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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