1. 1 1 In-Bound Logistics John H. Vande Vate Fall, 2002

2. 2 2 Exam 2

3. 3 3 Scores Average 81 Std. Dev: 11 Remember: All questions about grading submitted IN WRITING Probably made mistakes Happy to review grading Unwilling to discuss grading

4. 4 4 Causes of Bullwhip Today Product Proliferation/Mass Customization –More varieties of products Build-to-Order –Prohibits pooling orders to smooth requirements Lean –Prevents pooling releases to smooth demand on the supply chain

5. 5 5 Why Lean (Just-In-Time)? Reduces inventory –Capital requirements –Etc Reduces handling –Direct-to-Line Improves Quality –See problems quickly Increases launch speed

6. 6 6 Why Not Lean? Capacity Changes in requirements create upstream inventory Changes in requirements raise transport costs Reliability Distant supplies subject to disruption

7. 7 7 Lean Works When Total volume is relatively constant Product variety is limited Changeovers are fast and cheap Suppliers are nearby

8. 8 8 US Auto Industry Total volume is relatively constant Product variety is limited Changeovers are fast and cheap Suppliers are nearby

9. 9 9 How Lean Works Manufacturer has standing PO with supplier –Releases permission to supply against that PO –Daily quantities or even more frequent to match planned production

10. 10 How Lean Works

11. 11 A Financial Model Cash Acct From Revenues Cash Expenses

12. 12 Invest Sell Assets A Financial Model Cash Acct From Revenues Cash Expenses

13. 13 Controls When Cash balance reaches here Invest enough to bring it to here

14. 14 Controls When Cash balance falls to here Sell assets to bring it to here

15. 15 Controls T b t

16. 16 Trade-offs Opportunity cost of Cash Balance Transaction costs of investing and selling assets Set the controls, T, t and b to balance these costs

17. 17 Inventory Analogy Cash Expenses Daily Production reqs. From Revenue Constant supplies Sell Assets Expedited order Invest Excess Curtailed order

18. 18 Trade-offs Opportunity cost of Cash Balance Transaction costs of investing and selling assets Cost of holding Inventory Supply chain costs of expediting and curtailing orders Set the controls, T, t and b to balance these costs

19. 19 Cost Components Cost of Inventory H: an interest rate on the value of the goods ($/item/year) Cost of Expediting E: extra transport costs (above std) ($/event) Cost of Curtailing C: disruption costs - savings over std transport ($/event)

20. 20 Total Cost Minimize  H*Average Inventory Level  E*Expected number of times we expedite  C*Expected number of times we curtail

21. 21 Simple Model Simple model of production requirements Probability Avg demand: a units/day1 – 2p Above avg demand: a +  units p Below avg demand: a -  units p Standard Supply: a units/day

22. 22 A Markov Process 0  22 bb TT tt … … … “Cash Balance” p p p p p p p pp p 1 1 1-2p

23. 23 Transition Matrix Drop . So k refers to k  Instead of b, t, T we find b , t , T 

24. 24 Steady State Probabilities  (i): Steady state probability “cash balance is i*   (0) = p  (1) 2p  (1) = p  (2) 2p  (i) = p  (i-1)+p  (i+1), i=2, 3, …, b-1 2p  (b) =  (0) + p  (b-1)+p  (b+1) 2p  (i) = p  (i-1)+p  (i+1), i=b+1, b+2,…, t 2p  (t) =  (T) + p  (t-1)+p  (t+1) 2p  (i) = p  (i-1)+p  (i+1), i=t+1, t+2,…, T-2 2p  (T-1) = p  (T-2)  (T) = p  (T-1)

25. 25 Steady State Probabilities  (i): Steady state probability “cash balance is i*   (0) = p  (1) =>  (1) =  (0)/p 2p  (1) = p  (2) =>  (2) = 2  (1) = 2  (0)/p 2p  (i) = p  (i-1)+p  (i+1), i=2, 3, …, b-1 2p  (b) =  (0) + p  (b-1)+p  (b+1) 2p  (i) = p  (i-1)+p  (i+1), i=b+1, b+2,…, t 2p  (t) =  (T) + p  (t-1)+p  (t+1) 2p  (i) = p  (i-1)+p  (i+1), i=t+1, t+2,…, T-2 2p  (T-1) = p  (T-2) ==  (T-2) = 2  (T-1) = 2  (T)/p  (T) = p  (T-1) ==  (T-1) =  (T)/p

