Re-Order Point Problems Set 3: Advanced

Problem 1: Problem 7.2 – Average Inventory With Isafety Weekly demand for DVD-Rs at a retailer is normally distributed with a mean of 1,000 boxes and a standard deviation of 150. Currently, the store places orders via paper that is faxed to the supplier. Assume 50 working weeks in a year. Lead time for replenishment of an order is 4 weeks. Fixed cost (ordering and transportation) per order is $100. Each box of DVD-Rs costs $1. Annual holding cost is 25% of average inventory value. The retailer currently orders 20,000 DVD-Rs when stock on hand reaches 4,200. R/week = N(1000,150) L = 2 week Q = 20,000 ROP = 4,200 50 weeks /yr Ave. R /year = 50,000 S =100 C = 1 H=0.25C = $0.25 a.1) How long, on average, does a box of DVD-R spend in the store? Icycle =Q/2 = 20000/2 Isafety = ROP – LTD LTD = L×R = 4×1000 = 4000

Problem 1: Problem 7.2 – Average Inventory With Isafety Average inventory = Icycle + Isafety I = Average Inventory = 20000/2 + 200 = 10200 RT = I =  1000T = 10200  T = 10.2 weeks R/week = N(1000,150) L = 2 week Q = 20,000 ROP = 4,200 50 weeks /yr Ave. R /year = 50,000 S =100 C = 1 H=0.25C = $0.25 a.2) What is the annual ordering and holding cost. Number of orders R/Q R/Q = 50,000/20,000 = 2.5 Ordering cost = 100(2.5) = 250 Annual holding cost = 0.25% ×1 ×10200 =2550 Total inventory system cost excluding purchasing = 250+2550 = 2800. OC ≠ CC for two reason, (i) Not EOQ, (ii) Isafety

Problem 1: Problem 7.2 – Average Inventory With Isafety b.1) Assuming that the retailer wants the probability of stocking out in a cycle to be no more than 5%, recommend an optimal inventory policy: Q and R policy. =6325 Z(95%) = 1.65 Z(95%) = 1.65 Isafety = zσLTD =1.65(300) = 495 ROP = 4000+195 = 4495 Optimal Q and R Policy: Order 6325 whenever inventory on hand is 4495 b.2) Under your recommended policy, how long, on average, would a box of DVD-Rs spend in the store?

Problem 1: Problem 7.2 – Average Inventory With Isafety Average inventory = Icycle+Isafety = Q/2 + Isafety Average inventory = I = 6325/2 + 495 = 3658 RT = I  1000T = 3658  T = 3.66 weeks c) Reduce lead time form 4 to 1. What is the impact on cost and flow time? Safety stock reduces from 495 to 247.5 247.5 units reduction That is 0.25(247.5) = $62 saving Average inventory = Icycle + Isafey = Q/2 + Isafety Average Inventory = I = (6,325/2) + 247.5 = 3,410. Average time in store  I = RT 3410 = 1000T  T = 3.41 weeks

Problem 2: Problem 7.8 – Centralization R/week in each warehouse follows Normal distribution with mean of 10,000, and Standard deviation of 2,000. Compute mean and standard deviation of weekly demand in all the warehouse together – considered as a single central warehouse. Mean (central) = 4(10000) = 40,000 per week Variance (central) = SUM(variance at each warehouse) Variance (central) = 4(variance at each warehouse) Variance at each warehouse = (2000)2 = 4,000,000 Variance (central) = 4(4,000,000) = 16,000,000 Standard deviation (central) = =40000 R (central) = Normal(40,000, 4,000)

Problem 2: Problem 7.8 – Centralization The replenishment lead time (L) = 1 week. Standard deviation of demand during lead time in the centralized system Safety stock at each store for 95% level of service Isafety (central) = 1.65 x 4,000 = 6,600. Average inventory in the centralized system I = Icycle +Isafety I = Q/2 +Isafety 40000/2 +6600 =26600 Average Centralized Inventory = 26600 Average Decentralized Inventory = 53200 Average time spend in inventory (Centralized): RT =I 4000T = 26600  T = 0.67 weeks

