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Solutions to end-of-chapter problems Engineering Economy, 7th edition Leland Blank and Anthony Tarquin

Chapter 5 Present Worth Analysis 5.1 Mutually exclusive alternatives accomplish the same thing. Therefore, only one is to be selected, so they are compared against each other. Independent projects accomplish different things. Therefore, none, one, more than one, or all of them can be selected as they are only compared against the do-nothing alternative. 5.2 (a) The do-nothing alternative means that the status-quo should be maintained. That is, If none of the alternatives under consideration are economically attractive, all of them should be rejected. (b) Do-nothing is not an option when the alternatives being evaluated are cost alternatives, which means that one of them must be selected. 5.3

(a) Number of alternatives = 24 = 16 (b) Possibilities: DN, W, X, Y, Z, WX, WY, WZ, XY, XZ, YZ, WXY, WXZ, WYZ, XYZ, WXYZ

5.4 Revenue alternatives have cash inflows and outflows, while cost alternatives have only costs. 5.5 Equal service means that alternatives must provide service for the same period of time, and therefore, end at the same time. 5.6 Equal service can be satisfied by using a specified planning period or by using the least common multiple between the lives of the alternatives. 5.7 PWIn-house = -30 + (14 – 5)(P/A,10%,5) + 2(P/F,10%,5) = -30 + (14 – 5)(3.7908) + 2(0.6209) = $5.359 ($5,359,000) PWContract = (3.1 – 2)(P/A,10%,5) = (3.1 – 2)(3.7908) = $4.170 ($4,170,000) Select In-house production. 5.8 PWA = -42,000 – 28,000(P/A,10%,4) = -42,000 – 28,000(3.1699) = $-130,757 1

PWB = -51,000 – 17,000(P/A,10%,4) = -51,000 – 17,000(3.1699) = $-104,888 Select Machine B 5.9 (a) PWX = -15,000 – 9000(P/A,12%,5) + 2000(P/F,12%,5) = -15,000 – 9000(3.6048) + 2000(0.5674) = $-46,308 PWY = -35,000 – 7000(P/A,12%,5) + 20,000(P/F,12%,5) = -35,000 – 7000(3.6048) + 20,000(0.5674) = $-48,886 Select Material X (b) Let first cost of Y be XY. Set PWY = -46,308 -46,308 = -XY - 7000(P/A,12%,5) + 20,000(P/F,12%,5) = -XY - 7000(3.6048) + 20,000(0.5674) Xy = $32,422 Select Y if first cost is ≤ $32,422 5.10 Find Pg for each stock and select higher one. PgA = 30,000{1 – [(1 + 0.06)/(1 + 0.08)]5}/(0.08 – 0.06) = $133,839 PgB = 20,000{1 – [(1 + 0.12)/(1 + 0.08)]5}/(0.08 – 0.12) = $99,710 Select Class A stock 5.11 PWA = -952,000 - 1,300,000 - 126,000(P/A,6%,50) = -952,000 - 1,300,000 - 126,000(15.7619) = $-4,238,000 PWB = -5(366,000) -9000(151.18) - 340,000 - 81,500 + 500,000(P/F,6%,5) = -3,612,120 + 500,000(0.7473) = $-3,238,470 Select Plan B

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5.12 PWNo drains = -1500(P/A,4%,12) = -1500(9.3851) = $-14,078 PWCorrugated = -3(7000) +4000(P/F,4%,12) = -21,000 +4000(0.6246) = $-18,502 Do not install corrugated pipe 5.13 PW250 = -155,000 – 3000(P/A,10%,30) = -155,000 – 3000(9.4269) = $-183,281 PW300 = $-210,000 Install the 250 mm pipe 5.14 PWGaseous = -8000 - (650 + 800)(P/A,10%,5) = -8000 - (1450)(3.7908) = $-13,497 PWDry = - (1000 + 1900)(P/A,10%,5) = - (2900)(3.7908) = $-10,993 Add dry chlorine 5.15 PWVolt = -35,000 + 15,000(P/F,0.75%,60) = -35,000 + 15,000(0.6387) = $-25,420 PWLeaf = -1500 – 349(P/A,0.75%,60) = -1500 – 349(48.1734) = $-18,313 Select the Nissan Leaf 5.16 In $ million units, PWLand = -215 – 22(P/A,15%,50) – 30(P/F,15%,25) = -215 – 22(6.6605) – 30(0.0304) = $-362.443 ($-362,443,000)

