5 Chapter Present Worth Analysis
Solutions to Problems 5.1 A service alternative is one that has only costs (no revenues). 5.2 (a) For independent projects, select all that have PW 0; (b) For mutually exclusive projects, select the one that has the highest numerical value. 5.3 (a) Service; (b) Revenue; (c) Revenue; (d) Service; (e) Revenue; (f) Service 5.4 (a) Total possible = 25 = 32 (b) Because of restrictions, cannot have any combinations of 3,4, or 5. Only 12 are acceptable: DN, 1, 2, 3, 4, 5, 1&3, 1&4, 1&5, 2&3, 2&4, and 2&5. 5.5 Equal service means that the alternatives 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 of the lives of the alternatives. 5.7 Capitalized cost represents the present worth of service for an infinite time. Real world examples that might be analyzed using CC would be Yellowstone National Park, Golden Gate Bridge, Hoover Dam, etc. 5.8 PWold = -1200(3.50)(P/A,15%,5) = -4200(3.3522) = $-14,079 PWnew = -14,000 1200(1.20)(P/A,15%,5) = -14,000 1440(3.3522) = $-18,827 Keep old brackets 5.9 PWA = -80,000 30,000(P/A,12%,3) + 15,000(P/F,12%,3) = -80,000 30,000(2.4018) + 15,000(0.7118) = $-141,377
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PWB = -120,000 8,000(P/A,12%,3) + 40,000(P/F,12%,3) = -120,000 8,000(2.4018) + 40,000(0.7118) = $-110,742 Select Method B 5.10 Bottled water: Cost/mo = -(2)(0.40)(30) = $24.00 PW = -24.00(P/A,0.5%,12) = -24.00(11.6189) = $-278.85 Municipal water: Cost/mo = -5(30)(2.10)/1000 = $0.315 PW = -0.315(P/A,0.5%,12) = -0.315(11.6189) = $-3.66 5.11 PWsingle = -4000 - 4000(P/A,12%,4) = -4000 - 4000(3.0373) = $-16,149 PWsite = $-15,000 Buy the site license 5.12 PWvariable = -250,000 231,000(P/A,15%,6) 140,000(P/F,15%,4) + 50,000(P/F,15%,6) = -250,000 231,000(3.7845) 140,000(0.5718) + 50,000(0.4323) = $-1,182,656 PWdual = -224,000 235,000(P/A,15%,6) 26,000(P/F,15%,3) + 10,000(P/F,15%,6) = -224,000 235,000(3.7845) 26,000(0.6575) + 10,000(0.4323) = $-1,126,130 Select dual speed machine 5.13 PWJX = -205,000 29,000(P/A,10%,4) 203,000(P/F,10%,2) + 2000(P/F,10%,4) = -205,000 29,000(3.1699) 203,000(0.8264) + 2000(0.6830) = $-463,320 PWKZ = -235,000 27,000(P/A,10%,4) + 20,000(P/F,10%,4) = -235,000 27,000(3.1699) + 20,000(0.6830) = $-306,927 Select material KZ
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5.14
PWK = -160,000 7000(P/A,2%,16) 120,000(P/F,2%,8) + 40,000(P/F,2%,16) = -160,000 7000(13.5777) 120,000(0.8535) + 40,000(0.7284) = $-328,328 PWL = -210,000 5000(P/A,2%,16) + 26,000(P/F,2%,16) = -210,000 5000(13.5777) + 26,000(0.7284) = $-258,950 Select process L
5.15
PWplastic = -75,000 - 27,000(P/A,10%,6) - 75,000(P/F,10%,2) - 75,000(P/F,10%,4) = -75,000 - 27,000(4.3553) - 75,000(0.8264) - 75,000(0.6830) = $-305,798 PWaluminum = -125,000 12,000(P/A,10%,6) 95,000(P/F,10%,3) + 30,000(P/F,10%,6) = -125,000 12,000(4.3553) 95,000(0.7513) + 30,000(0.5645) = $-231,702 Use aluminum case
5.16
