Engineering Economy, 8th edition
Leland Blank and Anthony Tarquin
Chapter 11
Replacement and Retention Decisions
Foundations of Replacement
11.3 The defender’s value of P is its fair market value. If the asset must be updated or
11.5 (a) Yes, as long as assets similar to the ones under comparison (including the in–
11.6 P = market value = 85,000 – 10,000(1)
11.7 Defender:
Challenger:
P = $170,000
11.8 P is market value after 2 years: P = 40,000 – 2(3000) = $34,000
11.9 (a) P = 90,000 – 8000(2) = $74,000
Economic Service Life
11.10 (a) The ESL of the defender is 3 years with the lowest AW of $–85,000
11.11 ESL occurs when the AW is the lowest cost.
11.12 Add AW amounts; select lowest AW
Years Retained Total AW
Years Retained Total AW, $
(b) 14,968 = S(A/F,10%,2)
11.14 (a) By hand:
11.15 (a) By hand:
AW1 = -180,000(A/P,15%,1) – 84,000 + 45,000(A/F,15%,1)
= -180,000(1.1500) – 84,000 + 45,000(1.0000)
(b) By spreadsheet: Graph shows CR, AOC and total AW
ESL is 5 years with an AW = $-135,934 per year
11.16 AW1 = -70,000(A/P,12%,1) – 75,000 + 59,500(A/F,12%,1) = $-93,900
11.17 (a) Solution by hand using regular AW computations
Year
Salvage
Value, $
AOC, $
per year
AW3 = $-127,489
(b) Spreadsheet below utilizes the annual marginal costs to determine that ESL is 3 years
with AW = $-127,489.
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.
Results are significantly different. ESL is now 8 or 9 years, with a flat AW curve
for several years.
(b) ESL has decreased from 12 to 8 or 9 years (about a 25 to 33% decrease); AW of costs
Replacement Study
11.19 In chapter 6, neither asset is currently owned. Here, one is presently in place.
11.20 Determine the ESL and AW for the new challenger. If defender estimates changed,
11.23 (a) Purchase the challenger at the end of year 2, when its AW will be lower than that of
11.24 (a) AWD = – (12,000 + 25,000)(A/P,10%,3) – 48,000 + 19,000(A/F,10%,3)
11.25 AWC = -80,000(A/P,12%,3) – 19,000 + 10,000(A/F,12%,3)
AWD = -20,000(A/P,12%,3) -15,000 – 31,000 + 9000(A/F,12%,3)
= -20,000(0.41635) -15,000 – 31,000 + 9000(0.29635)
11.26 (a) AWD = -(7000 + 22,000)(A/P,10%,3) – 27,000 + 12,000(A/F,10%,3)
(b) Find AWC over 2 years and compare with AWD over 3 years
AWC = -65,000(A/P,10%,2) – 14,000 + 23,000(A/F,10%,2)
11.27 Find defender ESL; compare with AWC = $-97,000
AWD,1 = -37,000(A/P,10%,1) – 85,000 + 30,000(A/F,10%,1) –
11.28 No option for retention of defender D beyond one year
AWD = -9000(A/P,10%,1) – 192,000
= -320,000(0.31547) – 68,000 + 50,000(0.21547)
11.29 AWD = -25,000(A/P,15%,3) – 190,000
Select the challenger
11.30 Determine the ESL for the defender and compare to the AWC for the same number of
years.
By hand:
AWD,1 = -32,000(A/P,10%,1) – 24,000 + 25,000(A/F,10%,1)
= -32,000(1.10) –24,000 + 25,000
ESL for defender is n = 1 year with AWD = $-34,200
By spreadsheet: Defender ESL is n = 1; select challenger now
11.31 Find ESL of the defender; compare with AWC over 5 years.
AWD,3 = -8000(A/P,15%,3) – [50,000(P/F,15%,1) +
The ESL is 1 year with AWD,1 = $-53,200
Replacement Study over a Specified Study Period
11.32 Compare cost of replacement (challenger) with cost of keeping defender (in–place) for 3
more years
Replace in–place with challenger now, since AWC < AWD
11.33 Option 1 (lease) is defender; option 2 is challenger. Find AW of lease over 5–year period
Select option 1, continue to lease the land
11.34 (a) Option 1: Keep defender for 2 years and then replace it with challenger for 1 year
Option 2: Replace now with challenger for all 3 years
AW2 = $-75,000
Keep the machine 2 years and then replace with the challenger
(b) The annual worth will be AW1 = $–72,880
11.35 By hand:
There are three options for a 2–year study period
AWX,2 = -82,000(A/P,12%,2) – 30,000 + 42,000(A/F,12%,2)
AWY,1 = -97,000(A/P,12%,1) – 27,000 + 66,000(A/F,12%,1)
= $–69,640
AWY,2 = -97,000(A/P,12%,2) – 27,000 + 56,000(A/F,12%,2)
= $-57,980
Option Years for X Years for Y
AW cash flows, $ per year
Option AW,
$ per year
Year 1
Year 2
A
0
2
-57,980
-57,980
-57,980
B
1
1
-71,840
–69,640
-70,802
C
2
0
-58,708
-58,708
-58,708
Option
Keep defender X
Use challenger Y