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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 4-35 P = \$3,000 (P/A, 20%, 10) + \$7,000 (P/A, 20%, 5) = \$3,000 (4.1925) + \$7,000 (2.9906) = \$33,511.70 You can afford to pay as much as \$33,511.70 for the equipment. 4-42 Number of monthly deposits = (5 years)(12 months/yr) = 60 \$400,000 = \$200,000(F/P, i' / month, 60) + \$676(F/A, i' / month, 60) Try i' / month = 0.75%: \$400,000 > \$364,126.69, i' / month > 0.75% Try i' / month = 1%: \$400,000 < \$418,548.72, i' / month < 1% Using linear interpolation: i month '/ . \$400, \$364, . . \$418, . \$364, . - - = - - 0 75% 000 126 69 1% 0 75% 548 72 126 69 ; i' / month = 0.9148% Therefore, i' / year = (1.009148) 12 - 1 = 0.1155 or 11.55% per year 4-44 F 5 = \$100 (F/A, 8%, 4) (F/P, 8%, 1) + \$100 (P/A, 8%, 2)

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

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