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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