Solutions to end–of–chapter problems
Engineering Economy, 7th edition
Leland Blank and Anthony Tarquin
Chapter 6
Annual Worth Analysis
6.2 Three assumptions in the AW method are:
(2) The selected alternative will be repeated in succeeding life cycles
6.3 The AW over one life cycle of each alternative can be used to compare them because their
6.5 AW4 = -20,000(A/P,10%,4) – 12,000 + 4000(A/F,10%,4)
6.6 AW = -130,000(A/P,8%,50) – 290
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6.7 Find PW and convert to AW
PW = -13,000 – 13,000(P/A,8%,9) – 290(P/A,8%,50)
6.8 AW = -115,000(A/P,8%,8) – 10,500 – 3600(P/F,8%,4)(A/P,8%,8) + 45,000(A/F,8%,8)
6.9 AW = -2000(P/F,8%,5)(A/P,8%,8) – 800(A/F,8%,2)
6.10 (a) CR = -285,000(A/P,12%,10) + 50,000(A/F,12%,10)
(b) AW = -285,000(A/P,12%,10) + 50,000(A/F,12%,10) + 52,000 – 10,000
6.11 (a) CR = -500,000(A/P,8%,20) + (0.9)500,000(A/F,8%,20)
6.12 CR = -750,000(A/P,24%,5) + 75,000(A/F,24%,5)
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Cash flow diagrams are shown here.
6.13 AWX = -75,000(A/P,10%,4) – 32,000 + 9000(A/F,10%,4)
AWY = -140,000(A/P,10%,4) – 24,000 + 19,000(A/F,10%,4)
6.14 AWBuy = [-32,780 – 2200 + 7500 + 0.5(2200)](A/P,10%,3) + 0.40(32,780)(A/F,10%,3)
6.15 AWSingle = -6000(A/P,10%,4) – 6000(P/A,10%,3)(A/P,10%,4)
CR = $263,869 per year
0 1 2 3 4 5
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6.16 AWpermanent = -3,800,000(A/P,6%,20)
6.17 (a) AWSolar = -16,600(A/P,10%,5) – 2400
6.18 AWMF = -33,000(A/P,10%,3) – 8000 + 4000(A/F,10%,3)
6.19 (a) AWJoe = -85,000(A/P,8%,3) – 30,000 + 40,000(A/F,8%,3)
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(b) Spreadsheet and Goal Seek indicate that Watcheye’s first cost must be ≤ $-116,935.
6.20 AWR = -250,000(A/P,10%,3) – 40,000 + 20,000(A/F,10%,3)
6.21 AW4 yrs = -39,000(A/P,12%,4) – [17,000 + 1200(A/G,12%,4)] + 23,000(A/F,12%,4)
6.22 (a) CRSemi2 = -80,000(A/P,10%,5) + 13,000(A/F,10%,5)
Found using Goal Seek
when cell C9 was set equal
to cell B9 at $-50,662
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(b) AWSemi2 = -80,000(A/P,10%,5) – [21,000 + 500(A/G,10%,5)] + 13,000(A/F,10%,5)
6.23 AW = -200,000(0.10) – 100,000(A/F,10%,7)
6.24 AW = -5M(0.10) – 2M(P/F,10%,10)(0.10) – [(100,000/0.10)(P/F,10%,10)](0.10)
6.25 First find PW for years 1 through 10 and convert to AW.
PW = -[150,000(P/A,10%,4) + 25,000(P/G,10%,4)](P/F,10%,2)
6.26 AWCondi = -25,000(A/P,10%,3) – 9000 + 3000(A/F,10%,3)
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6.27 AW = -30,000,000(0.10) – 50,000 – 1,000,000(A/F,10%,5)
6.28 (a) AWX = -90,000(A/P,10%,3) – 40,000 + 7000(A/F,10%,3)
(b) Goal Seek (right figure, row 2) finds the required first costs for Y = $-341,912 and
6.29 The alternatives are A1, A2, B1, B2 and C. Use a + sign for costs.
AWA1 = {100,000+ [190,000 + 60,000(P/A,10%,9)](P/F,10%,1) }(A/P,10%,10)
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6.30 First find the present worth of all costs and then convert to annual worth over 20 years.
PW = -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)
6.31 First find the present worth of all costs and then convert to annual worth over 20 years.
PW = – 2.6(P/F,6%,1) – 2.0(P/F,6%,2) – 7.5(P/F,6%,3) – 10.0(P/F,6%,4)
6.32 Annual LCCA = -750,000(A/P,6%,20) – 72,000 – 24,000
– 150,000[(P/F,6%,5) + (P/F,6%,10) + (P/F,6%,15)](A/P,6%,20)
6.33 PWM = -250,000 – 150,000(P/A,8%,4) – 45,000 – 35,000(P/A,8%,2)
6.40 AW2 = -550,000(A/P,6%,15) +100,000
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6.45 AW = -40,000(A/P,15%,4) – 5000 + 32,000(A/F,15%,4)
6.46 AW = -50,000(0.12) – [(20,000/0.12)](P/F,12%,15)(0.12)
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Solution to Case Study, Chapter 6
There is not always a definitive answer to case study exercises. Here are example responses
THE CHANGING SCENE OF AN ANNUAL WORTH ANALYSIS
1. Spreadsheet and chart are below. Revised costs and savings are in columns F-H.
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2. In cell G18, the new AW = $17,904. This is only slightly larger than the PowrUp