Session 9 ch11 class.ppt

Based on the remaining useful life of the defender a

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Based on the remaining useful life of the defender , a 5-year planning horizon is used. After 5 years, the challenger will have a market value of \$200,000. With a BTMARR of 16.67%, using an opportunity cost approach , should the defender be replaced?

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Principles of Engineering Economic Analysis , 5th edition Solution to Example 11.9 Based on the incremental cash flows shown below and an EUAC ( equivalent uniform annual cash ) analysis, the old SMP machine ( defender ) should be retained. EUAC 2-1 (16.67%) = \$450,000( A|P 16.67%,5) - \$200,000( A|F 16.67%,5) - \$110,000 = \$890.00 > \$0* =PMT(16.67%,5,-450000,200000)-110000 = \$889.06 * CF(2) – CF(1) identical to cash flow approach *
Principles of Engineering Economic Analysis , 5th edition Solution to Example 11.9 Based on the incremental cash flows shown below and an EUAC analysis, the old SMP machine ( defender ) should be retained. EUAC 2-1 (16.67%) = \$450,000( A|P 16.67%,5) - \$200,000( A|F 16.67%,5) - \$110,000 = \$890.00 > \$0* =PMT(16.67%,5,-450000,200000)-110000 = \$889.06 * CF(2) – CF(1) identical to cash flow approach * Since EUAC 2-1 (16.67%) > \$0, Do Not Replace! Since EUAC 2-1 (16.67%) > \$0, Do Not Replace!

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Principles of Engineering Economic Analysis , 5th edition A filter press was purchased 3 yrs ago for \$30,000. O&M costs are expected to be \$7,000 next year if the filter press is kept; they will increase by \$1,000 per year. In 5 yrs, the filter press can be sold for \$2,000. Its book value is \$12,600. A new press can be purchased for \$36,000. If purchased, the old filter press can be sold for \$9,000. O& M costs will follow a \$1,000 gradient series, with no cost in the first year. At the end of 5 yrs, the new press will have a \$12,000 salvage value. With a 15% MARR , using an opportunity cost approach , what should be done? Example 11.10
Principles of Engineering Economic Analysis , 5th edition Solution to Example 11.10 EOY CF(1) CF(2) CF(2) - CF(1) 0 -\$9,000.00 -\$36,000.00 -\$27,000.00 1 -\$7,000.00 \$0.00 \$7,000.00 2 -\$8,000.00 -\$1,000.00 \$7,000.00 3 -\$9,000.00 -\$2,000.00 \$7,000.00 4 -\$10,000.00 -\$3,000.00 \$7,000.00 5 -\$9,000.00 \$8,000.00 \$17,000.00 Given the incremental cash flows shown below, based on an EUAC analysis, the old filter press should be replaced. EUAC 2-1 (15%) = \$27,000( A|P 15%,5) - \$10,000( A|F 15%,5) - \$7000 = \$27,000(0.29832) - \$10,000(0.14832) - \$7000 = -\$428.56/yr < \$0 =PMT(15%,5,-27000,10000)-7000 = -\$428.64/yr < \$0 The challenger is preferred; replace the old filter press.

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Principles of Engineering Economic Analysis , 5th edition Alternate Solution to Example 11.10 Alternative 1: keep the old press EUAC 1 (15%) = \$9,000(A|P 15%,5) + \$7,000 + \$1,000(A|G 15%,5) - \$2,000 (A|F 15%,5) EUAC 1 (15%) = \$9,000(0.29832) + \$7,000 + \$1,000(1.72281) - \$2,000(0.14832) EUAC 1 (15%) = \$11,111.05/yr Alternative 2: replace the old press EUAC 2 (15%) = \$36,000(A|P 15%,5) + \$1,000(A|G 15%,5) - \$12,000(A|F 15%,5) EUAC 2 (15%) = \$36,000(0.29832) + \$1,000(1.72281) - \$12,000(0.14832) EUAC 2 (15%) = \$10,682.49/yr Replace the Filter Press! EUAC 2 (15%) < EUAC 1 (15%)
Principles of Engineering Economic Analysis , 5th edition Example 11.11 In Example 11.10, suppose the equipment supplier offers a \$10,000 trade-in for the old press. Also, suppose two new alternatives are considered: a \$40,000 filter press having a \$13,000 salvage value in 5 yrs, O&M equal to a \$500 gradient series on a \$500 base, and a \$12,000 trade- in for the old press; and a leased press, with \$7,500 beginning-of-year lease costs and end-of-year O%M cost given by an \$800 gradient series. If leasing is pursued, the old press will be sold for \$9,000.

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Principles of Engineering Economic Analysis , 5th edition Solution to Example 11.11 EUAC 1 (15%) = \$9,000(A|P 15%,5) + \$7,000 + \$1,000( A|G 15%,5) - \$2,000( A|F 15%,5) = \$9,000(0.29832) + \$7,000 + \$1,000(1.72281) - \$2,000(0.14832) = \$11,111.05/yr ( \$11,111.02 with Excel) EUAC 2 (15%) = \$35,000( A|P 15%,5) + \$1,000( A|G 15%,5) - \$12,000( A|F 15%,5) = \$35,000(0.29832) + \$1,000(1.72281) - \$12,000(0.14832) = \$10,384.17/yr ( \$10,384.07 with Excel) EUAC 3 (15%) = \$37,000( A|P 15%,5) + \$500 +\$500( A|G 15%,5) = \$37,000(0.29832) + \$500 + \$500(1.72281) = \$10,471.09/yr ( \$10,470.98 with Excel) EUAC 4 (15%) = \$7,500( F|P 15%,1) + \$800( A|G 15%,5) = \$7,500(1.15000) + \$800(1.72281) = \$10,003.25/yr ( \$10,003.25 with Excel) Lease a New Filter Press!
Principles of Engineering Economic Analysis , 5th edition Optimum Replacement Interval

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Principles of Engineering Economic Analysis , 5th edition As equipment ages, O&M costs increase and CR cost decreases. As a result, the EUAC is often a
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• Fall '17
• Mike Heny
• Economics, Generally Accepted Accounting Principles, Replacements

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