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PWSea = -350 – 2(P/A,15%,50) – 70(P/F,15%,25) = -350 – 2(6.6605) – 70(0.0304) = $-365.449 ($-365,449,000) Select land route by a PW margin of only $3 million 5.17 PWA = -40,000[1+ (P/F,10%,2) + (P/F,10%,4) + (P/F,10%,6)] – 9000(P/A,10%,8) = -40,000 [1 + 0.8264 + 0.6830 + 0.5645] – 9000(5.3349) = $-170,970 PWB = -80,000[1 + (P/F,10%,4)] – 6000(P/A,10%,8) = -80,000[1 + 0.6830] – 6000(5.3349) = $-166,649 PWC = -130,000 – 4000(P/A,10%,8) + 12,000(P/F,10%,8) = -130,000 – 4000(5.3349) + 12,000(0.4665) = $-145,742 Select Method C 5.18 PWA = -5,000,000 – 5,500,000(P/A,10%,10) = -5,000,000 – 5,500,000(6.1446) = $-38,795,300 PWB = -5,000,000 – 25,000,000(P/F,10%,2) - 30,000,000(P/F,10%,7) = -5,000,000 – 25,000,000(0.8264) - 30,000,000(0.5132) = $-41,056,000 Select Plan A 5.19 (a) PWX = -250,000 – 60,000(P/A,10%,6) - 180,000(P/F,10%,3) + 70,000(P/F,10%,6) = -250,000 – 60,000(4.3553) - 180,000(0.7513) + 70,000(0.5645) = $-607,037 PWY = -430,000 – 40,000(P/A,10%,6) + 95,000(P/F,10%,6) = -430,000 – 40,000(4.3553) + 95,000(0.5645) = $-550,585 Select Machine Y

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(b) Spreadsheet solution

5.20 Set the PWS relation equal to $-33.16, and solve for the first cost XS ( a positive number) with repurchase in year 5. In $1 million units, -33.16 = -XS[1 + (P/F,12%,5)] - 1.94(P/A,12%,10) + 0.05XS[(P/F,12%,5) + (P/F,12%,10)] = -1.5674XS - 1.94(5.6502) + 0.0445XS 1.5229XS = -10.9614 + 33.16 XS = $14.576

($14,576,000)

Select seawater option for any first cost ≤ $14.576 million 5.21 PW1 = -26,000 – 5000(P/A,10%,6) - 26,000(P/F,10%,3) = -26,000 – 5000(4.3553) - 26,000(0.7513) = $-67,310 PW2 = -83,000 – 1400(P/A,10%,6) - 2500(P/F,10%,3) = -83,000 – 1400(4.3553) - 2500(0.7513) = $-90,976 Select Plan 1 5.22 Compare PW of costs over 30 years. PWPlastic = -(0.90)(110)(43,560) - [(0.90)(110)(43,560) + 500,000](P/F,8%,15) = -4,312,440 - [4,312,440 + 500,000](0.3152) = $-5,829,321 PWRubberized = -(2.20)(110)(43,560) = $-10,541,520 Select plastic liner 5