i/year = (1 + 0.03)2 1 = 6.09% PWA = -1,000,000 - 1,000,000(P/A,6.09%,5) = -1,000,000 - 1,000,000(4.2021) (by equation) = $-5,202,100 PWB = -600,000 600,000(P/A,3%,11) = -600,000 600,000(9.2526) = $-6,151,560 PWC = -1,500,000 500,000(P/F,3%,4) 1,500,000(P/F,3%,6) - 500,000(P/F,3%,10) = -1,500,000 500,000(0.8885) 1,500,000(0.8375) 500,000(0.7441) = $-3,572,550 Select plan C
5.17
FWsolar = -12,600(F/P,10%,4) 1400(F/A,10%,4) = -12,600(1.4641) 1400(4.6410) = $-24,945 FWline = -11,000(F/P,10%,4) 800(F/P,10%,4) = -11,000(1.4641) 800(4.6410) = $-19,818 Install power line
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5.18
FW20% = -100(F/P,10%,15) 80(F/A,10%,15) = -100(4.1772) 80(31.7725) = $-2959.52 FW35% = -240(F/P,10%,15) 65(F/A,10%,15) = -240(4.1772) 65(31.7725) = $-3067.74 20% standard is slightly more economical
5.19
FWpurchase = -150,000(F/P,15%,6) + 12,000(F/A,15%,6) + 65,000 = -150,000(2.3131) + 12,000(8.7537) + 65,000 = $-176,921 FWlease = -30,000(F/A,15%,6)(F/P,15%,1) = - 30,000(8.7537)(1.15) = $-302,003 Purchase the clamshell
5.20
FWHSS = -3500(F/P,1%,6) 2000(F/A,1%,6) 3500(F/P,1%,3) = -3500(1.0615) 2000(6.1520) 3500(1.0303) = $-19,625 FWgold = -6500(F/P,1%,6) 1500(F/A,1%,6) = -6500(1.0615) 1500(6.1520) = $-16,128 FWtitanium = -7000(F/P,1%,6) 1200(F/A,1%,6) = -7000(1.0615) 1200(6.1520) = $-14,813 Use titanium nitride bits
5.21
FWA = -300,000(F/P,12%,10) 900,000(F/A,12%,10) = -300,000(3.1058) 900,000(17.5487) = $-16,725,570 FWB = -1,200,000(F/P,12%,10) 200,000(F/A,12%,10) 150,000(F/A,12%,10) = -1,200,000(3.1058) 200,000(17.5487) 150,000(17.5487) = $-9,869,005 Select Plan B
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5.22
CC = -400,000 400,000(A/F,6%,2)/0.06 = -400,000 400,000(0.48544)/0.06 =$-3,636,267 CC = -1,700,000 350,000(A/F,6%,3)/0.06 = - 1,700,000 350,000(0.31411)/0.06 = $-3,532,308 CC = -200,000 25,000(P/A,12%,4)(P/F,12%,1) [40,000/0.12])P/F,12%,5) = -200,000 25,000(3.0373)(0.8929) [40,000/0.12])(0.5674) = $-456,933 CC = -250,000,000 800,000/0.08 [950,000(A/F,8%,10)]/0.08 - 75,000(A/F,8%,5)/0.08 = -250,000,000 800,000/0.08 [950,000(0.06903)]/0.08 -75,000(0.17046)/0.08 = $-251,979,538 Find AW and then divide by i. AW = [-82,000(A/P,12%,4) 9000 +15,000(A/F,12%,4)] = [-82,000(0.32923) 9000 +15,000(0.20923)]/0.12 = $-32,858.41 CC = -32,858.41/0.12 = $-273,820
5.23
5.24
5.25
5.26
5.27
(a) P29 = 80,000/0.08 = $1,000,000 (b) P0 = 1,000,000(P/F,8%,29) = 1,000,000(0.1073) = $107,300
5.28
Find AW of each plan, then take difference, and divide by i. AWA = -50,000(A/F,10%,5) = -50,000(0.16380) = $-8190 AWB = -100,000(A/F,10%,10) = -100,000(0.06275) = $-6275 CC of difference = (8190 - 6275)/0.10 = $19,150
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5.29
CC = -3,000,000 50,000(P/A,1%,12) 100,000(P/A,1%,13)(P/F,1%,12) - [50,000/0.01](P/F,1%,25) = -3,000,000 50,000(11.2551) 100,000(12.1337)(0.8874) - [50,000/0.01](0.7798) = $-8,538,500 CCpetroleum = [-250,000(A/P,10%,6) 130,000 + 400,000 + 50,000(A/F,10%,6)]/0.10 = [-250,000(0.22961) 130,000 + 400,000 + 50,000(0.12961)]/0.10 = $2,190,780 CCinorganic = [-110,000(A/P,10%,4) 65,000 + 270,000 + 20,000(A/F,10%,4)]/0.10 = [-110,000(0.31547) 65,000 + 270,000 + 20,000(0.21547)]/0.10 = $1,746,077 Petroleum-based alternative has a larger profit.
