存货管理习题解答

存货管理习题解答内容摘要：

g costs: a. cost of shortages (finished goods inventory) b. cost of failures (workinprocess inventory) c. cost of supplier reliability d. cost of ordering e. cost of setups f. cost of quantity discounts g. cost of smoothing production for seasonal products. 2. The possible advantages of using a single supplier include: a. obtaining a discount due to additional volume purchased from the supplier b. building trust and working with the supplier so that the material will be delivered in a timely fashion to avoid stockouts and excess inventory. The possible advantages of using multiple suppliers include: a. the adverse effect of tardiness will be felt much less when there are multiple sources for the materials. Operations Management, 9/e 306 The main advantage of using the fixed order interval model is the reduction in ordering cost because orders for different parts are aggregated during the order interval. The main disadvantage of using the fixed order interval model is that the pany faces the risk of experiencing shortages during the fixed interval. In this particular situation, the advantages of using a single supplier may be offset by using the fixed order interval model. Instructor’s Manual, Chapter 12 307 Solutions 1. Item Usage Unit Cost Usage x Unit Cost Category 4021 50 $1,400 $70,000 A 9402 300 12 3,600 C 4066 40 700 28,000 B 6500 150 20 3,000 C 9280 10 1,020 10,200 C 4050 80 140 11,200 C 6850 2,000 15 30,000 B 3010 400 20 8,000 C 4400 7,000 5 35,000 B In descending order: Item Usage x Cost Category 4021 $70,000 A 4400 35,000 B 6850 30,000 4066 28,000 4050 11,200 C 9280 10,200 3010 8,000 9402 3,600 6500 3,000 Operations Management, 9/e 308 Solutions (continued) 2. The following table contains figures on the monthly volume and unit costs for a random sample of 16 items for a list of 2,000 inventory items. Dollar Item Unit Cost Usage Usage Category K34 10 200 2,000 C K35 25 600 15,000 A K36 36 150 5,400 B M10 16 25 400 C M20 20 80 1,600 C Z45 80 250 16,000 A F14 20 300 6,000 B F95 30 800 24,000 A F99 20 60 1,200 C D45 10 550 5,500 B D48 12 90 1,080 C D52 15 110 1,650 C D57 40 120 4,800 B N08 30 40 1,200 C P05 16 500 8,000 B P09 10 30 300 C a. Develop an ABC classification for these items. [See table.] b. How could the manager use this information? To allocate control efforts. c. Suppose after reviewing your classification scheme, the manager decides to place item P05 into the “A” category. What would some possible explanations be for that decision? It might be important for some reason other than dollar usage, such as cost of a stockout, usage highly correlated to an A item, etc. 3. D = 4,860 bags/yr. S = $10 H = $75 a. b a g s 3675 10)8 6 0,4(2HDS2Q  b. Q/2 = 36/2 = 18 bags c. o r d e r s 135o r d e r s/b a g s 36 b a g s 860,4QD  Instructor’s Manual, Chapter 12 309 Solutions (continued) d. SQDH2/QTC  700,2$350,1350,1)10(36860,4)75(236  e. Using S = $5, Q = 75 )11)(860,4(2  3 1,2$ 1 5, 1 5,1)11(7 5 8 6 0,4)75(27 5  Increase by [$2, – $2,700] = $ 4. D = 40/day x 260 days/yr. = 10,400 packages S = $60 H = $30 a. o x e sb 2 0 0 330 60)4 0 0,10(2HDS2Q 0  b. SQDH2QTC  1 8,6$ 5 8,30 6 0,3)60(2 0 44 0 0,10)30(22 0 4  c. Yes d. )60(202000,10)30(2200TC 200  TC200 = 3,000 + 3,120 = $6,120 6,120 – 6, (only $ higher than with EOQ, so 200 is acceptable.) 5. D = 750 pots/mo. x 12 mo./yr. = 9,000 pots/yr. Price = $2/pot S = $20 P = $50 H = ($2)(.30) = $.60/unit/year a. 7 7 7 460. 20)0 0 0,9(2HDS2Q 0  )20( 7 40 0 0,9)60(.2 7 4TC  TC = + = If Q = 1500 Operations Management, 9/e 310 Solutions (continued) )20(500,1 000,9 )6(.2500,1TC  TC = 450 + 120 = $570 Therefore the additional cost of staying with the order size of 1,500 is: $570 – $ = $ b. Only about one half of the storage space would be needed. 6. u = 800/month, so D = 12(800) = 9,600 crates/yr. H = .35P = .35($10) = $ S = $28 736,1$)28(800600,9)(2800:TCP r e s e n t  a. ]392 t or o u n d[ $ 28)$600,9(2HDS2Q 0  TC at EOQ: . 7 1,1$)28(3 9 26 0 0,9)(23 9 2  Savings approx. $ per year. 7. H = $2/month S = $55 D1 = 100/month (months 1–6) D2 = 150/month (months 7–12) a. 2 55)100(2Q:D HDS2Q 010  2 55)1 5 0(2Q:D 02  b. The EOQ model requires this. c. Discount of $10/order is equivalent to S – 10 = $45 (revised ordering cost) 1–6 TC74 = $ 180$)45(150100)2(2150TC145$)45(100100)2(2100TC*140$)45(50100)2(250TC15010050 Instructor’s Manual, Chapter 12 311 Solutions (continued) 7–12 TC91 = $ 1 9 5$)45(1 5 01 5 0)2(21 5 0TC* 6 7$)45(1 0 01 5 0)2(21 0 0TC1 8 5$)45(501 5 0)2(250TC1 5 01 0 050 8. D = 27,000 jars/month H = $.18/month S = $60 a. .243,418. 60)000,27(2 HDS2Q  TC= SQDH2Q  TC4,000 = $ TC4,243 = $ $ TC4000 = 7 6 5)60(0 0 0,4 0 0 0,27)18(.20 0 0,4  TC4243 = 6 3602 4 3,4 0 0 0,27)18(.22 4 3,4  b. Current: 000,4 000,27QD  ForQDto equal 10, Q must be 2,700 . 1 82 ( 2 7 , 0 0 0 ) S2 , 7 0 0 So HDS2Q  Solving, S = $ c. the carrying cost happened to increase rather dramatically from $.18 to approximately $.3705. H 02 ( 2 7 , 0 0 0 ) 57 0 0,2 HDS2Q  Solving, H = $.3705 D。