Solutions to end–of–chapter problems
Engineering Economy, 8th edition
Leland Blank and Anthony Tarquin
Chapter 6
Annual Worth Analysis
Annual Worth and Capital Recovery Calculations
6.1 Multiply the FW values by (A/F,i%,n), where n is equal to the LCM or stated study period.
6.2 The estimate obtained from the three-year AW would not be valid, because the AW
Spreadsheet: The answer is the G3 display of $3656
6.4 (a) AW4 = -30,000(A/P,10%,6) – 12,000 + 4000(A/F,10%,6)
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(b) S = $22,923 is very high for a used delivery car; this market value is over 5.7 times the
estimated salvage of $4000.
6.5 CR1 = -50,000(A/P,10%,5) + 5,000(A/F,10%,5)
6.6 (a) Capital recovery = P/n = 6000/3
= $2000 per year
(b) A = 6,000(A/P,10%,3)
6.7 (a) CR = -750,000(A/P,25%,5) + 125,000(A/F,25%,5)
(b) AW = CR + AOC = –263,656 – 80,000
6.8 AW = -638,000(A/P,25%,9) – 240,000 + 184,000(A/F,25%,9)
6.9 In $ billion units
6.10 Find the CR; in $ million units
6.11 Factors: In $ million units
6.12 AW = -800,000(A/P,10%,4) – 300,000(P/F,10%,2)(A/P/10%,4) – 950,000
Alternative Comparison
6.13 (a) In $1000 units
AWPush = -2250(A/P,10%,8) – 600 + 50(A/F,10%,8)
(b) Set AWPush equal to $-972,416, and solve for SPush. In $1000 units
6.14 AWA = -2,000,000(A/P,12%,3) – 60,000 + 2,000,000(0.10)(A/F,12%,3)
Select Plan B
6.15 AW1 = -550,000(A/P,10%,3) – 160,000 + 125,000(A/F,10%,3)
AW2 = -830,000(A/P,10%,3) – 120,000 + 240,000(1.35)(A/F,10%,3)
6.16 Factors:
AWResturant = –26,000(A/P,1%,60) -2200 –3700 + 14,100 + 26,000(0.10)(A/F,1%,60)
Truck: = – PMT(1%,60,-17900,6265) + 5300 displays $+4978.54
6.17 (a) AWGM = -36,000(A/P,15%,3) – 4000 + 15,000(A/F,15%,3)
Purchase the Ford SUV
(b) PWGM = -15,448(P/A,15%,12)
6.18 (a) Factors
AWRound = -250,000(A/P,10%,6) – 31,000 + 40,000(A/F,10%,6)
Select Round Knife
(b) Round knife: = – PMT(10%,6,-250000,40000) – 31000 displays $-83,218
6.19 AWCart = -300,000(A/P,10%,4) – 60,000 + 70,000(A/F,10%,4)
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= $-126,420
Select articulated robot
6.20 (a) AWC = -40,000(A/P,15%,3) – 10,000 + 12,000(A/F,15%,3)
(b) Factors: Set AWD = AWC with nD as an unknown.
6.21 (a) AWA = -25,000(A/P,12%,2) – 4000
(b) Use the LCM of 6 years
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FWA =AWA(F/A,12%,6) = -18,793(8.1152) = $-152,509
FWB =AWB(F/A,12%,6) = -22,804(8.1152) = $-185060
Select plan A
6.22 (a) AWLand = –150,000(A/P,10%,4) – 95,000 + 25,000(A/F,10%,4)
6.23 (a) AWForklift = CR – AOC – salary – AW of pallets
(b) Functions shown. Select forklift.
6.24 AWQ = –42,000(A/P,10%,2) – 6000
6.25 AWsmall = -1,700,000(A/P,1%,120) – 12,000 + 170,000(A/F,1%,120)
Permanent Investments
6.26 AW = P(i) = CC(i)
6.27 Find P in year 29 using (P/F,i%,20) to move back to year 9, and then use (A/F,i%,10)
factor to find A
(a) A = [80,000/0.10](P/F,10%,20)(A/F,10%,10)
6.28 AW = -300,000(0.10) – 100,000(A/F,10%,5)
6.29 Find PW in year -1; multiply by i
PW-1 = 5,000,000(P/F,10%,1) + 2,000,000(P/F,10%,11) + (100,000/0.10)(P/F,10%,11]
Difference is $478
(b) AW50 = 100,000(A/P,10%,50)
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= $10,086
AW∞ = Pi = 100,000(0.10)
= $10,000
Difference is $86, a lot smaller than at 5% per year
6.31 AWcontract = –1 + 2.5
6.32 (a) Perpetual AW is equal to AW over one life cycle
AW = -[6000(P/A,8%,28) + 1000(P/G,8%,28)](P/F,8%,2)(A/P,8%,30)
(b) With no function to directly accommodate gradients, list cash flows in cells B2 through
B32 and use PMT function to verify.
6.33 AWbrush = -400,000(A/P,8%,10) + 50,000(A/F,8%,10)
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= $-96,804
AWblast = -400,000(0.08) – 70,000
= $-102,000
Select brush alternative
Life Cycle Cost
6.34 PW of LCC = –6.6 – 3.5(P/F,7%,1) – 2.5(P/F,7%,2) – 9.1(P/F,7%,3) – 18.6(P/F,7%,4)
– 21.6(P/F,7%,5) – 17(P/A,7%,5)(P/F,7%,5) – 14.2(P/A,7%,10)(P/F,7%,10)
AW of LCC = -151,710,860(A/P,7%,20)
6.35 PW of LCC = – 2.6(P/F,6%,1) – 2.0(P/F,6%,2) – 7.5(P/F,6%,3) – 10.0(P/F,6%,4)
6.36 PW of LCCA = -250,000 – 150,000(P/A,8%,4) – 45,000 – 35,000(P/A,8%,2)
AW of LCCB = –937,525(A/P,8%,10)
AW of LCCC = -1,174,268(A/P,8%,10)
= -1,174,268(0.14903)
= $-175,001
Select Alternative B (adapted system)
Spreadsheet Exercises
6.38 (a) Spreadsheet details omitted intentionally; resulting chart shown.
6.39 Spreadsheet functions omitted
6.40 Testing stage: Equipment first cost increases to $-30,000,000; S remains at $600,000
-$400,000
-$250,000
-$200,000
60% 70% 80% 90% 100% 110% 120% 130% 140%
AW of AW of
stage is entered.
Additional Problems and FE Exam Review Questions
6.41 Answer is (b)
6.47 Find PW in year 0 and then multiply by i
6.48 A = [40,000/0.08](P/F,8%,2)(A/F,8%,3)
6.50 Answer is (a)
Solution to Case Study, Chapter 6
ANNUAL WORTH ANALYSIS – THEN AND NOW
1. Spreadsheet and graph are below. Revised costs and savings are in columns F-H.
Maintenance costs increase; repair savings decrease.
2. In cell G18, the new AW = $17,904. This is only slightly larger than the PowrUp
AW = $17,558. Lloyd’s would have been selected, but only by a small margin.
3. New CR is $7173 per year (cell E17), an increase from $7025 previously determined when
the salvage estimate was $3000 after 10 years.