5.30
5.31
CC = 100,000 + 100,000/0.08 = $1,350,000 CCpipe = -225,000,000 10,000,000/0.10 [50,000,000(A/F,10%,40)]/0.10 = -225,000,000 10,000,000/0.10 [50,000,000(0.00226)]/0.10 = $-326,130,000 CCcanal = -350,000,000 500,000/0.10 = $-355,000,000 Build the pipeline
5.32
5.33 CCE = [-200,000(A/P,3%,8) + 30,000 + 50,000(A/F,3%,8)]/0.03 = [-200,000(0.14246) + 30,000 + 50,000(0.11246)]/0.03 = $237,700 CCF = [-300,000(A/P,3%,16) + 10,000 + 70,000(A/F,3%,16)]/0.03 = [-300,000(0.07961) + 10,000 + 70,000(0.04961)]/0.03 = $-347,010 CCG = -900,000 + 40,000/0.03 = $433,333 Select alternative G.
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5.34 No-return payback refers to the time required to recover an investment at i = 0%. 5.35 The alternatives that have large cash flows beyond the date where other alternatives recover their investment might actually be more attractive over the entire lives of the alternatives (based on PW, AW, or other evaluation methods). 5.36 0 = - 40,000 + 6000(P/A,8%,n) + 8000(P/F,8%,n) Try n = 9: 0 +1483 Try n = 8: 0 -1198 n is between 8 and 9 years 5.37 0 = -22,000 + (3500 2000)(P/A,4%,n) (P/A,4%,n) = 14.6667 n is between 22 and 23 quarters or 5.75 years 5.38 0 = -70,000 + (14,000 1850)(P/A,10%,n) (P/A,10%,n) = 5.76132 n is between 9 and 10; therefore, it would take 10 years. 5.39 (a) n = 35,000/(22,000 17,000) = 7 years (b) 0 = -35,000 + (22,000 17,000)(P/A,10%,n) (P/A,10%,n) = 7.0000 n is between 12 and 13; therefore, n = 13 years. 5.40 250,000 500n + 250,000(1 + 0.02)n = 100,000 Try n = 18: 98,062 < 100,000 Try n = 19: 104,703 > 100,000 n is 18.3 months or 1.6 years. 5.41 Payback: Alt A: 0 = -300,000 + 60,000(P/A,8%,n) (P/A,8%,n) = 5.0000 n is between 6 and 7 years Alt B: 0 = -300,000 + 10,000(P/A,8%,n) + 15,000(P/G,8%,n) Try n = 7: 0 > -37,573 Try n = 8: 0 < +24,558 n is between 7 and 8 years Select A
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PW for 10 yrs: Alt A: PW = -300,000 + 60,000(P/A,8%,10) = - 300,000 + 60,000(6.7101) = $102,606 Alt B: PW = -300,000 + 10,000(P/A,8%,10) + 15,000(P/G,8%,10) = -300,000 + 10,000(6.7101) + 15,000(25.9768) = $156,753 Select B Income for Alt B increases rapidly in later years, which is not accounted for in payback analysis. 5.42 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) - 2.7(P/A,7%,3)(P/F,7%,17) = -6.6 3.5(0.9346) 2.5(0.8734) 9.1(0.8163) 18.6(0.7629) - 21.6(0.7130) - 17(4.1002)(0.7130) 14.2(7.0236)(0.5083) - 2.7(2.6243)(0.3166) = $-151,710,860 5.43 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) -6.3(P/F,6%,5) 1.36(P/A,6%,15)(P/F,6%,5) -3.0(P/F,6%,10) - 3.7(P/F,6%,18) = 2.6(0.9434) 2.0(0.8900) 7.5(0.8396) 10.0(0.7921) -6.3(0.7473) 1.36(9.7122)(0.7473) -3.0(0.5584) - 3.7(0.3503) = $-36,000,921 5.44 LCCA = -750,000 (6000 + 2000)(P/A,0.5%,240) 150,000[(P/F,0.5%,60) + (P/F,0.5%,120) + (P/F,0.5%,180)] = -750,000 (8000)(139.5808) 150,000[(0.7414) + (0.5496) + (0.4075)] = $-2,121,421 LCCB = -1.1 (3000 + 1000)(P/A,0.5%,240) = -1.1 (4000)(139.5808) = $-1,658,323 Select proposal B. 5.45 LCCA = -250,000 150,000(P/A,8%,4) 45,000 35,000(P/A,8%,2) -50,000(P/A,8%,10) 30,000(P/A,8%,5) = -250,000 150,000(3.3121) 45,000 35,000(1.7833) -50,000(6.7101) 30,000(3.9927) = $-1,309,517
