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.