Solutions to Homework _4

Solutions to Homework _4 - Solutions to HW#4 IE 1040 Fall...

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Solutions to HW #4 IE 1040 Fall, 2005 5-5 PW(15%) = -\$10,000 + (\$8,000 - \$4000)(P/A,15%,5) - \$1,000 (P/F,15%,5) = -\$10,000 + \$4000(3.3522) - \$1,000 (0.4972) = \$2,911.60 FW(15%) = -\$10,000 (F/P,15%,5) + (\$8,000 - \$4000)(F/A,15%,5) - \$1,000 = -\$10,000 (2.0114) + \$4000(6.7424) - \$1,000 = \$5,855.60 AW(15%) = \$8,000 - \$4000 - [\$10,000(A/P,15%,5) - (- \$1,000)(A/F,15%,5)] = \$4000 - [\$10,000(0.2983) + \$1,000(0.1483)] = \$868.70 5.7 Your Point of View: + 0 1 2 3 4 5 6 (R-E) = \$1,300 (yr. 1) = \$1,300 + (k-1)(\$100) for 1 ≤ k ≤ 6 PW(15%) = -\$10,000 + \$1,300 (P/A, 15%, 6) + \$100 (P/G, 15%, 6) + \$1,200 (P/F, 15%, 6) = -\$10,000 + \$1,300 (3.7845) + \$100 (7.937) + \$12,000 (0.4323) = \$901.15 > 0; the investment appears to be a good one if the risk is low 5.11 Desired yield per year = 10% V N = \$1,000 (P/F, 10%, 10) + 0.14 (\$1,000) (P/A,10%,10) = \$1,000 (.3855) + \$140 (6.1446) = \$1,245.74 \$10,000 EOY \$1,300 \$1,400 \$1,500 \$1,600 \$1,700 \$1,800 \$12k

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Solutions to HW #4 IE 1040 Fall, 2005 5-14 Interest payments from bond = \$100,000 (0.0725) = \$7,250 every six months
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Solutions to Homework _4 - Solutions to HW#4 IE 1040 Fall...

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