Solutions to Homework _3

Solutions to Homework _3 - Solutions to HW #3 4-8 F =...

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Solutions to HW #3 IE 1040 Fall, 2005 4-8 F = $10,000 (F/P, 6%, 5) = $10,000 (1.3382) = $13,382 4-11 $240,000 = $189,500 (1+i) 11 1.26649 = (1+i) 11 11 26649 . 1 - 1 = i, or i = 2.17% per year 4-14 A = $150,000 (A/F, 9%, 20) = $150,000 (0.0195) = $2,925 4-15 A = $5,000 (A/F, 8%, 15) = $5,000 (0.0368) = $184.00 F = $5,000 i = 8% / yr 0 1 2 3 14 15 A = $184 4-21 A = $22,000 0 1 2 3 4 5 i = 15% / yr P = $22,000 (P/A, 15%, 5) = $22,000 (3.3522) = $73,748.40 The company can justify spending up to $73,748.40 for this piece of equipment. 4-31 The value of N (years) at which you become a millionaire is found as follows: $1,000,000 = $10,000 (F/P, 10%, N); 100 = (F/P, 10%, N); 100 = (1.1) N By logarithms, or trial and error, N = 48.317 years = 49 years (to nearest whole year). EOY P = ?
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Solutions to HW #3 IE 1040 Fall, 2005 4-33 P 0 = -$1,000 (P/A, 10%, 5) - $10,000 (P/F, 10%, 15) - $10,000 (P/F, 10%, 30) = -$1,000 (3.7908) - $10,000 (0.2394) - $10,000 (0.0573) = -$6,757.80 A = -$6,757.80 (A/P, 10%, 50) = -$6,757.80 (0.1009) = -$681.86
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This note was uploaded on 06/22/2008 for the course IE 1040 taught by Professor Karenbursic during the Spring '08 term at Pittsburgh.

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Solutions to Homework _3 - Solutions to HW #3 4-8 F =...

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