24 months
AWK,3 = -160,000(A/P,1%,36) – 7000
11.38 Use the market value estimates in Example 11.3 (Figure 11–3) to calculate CR for n = 6
= $-9.36 ($–9.36 million)
n = 12 years: CR = –38(A/P,15%,12) + 1.06(A/F,15%,12)
Replacement Value
11.40 (a) Relations to determine RV or minimum trade–in value
11.41 (a) By hand:
–RV(A/P,8%,3) – 60,000 + 15,000(A/F,8%,3) = -80,000(A/P,8%,5)
(b) By spreadsheet:
11.42 By hand:
By spreadsheet:
Note that the RV value found by Goal Seek is positive, indicating that the trade–in value
is in favor of the seller of machine B, that is, machine A owner should pay the $21,950 to
get rid of A to purchase B.
11.43 –RV(A/P,15%,1) – 53,000 = -226,000(A/P,15%,10) – 48,000 + 60,000(A/F,15%,10)
18
Spreadsheet Exercises
11.44 (a) ESL is n = 5 years with an AW = $-45,363 per year
(b) ESL is now n = 7 years and the total AW curve is very flat between about n = 6 to 8,
making the ESL less sensitive to changes in estimates.
11.45 Spreadsheet verification below: ESL for defender is n = 1; select defender over
challenger with AWC = $-60,806
(a) Write the challenger PMT function in cell–reference format in cell F11
(b) Use Goal seek to force AWD,1 = $-60,806 (cell F5). Required trade-in/market value is
now $14,614.
11.46 Process M (upgrade current machinery) is economically better (cell B11)
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of McGraw–Hill Education.
20
Additional Problems and FE Exam Review Questions
11.47 Answer is (a)
11.48 Answer is (d)
Years Retained Total AW
11.55 -RV(A/P,10%,5) – 120,000 + 40,000(A/F,10%,5) = -670,000(A/P,10%,10) – 94,000
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of McGraw–Hill Education.
21
-RV(0.26380) – 120,000 + 40,000(0.16380) = -670,000(0.16275) – 94,000
+ 60,000(0.06275)
RV = $325,358
Answer is (c)
11.56 AW2 = -32,000(A/P,10%,2) – 24,000 + 14,000(A/F,10%,2)
11.57 Answer is (a)
Solution to Case Study, Chapter 11
Sometimes, there is not a definitive answer to a case study exercise. Here are example responses.
A PUMPER SYSTEM WITH AN ESL PROBLEM
1. The ESL is 13 years. Year 13 is predicted to require the 4th rebuild; the pump will not be used
beyond 13 years anyway.
2. Required MV = $1,420,983 found using Solver with F12 the target cell and B12 the
changing cell. This MV is well above the first cost of $800,000.
3. Solver yields the base AOC = $-201,983 in year 1 with increases of 15% per year.
The rebuild cost in year 4 (after 6000 hours) is $150,000. This AOC series is huge compared
to the estimated AOC of $25,000 (years 1 – 4).
4. Compare the results in #2 and #3 with that in #1 and comment on them.