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LCCB = -10,000 45,000 - 30,000(P/A,8%,3) 80,000(P/A,8%,10) - 40,000(P/A,8%,10) = -10,000 45,000 - 30,000(2.5771) 80,000(6.7101) - 40,000(6.7101) = $-937,525 LCCC = -175,000(P/A,8%,10) = -175,000(6.7101) = $-1,174,268 Select alternative B. 5.46 I = 10,000(0.06)/4 = $150 every 3 months 5.47 800 = (V)(0.04)/2 V = $40,000 5.48 1500 = (20,000)(b)/2 b = 15% 5.49 Bond interest rate and market interest rate are the same. Therefore, PW = face value = $50,000. 5.50 I = (50,000)(0.04)/4 = $500 every 3 months PW = 500(P/A,2%,60) + 50,000(P/F,2%,60) = 500(34.7609) + 50,000(0.3048) = $32,620 5.51 There are 17 years or 34 semiannual periods remaining in the life of the bond. I = 5000(0.08)/2 = $200 every 6 months PW = 200(P/A,5%,34) + 5000(P/F,5%,34) = 200(16.1929) + 5000(0.1904) = $4190.58 5.52 I = (V)(0.07)/2 201,000,000 = I(P/A,4%,60) + V(P/F,4%,60) Try V = 226,000,000: 201,000,000 > 200,444,485 Try V = 227,000,000: 201,000,000 < 201,331,408 By interpolation, V = $226,626,340
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5.53 (a) I = (50,000)(0.12)/4 = $1500 Five years from now there will be 15(4) = 60 payments left. PW5 then is: PW5 = 1500(P/A,2%,60) + 50,000(P/F,2%,60) = 1500(34.7609) + 50,000(0.3048) = $67,381 (b) Total = 1500(F/A,3%,20) + 67,381 = 1500(26.8704) + 67,381 = $107,687 [PW in year 5 from (a)]
FE Review Solutions 5.54 Answer is (b) 5.55 PW = 50,000 + 10,000(P/A,10%,15) + [20,000/0.10](P/F,10%,15) = $173,941 Answer is (c) 5.56 CC = [40,000/0.10](P/F,10%,4) = $273,200 Answer is (c) 5.57 CC = [50,000/0.10](P/F,10%,20)(A/F,10%,10) = $4662.33 Answer is (b) 5.58 PWX = -66,000 10,000(P/A,10%,6) + 10,000(P/F,10%,6) = -66,000 10,000(4.3553) + 10,000(0.5645) = $-103,908 Answer is (c) 5.59 PWY = -46,000 15,000(P/A,10%,6) - 22,000(P/F,10%,3) + 24,000(P/F,10%,6) = -46,000 15,000(4.3553) - 22,000(0.7513) + 24,000(0.5645) = $-114,310 Answer is (d) 5.60 CCX = [-66,000(A/P,10%,6) 10,000 + 10,000(A/F,10%,6)]/0.10 = [-66,000(0.22961) 10,000 + 10,000(0.12961)]/0.10 = $-238,582 Answer is (d)
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5.61 CC = -10,000(A/P,10%,5)/0.10 = -10,000(0.26380)/0.10 = $-26,380 Answer is (b) 5.62 Answer is (c) 5.63 Answer is (b) 5.64 Answer is (a) 5.65 Answer is (b)
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Extended Exercise Solution Questions 1, 3 and 4:
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Question 2:
