# Solution - Chapter 6: Annual Cash Flow Analysis 6-1 C = \$15...

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Unformatted text preview: Chapter 6: Annual Cash Flow Analysis 6-1 C = \$15 + \$15 (A/G, 10%, 4) = \$15 + \$15 (1.381) = \$35.72 6-2 B = [\$100 + \$100 (F/P, 15%, 4)] (A/F, 15%, 5) = [\$100 + \$100 (1.749)] (0.1483) = \$40.77 6-3 E = \$60 - \$15 (A/G, 12%, 4) = \$60 - \$15 (1.359) = \$39.62 \$15 \$30 \$45 \$60 C C C B B B B \$10 \$10 \$60 \$45 \$30 \$15 E E E 6-4 D = [\$100 (F/P, 6%, 2) + \$200 (F/P, 6%, 4)] (A/F, 6%, 6) = [\$100 (1.124) + \$200 (1.262)] (0.1434) = \$52.31 6-5 \$500 = D (F/A, 12%, 3) + 0.5D + D (P/A, 12%, 2) = D (3.374 + 0.5 + 1.690) D = \$500/5.564 = \$89.86 6-6 x = \$40 + \$10 (P/A, 10%, 4) + \$20 (P/F, 10%, 1) + \$10 (P/F, 10%, 2) D D D D D \$200 \$100 D D D D 1.5D \$50 \$40 \$30 \$20 \$10 x C C C C = \$40 + \$10 (3.170) + \$20 (0.9091) + \$10 (0.8264) = \$98.15 C = \$98.15 (A/P, 10%, 4) = \$98.15 (0.3155) = \$30.97 6-7 P = \$40 (P/A, 10%, 4) - \$10 (P/G, 10%, 4) + [\$20 (P/A, 10%, 3) + \$10 (P/G, 10%, 3)] (P/F, 10%, 4 = \$40 (3.170) - \$10 (4.378) + [\$20 (2.487) + \$10 (2.329)] (0.6830) = \$132.90 A = \$132.90 (A/P, 10%, 7) = \$132.90 (0.2054) = \$27.30 6-8 There is a repeating series:; 100 200 300 200. Solving this series for A gives us the A for the infinite series. A = \$100 + [\$100 (P/F, 10%, 2) + \$200 (P/F, 10%, 3) \$40 \$30 \$20 \$10 \$20 \$30 \$40 P . n = \$10 \$20 \$300 \$20 \$10 \$20 \$30 \$200 A Pattern repeats infinitely + \$100 (P/F, 10%, 4)] (A/P, 10%, 4) = \$100 + [\$100 (0.8254) + \$200 (0.7513) + \$100 (0.6830)] (0.3155) = \$100 + [\$301.20] (0.3155) = \$195.03 6-9 A = \$100 (A/P, 3.5%, 3) = \$100 (0.3569) = \$35.69 6-10 EUAC = \$60,000 (0.10) + \$3,000 + \$1,000 (P/F, 10%, 1) (A/P, 10%, 4) = \$6,000 + \$3,000 + \$1,000 (0.9091) (0.3155) = \$9,287 This is the relatively unusual situation where Cost = Salvage Value. In this situation the annual capital recovery cost equals interest on the investment. If anyone doubts this, they should compute: \$60,000 (A/P, 10%, 4) - \$60,000 (A/F, 10%, 2). This equals P*i = \$60,000 (0.10) = \$6,000. 6-11 Prospective Cash Flow: Year Cash Flow-\$30,000 1-8 +A 8 +\$35,000 EUAC = EUAB \$30,000 (A/P, 15%, 8) = A + \$35,000 (A/F, 15%, 8) \$100 A A A \$30,000 (0.2229) = A + \$35,000 (0.0729) \$6,687 = A + \$2,551.50 A = \$4,135.50 6-12 This problem is much harder than it looks! EUAC = {\$600 (P/A, 8%, 5) + \$100 (P/G, 8%, 5) + [\$900 (P/A, 8%, 5) \$100 (P/G, 8%, 5)][(P/F, 8%, 5)]}{(A/P, 8%, 10)} = {\$600 (3.993) + \$100 (7.372) + [\$900 (3.993) - \$100 (7.372)][0.6806]}{0.1490} = \$756.49 6-13 EUAC = \$30,000 (A/P, 8%, 8) - \$1,000 - \$40,000 (A/F, 8%, 8) = \$30,000 (0.1740) - \$1,000 - \$40,000 (0.0940) = \$460 The equipment has an annual cost that is \$460 greater than the benefits. The equipment purchase did not turn out to be desirable. \$60 \$70 \$80 \$90 \$1,000 \$90 \$80 \$70 \$60 \$50 A = ? 6-14 The 1.5% interest table does not contain n = 516. The problem must be segmented to use the 1.5% table....
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## This note was uploaded on 04/18/2011 for the course ENGINEER 300 taught by Professor Kashayar during the Spring '11 term at California State University Los Angeles .

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Solution - Chapter 6: Annual Cash Flow Analysis 6-1 C = \$15...

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