1. Stochastic Models Inventory Control

2. Inventory Forms u Form u Raw Materials u Work-in-Process u Finished Goods

3. Inventory Function u Safety Stock u inventory held to offset the risk of unplanned demand or production stoppages u Decoupling Inventory u buffer inventory required between adjacent processes with differing production rates u Synchronized Production u In-transit (pipeline) u materials moving forward through the value chain u order but not yet produced/received

4. Inventory Function u Cycle Inventory u orders in lot size not equal to demand requirements to lower per unit purchase costs u Decoupling Inventory u buffer inventory required between adjacent processes with differing production rates u Synchronized Production u In-transit (pipeline) u materials moving forward through the value chain u order but not yet produced/received

5. Inventory Function u Seasonal Inventory u produce in low demand periods to meet the needs in high demand periods u anticipatory - produce ahead of planned downtime u In-transit (pipeline) u materials moving forward through the value chain u order but not yet produced/received u Cycle Inventory u orders in lot size not equal to demand requirements to lower per unit purchase costs

6. Inventory Costs u Item Cost (C) u Order Cost (S) u Process Setup Costs u Holding Costs (H) u Function of time in inventory, average inventory level, material handling, utilities, overhead,... u Often calculated as a % rate of inventory cost (iC) u Stockout or Shortage Costs (s) u reflects costs associated with lost opportunity

7. Economic Order Quantity Q r time L T 1 D order arrives Q = reorder quantity r = reorder point D = demand rate L = leadtime T = inventory cycle

8. Economic Order Quantity Q = reorder quantity r = reorder point D = demand rate L = leadtime T = inventory cycle Q r time L T Avg. Inventory

9. 2 Inventory Cost TRC= Total Relevant Cost = Order Cost + Holding Costs = cost per cycle TCU = Total Relevant Cost per Unit Time = TRC/T  SHQT 1 2 S T  HQ 1

10. Cost per unit Time But, T= length of a cycle  Q D TCU SD Q HQ  1 2

11. Cost per Unit Time

12. Note: that minimal per unit cost occurs when holding cost = order cost (per unit time)

13. Economic Order Quantity Find min TCU     TCU QQ SD Q HQ  0 1 2 () -SDQ0 2 1 2   H Q*Q* SD iC  SD H 22

14. Example The monthly demand for a product is 50 units. The cost of each unit is $500 and the holding cost per month is estimated at 10% of cost. It costs $50 for each order made. Compute the EOQ. Q*Q* *50*50.1*500  2 = 10 Sol:

15. Optimal Inventory Cost Recall TCU SD Q HQ  1 2 TCU * SD Q*Q* HQ *  1 2 TCU * SD   H 2 H 1 2 H 2

16. Optimal Inventory Cost TCUHSDiCSD *  22Example: 2*.1*500*50*50TCU *  = $500 per month

17. Orders per year  HD N D Q S  * 2 N = number of orders per year Example: Example: D = 50 / month, Q * = 10 D = 50 x 12 = 600 / yr. H =.1x12x500 = $600 / unit-yr N  600*600 2*50 = 60

18. Cycle Time  T Q D S HD  * 2 T = cycle time Example: Example: D = 50 units/month, Q * = 10  T 2*50 50*50  10 50 =.2 months = 6 days

19. Reorder Point L= lead time r = reorder point = inventory depleted in time L = L*D Example: Example: Lead time for company is 2 days. Demand is 50 units per month or 1.67 units/day. r= 2*1.67 = 3.33 Reorder at 4 units

20. Lead Time Example 2: Example 2: Suppose our lead time is closer to 8 days. r = 8*1.67 = 13.33 but, recall we only order 10 units at a time r = 13.33 - 10 = 3.33

21. Example (cont.) Reorder at 4 units 1 cycle ahead. 10 4 time L T order arrives reorder

22. Sensitivity Q*Q*  SD H 2 Recall that Now suppose we deviate by p amount so that Q = Q * (1+p). What affect does this have on total cost? Let PCP = Percentage Cost Penalty

23. Sensitivity Q*Q*  SD H 2 Recall that Now suppose we deviate by p amount so that Q = Q * (1+p). What affect does this have on total cost? Let PCP = Percentage Cost Penalty PCP TCUQ Q Q x   ()() () * * 100

24. Senstivity (cont.) TCUQ SD Qp HQp() () () * *    1 1 2 1TCU SD Q HQ  1 2 Recall Miracle 1 Occurs 2TCUQSDHp p ()()         1 2 1 1 1

25. Sensitivity (cont.) Recall TCUHSD *  2 2SDHp p ()         1 2 1 1 1 PCP = HSD2 2 x 100 = 50 p p ()         1 1 1 100

26. Sensitivity (cont.) PCP = 50 p p ()         1 1 1 100 Miracle 2 Occurs PCP p p         50 1 2

27. Example Recall that Q * = 10. Suppose now that a minimum order of 15 is introduced. Compute the percentage cost penalty (increase). p    1510 5. PCP    505 15 83 2 (.).. Total relevant costs increase 8.3%

28.  Example 2 Suppose demand forecast increases by 25% so that D = 50(1.25) = 62.5. Then TCU * *.**2  62550 559 or TCU * increases by 11.8%

29. Shortages ImIm r time L T T1T1 T2T2 Q Q-R T 1 = time inventory carriedH = holding cost T 2 = time of stockoutS = order cost I m = max inventory level p = cost per unit short per unit time

30. Inventory Costs TCR = order + holding + shortages  SHI m TpQImIm T 1 2 1 2 12 () Miracle 3 Occurs  HDS Q H p p *  2  H R H p p * 2 +

