Solutions to Homework _5

# Solutions to - 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

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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. 6-11 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 ∆ (C-A) ∆ (B-A)...
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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 - 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

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