7 Solutions

 

CHAPTER CHAP TER 7 Solutions Manual For 

Basics of Engineering Economy, 1e Leland Blank, PhD, PE Texas A&M University and American University of Sharjah, UAE

Anthony Tarquin, PhD, PE University of Texas at El Paso

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Chapter 7 7.1 The primary purpose of public sector projects is to provide services for the  public good good at no profit. profit. 7.2 eBay - private, farmer’s market market - private, state police department – public, car racing facility - private, social security – public, EMS - public, ATM - private, travel agency- private, Six Flags amusement park - private, gambling casino  private,, swap meet  private meet - private. private. 7.3 Large initial investment investment - public, short life project projectss - private, private, profit - private, disbenefits - public, tax-free bonds - public, subsidized loans – public, low interest rate - public, income tax – private, water quality standards – public. 7.4 Disbenefits Disbenefits are consequences consequences to people while costs are consequences to governme government; nt; disbenefits are considered in the numerator of a B/C ratio while costs are in the denominator. 7.5 (a) $400,000 $400,000 annual income to local businesses businesses because of tourism tourism created created by park benefit (b)new Costnational of fish from fr om -hatchery hatche ry to stock lake at state state park - cost (c) Less tire tire wear because because of smoother smoother road road surface - benefit benefit (d) Decrease Decrease in property values due to closure closure of gov’t research lab - disbenefi disbenefitt (e) School overcrowding overcrowding because because of military military base expansion expansion - disbenefit disbenefit (f) Additional revenue to local motels because of extended national  park season season – benefit benefit 7.6 A BOT project is one that involves a public-private partnership wherein the private  partner is is responsible responsible for for building, building, operating, operating, and subsequently subsequently transferr transferring ing the  project to to a governmental governmental entity. entity. 7.7 Benefits and disbenefits might exactly offset each other because benefits to one part of the general population might represent disbenefits to a different part of the  population.  populat ion. For example, example, reduced reduced traffic traffic accidents accidents are are good for for motorists motorists but but bad for  tow- truck owners 7.8 The salvage value placed placed in the denominator because because it is a recovery of cost, which is a consequence to the government. 7.9 A modified B/C analysis analysis has only the initial initial investment investment cost in the denominator. denominator.

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7.10 B = $285,000 C = 12,000,000(A/P,6%,40) + 1,100,000 = 12,000,000(0.06646) + 1,100,000 = 797,520 + 1,100,000 = $1,897,520 B/C = 285,000/1,897,520 = 0.15 7.11 Convert all cash flows to AW B = 3,800,000(A/P,8%,20) = 3,800,000(0.10185) = $387,030 D = $65,000 C = 1,200,000(A/P,8%,20) + 300,000 = 1,200,000(0.10185) + 300,000 = $422,220   B/C = (387,030 – 65,000)/ 422,220 = 0.76 7.12 The total cost (TC) has to be $800,000. TC = P(A/P,6%,10) + 600,000 800,000 = P(0.13587) + 600,000 P = $1,471,995 7.13

B = $600,000 D = $190,000 C = 650,000(A/P,6%,20) + 150,000 = 650,000(0.08718) + 150,000 = $206,667 B/C == 1.98 (600,000 – 190,000)/206,667

7.14 (a) B = $340,000 D = $40,000 C = 2,300,000(0.06) + 120,000 = $258,000 B/C = (340,000 – 40,000)/258,000 = 1.16 (b) Modified B/C = (340,000 – 40,000 – 120,000)/138,000 = 1.30

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7.15 Use present worth, since since most of the cash flows are already present dollars. (a) Conventional B/C ratio B = 200,000 + 100,000(P/A,6%,40) = 200,000 + 100,000(15.0463) = $1,704,630 D = 18,000(P/A,6%,40) = $270,833 C = 1,200,000 + 200,000(P/F,6%,3) = 1,200,000 + 200,000(0.8396) = $1,367,920 S = 90,000(P/F,6%,5) = 90,000(0.7473) = $67,257 B/C = (1,704,630 – 270,833)/(1,367,920 – 67,257) = 1.10   (b) Modified B/C ratio = (B – D + S)/C = (1,704,630 – 270,833 + 67,257)/1,367,920 = 1.10 7.16 Use annual worth, since most most of the cash flows are in annual dollars. (a) Conventional B/C ratio B = 300,000(0.06) + 100,000 = 18,000 + 100,000 = $118,000 D = $40,000 C = 1,500,000(0.06) + 200,000(P/F,6%,3)(0.06) = 90,000 + 200,000(0.8396)(0.06) = $100,075 S = 70,000 B/C = (118,000 – 40,000)/(100,075 – 70,000) = 2.59   (b) Modified B/C ratio = (B – D + S)/C = (118,000 – 40,000 + 70,000)/100,075 = 1.48 7.17

