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Unformatted text preview: Solutions to HW #5 IE 1040 Fall, 2005 6.4 Assume all units are produced and sold each year. AW A (20%) = $30,000 (A/P,20%,10) + 15,000 ($3.10  $1.00)  $15,000 + $10,000 (A/F,20%,10) = $9,730 AW B (20%) = $6,000 (A/P,20%,10) + 20,000 ($4.40  $1.40)  $30,000 + $10,000 (A/F,20%,10) = $16,075 AW C (20%) = $40,000 (A/P,20%,10) + 18,000 ($3.70  $0.90)  $25,000 + $10,000 (A/F,20%,10) = $16,245 Select Design C to minimize the annual worth. 611 The ranking of the alternatives by increasing capital investment is A, C, and B. Is Alternative A an acceptable base alternative? PW A (i'%) = $200,000 + $90,000 (P/A, i'%,6) = 0 By trial and error, i' = IRR A = 38.7% > MARR, ∴ acceptable as base alternative. Or, PW A (15%) = $200,000 + $90,000 (P/A,15%,6)= $140,605 which is greater than zero, ∴ acceptable as base alternative. PW B (15%) = $230,000 + $108,000 (P/A,15%,6)= $178,726 PW B (15%) = $212,500 + $15,000(P/F,15%,1) + $122,500(P/A,15%,5)(P/F, 15%,1)= $131,552.46 ∆ (CA) ∆ (BA)...
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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.
 Spring '08
 KarenBursic

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