26. 26 Steady State Probabilities  (i): Steady state probability “cash balance is i*   (0) = p  (1) =>  (1) =  (0)/p 2p  (1) = p  (2) =>  (2) = 2  (1) = 2  (0)/p 2p  (i) = p  (i-1)+p  (i+1), i=2, 3, …, b-1 2p  (2) = p  (1)+p  (3) == 4  (0) =  (0)+p  (3)  (3) = 3  (0)/p 2p  (i) = p  (i-1)+p  (i+1) 2i  (0) = (i-1)  (0)+p  (i+1)  (i+1) = (i+1)  (0)/p

27. 27 Steady State Probabilities  (i): Steady state probability “cash balance is i*   (T) = p  (T-1) =>  (T-1) =  (T)/p 2p  (T-1) = p  (T-2) =>  (T-2) = 2  (T-1) = 2  (T)/p 2p  (i) = p  (i-1)+p  (i+1), i=t+1, t+2,…, T-2 2p  (T-2) = p  (T-1)+p  (T-3) == 4  (T) =  (T)+p  (T-3)  (T-3) = 3  (T)/p 2p  (T-i) = p  (T-i+1)+p  (T-i+1) 2i  (T) = p  (T-i-1)+(i-1)  (T) (i+1)  (T) = p  (T-i-1)  (T-i-1) = (i+1)  (T)/p

28. 28 Summary  (i) = i  (0)/p, i=1, 2, …, b  (T-i) = i  (T)/p, i=1, 2, …, T-t Still have to solve 2p  (b) =  (0) + p  (b-1)+p  (b+1) 2p  (i) = p  (i-1)+p  (i+1), i=b+1, b+2,…, t 2p  (t) =  (T) + p  (t-1)+p  (t+1)

29. 29 Between b and t 2p  (b) =  (0) + p  (b-1)+p  (b+1) 2b  (0) =  (0) + (b-1)  (0)+p  (b+1) b  (0) = p  (b+1)  (b+1) = b  (0)/p

30. 30 Summary  (i) = i  (0)/p, i=1, 2, …, b  (T-i) = i  (T)/p, i=1, 2, …, T-t We just solved 2p  (b) =  (0) + p  (b-1)+p  (b+1)  (b+1) = b  (0)/p Now we have to solve… 2p  (i) = p  (i-1)+p  (i+1), i=b+1, b+2,…, t 2p  (t) =  (T) + p  (t-1)+p  (t+1)

31. 31 Between b and t 2p  (t) =  (T) + p  (t-1)+p  (t+1) 2(T-t)  (T) =  (T) + p  (t-1)+(T-t-1)  (T) (T-t)  (T) = p  (t-1)  (t-1) = (T-t)  (T)/p

32. 32 Summary  (i) = i  (0)/p, i=1, 2, …, b  (T-i) = i  (T)/p, i=1, 2, …, T-t  (b+1) = b  (0)/p We just solved 2p  (t) =  (T) + p  (t-1)+p  (t+1) to get  (t-1) = (T-t)  (T)/p Now we have to solve… 2p  (i) = p  (i-1)+p  (i+1), i=b+1, b+2,…, t

33. 33 Between b and t 2p  (i) = p  (i-1)+p  (i+1), i=b+1, b+2,…, t 2b  (0) = b  (0)+p  (i+1)  (i+1) = b  (0)/p

34. 34 Summary  (i) = i  (0)/p, i=1, 2, …, b  (T-i) = i  (T)/p, i=1, 2, …, T-t  (i) = b  (0)/p, i = b, b+1, …, t And  (t-1) = (T-t)  (T)/p So b  (0)/p = (T-t)  (T)/p  (T) = b  (0)/ (T-t) and  (T-i) = i  (T)/p, i=1, 2, …, T-t becomes  (T-i) = [ib/(T-t)]  (0)/p

