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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
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
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
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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