【正文】 day Total time per batch = + = 3 days Since the time of production for the second ponent is 4 days, total time required for both ponents is 7 days (3 + 4). Since we have to make batches of the heating element per year, we need ( batches) x (7 days) = days per year. days exceed the number of working days of 250, therefore we can conclude that there is not sufficient time to do the new ponent (job) between production of batches of heating elements. An alternative approach for part d is: The max inventory of 1,250 will last 1250/300 = days – .50 day for setup = days. Since is less than 4 days, there is not enough time. 13. D = 18,000 boxes/yr. S = $96 H = $.60/box per yr. a. Qo = b o x e s 400,260. 96)000,18(2HDS2 ?? Since this quantity is feasible in the range 2020 to 4,999, its total cost and the total cost of all lower price breaks (., 5,000 and 10,000) must be pared to see which is lowest. TC2,400 = 040,23$)000,18($)96($400,2 000,18)60(.2400,2 ??? TC5,000 = ]l o w e st[ ,22$)000,18($)96($000,5 000,18)60(.2020,5 ??? TC10,000 = 7 2,22$)0 0 0,18($)96($0 0 0,10 0 0 0,18)60(.20 0 0,10 ??? Instructor’s Manual, Chapter 12 315 Solutions (continued) Hence, the best order quantity would be 5,000 boxes. b. y e a rp e r r d e r so 000,5 000,18QD ?? 14. a. S = $48 D = 25 stones/day x 200 days/yr. = 5,000 stones/yr. Quantity Unit Price a. H = $2 1 – 399 $10 400 – 599 9 8 92 48)0 0 0,5(2HDS2Q ??? 600 + 8 TC490 = 490 2 + 5,000 48 + 9 (5,000) = $45,980 2 490 TC600 = 600 2 + 5,000 48 + 8 (5,000) = $41,000 2 600 ? 600 is optimum. b. H = .30P 422)9(30.48)000,5(2E O Q$ 8 / s t o n e)at f eas i b l eN o t (NF 447)8(30.48)000,5(2E O Q$98$???? 2,400 5,000 10,000 Quantity TC ? ? ? Lowest TC 422 447 600 TC Quantity Operations Management, 9/e 316 Solutions (continued) (Feasible) Compare total costs of the EOQ at $9 and lower curve’s price break: TC = Q (.30P) + D (S) +PD 2 Q TC422 = 422 [.30($9)] + 5,000 ($48) + $9(5,000) = $46,139 2 422 TC600 = 600 [.30($8)] + 5,000 ($48) + $8(5,000) = $41,120 2 600 Since an order quantity of 600 would have a lower cost than 422, 600 stones is the optimum order size. c. ROP = 25 stones/day (6 days) = 150 stones. 15. Range P H Q D = 4,900 seats/yr. 0–999 $ $ 495 H = .4P 1,000–3,999 497 NF S = $50 4,000–5,999 500 NF 6,000+ 503 NF Compare TC495 with TC for all lower price breaks: TC495 = 495 ($2) + 4,900 ($50) + $(4,900) = $25,490 2 495 TC1,000 = 1,000 ($) + 4,900 ($50) + $(4,900) = $25,490 2 1,000 TC4,000 = 4,000 ($) + 4,900 ($50) + $(4,900) = $27,991 2 4,000 TC6,000 = 6,000 ($) + 4,900 ($50) + $(4,900) = $29,626 2 6,000 Hence, one would be indifferent between 495 or 1,000 units ? ? ? ? 495 497 500 503 1,000 4,000 6,000 Quantity TC Instructor’s Manual, Chapter 12 317 Solutions (continued) 16. D = (800) x (12) = 9600 units S = $40 H = (25%) x P For Supplier A: *178,132$TC560,130850768TC)]600,9)([()40(500600,9))(25(.2500TC,134$TC480,)]600,9)([()40(600,92TC))(25(.)40)(600,9(2Q)f e a s i b l e not( ))(25(.)40)(600,9(2Q500500500?????????????????????????????? For Supplier B: ,133$TC520,)]600,9)([()40(600,9))(25(.2TC))(25(.)40)(600,9(2Q??????????????? Since $132,178 $133, choose supplier A. The optimal order quantity is 500 units. Operations Management, 9/e 318 Solution (continued) 17. D = 3600 boxes per year Q = 800 boxes (remended) S = $80 /order H = $10 /order If the firm decides to order 800, the total cost is puted as follows: 3 2 0,89 6 0,33 6 00 0 0,4TC) 0 0,3(80$8 0 06 0 0,310$28 0 0TC)D*P(SQDH2QTC8 0 0Q8 0 0Q?????????????????????????????????????? Even though the inventory total cost curve is fairly flat around its minimum, when there are quantity discounts, there are multiple U shaped total inventory cost curves for each unit price depending on the unit price. Therefore when the quantity changes from 800 to 801, we shift to a different total cost curve. If we take advantage of the quantity discount and order 801 units, the total cost is puted as follows: 6 4,76 0 0, 5 90 0 5,4TC) 0 0,3(80$8 0 16 0 0,310$28 0 1TC)D*P(SQDH2QTC8 0 1Q8 0 1Q?????????????????????????????????????? The order quantity of 801 is preferred to order quantity of 800 because TCQ=801 TCQ=800 or 8320. 360,6960,3200,1200,1TC),3(80$240600,310$2240TC)D*P(SQDH2QTCb o x es 24010)80)(600,3(2HDS2E O QE O QE O QE O Q??????????????????????????????????????? The order quantity of 800 is not around the flat portion of the curve because the optimal order quantity (EOQ) is much lower than the suggested order quantity of 800. Since the EOQ of 240 boxes provides the lowest total cost, it is the remended order size. Instructor’s Manual, Chapter 12 319 Solution (continued) 18. Daily usage = 800 ft./day Lead time = 6 days Service level desired: 95 percent. Hence, risk should be – .95 = .05 This requires a safety stock of 1,800 feet. ROP = expected usage + safety stock = 800 ft./day x 6 days + 1,800 ft. = 6,600 ft. 19. expected demand during LT = 300 units ?dLT = 30 units a. Z = , ROP = exp. demand + Z?d LT 300 + (30) = ? 370 units b. 70 units (from a.) c. smaller Z ? less safe