B = 41,000(33 –18) = $615,000 D = 1100(85) = $93,500 C = 750,000(A/P,0.5%,36) = 750,000(0.03042) = $22,815

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B/C = (615,000 – 93,500)/22,815 = 22.86 7.18

B = $194,000 per year  C = 45(172,000)(A/P,6%,10) = 7,740,000(0.13587) = $1,051,634 S = $80,000 B/C = 194,000/(1,051,634 – 80,000) = 0.20

7.19

B = $500,000 per year  D = $45,000 per year  C = P(A/P,7%,10) + 150,000 = 0.14238P + 150,000 2.3 = (500,000 – 45,000)/ (0.14238P + 150,000) 0.3275P = 455,000 - 2.3(150,000) P = $335,880

7.20

B = 1250(140) = $175,000 per year  C = 595,000(A/P,6%,5) + 56,000 + 19,000 = 595,000(0.23740) + 75,000 = $216,253 B/C = 0.81

Therefore, the the utility should send the samples out.

7.21 The alternatives are usage alternatives; only compare against each other. PW of Incremental benefits, B = 950,000 – 250,000 = $700,000 PW of cost alt 1 = 600,000 + 50,000(P/A,8%,20) = 600,000 + 50,000(9.8181) = $1,090,905 PW of cost alt 2 = 1,100,000 + 70,000(P/A,8%,20) = 1,100,000 + 70,000(9.8181) = $1,787,267 PW incremental cost = 1,787,267 - 1,090,905 = $696,362  

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B/C = 700,000/696,362 = 1.01 Select alternative 2. 7.22

Compare EC vs DN B = $110,000 per year  D = $26,000 per year  C = 38,000(A/P,7%,10) + 49,000 = 38,000(0.14238) + 49,000 = $54,410 B/C = (110,000 – 26,000)/54,410 = 1.54 Eliminate DN Compare EC vs NS Incremental B = 160,000 – 110,000 = $50,000 Incremental D = 0 – 26,000 = $-26,000 Cost EC = $54,410 (from above) Cost NS = 87,000(A/P,7%,10) + 64,000 = 87,000(0.14238) + 64,000 = $76,387 Incremental C = 76,387 - 54,410 = $21,977 Incremental B/C = [50,000 -(-26,000)]/21,977 = 3.46 Eliminate Eliminate EC Select alternative NS.

7.23 Cost for method A = 14,100 + 6000 + 4300 + 2600 = $27,000 Cost for method B = $5200 + 1400 + 2600 + 1200 = $10,400 Incremental costs for method A = 27,000 – 10,400 = $16,600 Benefits for method A = 600 (P/A,7%,20) = 600(10.5940) = $6356.40 B/C = 6356/16,600 = 0.38 Select method B.

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7.24

(a) Program 1 vs DN B = 1.25 per month C = 60(A/P,0.5%,60) = 60(0.01933) = $1.16 B/C = 1.25/1.16 = 1.08Elimina 1.08 Eliminate te DN Program 1 vs 2 Incremental B = 8.00 – 1.25 = $6.75 Incremental C = (500 – 60)(A/P,0.5%,60) = 440(0.01933) = $8.51 Incremental B/C = 6.75/8.51 = 0.79 Select program 1. (b) B/C for program 1 = 1.08 (from above) Therefore, program 1 O.K. B for program 2 = 8.00 per month C for program 2 = 500(A/P,0.5%,60) = 500(0.01933) = $9.67 B/C for program 2 = 8.00/9.67 = 0.83 Therefore, reject program 2

7.25 Cost conventional = 200,000 – 10,000(P/F,7%,5) = 200,000 – 10,000(0.7130) = $192,870 Cost Solar = 1,300,000 – 150,000(P/F,7%,5) = 1,300,000 – 150,000(0.7130) = $1,193,050 Incremental cost for solar = 1,193,050 – 192,870 = $1,000,180 Incremental benefits solar = (80,000 – 9000)(P/A,7%,5)

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= 71,000(4.1002) = $291,114 Incremental B/C = 291,114/1,000,180 = 0.29 Therefore, select conventional 7.26

