MSCI 432/633: Production and Service Operations Management, Winter MSCI /633: Assignment # 3; MSCI : Assignment # 2

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1 MSCI 43/633: Production and Service Operations Management, Winter 010 Q1. Discounting schemes. MSCI /633: Assignment # 3; MSCI 43.00: Assignment # a. Find the optimal ordering quantities for each of the items and the corresponding annual total cost. [3 marks X3] b. What is the primary difference between the ordering decisions made in a? Illustrate your answer graphically. [ marks]

2 Q. Limit on shelf life. Simply find T EOQ EOQ/ and if T EOQ >SL then Q SL SL*λ Q3. Uncertain demand. a. Assume that the firm has centralized all inventories in a single warehouse and that the probability of stocking out in a cycle can still be no more than 5%. Ideally, how much average inventory can the company now expect to hold, and at what cost? In this case, how long will a unit spend in the warehouse before being sold? [5 marks]

3 Solution: (Since pipeline inventory is not influenced, we do not include it in the calculations below.) (a) To determine the optimal order quantity, we will use the EOQ formula. Observe that H $10 * 5%/yr $.5/yr, d 10,000 /wk, (i.e., annual demand, D 500,000/yr), and S $1000. Optimal order quantity at each warehouse DS 500, Q H. 5 0,000. Demand is normally distributed with: Mean weekly demand 10,000 units Std dev. 000 units Since replenishment lead time is 1 week: LT 1 week And, the standard deviation of demand during lead time at each warehouse: σ LT LT*σ d,000. For a 95% desired service level, z Safety stock (SS) at each warehouse for 95% level of service z*σ LT 1.65*,000 3,300. Average lead time demand (σ LT ) 10,000. Then, the reorder point (ROP) σ LT* z+ SS 13,300. (Not necessary to calculate ROP in this question.) Average inventory at each warehouse (I) Q/ + SS (0,000/) + 3,300 13,300 units. Average inventory holding cost per warehouse H*I $.5*13,300 $33,50. Number of orders per year at each warehouse D/Q 500,000/0,000 5 order/year. Annual Order cost per warehouse S*R/Q $1,000 * 5 $5,000. Average time unit spends in warehouse (by Little s Law) I/R 13,300/10, weeks. Since each warehouse is identical, the total average inventory across four warehouses 4* Average inventory in each warehouse 4*(13,300) 53,00. Annual order cost for all four warehouses 4* Annual order cost per warehouse 4* $5,000 $100,000. Annual holding cost for all four warehouses 4 * Average annual holding cost per warehouse 4* $33,50 $133,000. [b] The total average demand that the centralized warehouses faces 4* Average demand per warehouse 4 * 10,000 40,000 units per week. We assume that the centralized warehouse has the same cost structure as the individual warehouses. The optimal order quantity for the centralized warehouse is determined using the EOQ formula with d 40,000 per week (i.e., Annual Demand D 50*40,000,000,000) and the remaining parameters as in [a]. This gives Q 40,000 units. Standard deviation of weekly demand at the central warehouse SQRT(# of warehouses) * std dev of weekly demand at one warehouse SQRT(4)* 000 4,000 / week. With a replenishment lead time of 1 week, the standard deviation of lead time demand at central warehouse (σ LT ) SQRT(LT)*( σ d ) 4,000. Safety stock (SS) at central warehouse for 95% level of service

4 z*σ LT 1.65*4,000 6,600. Average lead time demand (σ LT ) 40,000 units. Reorder point (ROP) σ LT + SS 46,600 units. (Not necessary to calculate ROP in this question.) Average inventory in central warehouse Q/ + SS (40,000/) + 6,600 6,600. Number of orders per year the central places D/Q 50*40,000/40, orders/year. Annual order cost for central warehouse 50*$1,000 $50,000. Annual holding cost for central warehouse H*I $.5 * 6,600 $66,500. Average time unit spends in warehouse I/R 6,600/40, weeks. Q4 is a bonus question! Q4. Order quantity under inflation.