5.23 (a) PWFan X = -130,000 - 290(P/A,8%,50) = -130,000 - 290(12.2335) = $-133,548 PWFan Y = -290(P/A,8%,50) - 20(P/G,8%,50) = -290(12.2335) - 20(139.5928) = $-6,340 Fan Y made the far better deal (unless fan X’s seats are much better!!) (b) Let MX = ‘mortgage’ cost for fan X for equivalence of plans -6340 = -MX - 290(P/A,8%,50) Mx = 6340 - 290(12.2335) = $2792 Fan X should pay only $2792, not $130,000 5.24 (a) PWLand = -130,000 – 95,000(P/A,10%,6) – 105,000(P/F,10%,3) + 25,000(P/F,10%,6) = -130,000 – 95,000(4.3553) – 105,000(0.7513) + 25,000(0.5645) = $-608,528 PWIncin = -900,000 – 60,000(P/A,10%,6) + 300,000(P/F,10%,6) = -900,000 – 60,000(4.3553) + 300,000(0.5645) = $-991,968 PWContract = –120,000(P/A,10%,6) = –120,000(4.3553) = $-522,636 Select private disposal contract (b) Recalculate PW for the contract alternative with 20% increases each 2 years. PWContract = -120,000(P/A,10%,2) - 120,000(1.20)(P/A,10%,2)(P/F,10%,2) - 120,000(1.2)2(P/A,10%,2)(P/F,10%,4) = -120,000(1.7355) - 144,000(1.7355)(0.8264) - 172,800(1.7355)(0.6830) = $-619,615 Select land application; the selection changed 5.25 (a) Use LCM of 12 years and select L. (b) Use PW over life of each alternative and select I, J and L with PW > 0.

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5.26 FWX = -80,000(F/P,15%,3) – 30,000(F/A,15%,3) + 40,000 = -80,000(1.5209) – 30,000(3.4725) + 40,000 = $-185,847 FWY = -97,000(F/P,15%,3) – 27,000(F/A,15%,3) + 50,000 = -97,000(1.5209) – 27,000(3.4725) + 50,000 = $-191,285 Select robot X 5.27 FWT = -750,000(F/P,12%,4) – 60,000(F/A,12%,4) – 670,000(F/P,12%,2) + 80,000 = -750,000(1.5735) – 60,000(4.7793) – 670,000(1.2544) + 80,000 = $-2,227,331 FWW = -1,350,000(F/P,12%,4) - 25,000(F/A,12%,4) - 90,000(F/P,12%,2) + 120,000 = -1,350,000(1.5735) – 25,000(4.7793) - 90,000(1.2544) + 120,000 = $-2,236,604 Select process T, by a small margin of only $9273 in FW. 5.28 FWP = -23,000(F/P,8%,6) – 4000(F/A,8%,6) – 20,000(F/P,8%,3) + 3000 = -23,000(1.5869) – 4000(7.3359) – 20,000(1.2597) + 3000 = $-88,036 FWQ = -30,000(F/P,8%,6) – 2500(F/A,8%,6) + 1000 = -30,000(1.5869) – 2500(7.3359) + 1000 = $-64,947 Select alternative Q 5.29 FWK = -1,600,000(F/P,12%,8) – 70,000(F/A,12%,8) – 1,200,000(F/P,12%,4) + 400,000 = -1,600,000(2.4760) – 70,000(12.2997) – 1,200,000(1.5735) + 400,000 = $-6,310,780 FWL = -2,100,000(F/P,12%,8) – 50,000(F/A,12%,8) - 3000(P/G,12%,8)(F/P,12%,8) = -2,100,000(2.4760) – 50,000(12.2997) - 3000(14.4714)(2.4760) = $-5,922,079 Select system L 5.30 FWOld = -1,300,000(F/P,10%,5) – 100,000,000(F/P,10%,4) = -1,300,000(1.6105) – 100,000,000(1.4641) = $-148,503,650

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FWNew = -1,300,000(F/P,10%,6) – 100,000,000 = -1,300,000(1.7716) – 100,000,000 = $-102,303,080 Difference = 148,503,650 - 102,303,080 = $46,200,570 (higher cost for old contract) 5.31 CC = (-100,000/0.08)(P/F,8%,5) = -1,250,000(0.6806) = $-850,750 5.32 (a)

CC = -10,000(A/F,3%,5)/0.03 = -10,000(0.18835)/0.03 = $-62,783

(b)