Rank alternatives according to increasing cost: cost: DN, 1, 2, 3, 4 DN vs 1: B/C = (16,000 – 10,000)/15,000(A/P,10%10) = 6000/15,000(0.16275) = 2.46 Eliminate DN 1 vs 2: Incremental B = (20,000 – 16,000) – (12,000 – 10,000) = $2000 Incremental C = (19,000 – 15,000)(A/P,10%,10) = 4000(0.16275) = $651 Incremental B/C = 2000/651 = 3.07 Eliminate 1 2 vs 3: Incremental B = (19,000 – 20,000) – (9,000 – 12,000) = $2000 Incremental C = (25,000 – 19,000)(A/P,10%,10) = 6000(0.16275) = $976.50 Incremental B/C = 2000/976.50 = 2.05 Eliminate 2

3 vs 4: Incremental B = (22,000 – 19,000) – (11,000 – 9,000) = $1000 Incremental C = (33,000 – 25,000)(A/P,10%,10) = 8000(0.16275) = $1302 Incremental B/C = 1000/1302 = 0.77 Eliminate 4 Select method 3.

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7.27 Cost of S = 50,000,000(0.06) + 150,000 = $3,150,000 Cost of D = 75,000,000(0.06) + 130,000 = $4,630,000 Cost of N = 60,000,000(0.06) + 140,000 = $3,740,000   Rank alternatives according to increasing cost: S, N, D S vs N: Incremental B = 7,600,000 – 5,900,000 = $1,700,000 Incremental C = 3,740,000 – 3,150,000 = $590,000 Incremental B/C = 1,700,000/590,000 = 2.88 Eliminate S N vs D: Incremental B = 5,900,000 – 4,100,000 = $1,800,000 Incremental C = 4,630,000 - 3,740,000 = $890,000 Incremental B/C = 1,800,000/890,000 = 2.02 Eliminate N Select alternative D. 7.28 Cost of Project Good = 10,000 – 1500 = $8500 Cost of Project Better = 8000 – 2000 = $6000 Cost of Project Best = 20,000 – 16,000 = $4000 Cost of Project Best of All = 14,000 – 3000 = $11,000 Rank alternatives according to increasing cost: DN, Best, Better, Good, Best of All   DN vs Best: Incremental B = 25,000 – 20,000 = $5000 Incremental C = 4000 – 0 = $4000 Incremental B/C = 5000 /4000 = 1.25 Eliminate DN Best vs Better: Incremental B = [(11,000 – 1,000) – (25,000 – 20,000) = $5000 Incremental C = 6000 – 4000 = $2000 Incremental B/C = 5000 /2000 = 2.50 Eliminate Best Better vs Good: Incremental B = [(15,000 – 6000) – (11,000 – 1000) = $-1000 Eliminate Good Better vs Best of All: Incremental B = [(42,000 – 31,000) – (11,000 – 1000) = $1000 Incremental C = 11,000 – 6000 = $5000 Incremental B/C = 1000 /5000 = 0.20 Eliminate Best of All Select project Better 

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7.29

Rank alternatives according to increasing cost: DN, A, F, E, C, D, B Eliminate A based on B/C vs DN < 1.0 F vs DN = 1.08 Eliminate DN F vs E = 1.33 Eliminate F E vs C = 4.00 Eliminate E C vs D = 1.17 Eliminate C D vs B = 0.33 Eliminate B Select alternative D

7.30

Rank alternatives according to increasing cost: DN, G, J, H, I, L, K  Eliminate H and K based on B/C vs DN < 1.0 G vs DN = 1.15 Eliminate DN G vs J = 1.07 Eliminate G J vs I = 1.07 Eliminate J I vs L is not done Must compare I vs L incrementally; survivor is best alternative.

7.31

Rank alternatives according to increasing cost: DN, X, Y, Z, ZZ Eliminate X based on B/C vs DN < 1.0 Y vs DN = 1.07 Eliminate DN Y vs Z = 1.40 Eliminate Y Z vs ZZ = 1.00 Eliminate Eliminate Z

  Select alternative ZZ Problems for Test Review and FE Exam Practice

7.32 Answer is (d) 7.33 Answer is (b) 7.34 Answer is (b) 7.35 Can use either PW, AW, or FW values; For PW, B/C = (245,784 – 30,723)/(100,000 + 61,446) = 1.33 Answer is (b)

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7.36

Alternatives are ranked according to cost; eliminate K and M since B/C