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Case Study Solution 1. Set first cost of toilet equal to monthly savings and solve for n: (115.83 76.12) + 50 (A/P,0.75%,n) = 2.1(0.76 + 0.62) 89.71(A/P,0.75%,n) = 2.898 (A/P,0.75%,n) = 0.03230 From 0.75% interest table, n is between 30 and 36 months By interpolation, n = 35 months or 2.9 years 2. If the toilet life were to decrease by 50% to 2.5 years, then the homeowner would not breakeven at any interest rate (2.6 years is required at 0% and longer times would be required for i > 0%). If the interest rate were to increase by more than 50% (say from 9% to 15%), the payback period would increase from 2.9 years (per above solution) to a little less than 3.3 years (from 1.25% interest table). Therefore, the payback period is much more sensitive to the toilet life than to the interest rate. 3. cost/month = 76.12 (A/P,0.5%,60) = 76.12 (0.01933) = $1.47 CCF/month = 2.1 + 2.1 = 4.2 cost/CCF = 1.47/4.2 = $0.35/CCF or $0.47/1000 gallons (vs $0.40/1000 gallons at 0% interest) 4. (a) If 100% of the $115.83 cost of the toilet is rebated, the cost to the city at 0% interest is c= 115.83 (2.1 + 2.1) (12) (5)
= $0.46/CCF or $0.61/1000 gal (vs $0.40/1000 gal at 75% rebate) This is still far below the city's cost of $1.10/1000 gallons. Therefore, the success of the program is not sensitive to the percentage of cost rebated. (b) Use the same relation for cost/month as in question 3 above, except with varying interest rates, the values shown in the table below are obtained for n = 5 years.
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Interest Rate, % $ / CCF $/1000 gal
4 6 0.33 0.35 0.45 0.47
8 0.37 0.49
10 0.39 0.51
12 0.40 0.54
15 0.43 0.58
The results indicate that even at an interest rate of 15% per year, the cost at $0.58/1000 gallons is significantly below the city's cost of $1.10/1000 gallons. Therefore, the program's success is not sensitive to interest rates. (c) Use the same equation as in question 3 above with i = 0.5% per month and varying life values. Life, years 2 3 $ / CCF 0.80 0.55 $/1000 gal 1.07 0.74 4 0.43 0.57 5 0.35 0.47 6 0.30 0.40 8 0.24 0.32 10 0.20 0.27 15 0.15 0.20 20 0.13 0.17
For a 2-year life and an interest rate of a nominal 6% per year, compounded monthly, the cost of the program is $1.07/1000 gallons, which is very close to the savings of $1.10/1000 gallons. But the cost decreases rapidly as life increases. If further sensitivity analysis is performed, the following results are obtained. At an interest rate of 8% per year, the costs and savings are equal. Above 8% per year, the program would not be cost effective for a 2-year toilet life at the 75% rebate level. When the rebate is increased to 100%, the cost of the program exceeds the savings at all interest rates above 4.5% per year for a toilet life of 3 years. These calculations reveal that at very short toilet lives (2-3 years), there are some conditions under which the program will not be financially successful. Therefore, it can be concluded that the program's success is mildly sensitive to toilet life.
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