CC = -10,000(A/F,8%,5)/0.08 = -10,000(0.17046)/0.08 = $-21,308

(c) When money earns at the lower 3% rate, it is necessary to start with more. 5.33 CC = -300,000 – 35,000/0.12 – 75,000(A/F,12%,5)/0.12 = -300,000 – 291,667 – 75,000(0.15741)/0.12 = $-690,048 5.34 Use C to identify the contractor option. (a) CCC = -5 million/0.12 = $-41.67 million Between the three options, select the contractor (b) Find Pg and A of the geometric gradient (g = 2%), then CC. Pg = -5,000,000[1 - (1.02/1.12) 50]/(0.12 – 0.02) = -5,000,000[9.9069] = $-49.53 million A = Pg(A/P,12%,50) = -49.53 million(0.12042) = $-5.96 million per year CCC = A/i = -5.96 million/0.12 = $-49.70 million Now, select groundwater (CCG = $-48.91) source by a relatively small margin.

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5.35 For M, first find AW and then divide by i to find CC. AWM = -150,000(A/P,10%,5) – 50,000 + 8000(A/F,10%,5) = -150,000(0.26380) – 50,000 + 8000(0.16380) = $-88,260 CCM = -88,260/0.10 = $-882,600 CCN = - 800,000 – 12,000/0.10 = $-920,000 Select alternative M 5.36 CC = -1000/0.10 – 5000(A/F,10%,4)/0.10 = -1000/0.10 – 5000(0.21547)/0.10 = $-20,774 5.37 CC = (-40,000/0.08)(P/F,8%,11) = (-40,000/0.08)(0.4289) = $-214,450 5.38 CC = -150,000 – 5000/0.06 – 20,000(P/F,6%,2) = -150,000 – 5000/0.06 – 20,000(0.8900) = $-251,133 5.39 Answer is (c) 5.40 Answer is (a) 5.41 Answer is (d) 5.42 Answer is (c) 5.43 FWP = -23,000(F/P,8%,6) -20,000(F/P,8%,3) - 4,000(F/A,8%,6) + 3000 = -23,000(1.5869) -20,000(1.2597) - 4,000(7.3359) + 3000 = $-88,036 Answer is (a) 5.44 CC = -50,000 – 10,000(P/A,10%,15) – (20,000/0.10)(P/F,10%,15) = -50,000 – 10,000(7.6061) – (20,000/0.10)(0.2394) = $-173,941 Answer is (c)

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5.45 CC = (-40,000/0.10)(P/F,10%,4) = (-40,000/0.10)(0.6830) = $-273,200 Answer is (c) 5.46 Answer is (b) 5.47 Answer is (b) 5.48 Answer is (d) 5.49 Answer is (d) 5.50 PWY = -95,000 – 15,000(P/A,10%,4) + 30,000(P/F,10%,4) = -95,000 – 15,000(3.1699) + 30,000(0.6830) = $-122,059 Answer is (b) 5.51 CC = -10,000 - [10,000(A/F,10%,5)]/0.10 = -10,000 - [10,000(0.16380)]/0.10 = $-26,380 Answer is (c) 5.52 CC = -10,000 – 5000(P/A,10%,5) – (1000/0.10)(P/F,10%,5) = -10,000 – 5000(3.7908) – (1000/0.10)(0.6209) = $-35,163 Answer is (b)

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Solution to Case Study, Chapter 5 There is not always a definitive answer to case study exercises. Here are example responses

COMPARING SOCIAL SECURITY BENEFITS 1. Total payments are shown in row 30 of the spreadsheet. 2. Future worth values at 6% per year are shown in row 29.

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3. Plots of FW values by year are shown in the (x-y scatter) graph below.

4. Develop all feasible plans for the couple and use the summed FW values to determine which is the largest. Spouse #1 A A A B B B C

Spouse #2 A B C B C D C

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FW, $ 1,707,404 1,663,942 1,617,716 1,620,480 1,574,254 1,671,304 1,528,028