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 *
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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 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
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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%)
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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!
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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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