Inventory Control with Time-Varying Demand

1. Inventory Control with Time-Varying Demand

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

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

4. Characteristics • 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

5. Big versus Small Buckets • 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

6. The Lot Sizing Model

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

10. Example

11. The Lot for Lot Solution

12. The Fixed Order Quantity Solution

13. The Fixed Order Period Solution

14. A Mixed Integer Linear Program (MILP) Formulation

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

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. Example

20. Example • Step 1: Obviously, just satisfy D1 (note we are neglecting production cost, since it is fixed). • Step 2: Two choices, either j2* = 1 or j2* = 2.

21. Example (Continued…) • Step3: Three choices, j3* = 1, 2, 3.

22. Example (Continued…) • Step 4: Four choices, j4* = 1, 2, 3, 4.

23. Property 2 If jk*=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+1.

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. Example (Continued…) • Step 5: Only two choices, j5* = 4, 5. • Step 6: Three choices, j6* = 4, 5, 6. And so on.

26. Example Solution

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. 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. The Network 1 6 2 5 3 4 Node 6 is a pseudo node representing the “end” of the problem.

30. Example Path 1 6 5 2 3 4 Interpretation: Order (or produce) in periods 1, 3, and 4 so that Q1 = D1 + D2 ; Q2 = 0; Q3 = D3; Q4 = D4 + D5 ; and Q5 = 0.

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

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. A Dynamic Programming Algorithm to Find the Least Costly Path Step 1:t = 1, zt* = 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. 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. Example 1 6 2 5 3 4

36. Example (Continued…) 1 6 2 5 3 4

37. Example (Continued…) 1 6 2 5 3 4

38. Example (Continued…) 1 6 2 5 3 4

39. Example (Continued…) 1 6 2 5 3 4

40. Example (Continued…) 1 6 2 5 3 4

41. Example (Continued…) 1 6 2 5 3 4

42. Example (Continued…) 1 6 2 5 3 4

43. Example (Continued…) 1 6 2 5 3 4

44. Example t Dt At ct ht 1 10 40 2 1 2 2 40 2 1 3 12 40 2 1 4 4 40 2 1 5 14 40 2 1

45. Example z1*= 0 c1,2= 40 + 20 = 60 z2*=z1*+ c1,2 = 0+60=60 p2*= 1 c1,3= 40 + 24+2 = 66 c2,3= 40 + 4 = 44 z3* = min (z1*+ c1,3, z2*+ c2,3) = min (66, 104) = 66 p3*= 1

46. Example c1,4= 114, c2,4= 80, c3,4= 64 z4* = min (z1*+ c1,4, z2*+ c2,4, z3*+ c3,4) = min (114, 140, 130) = 114 p4*= 1 c1,5= 134, c2,5 = 96, c3,4= 76, c4,5= 48, z5* = min (134, 156, 142, 162) = 134 p5*= 1 c1,6= 218, c2,6 = 166, c3,6= 132, c4,6= 86, c5,6= 68 z6* = min (218, 226, 198, 200, 202) = 198 p6*= 3

47. Heuristics • 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.

48. Example Heuristics • 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.

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

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: