4 Solutions

 

CHAPTER CHAP TER 4 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

PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be

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Chapter 4 4.1 Do-nothing (DN) represents the status quo and is understood to always be an option. If one of the projects absolutely must be selected, the DN alternative is present, it simply will not be selected when the final selection is made. 4.2 (a) Do-nothing, Do-nothing, which is to leave in place the existing equipment. Annual costs for  equipment used now, its estimated remaining life, and an interest rate at which the evaluation will be performed. (b) Cost series, since all estimates are cost cash flows. (c) Conversion to a revenue series requires annual revenue or savings estimates so that net cash flows can be calculated. 4.3

Independent projects are compared against the M MARR, ARR, not each each other; mutually exclusive alternatives compete with each other for selection. Additionally, each independent project usually accomplishes a different objective, whereas mutually exclusive alternatives are different ways to accomplish the same objective.

4.4 (a) A, A, B and C are mutually exclusive; D and E are independent. (b) X is in all bundles as the mutually exclusive selection. The two independent  projectss have 22 = 4 bundles. The 4 viable options are:  project X only

XD

XE

XDE

4.5 Of the 24 = 16 bundles possible, there are 12 acceptable bundles. DN 12 123

1 13 124

2 14

3 23

4 24

4.6 PW = -200,000 + (40,000 – 5000)(P/A,10%,5) + 10,000(P/G,10%,5) = -200,000 +35,000(3.7908) + (10,000(6.8618) = $+1296 Excel: enter cash flows for years 0 through 5 and use the NPV function. 4.7 12% per year compounded monthly is 1% per month. Since PW > 0, the service is financially justified. PW = -600 + 30(P/A,1%,24) = -600 + 30(21.2434) = $+37.30

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4.8 Determine if the deposit’s F value in year 10 equals the $20 million target target amount. First use Equation [2.7] to find P with g = 0.1 and i = 0.0525. all monetary terms are in $ million. P = 1{[1- (1.1/1.0525)10]/-0.0475} = 1{-0.5549/-0.0475} = 11.68236 F = 11.68236(F/P,5.25%,10) = 11,68236(1.66810) = $19.4873 million The deposits fall short of the target by $512,700. A spreadsheet solution follows.

4.9 Subscripts are C for contract service and B for Burling Coop installed. PWC = -75,000(P/A,6%,3) – 100,000(P/A,6%,2)(P/F,6%,3) = -75,000(2.6730) – 100,000(1.8334)(0.8396) = $- 354,407 PWB = -150,000 – 60,000(P/A,6%,5) = $-402,744 The contract service is a better deal with a smaller PW of costs. 4.10 Semiannual bond dividend is 1000(0.05)/2 = $25 $25 per 6 months. Se Semiannual miannual interest rate is 5%/2 = 2.5%. PW = -825 + 25(P/A,2.5%,16) + 800(P/F,2.5%,16) = -825 + 25(13.0550) + 800(0.6736) = $+40.26 Yes, the bond investment does make over the target rate since PW > 0. A spreadsheet solution follows.

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4.11 Semiannual bond dividend is 10,000(0.05)/2 = $250 per 6 months. Semiannual Semiannual interest rate expected is 6%/2 = 3%. PW = -9000 + 250(P/A,3%,40) + 10,000(P/F,3%,40) = -9000 + 250(23.1148) + 10,000(0.3066) = $-155  No, the bond bond investment investment does does not make make the target target rate since since PW < 0. A spreads spreadsheet heet solution follows.

4.12 (a) Calculate the face value V using Equation [4.1]. C Carla arla gets this amount. 1000 = V(0.04)/4 V = $100,000 (b) Based on dividends paid quarterly at 4% per year, and V = $100,000, solve the relation PW = 0 for the number of quarters n. 0 = -100,000 + 1,000(P/A,1%,n) + 100,000(P/F,1%,n) For n = 60: -100,000 + 1,000(44.9550) + 100,000(0.5504) = $0. Maturity is 60/4 = 15 years. By spreadsheet enter = -PV(1%,60,1000,100000)-100000 into a single cell to display $0.00, indicating that n = 60 quarters.

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(c) Purchase price was $95,000. The quarterly rate earned is i% in the PW relation. PW = -95,000 + 1,000(P/A,i ,60) + 100,000(P/F,i,60) Solve manually by trial and error using the interest factors. i = 1%: PW = $+4995 i = 1.25%: PW = $-5505 By interpolation, the earned quarterly rate is 1.11%. Use Equation [3.2]. Effective annual rate = (1.0111)4 -1 = 4.51% per year  For a rapid spreadsheet solution, enter into single cells, =RATE(60,1000,-95000,100000) to display the nominal rate of 1.11% as the quarterly rate, then enter =EFFECT(4.44%,4) to display 4.51%. (This does not use the NPV function; it uses the rate of return function to find i.) 4.13 

Bottled: Cost/mo = -1($1/gal)(30 days/mo) = $30 PW = -30(P/A,0.5%,12) = -30(11.6189) = $-348.57 Filtered: Cost/mo = -2($0.27/gal)(30 days/mo) = $16.20 PW = -16.20(P/A,0.5%,12) = $-188.23 Tap:

Cost/mo = -5($2.75/1000 gal)(30 days/mo) = $0.4125 PW = -0.4125(P/A,0.5%,12) = $-4.79

4.14 Monetary units are in $1000. Calculate PW values to to select the pull system.  PW pull == -1500 + 100(P/F,10%,8) -1500 –– 700(P/A,10%,8) 700(5.3349) + 100(0.4665) = $-5187.780 ($-5,187,780) PW push = -2250 – 600(P/A,10%,8) + 50(P/F,10%,8) – 500(P/F,10%,3) = -2250 – 600(5.3349) + 50(0.4665) – 500(0.7513) = $-5803.265 ($-5,803,265) By spreadsheet, enter the following into single cells to display the PW values. PW pull: =-PV(10%,8,-700000,100000)-1500000 PW push: =-PV(10%,8,-600000,50000)-2250000-PV(10%,3,,-500000)

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4.15 Monetary units are in $1000. Calculate PW values to select Siemens. PWDel = -250 – 231(P/A,5%,6) -140(P/F,5%,4) + 50(P/F,5%,6) = -250 -231(5.0757) – 140(0.8227) + 50(0.7462) = $-1500.355 ($-1,500,355) PWSie = -224 – 235(P/A,5%,6) - 26(P/F,5%,3) + 10(P/F,5%,6) = -224 -235(5.0757) – 26(0.8638) + 10(0.7462) = $-1431.786 ($-1,431,786) By spreadsheet, enter the following into single cells to display the PW values. PWDel: =-PV(5%,6,-231000,50000)-250000 =-PV(5%,6,-231000,50000)-25 0000 - PV(5%,4,,-140000) PWSie: =-PV(5%,6,-235000,10000)-224000 =-PV(5%,6,-235000,10000)-22 4000 - PV(5%,3,,-26000) 4.16 Monetary units are in $ million. Calculate PW values to select alternative B. PWA = -200 – 450(P/A,12%,20) + 75(P/F,12%,20) = -200 - 450(7.4694) + 75(0.1037) = $-3553.4525 ($-3.55 billion) PWB = -350 - 275(P/A,12%,20) + 50(P/F,12%,20) = -350 - 275(7.4694) + 50(0.1037) = $-2398.90 ($-2.40 billion) PWC = -475 – 400(P/A,12%,20) + 90(P/F,12%,20) = -475 - 400(7.4694) + 90(0.1037) = $-3453.427 ($-3.45 billion) By spreadsheet, enter the following to display the PW values in $ million units. PWA:: = = -PV(12%,20,-275,50)-350 -PV(12%,20,-450,75)-200 PW B PWC: = -PV(12%,20,-400,90)-475 4.17 PWProf  = -52,000 - 1000(P/A,9%,4) – 500(P/G,9%,4) = -52,000 - 1000(3.2397) – 500(4.5113) = $-57,495 PWExec = -62,000 - 5000(P/A,9%,4) + 500(P/G,9%,4) = -62,000 - 5000(3.2397) + 500(4.5113) = $-75,943 Select the Professional Plan. A spreadsheet solution follows.

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4.18 For Harold (H), use i = 1% per month and n = 60 months to calculate the PW. PWH = -40,000 - 5000(P/A,1%,60) + 10,000(P/F,1%,60) = -40,000 - 5000(44.9550) + 10,000(0.5504) = $-259,271 For Gwendelyn (G), use effective semiannual i and n = 10 to calculate the PW. Effective i = (1.01)6 -1 = 6.152% PWG = -60,000 – 13,000(P/A,6.152%,10) + 8,000(P/F,6.152%,10) = -60,000 – 13,000(7.30737) + 8,000(0.55045) = $-150,592 Select Gwendelyn’s plan. A spreadsheet solution follows.

4.19 (a) Use LCM = 6 years for PW analysis to select method A. PWA = -100,000[1+(P/F,10%,3)] -30,000(P/A,10%,6) -5,000[(P/G,10%,3) +(P/G,10%,3)(P/F,10%,3)] = -100,000[1.7513] -30,000(4.3553) -5,000[2.3291+(2.3291)(0.7513)] = $-326,184

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PWB = -250,000 – 20,000(P/A,10%,6) = -250,000 – 20,000(4.3553) = $-337,106 (b) Use n = 3 in all calculations and do not repurchase A. Still select method A, now by a larger margin. PWA = -100,000 -30,000(P/A,10%,3) -5,000(P/G,10%,3) = -100,000 -30,000(2.4869) -5,000(2.3291) = $-186,253 PWB = -250,000 – 20,000(P/A,10%,3) = -250,000 – 20,000(2.4869) = $-299,738 A spreadsheet solution for parts (a) and (b) follows.

4.20

PWA = -100,000 -150,000(P/F,10%,3) -30,000(P/A,10%,3) - 5,000(P/G,10%,3) - [40,000(P/A,10%,3) + 5,000(P/G,10%,3)](P/F,10%,3) = -100,000 – 150,000(0.7513) -30,000(2.4869) - 5,000(2.3291) - [40,000(2.4869) + 5,000(2.3291)](0.7513) = $-382,433 PWB = -250,000 – 20,000(P/A,10%,6) = -250,000 – 20,000(4.3553) = $-337,106 Select method B, which is a change from the previous choice.

4.21 (a) PWC = -375,000 -200(P/A,8%,13)(P/F,8%,7) = -375,000 – 200(7.9038)(0.5835) = $-375,922 For asphalt, repave after 10 years and re-start maintenance charge in year 12.

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PWA = -250,000[1+(P/F,8%,10)] – 2500(P/A,8%,9)[(P/F,8%,1)+(P/F,8%,11)] = -250,000[1.4632] – 2500(6.2469)[0.9259 + 0.4289] = $-386,958 Select the concrete option, with a marginal advantage. (b) Maintenance costs are incurred over 5 years only; there are none for concrete.  Now select select the asphalt asphalt option option by a large margin. margin. PWC = $-375,000 PWA = -250,000 – 2500(P/A,8%,4)(P/F,8%,1) = -250,000 – 2500(3.3121)(0.9259) = $-257,667 A spreadsheet solution for parts (a) and (b) follows.

4.22 (a) LCM is 6 years for R and T evaluation. Select vendor T. PWR  = -75,000[1+(P/F,10%,2)+(P/F,10%,4)] – 27,000(P/A,10%,6) = -75,000[1 + 0.8264 + 0.6830] – 27,000(4.3553) = $-305,798 PWT = -125,000[1+(P/F,10%,3)] + 30,000[(P/F,10%,3)+(P/F,10%,6)] -12,000(P/A,10%,6) = -125,000[1 + 0.7513] + 30,000[0.7513+0.5645] – 12,000(4.3553) = $-231,702 (b) Re-purchase R after 2 years. Rental is paid at the end of each year. PWR  = -75,000[1+(P/F,10%,2)] – 27,000(P/A,10%,3) = - 75,000[1 + 0.8264)] – 27,000(2.4869) = $-204,126 4- 9

 

PWT = -125,000 + 30,000(P/F,10%,3) -12,000(P/A,10%,3) = -125,000 + 30,000(0.7513) – 12,000(2.4869) = $-132,304 PWrent = -50,000(P/A,10%,3) =$-124,345 Rental option is the cheapest for a 3-year period. A spreadsheet solution follows.

4.23 (a) PW analysis analysis requires an LCM of 12 years. Select machine D. PWD = -62,000[1+(P/F,15%,4)+(P/F,15%,8)] – 15,000(P/A,15%,12) + 8,000[(P/F,15%,4)+(P/F,15%,8)+(P/F,15%,12)] = -62,000[1 + 0.5718 + 0.3269] – 15,000 (5.4206) + 8,000[0.5718 + 0.3269 + 0.1869] = $-190,344 PWE = -77,000[1+(P/F,15%,6)] – 21,000(P/A,15%,12) + 10,000[(P/F,15%,6)+P/F,15%,12)] = -77,000[1 + 0.4323] – 21,000(5.4206) + 10,000[0.4323 + 0.1869] = $-217,928 (b) Calculate the FW from PW values over the 12-year LCM, or set up FW relations directly from cash flow estimates. Select machine D. FWD = PWD(F/P,15%,12) = -190,344(5.3503) = $-1,018,398 FWE = PWE(F/P,15%,12) = -217,928(5.3503) = $-1,165,980 A spreadsheet solution for parts (a) and (b) follows.

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4.24

PWD = -62,000 - 30,000(P/F,15%,4) -15,000(P/A,15%,5) = -62,000 - 30,000(0.5718) -15,000(3.3522) = $-129,437 PWE = -77,000 - 21,000(P/A,15%,5) + 10,000(P/F,15%,5) = -77,000 - 21,000(3.3522) + 10,000(0.4972) = $-142,424 Select machine D. This is the same decision previously determined.

4.25  Calculate FW in 15 years to select the 35% more efficient freezer. FW20% = -150(F/P,10%,15) - 115(F/A,10%,15) = -150(4.1772) - 115(31.7725) = $-4280 FW35% = -340(F/P,10%,15) - 80(F/A,10%,15) = -340(4.1772) - 80(31.7725) = $-3962 4.26

FWC = -80,000[(F/P,12%,15)+(F/P,12%,10)+(F/P,12%,5)+1] = -80,000[5.4736 + 3.1058 + 1.7623 + 1] = $-907,336

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FWM = -200,000(F/P,12%,15) – 500[(F/P,12%,12)+(F/P,12%,9)+(F/P,12%,6) +(F/P,12%,3)+1] +25,000 = -200,000(5.4736) - 500[3.8969 + 2.7731 + 1.9738 + 1.4049 + 1] + 25,000 = $-1,075,244 Select the concrete exterior by a future worth amount of approximately $168,000.

4.27 Calculate the LCC to select alternative A. All monetary terms are in $ million units. LCCT = -250 - 150(P/A,8%,3) - 45 - 35(P/A,8%,2) -50(P/A,8%,10) - 30(P/A,8%,4) = -250 – 150(2.5771) – 45 – 35(1.7833) -50(6.7101) – 30(3.3121) = $-1178.8485 ($1,178,848,500) ($1,178,848,5 00) LCCA = -10 - 45 - 30(P/A,8%,3) - 100(P/A,8%,10) - 40(P/A,8%,10) = -10 – 45 - 30(2.5771) - 100(6.7101) - 40(6.7101) = $-1071.7270 ($-1,071,727,000) LCCC = -190(P/A,8%,10) = $-1274.9 -190(6.7101) = $-1 274.919 19

($-1,27 ($-1,274,91 4,919,00 9,000) 0)

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4.28 LCC is determined by spreadsheet much easier than than by hand calculator. Select the UPMOST system with the lower LCC.

To use factors, set up LCC relations to select UPMOST. Monetary units are $1000. LCCEMOST = -25 -57(P/A,20%,10) = -25 – 57(4.1925) = $-263.972 ($-263,972) LCCUPMOST= -150[1+(P/F,20%,1)] - 120(P/A,20%,2) - 20(P/A,20%,3) -130(P/A,20%,8)(P/F,20%,2) + [150(P/A,20%,8) + 50(P/G,20%,8)]((P/F,20%,2) = -150[1 + 0.8333] – 120(1.5278) – 20(2.1065) – 130(3.8372)(0.6944) + [150(3.8372) + 50(9.8831)](0.6944) = -846.853 + 742.824 = $-104.029 ($-104,029) 4.29 (a) Earning rate is 0.5% per month. Amount Amount to accumulate by month 20(12) = 240 is the capitalized cost (CC) amount for $5000, which is an F value used to calculate the monthly deposit A. CC = A/i = 5000/0.005 = $1 million A = F(A/F,0.5%,240) = 1,000,000(0.00216) = $2160 per month By spreadsheet, enter =-PMT(0.5%,240,,1000000) to display $2164 per month.

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(b) Effective annual i = [(1.005)12 – 1](100%) = 6.1678% per year  A = 1,000,000(A/F,6.1678%,20) = 1,000,000(0.026698) = $26,698 per year  By spreadsheet, =-PMT(6.1678%,20,,1000000) displays the same amount. 4.30 Monetary terms are in $1000 units. CC = -200 - 300(P/F,6%,4) - 50(A/F,6%,5)/0.06 - (8/0.06)(P/F,6%,14) = -200 - 300(0.7921) - 50(0.1774)/0.06 - (8/0.06)(0.4423) = $-644.437 ($-644,437) 4.31 (a) CC = -85,000,000 – 550,000(A/F,8%,3) + 18,500,000 0.08 0.08 = -85,000,000 – 6,875,000(0.30803) + 231,250,000 = $+144,132,294 (b) Use Equation [4.3] to calculate A. A = CC(i) = 144,132,294(0.08) = $11,530,584 The A means that the toll road should have an equivalent annual cash flow of approximately $11.53 million for the foreseeable future. 4.32 Find AW and then divide by i, according to Equation [4.2]. AW = -97,000(A/P,12%,4) - 10,000 + 20,000(A/F,12%,4) = -97,000(0.32923) - 10,000 + 20,000(0.20923) = $-37,751 CC = -37,751/0.12 = $-314,589   The CC value is the present worth of the computer system cost assuming its need will last ‘forever’. That is, the function will be provided at the same costs as those used here. This assumes that all costs will change the same as inflation and deflation over  time. 4.33 CC = 100,000 100,000 + 100,000/0.05 = $2,100,000

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4.34 Monthly rate is 0.18/12 = 0.015. Calculate total debt allowed iin n month n, then determine the time to accumulate this amount. CC = A/i = 500/0.015 = $33,333 1000(F/A,1.5%,n) = 33,333 By spreadsheet, enter =NPER(1.5%,-1000,,33333) to display n = 27.2 months. By trial and error, n = 27.2 (interpolated) using the following. For n = 25: 1000(F/A,1.5%,25) = $30,063 For n = 28: 1000(F/A,1.5%,28) = $34,482   Tom can spend at the $1000 per month rate for only 2 years and 3 months before obligating himself to a $500 per month debt payment for the rest of his life! 4.35 Monetary terms are $ million units. Determine Determine CC values to select select undersea route.

 

CCland == -225 -225 -- 20/0.10 20/0.10 -- [50(A/F,10%,40)]/0.10 [50(0.00226)]/0.10 = $-426.13 ($-426,130,000) CCsea = -350 - 2/0.10 = $-370.0 ($-370,000,000)

4.36 Monetary terms are $1000 units. Determine CC values of revenues minus costs to select plan 2. (Note: Revenues do not come close to covering costs.) CC1 = +190(P/A,6%,5) + 20(P/G,6%,5) + 270((P/A,6%,5)(P/F,6%,5) + [350/0.06](P/F,6%,10) - 40,000 – 8,000(P/F,6%,10) – 250/0.06 = +190(4.2124) + 20(7.9345) + 270(4.2124)(0.7473) [5833.33](0.5584) - 40,000 – 8,000(0.5584) – 4166.67 =+ +5066.32 – 48,633.87 = $-43, $-43,56 567. 7.55 552 2 ($-4 ($-43, 3,56 567, 7,55 552) 2) CC2 = +260(P/A,6%,7) + 30(P/G,6%,7) + [440/0.06](P/F,6%,7) - 42,000 - 300/0.06 = +260(5.5824) + 30(15.4497) + 7333.33(0.6651) – 42,000 – 5,000 = +6792.315 – 47,000 = $$-40 40,2 ,207 07.6 .685 85 ($-4 ($-40, 0,20 207, 7,68 685) 5) 4.37 Monetary terms are $1000 units. Determine CC values of revenues minus costs to select design B. Use Equation [4.2], CC = AW/i. CCA = [+800 + 50(A/F,10%,6) - 2,500(A/P,10%,6) - 130]/0.10 = [+800 + 50(0.12961) - 2,500(0.22961) –130]/ 0.10 4- 15

 

CCA = $+102.4555/0/10 = $+1024.555

($+1,024,555)

CCB = [+625 + 20(A/F,10%,4) -1,100(A/P,10%,4) – 65]/0.10 = [+625 + 20(0.21547) -1,100(0.31547) – 65]/0.10 = 217.292/0.10 = +$21 217 72.9 2.924 ($+2, $+2,1 172,9 2,924 24)) A spreadsheet solution follows.

4.38 Monetary terms are $1000 units. Determine CC values of to select select wells. CCdam = -10,000 – 25/0.05 = $-10,500 ($-10,500,000) CCwells = AW/i = (-1,500(A/P,5%,10) – 120]/0.05 = -314.25/0.05 = $-628 -6285. 5.0 00 ($-6 $-6,285 ,285,0 ,000 00)) By spreadsheet, enter the following into single cells to display the CC values. Select wells alternative. Dam: = -10000000 -25000/0.05 Wells: = (-PMT(5%,10,-1500000) -120000)/0.05

Display: CC = $-10,500,000 Display: CC = $-6,285,137

4.39 Quarterly interest rate is 12/4 = 3% with 4 quarters per year. Use Equation [4.2], CC = AW/i; select alternative E. Monetary values are in $1000 units. CCE = [-2000(A/P,3%,16) + 300 + 50(A/F,3%,16)]/0.03 = [-2000(0.07961) + 300 + 50(0.04961)]/0.03 = $+4775.35 ($+4,775,350)   CCF = [-3000(A/P,3%,32) + 100 + 70(A/F,3%,32)]/0.03 = [-3000(0.04905) + 100 + 70(0.01905)]/0.03 = $-1527. 27.217 ($-1,527,217)

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CCG = -10,000 + 400/0.03 = $+33 3333 33..33 333 3 ($+3 $+3,333 ,333,3 ,333 33)) By spreadsheet, enter the following into single cells to display the CC values ans select alternative E. E: = (-PMT(3%,1 (-PMT(3%,16,-20000 6,-2000000,50000) 00,50000)+300000 +300000)/0.03 )/0.03 Display: Display: $+4,775,295 $+4,775,295 F: = (-PMT(3%,32, (-PMT(3%,32,-300000 -3000000,70000)+ 0,70000)+100000 100000)/0.03 )/0.03 Display: Display: $-1,526,886 $-1,526,886 G: = -10000000+400000/0.03 Display: $+3,333,333 4.40 Independent projects are considered one-time investments and are compared to the the MARR (not each other), so they are evaluated over their respective lives. Mutually exclusive alternatives are assumed to be needed over multiple life cycles and are compared to each other, thus the need for equal-service comparison. 4.41 (a) Select 2, 3 and 4 with PW > 0 at 12%. (b) Of 24 = 16 bundles, list acceptable bundles and PW values. Select projects 2 and 3 with largest PW = $9000. Bundle DN 2 3 4 23 34

Investment 0 $-15,000 -20,000 -40,000 -45,000 -60,000

PW 0 $ 8,500 500 7,600 9,000 8,100

4.42 Determine the PW for each project. PWA = -1,500,000 + 360,000(P/A,10%,8) = $420,564 PWB = -3,000,000 + 600,000(P/A,10%,10) = $686,760 PWC = -1,800,000 + 620,000(P/A,10%,5) = $550,296 PWD = -2,000,000 + 630,000(P/A,10%,4) = $-2,963 (not acceptable) By spreadsheet, enter the following to display the project PW values. A: = -PV( -PV(10 10%, %,8,3 8,360 6000 000) 0)-15 -1500 0000 000 0 B: = -PV(10 -PV(10%,10 %,10,600 ,600000 000)-30 )-30000 00000 00 C: = -P -PV( V(10 10%, %,5,6 5,6200 20000 00)-1 )-1800 80000 000 0 D: = -PV -PV(10 (10%, %,4,6 4,630 3000 000) 0)-20 -2000 0000 000 0

Di Displ splay: ay: $420 $420,57 ,573 3 Display Display:: $686,740 $686,740 Di Displ splay: ay: $5 $550 50,28 ,288 8 Di Displ splay; ay; $-2, $-2,98 985 5

Formulate acceptable bundles from the 24 = 16 possibilities, without both B and C and select projects with largest total PW of a bundle. (a) With, b = $4 million, select projects A and C with PW = $970,860.

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Bundle DN A B C AC

Investment, $ million 0 -1.5 -3.0 -1.8 -3.3

PW, $ 0 420,564 686,760 550,296 970,860

(b) With, b = $5.5 million, select projects A and B with PW = $1,107,313.

Bundle DN A B C AB AC

Investment, $ million 0 -1.5 -3.0 -1.8 -4.5 -3.3

PW, $ 0 420,564 686,760 550,296 1,107,313 970,860

(c) With no-limit, select all with PW > 0. Select projects A, B and C. 4.43 Hand calculate each project’s PW using P/F factors since all cash flows are different each year. Or, use a spreadsheet as follows.

Use b = $20,000 to formulate bundles from the 24 = 16 possibilities. Select projects X and Z with PW = $4,856.

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Bundle DN W X Y Z WX WY WZ XY XZ YZ

Investment, $ 0 -5,000 -8,000 -8,000 -10,000 -13,000 -13,000 -15,000 -16,000 -18,000 -18,000

PW, $ 0 2,011 2,360 1,038 2,496 4,371 3,049 4,507 3,398 4,856 3,534

4.44 Determine PW values at 0.5% per m month onth by spreadsheet using the PV PV function =-PV(0.5%,36,revenue)-cost, or by hand, as follows. PWauto = -4500 + 220(P/A,0.5%,36) = $2732 PWweb = -3000 + 200(P/A,0.5%,36) = $3574 PWfast = -2200 + 140(P/A,0.5%,36) = $2402 $2402 With 23 = 8 bundles and b = $7000, select the last two features (web interface and fast search software) with PW = $5976. Bundle DN Auto Web Fast Auto/fast Web/fast

Investment, $ 0 -4500 -3000 -2200 -6700 -5200

PW, $ 0 2732 3574 2402 5134 5976

Problems for Test Review and FE Exam Practice

4.45 Answer is (b). 4.46 Answer is (c).   4.47 PW = -65,000[1+(P/F,12%,4)+(P/F,12%,8)] -65,000[1+(P/F,12%,4)+(P/F,12 %,8)] -15,000(P/A,12%,12) + 25,000[(P/F,12%,4)+(P/F,12%,8)+(P/F,12%,12)] = -65,000[1 + 0.6355 + 0.4039] -15,000(6.1944) + 25,000[0.6355 + 0.4039 + 0.2567] = $-193,075 Answer is (a).

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4.48 FW = -65,000[(F/P,12%,8)+(F/P,12%,4)] -65,000[(F/P,12%,8)+(F/P,12 %,4)] – 15,000(F/A,12%,8) + 25,000[(F/P,12%,4)+1] = -65,000[(2.4760) + (1.5735)] – 15,000(12.2997) + 25,000[(1.5735) + 1] = $-383,376 Answer is (a). 4.49 Answer is (d). 4.50 Answer is (b). 4.51 Answer is (c). 4.52 Answer is (a). 4.53 eff i = er  – 1 = e0.1 – 1 = 0.1052 or 10.52% and CC = AW/i Answer is (d). 4.54 PWB = CCB = -90,000 – 4000/0.08 = -140,000 Answer is (d). 4.55 CC in year 5 = 40,000/0.10 = $400,000 PW = 400,000(P/F,10%,4) = $273,200 Answer is (c). 4.56 Bond dividend = 50,000(0.06)/4 = $750 per quarter  Answer is (b). 4.57 Bond dividend = 20,000(0.04)/2 = $400 Answer is (a). 4.58 Bond dividend = 5000(0.05)/4 = $62.50 per quarter  PW = 62.50(P/A,1.5%,20) + 5000(P/F,1.5%,20) = $4786 Answer is (c).

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