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Park F Ch 6 FEE Problem

Park F Ch 6 FEE Problem - Note 1 Unless otherwise stated...

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Note 1 : Unless otherwise stated, all cash flows given in the problems represent after- tax cash flows in actual dollars. The MARR also represents a market interest rate, which considers any inflationary effects in the cash flows. Note 2 : Unless otherwise stated, all interest rates presented in this set of problems assume annual compounding. 6.1 Consider the following cash flows and compute the equivalent annual worth at i = 12%: A n n Investment Revenue 0 -$10,000 1 $2,000 2 $2,000 3 $3,000 4 $3,000 5 $1,000 6 $2,000 $500 Answer AE (12%) = [-$10,000 + $2,000(P/F, 12%, 1) + … +$2,500(P/F, 12%, 6)] (A/P, 12%, 6) = -$180.96 1
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6.2 (A) The following investment has a net present value of zero at i = 8%: X X X X $400 $400 0 1 2 3 4 5 6 Years $2,145 Which of the following is the net equivalent annual worth at 8% interest? (a) $400 (b) $0 (c) $500 (d) $450 Answer (b) NEAW = NPW * (A/P, 8, 6) 0 = -2145 + 400 * (P/F, 8, 1) + 400 * (P/F, 8, 4) + X * (P/A, 8, 2) (P/F, 8, 1) + X * (P/A, 8, 2) (P/F, 8, 4) = -2145 + 400 * [(P/F, 8, 1) + (P/F, 8, 4)] + X * (P/A, 8, 2) [(P/F, 8, 1) + (P/F, 8, 4)] = -2145 + 400 * [.9259 + .7350] + X * 1.7833 * [9259 + .7350] = -2145 + 400 * 1.6609 + X * 2.9619 1475.64 / 2.9619 = X = 498.21 2
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6.3 Consider the following sets of investment projects: Project’s Cash Flow n A B C D 0 -$2,000 -$4,000 -$3,000 -$9,000 1 $400 $3,000 -$2,000 $2,000 2 $500 $2,000 $4,000 $4,000 3 $600 $1,000 $2,000 $8,000 4 $700 $500 $4,000 $8,000 5 $800 $500 $2,000 $4,000 Compute the equivalent annual worth of each project at i = 10%, and determine the acceptability of each project. Answer AE (10%) A = -$2,000(A/P, 10%, 5) + $400 +$100(A/G, 10%, 5) = $53.42 (Accept) AE (10%) B = -$4,000(A/P, 10%, 5) + $500 + [$2,500(P/F, 10%, 1) + $1,500(P/F, 10%, 2) + $500(P/F, 10%, 3)] (A/P, 10%, 5) = $470.47 (Accept) AE (10%) C = [-$3,000 - $2,000(P/F, 10%, 1) + … + $2,000(P/F, 10%, 5)] (A/P, 10%, 5) = $1,045.73 (Accept) AE (10%) D = [-$9,000 + $2,000(P/F, 10%, 1) + … + $4,000(P/F, 10%, 5)] (A/P, 10%, 5) = $2,659.68 (Accept) 3
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6.4 (A) What is the annual-equivalence amount for the following infinite series at i = 12%? $1,200 $700 0 1 2 3 4 5 6 7 8 9 10 11 12 13 … Years (a) $950 (b) $866 (c) $926 (d) None of the above Answer (a) AE (12%) = $700 + [($500/0.12) (P/F, 12%, 6)] (0.12) = $700 + [($500/0.12) (.50663)] (0.12) = $953 NB AE (12%) = $700 +$500 * (P/F, 12%, 6) 4
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6.5 Consider the following sets of investment projects: Period Project’s Cash Flow (n) A B C D 0 -$3,500 -$3,000 -$3,000 -$3,600 1 $0 $1,500 $3,000 $1,800 2 $0 $1,800 $2,000 $1,800 3 $5,500 $2,100 $1,000 $1,800 Compute the equivalent annual worth of each project at i = 13%, and determine the acceptability of each project. Answer AE (13%) A = -$3,500(A/P, 13%, 3) + $5,500(A/F, 13%, 3) = $132.04, Accept AE (13%) B = -$3,000(A/P, 13%, 3) + $1,500 + $300(A/G, 13%, 3) = $505.05, Accept AE (13%) C = -$3,000(A/P, 13%, 3) + $3,000 - $1,000(A/G, 13%, 3) = $810.71, Accept AE (13%) D = -$3,600(A/P, 13%, 3) + $1,800 = $275.32, Accept 5
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6.6 (A) Consider an investment project with the following repeating cash flow pattern every four years for forever: $80 $80 $80 $80 $40 $40 $40 $40 0 1 2 3 4 5 6 7 8 Years What is the annual-equivalence amount of this project at an interest rate of 12%? Answer PW (12%) one-cycle = 80 * (P/F, 12, 1) + 80 * (P/ F, 12, 2) + 40 * (P/F, 12, 3) + 40 * (P/F, 12, 4) = $189.10 AE (12%) = 189.10(A/P, 12%, 4) = $62.25 (.3292) 6
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6.7 The owner of a business is considering investing $55,000 in new equipment. He estimates that the net cash flows will be $5,000 during the first year and will increase by $2,500 per year each year thereafter. The equipment is estimated to have a 10-year service life and a net salvage value at the end of this time of $6,000. The firm’s interest rate is 12%. (a) Determine the annual capital cost (ownership cost) for the equipment. (b) Determine the equivalent annual savings (revenues).
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