31. Example Suppose we allow backorders for our previous example. We estimate that the cost of a backorder is $1 per unit per day ($30 / month). Then  *50*50 Q 50 * 30  2 = 16.3 = 17 units

32. Production Model (ELS) ImIm time T1T1 1 P-D T2T2 T Q = batch size order quantity D= demand rate P = production rate P-D = replenish rate during T 1 S = setup costs H = holding cost /unit-time I m = max inventory level

33. Production Model (ELS) ImIm time T1T1 1 P-D T2T2 T T= cycle length = T 1 + T 2 = Q/D T 1 = length of production run = Q/P T 2 = depletion time = I m /D I m = max inventory level = (P-D)T 1 = (P-D)Q/P

34. Costs TC = total costs per cycle = order + holding   S HPDQT P 1 2 () TCU= Cost per unit time TCT  / S T HPDQ P   ()1 2 SD Q HPDQ P   () 2

35. Optimal Q* (ELS)   TCU Q SD Q HPD P   0 2 2 () Solving for Q, Q SDP HPD EOQ D P * ()     2 1

36. Max Inventory I m = max inventory = (P-D)T 1 = (P-D)Q*/P IQ D P ELS D P m               * 11

37. Summary ImIm time T1T1 1 P-D T2T2 T Q SDP HPD EOQ * ()    2 D P  1 IQ D P m         * 1 D P  1

38. Probabilistic Models ImIm time L R=B+LD B Q* D= demand rate B= buffer stock R= reorder point = B+LD D L = actual demand from time of order to time of arrival

39. Probabilistic Model Let  = max risk level for out of stock condition Idea: we want to set a buffer level B so that the probability of running out of stock is < .  > P{out of stock} = P{demand in D L > R} = P{D L > B+LD}

40. Example Prob: Suppose S=$100, H=$.02/day, L=2 days. D = daily demand N(100, 10). From EOQ model, Q* = 1,000 units Find: Buffer level, B, such that probability of out of stock <.05.

41. Solution D L = demand for 2 days = D 1 + D 2 Question: D 1 & D 2 are identically independently distributed normal variates with mean  and standard deviation  =10. What can we say about the distribution of D L ?

42. Prob. Review Suppose we have a random variable X L given by X = Y 1 + Y 2 Then E[X] = E[Y 1 ] + E[Y 2 ] If Y 1 & Y 2 are independent, then xx  2 1 2 2 2 12 2  cov(,)yy xx  2 1 2 2 2 

43. 200 Solution (cont.) Recall D L = demand for 2 days = D 1 + D 2 ~ N(  L,  L ) Then  L = E[D L ] = E[D 1 ] + E[D 2 ] = 200  LDD 22222 12 10 200   L  14.

44. Solution (cont.) D L ~ N(200, 14.14)  B + LD} = P{D L > B +  L }         P D B ll LL         PZ B L 

45. Solution (cont.) Recall for our problem that  =.05 and  L =14.14. Then,.. 05 14       PZ B 0 Z  =1.645  =.05 B 14 1645..  B = 23.3

46. Summary For D ~ N(100,10), L = 2 days, Q* = 1,000 units, and  = risk level =.05 D L ~ N(  L =200,  L =14.14) B = Z   L = 23.3 R = B + D L = B +  L = 23.3 + 200 = 224

47. Optional Replacement In the continuous review model, an order of Q* is made whenever inventory level reaches the reorder point R. We can also utilize periodic review systems with variable order quantities. The two most common are Optional Replacement (s,S) P system ImIm time L R=B+LD B

48. Optional Replacement At t=1, inventory level is above minimum stock level s, no order is made. At t =2, inventory level is below s, order up to S s = R = B+DL S = Q* S time s 12345

49. P System Order up to Target level T at each review interval P. Let D P+L = demand in review period + lead time  P+L = standard deviation of demand in period P+L  = level of risk associated with a stockout T = D P+L + Z   P+L T time 12345

50. Newsboy Problem Often inventory for a single product is met only once; e.g. News Stand (can’t sell day old papers) Pet Rocks Christmas Trees If Q > D, incur costs for Q but revenue only for D If Q < D, incur opportunity costs in form of lost sales

51. Newsboy (cont.) Objective: Determine best order quantity which maximizes expected profit Payoff Matrix: R ij = payoff for order quantity Q i and demand level D j P = profit per unit sold L = loss per unit not sold R PQifQD PDLQDifQD ij iij iijij       ()

52. Newsboy (cont.) Expected Payoff: EPQPR iD j m ij j ()    1 where EP(Qi) = expected payoff for order quantity Qi P = probability of demand level j Rij = payoff for order quantity Qi and demand level Dj D j

53. Example; Newsboy Boy Scout troop 53 plans to sell Christmas trees to earn money. Each tree costs the troop $10 and can be sold for $25. They place no value on lost sales due to lack of trees, L=0. Demand schedule is shown below. DemandP{demand} 1000.10 1200.15 1400.25 1600.25 1800.15 2000.10

54. Example (cont.) Payoff Matrix P = Profit = $25 - $10 per tree sold L = Loss = $-10 per tree not sold Demand Level Order Q100120140160180200 1001,500 1201,3001,800 1401,1001,6002,100 1609001,4001,9002,400 1807001,2001,7002,2002,700 2005001,0001,5002,0002,5003,000

55. Example (cont.) Expected Payoff: Demand Level 0.10.150.25 0.150.1Expected Order Q100120140160180200Payoff 100150225375 225150 1,500 120130270450 270180 1,750 140110240525 315210 1,925 16090210475600360240 1,975 18070180425550405270 1,900 20050150375500375300 1,750 Order quantity has largest expected payoff of $1,975 order 160 trees