Problem 2: Problem 7.8 – Centralization S(R/Q) = 1000(4)(500,000/40,000) = 50,000 H(Isafety + Icycle) = 2.5(26,000) = 66,500 Purchasing cost = CR = 10(500,000) = 5,000,000 We do not consider RC because it does not depend on the inventory policy. But you can always add it. Inventory system cost for four warehouse in centralized system 116,500 Inventory system cost for four warehouses in decentralized system 233,000

Problem 3: Problem 7.3- Lead Time vs Purchase Price Home and Garden (HG) chain of superstores imports decorative planters from Italy. Weekly demand for planters averages 1,500 with a standard deviation of 800. Each planter costs $10. HG incurs a holding cost of 25% per year to carry inventory. HG has an opportunity to set up a superstore in the Phoenix region. Each order shipped from Italy incurs a fixed transportation and delivery cost of $10,000. Consider 52 weeks in the year. R/week = N(1500,800) C = 10 h = 0.25 H =0.25(10) = 2.5 52 weeks /yr R = 78000/yr S =10000 L = 4 weeks SL = 90% a) Determine the optimal order quantity (EOQ). =24980 b) If the delivery lead time is 4 weeks and HG wants to provide a cycle service level of 90%, how much safety stock should it carry?

Problem 3: Problem 7.3- Lead Time vs Purchase Price c) Reduce L from 4 to 1, Increase C by 0.2 per unit. Yes or no? c.1) Rough computations. Purchasing cost increase  0.2(78000) = 15600 Safety stock decrease  from 2048 to 1024  1024 units reduction Safety stock cost saving = 2.5(1024) = 2560 15600-2560 = 13040 increase in cost

Problem 3: Problem 7.3- Lead Time vs Purchase Price c.1) Detailed computations. R/week = N(1500,800) C = 10 + 0.2 H = .25(10+.2) = 2.55 52 weeks /yr R = 78000 S =10000 L = 4 weeks SL = 90% For the original case of C = 10 and H =2.5 +CR +HIs Change in C  Change in H. Not only purchasing cost changes But also cost of EOQ and Is Changes Reduce L from 4 to 1 Purchasing cost increase = 0.2(78000) = 15600 Safety stock reduces from 2048@2.5 to 1024@2.55 Safety stock cost saving = 2.5(2048)-2.55(1024) = 2509

Problem 3: Problem 7.3- Lead Time vs Purchase Price Total inventory cost (ordering + carrying) will also increase R/week = N(1500,800) C = 10 + 0.2 H = .25(10+.2) = 2.55 52 weeks /yr R = 78000 S =10000 L = 4 weeks SL = 90% Total inventory cost (ordering + carrying) increase = 0.00995*62450 = 621 Total impact = +621+15600-2509 = 13712

Problem 3: Problem 7.3- Lead Time vs Purchase Price # of warehouses = 4 R/week at each warehouse N(10,000, 2,000) 50 weeks per year C = 10 H=0.25(10) = 2.5 /year S = 1000 L = 1 week SL = 0.95 =20000 z(0.95) = 1.65, σLTD = 2,000 Is = 1.65 x 2,000 = 3,300. ROP = LTD + Is = 1×10,000 + 3,300 = 13,300. Average inventory at each warehouse : Each time we order Q=20000  Icycle = Q/2 = 20000/2=10000 Average Inventory = Ic + Is =10000+ 3300 =13300 Average Decentralized inventory in 4 warehouses = I = 4(13300) = 53200 Demand in in 4 warehouses = 4(10000) =40000/w RT = I  40000T= 53200 T = 1.33 weeks

Problem 3: Problem 7.3- Lead Time vs Purchase Price OC = S(R/Q) = 1000[(50×10,000)]/Q = 25,000. CC = H (Icycle + Isafety) = H(Q/2+Isafety) = 2.5 (13300) =32,250. We do not consider RC because it does not depend on the inventory policy. But you can always add it. RC = 10 [(50×10,000)]= 5,000,0000 Inventory system cost for one warehouse excluding purchasing TC = 25,000 + 32,000 = 58,200 Inventory system cost for four warehouses = 4(58,250) Inventory system cost for four warehouses = 233,000 Inventory system cost for four warehouses including purchasing = 4(5,000,000) + 233,000 = 20,233,000