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4 Pages

### Solutions to Homework _5

Course: IE 1040, Spring 2008
School: Pittsburgh
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Word Count: 501

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to Solutions HW #5 IE 1040 Fall, 2005 6-4 Assume all units are produced and sold each year. AWA(20%) = -\$30,000 (A/P,20%,10) + 15,000 (\$3.10 - \$1.00) - \$15,000 + \$10,000 (A/F,20%,10) = \$9,730 AWB(20%) = -\$6,000 (A/P,20%,10) + 20,000 (\$4.40 - \$1.40) - \$30,000 + \$10,000 (A/F,20%,10) = \$16,075 AWC(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...

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to Solutions HW #5 IE 1040 Fall, 2005 6-4 Assume all units are produced and sold each year. AWA(20%) = -\$30,000 (A/P,20%,10) + 15,000 (\$3.10 - \$1.00) - \$15,000 + \$10,000 (A/F,20%,10) = \$9,730 AWB(20%) = -\$6,000 (A/P,20%,10) + 20,000 (\$4.40 - \$1.40) - \$30,000 + \$10,000 (A/F,20%,10) = \$16,075 AWC(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? PWA (i'%) = -\$200,000 + \$90,000 (P/A, i'%,6) = 0 By trial and error, i' = IRRA = 38.7% > MARR, acceptable as base alternative. Or, PWA (15%) = -\$200,000 + \$90,000 (P/A,15%,6)= \$140,605 which is greater than zero, acceptable as base alternative. PWB (15%) = -\$230,000 + \$108,000 (P/A,15%,6)= \$178,726 PWB (15%) = -\$212,500 + -\$15,000(P/F,15%,1) + \$122,500(P/A,15%,5)(P/F, 15%,1)= \$131,552.46 (C-A) - \$12,500 - 105,000 32,500 11.4% No A -\$9,068 No A (B-A) -\$30,000 18,000 18,000 55.8% Yes B \$38,121 Yes B Capital Investment Net cash flow (year 1) Net cash flow (years 2-6) IRR Is increment justified? Current best design PW (15%) Is increment justified? Current best design Both methods result in the decision to invest in design B. Solutions to HW #5 IE 1040 Fall, 2005 6-17 (a) Use AW to deal with different useful lives AWx(5%) = -\$6,000(A/P,5%,12) - \$2,500 = -\$3,176.80 AWy(5%) = -\$14,000(A/P,5%,18) + \$2,800(A/F,5%,18) - \$2,400 = -\$3,497.60 Select Alternative X (can also calculate PW over 36 years and compare) (b) PWx(5%) = -\$6,000 - \$2,500(P/A,5%,12) - \$8,000(P/A,5%,6)(P/F,5%,12) = -\$50,767.45 PWy(5%) = -\$3,497.60(P/A,5%,18) = -\$40,885.54 Select Alternative Y (can also calculate AW over 18 years and compare) 6-24 Assume repeatability. (a) Machine A: CR cost/yr = \$35,000(A/P,10%,10) - \$3,500(A/F,10%,10) = \$5,475.05 Maint. Cost/yr = \$1,000 \$5,475.05 \$1,000 CR and cost/part maintenance = = \$0.6475 10,000 parts \$16 / hr Labor cost/part = = \$5.3333 3 parts / hr Total cost/part (to nearest cent) = \$5.98 Machine B: CR cost/yr = \$150,000(A/P,10%,8) - \$15,000(A/F,10%,8) = \$26,799 Maint. Cost/yr = \$3,000 \$26,799 \$3,000 CR and maintenance cost/part = = \$2.98 10,000 parts \$20 / hr Labor cost/part = = \$3.33 6 parts / hr Total cost/part (to nearest cent) = \$6.31 Select Machine A to minimize total cost per part. Solutions to HW #5 IE 1040 Fall, 2005 (b) Machine A: \$8 / hr = \$2.67 3 parts / hr. (other costs remain unchanged) Total cost/part (to nearest cent) = \$3.31 Labor cost/part = Machine B: \$10 / hr = \$1.67 6 parts / hr. (other costs remain unchanged) Total cost/part (to nearest cent) = \$4.65 Labor cost/part = Select Machine A to minimize total cost per part. 6-25 (a) Assume repeatability. AWA = -\$20,000(A/P,20%,5) - \$5,500 + \$1,000(A/F,20%,5) = -\$12,053.60 AWB = -\$38,000(A/P,20%,10) - \$4,000 + \$4,200(A/F,20%,10) = -\$12,901.30 Select A (b) AWA(20%) = -\$12,053.60 AWB(20%) = -\$38,000(A/P,20%,5) - \$4,000 +\$15,000(A/F,20%,5) = -\$14,691.20 Select A 6-29 CWD1(10%) = [ \$50,000(A P ,10%,20) + \$10,000(A F,10%,20) - \$9,000] 0.10 = -\$147,000 [ \$120,000(A P ,10%,50) + \$20,000(A F,10%,50) - \$5,000] 0.10 = -\$170,900 Select Design D1 to minimize costs. CWD2(10%) = Solutions to HW #5 IE 1040 Fall, 2005 6-39 Assume repeatability. Rank order: DN, Komatsu, Caterpillar, Deere IRR on (Komatsu DN) = 27.2% > 15%, so select Komatsu. IRR on (Caterpillar Komatsu): Set AWK(i') AWC(i') and solve for i' -\$17,000 (A/P, i',5) + \$6,200 + \$3,500 (A/F, i',5) = -\$22,000 (A/P, i',4) + \$7,000 + \$4,000 (A/F, i',4) At i' = 0%, AWC(0%) AWK(0%) = -\$1,000 Therefore, i' < 0% << 15%, so select Komatsu. IRR on (Deere Komatsu): Set AWK(i') AWD(i') and solve for i' -\$17,000 (A/P, i',5) + \$6,200 + \$3,500 (A/F, i',5) = -\$26,200 (A/P, i',10) + \$7,500 + \$5,000 (A/F, i',10) i' 25.1% > 15%, so select Deere.
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