1. 1 Inventory Control with Time-Varying Demand

2. 2  Week 1Introduction to Production Planning and Inventory Control  Week 2Inventory Control – Deterministic Demand  Week 3Inventory Control – Stochastic Demand  Week 4Inventory Control – Stochastic Demand  Week 5Inventory Control – Stochastic Demand  Week 6Inventory Control – Time Varying Demand  Week 7Inventory Control – Multiple Echelons Lecture Topics

3. 3  Week 8Production Planning and Scheduling  Week 9Production Planning and Scheduling  Week 10Managing Manufacturing Operations  Week 11Managing Manufacturing Operations  Week 12 Managing Manufacturing Operations  Week 13Demand Forecasting  Week 14Demand Forecasting  Week 15Project Presentations Lecture Topics (Continued…)

4. 4  Demand varies from period to period  The demand for each period is exactly known  Costs may vary from period to period  Capacity may vary from period to period Characteristics

5. 5  Big time-bucket models  Items produced/ordered in a period can be used to satisfy the demand for that period  Small time-bucket models  Production/supply leadtimes can take multiple periods Big versus Small Buckets

6. 6 The Lot Sizing Model

7. 7 Assumptions for the Basic Model  Demand varies from period to period but is exactly known for each period.  Demand in each period must be satisfied during the same period (backordering is not allowed).  There are no limits on how much can be produced or ordered.  Items produced/ordered in a period are available to satisfy demand during the same period (big bucket model).  Setup/ordering, production/purchasing, and inventory holding costs can vary period to period.

8. 8 Objective Determine the optimal order quantity ( lot size ) in each period so that the demand in each period is met while the sum of ordering, purchasing, and inventory holding costs are minimized.

9. 9 Notation  t : a period (e.g., day, week, month); t = 1, …, T, where T represents the planning horizon  D t : demand in period t (number of units)  c t : unit purchasing/production cost  A t : ordering/setup cost associated with placing an order (or initiating production) in period t  h t : cost of holding one unit of inventory from period t to period t +1  Q t : the size of the order (or lot size ) in period t ; a decision variable

10. 10 Example

11. 11 The Lot for Lot Solution

12. 12 The Fixed Order Quantity Solution

13. 13 The Fixed Order Period Solution

14. 14 A Mixed Integer Linear Program (MILP) Formulation

15. 15 Solution Approach  Solve as a standard MILP (using for example a branch and bound algorithm); several commercial MILP solver software tools are available  Develop a customized solution that takes advantage of structural properties specific to the problem (e.g., the Wagner-Whitin algorithm)

16. 16 Property 1 Under an optimal lot-sizing policy either the inventory carried to period t+1 from the previous period will be zero or the production quantity in period t+1 will be zero.

17. 17 The Basic Idea of the Wagner- Whitin Algorithm  Using property 1, either Q t = 0 or Q t =D t +…+D k for some k.  If j k * = t = last period of production (or ordering) in a k period problem, then we will produce (or order) exactly D t + D t+ 1 …D k in period j k *.  We can then consider periods 1, …, j k * - 1 as if they are an independent j k * -1 period problem.

18. 18 The Basic Idea of the Wagner- Whitin Algorithm (Continued…)  Construct an algorithm where the decision is whether or not to order in a given period. If we order, then the order quantity should be just enough to cover demand until the next period in which we order.  Solve a series of smaller sub-problems (a one period problem, a two period, …., N period problem), where the solution to each sub-problem is used in solving the next subproblem.

19. 19 Example

20. 20 Example Step 1: Obviously, just satisfy D 1 (note we are neglecting production cost, since it is fixed). Step 2: Two choices, either j 2 * = 1 or j 2 * = 2.

21. 21 Example (Continued…) Step3: Three choices, j 3 * = 1, 2, 3.

22. 22 Example (Continued…) Step 4: Four choices, j 4 * = 1, 2, 3, 4.

23. 23 Property 2 If j k * =t, then the last period in which ordering/production occurs in an optimal k+1 period policy must be in the set t, t+ 1,…k+ .

24. 24 Property 2 (Continued…) In the Example:  We order in period 4 for period 4 of a 4 period problem.  We would never order in period 3 for period 5 in a 5 period problem.

25. 25 Example (Continued…) Step 5: Only two choices, j 5 * = 4, 5. Step 6: Three choices, j 6 * = 4, 5, 6. And so on.

26. 26 Example Solution

27. 27 Example Solution (Continued…) Optimal Policy:  Order in period 8 for 8, 9, 10 (40 + 20 + 30 = 90 units)  Order in period 4 for 4, 5, 6, 7 (50 + 50 + 10 + 20 = 130 units)  Order in period 1 for 1, 2, 3 (20 + 50 + 10 = 80 units) Note: we order in 7 for an 8 period problem, but this never comes into play in optimal solution.