35. 35 Summary  (i) = i  (0)/p, i=1, 2, …, b  (i) = b  (0)/p, i = b, b+1, …, t  (i) = [(T-i)/(T-t)]b  (0)/p, i=t, t+1, …, T-1  (T) = b  (0)/(T-t) 1 =  (i) = =  (0) +  (i  (0)/p: i = 1,.., b-1) +  (b  (0)/p: i = b,.., t) +  (ib  (0)/[p(T-t)]: i = 1,.., T-t-1) + b  (0)/[T-t] =  (0) + b(b-1)  (0)/2p + 2(t-b+1)b  (0)/2p + (T-t-1)b  (0)/2p + b  (0)/[T-t] =  (0) + [T+t-b]b  (0)/2p + b  (0)/[T-t] =  (0)[2p(T-t+b) + b(T-t)(T+t-b)]/[(T-t)2p]

36. 36 Calculating  (0) 1 =  (0)[2p(T-t+b) + b(T-t)(T+t-b)]/[(T-t)2p]  (0) = 2p(T-t)/[2p(T-t+b) + b(T-t)(T+t-b)]

37. 37 Costs Expected number of times we expedite  (0)*Number of “Days” in Year Expected number of times we curtail  (T)*Number of “Days” in Year Average Inventory Level  (i  (i): i = 0,.., T)

38. 38 Average Inventory  i  (i) = =  (i 2  (0)/p: i = 1,.., b-1) +  (ib  (0)/p: i = b,.., t-1) +  (i(T-i)b  (0)/[p(T-t)]: i = t,.., T-1) +  Tb  (0)/[T-t] =  b[3T(T-t) – 2(T-t) 2 + 3(t 2 – b 2 ) + 6b-4]  (0)/6p 5 points extra credit for first to find any errors

39. 39 Average Inventory  i  (i) =  b[3T(T-t) – 2(T-t) 2 + 3(t 2 – b 2 ) + 6b-4]  (0)/6p  (0) = 2p(T-t)/[2p(T-t+b) + b(T-t)(T+t-b)]  i  (i) =  b[3T(T-t) – 2(T-t) 2 + 3(t 2 – b 2 ) + 6b-4] (T-t)/ [6p(T-t+b) + 3b(T-t)(T+t-b)]

40. 40 Costs Expected number of times we expedite  (0)*Number of “Days” in Year 2p(T-t)N/[2p(T-t+b) + b(T-t)(T+t-b)] Expected number of times we curtail  (T)*Number of “Days” in Year 2pbN/[2p(T-t+b) + b(T-t)(T+t-b)] Average Inventory Level  (i  (i): i = 0,.., T)  b[3T(T-t) – 2(T-t) 2 + 3(t 2 – b 2 ) + 6b-4] (T-t)/ [6p(T-t+b) + 3b(T-t)(T+t-b)]

41. 41 Convert Variables Focus on the differences b, x, y T b t b x y

42. 42 Optimize Expected number of times we expedite E*  (0)*Number of “Days” in Year E2pyN/[2p(y+b) + by(y+2x+b)] Expected number of times we curtail C*  (T)*Number of “Days” in Year C2pbN/[2p(y+b) + by(y+2x+b)] Average Inventory Level H  (i  (i): i = 0,.., T) H  by[3Ty – 2y 2 + 3x(x+2b) + 6b-4]/ [6p(y+b) + 3by(y+2x+b)]

43. 43 Optimize Total Cost {6pN(Ey+Cb) + H  by[3(b+x+y)y – 2y 2 + 3x(x+2b) + 6b-4]}/ [6p(y+b) + 3by(y+2x+b)]

44. 44 Dell

45. 45 Additional Constraint Can’t curtail more than one days usage On average y  a

46. 46

47. 47 Differences Constant Stream of Releases punctuated by Expediting and Curtailing If supplier can see inventory and knows T, can anticipate and plan for coming expedited and curtailed orders Have to set a lower bound > 0 to protect against disruptions – safety stock Complicates the calculation of cost of Expediting

48. 48 Example: Shipments

49. 49 Example: Inventory