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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