28. 28 A Network Representation The lot sizing problem can be represented as a network, where each node t represents a period and an arc from node t’ to node t represents the fact that we order (or produce) in both periods t’ and t but not in periods in between.

29. 29 2 34 5 61 The Network Node 6 is a pseudo node representing the “end” of the problem.

30. 30 2 34 5 61 Example Path Interpretation: Order (or produce) in periods 1, 3, and 4 so that Q 1 = D 1 + D 2 ; Q 2 = 0; Q 3 = D 3 ; Q 4 = D 4 + D 5 ; and Q 5 = 0.

31. 31 Arc Costs The cost c t’,t of reaching node t from t ’ is the cost of ordering in t ’ but not in t ’+1, t ’+2, …, t -1:

32. 32 Key Insight Finding the minimum cost solution is equivalent to finding the least costly path (shortest path) in the network to go from node 1 to node T +1, where T is number of periods.

33. 33 A Dynamic Programming Algorithm to Find the Least Costly Path Step 1: t = 1, z t * = 0 Step 2: t = t +1. If t > T +1, stop. Otherwise go to step 3. Step 3: For all t’ = 1, 2, …, t - 1, Step 4: Compute

34. 34 Step 5: Compute (that is, choose the period t’ that minimizes ) Step 6: Go to to step 2. The optimal cost is given by The optimal set of periods in which ordering/production takes place can be obtained by backtracking from

35. 35 2 34 5 61 Example

36. 36 2 34 5 61 Example (Continued…)

37. 37 2 34 5 61 Example (Continued…)

38. 38 2 34 5 61 Example (Continued…)

39. 39 2 34 5 61 Example (Continued…)

40. 40 2 34 5 61 Example (Continued…)

41. 41 2 34 5 61 Example (Continued…)

42. 42 2 34 5 61 Example (Continued…)

43. 43 2 34 5 61 Example (Continued…)

44. 44 Example tD t A t c t h t 1104021 224021 3124021 444021 5144021

45. 45 Example z 1 *= 0 c 1,2 = 40 + 20 = 60 z 2 *= z 1 *+ c 1,2 = 0+60=60 p 2 *= 1 c 1,3 = 40 + 24+2 = 66 c 2,3 = 40 + 4 = 44 z 3 * = min ( z 1 *+ c 1,3, z 2 *+ c 2,3 ) = min (66, 104) = 66 p 3 *= 1

46. 46 Example c 1,4 = 114, c 2,4 = 80, c 3,4 = 64 z 4 * = min ( z 1 *+ c 1,4, z 2 *+ c 2,4, z 3 *+ c 3,4 ) = min (114, 140, 130) = 114 p 4 *= 1 c 1,5 = 134, c 2,5 = 96, c 3,4 = 76, c 4,5 = 48, z 5 * = min (134, 156, 142, 162) = 134 p 5 *= 1 c 1,6 = 218, c 2,6 = 166, c 3,6 = 132, c 4,6 = 86, c 5,6 = 68 z 6 * = min (218, 226, 198, 200, 202) = 198 p 6 *= 3

47. 47  Instead of solving the problem optimally, we could use a heuristic (a rule) that leads to reasonably good solutions but not necessarily optimal.  The advantage of heuristics is ease of implementation and lower computational effort to reach a solution. Heuristics

48. 48  Choose a fixed order quantity and order in multiples of this order quantity. Order again when demand in a period cannot be met from available inventory.  Choose a fixed order period P. Then, every P periods order all the demand for the next P periods.  Use a greedy heuristic such as the Silver-Meal heuristic. Example Heuristics

49. 49 The Silver-Meal Heuristic Starting with a period t, order for the next k periods if the resulting average cost per period z t,t+k is smaller than the average cost per period if we ordered only for the next ( k -1) periods.

50. 50 The Silver-Meal Algorithm Step 1: Set t = 1 Step 2: Step 3: t ’= t +1 Step 4: If t’ > T, go to step 7. Otherwise go to step 5 Step 5:

51. 51 Step 6: If z t,t’  z t,t’ -1, set t’ = t’ +1 and go to step 4. Otherwise go to step 7. Step 7: Set Q t = D t + D t+1 + … + D t’ -1 Step 8: Set t=t’. Step 9: If t > T, stop. Otherwise, go to step 2.

52. 52 Example tD t A t c t h t 1104021 224021 3124021 444021 5144021

53. 53 Example z 1,1 = 40 + 20 = 60 z 1,2 = (60 + 4 + 2)/2 = 66/2=33 < 60 z 1,3 = 114/3 = 38>33 So the heuristic sets Q 1 = 12. Next, z 3,3 = 64 z 3,4 = 76/2 = 36 < 64 z 3,5 = 132/3 = 44>36 So, Q 3 = 16. Next,

54. 54 z 5,5 = 68. Since we have reached the end of the planning horizon, the heuristic sets Q 5 = 14.