Selected Solutions - 5

Selected Solutions - 5.25 CC =-250,000,000 – 800,000/0.08 –[950,000(A/F,8,10/0.08 75,000(A/F,8,5/0.08 = $-251,979,538 5.28 Find AW of each plan

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SOLUTIONS TO SELECTED PROBLEMS Student: You should work the problem completely before referring to the solution. CHAPTER 5 Solutions included for problems 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58, 61, and 64 5.1 A service alternative is one that has only costs (no revenues). 5.4 (a) Total possible = 2 5 = 32 (b) Because of restrictions, cannot have any combinations of 3,4, or 5. Only 12 are 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. 5.10 Bottled water: Cost/mo = -(2)(0.40)(30) = $24.00 PW = -24.00(P/A,0.5%,12) = $-278.85 Municipal water: Cost/mo = -5(30)(2.10)/1000 = $0.315 PW = -0.315(P/A,0.5%,12) = $-3.66 5.13 PW JX = -205,000 – 29,000(P/A,10%,4) – 203,000(P/F,10%,2) + 2000(P/F,10%,4) = $-463,320 PW KZ = -235,000 – 27,000(P/A,10%,4) + 20,000(P/F,10%,4) = $-306,927 Select material KZ 5.16 i/year = (1 + 0.03) 2 – 1 = 6.09% PW A = -1,000,000 - 1,000,000(P/A,6.09%,5) = -1,000,000 - 1,000,000(4.2021) (by equation) = $-5,202,100 PW B = -600,000 – 600,000(P/A,3%,11) = $-6,151,560 Chapter 5 1
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PW C = -1,500,000 – 500,000(P/F,3%,4) – 1,500,000(P/F,3%,6) - 500,000(P/F,3%,10) = $-3,572,550 Select plan C 5.19 FW purchase = -150,000(F/P,15%,6) + 12,000(F/A,15%,6) + 65,000 = $-176,921 FW lease = -30,000(F/A,15%,6)(F/P,15%,1) = $-302,003 Purchase the clamshell 5.22 CC = -400,000 – 400,000(A/F,6%,2)/0.06 =$-3,636,267
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Unformatted text preview: 5.25 CC = -250,000,000 – 800,000/0.08 – [950,000(A/F,8%,10)]/0.08 - 75,000(A/F,8%,5)/0.08 = $-251,979,538 5.28 Find AW of each plan, then take difference, and divide by i. AW A = -50,000(A/F,10%,5) = $-8190 AW B = -100,000(A/F,10%,10) = $-6275 CC of difference = (8190 - 6275)/0.10 = $19,150 5.31 CC = 100,000 + 100,000/0.08 = $1,350,000 5.34 No-return payback refers to the time required to recover an investment at i = 0%. 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 5.40 –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 Chapter 5 2 n is 18.3 months or 1.6 years. 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) = $-36,000,921 5.46 I = 10,000(0.06)/4 = $150 every 3 months 5.49 Bond interest rate and market interest rate are the same. Therefore, PW = face value = $50,000. 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 5.55 PW = 50,000 + 10,000(P/A,10%,15) + [20,000/0.10](P/F,10%,15) = $173,941 Answer is (c) 5.58 PW X = -66,000 –10,000(P/A,10%,6) + 10,000(P/F,10%,6) = $-103,908 Answer is (c) 5.61 CC = -10,000(A/P,10%,5)/0.10 = $-26,380 Answer is (b) 5.64 Answer is (a) Chapter 5 3...
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This homework help was uploaded on 04/01/2008 for the course ISE 2014 taught by Professor Cpkoelling during the Spring '07 term at Virginia Tech.

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Selected Solutions - 5.25 CC =-250,000,000 – 800,000/0.08 –[950,000(A/F,8,10/0.08 75,000(A/F,8,5/0.08 = $-251,979,538 5.28 Find AW of each plan

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