5 Q5 and Q6 are required only from MSCI43.001/633 students Q5 [Nahmias 4.0] EPQ. (a) Calculate the number of JJ39877 filters that Filter Systems should produce in each production run of this particular part, so as to minimize annual holding and set-up costs? [ marks] (b) Assuming that it produces the optimal number of filters in each run, compute the maximum level of on-hand inventory of these filters that the firm has at any point in time. [1.5 marks] (c) Assuming that the policy in part (1a) is used, calculate the percentage of the working time that the company spends producing these particular filters. [1.5 marks] Solution a) h' I*c(1-λ/P)(.)(.50)[1 - (00*1)/(50*6*0*1)].5317 Setup cost, K (100+55)(1.5) Q* ()(3.5)(400) ,449 b) H Q (1 - λ/p) (1449)(1-400/7,000) 1,401. c) T Q/λ 1449/ years T 1 Q/P 1,449/7, years.001/ or 3.3% of each cycle is up-time. Q6.[based on Nahmias 4.]. (a) Which source should be used, and what is the size of the optimal order? (b) If the replenishment lead time for wafers is three months, determine the reorder point based on the onhand inventory of wafers.

6 (c) Assume that two years have passed, and the purchasing agent is reconsidering the optimal number of wafers to purchase and from which source to purchase them. Source A still has the same pricing policy. Source B now accepts orders of any size, but sells the wafers for $.55 each for orders of up to 3000 wafers, and $.5 each for the incremental amount ordered over 3000 wafers. Source C went out of business. Which source should now be used, and what is the new size of the optimal order? Solution: a) λ 0,000 All Units Discount K 100 I 0.0 c 0 $.50 c 1 $.40 c $.30 Q (0) Kλ 0 ()(100)(0,000) (.0)(.50) 88(realizable) Q (1) Kλ 1 ()(100)(0,000) (.)(.40) 887(not realizable) Q () Kλ ()(100)(0,000) (.)(.30) 949(not realizable) Only Q (0) is realizable. Cost at Q 4,000 (0,000)(.30) + $47,40 Cost at Q 3,000 (0,000)(.40) + $49, (.)(.30)(4,000) (100)(0,000) + (4,000) (.)(.40)(3,000) (100)(0,000) + 3,000 Cost at Q Q (0) 88 (0,000)(.50) + Kλ IC > 50,000 0 It follows that the optimal order size is Q 4,000. Hence, source C should be selected. b) T Q/λ 4000/ years; τ 3 months 0.5 years τ/t 0.5/ cycles > 1. Therefore, we use cycle for the purpose of calculating reorder point R. 0.5 cycle (0.5)(0.) 0.05 year R λτ (0,000)(0.05) 1000 units

7 c) Incremental schedule for source B. For Q < 3,000, c (price/unit) charged by source B ($.55) is greater than that charged by source A, it is, therefore, never optimal to use source B for Q < 3,000. Therefore, need only compare source A and source B for Q > we C(Q).50Q for Q 3,000 (Source A) C(Q) (.55)(3,000) +.5(Q - 3,000) for Q 3, Q Q for Q 3,000 C(Q)/Q.55 for Q 3, /Q +.5 for Q 3,000 (Source B) For source B for Q > 3000, * 0, Q G(Q) 0, Q Q Q 18,000,000,000,000 (.0)(.5) Q $45, 090 Q Q 0,000, Q + $45, 090 Q The minimizing Q occurs where d(c(q))/d(q) 0 0,000,000 Q* Cost 0,000,000 + (.5) (948) + $45, ,090 $49,33.64 Note: There is an easier way to solve this by noting the structure of cost function, C(Q). C(Q) Q for Q 3,000 This suggests that 900 is an additional setup cost under the new incremental discount price scheme. Therefore, K ' λ ()( )(0,000) Q* 948 (.0)(.5) Since the annual cost for source A exceeds $50,000 (as shown in part a), now source B should be used and the new size of optimal order is 948 units.