Fundamentals of supply chain theory by Larry Synder

Fundamentals of supply chain theory

1. Fundamentals of Supply Chain Theory Larry Snyder Z.-J. Max Shen Slides © 2014 Lawrence V. Snyder Chapter 9: Supply Uncertainty

2. 9.1 Introduction to Supply Uncertainty 2 © 2014 Lawrence V. Snyder

3. Introduction 3 We have dealt with demand uncertainty quite a bit, but have had little to say about supply uncertainty Types of supply uncertainty: Disruptions (due to, e.g., bad weather, natural disasters, labor strikes, supplier goes out of business) Yield uncertainty – production process produces defects, and less than 100% of production meets specifications Capacity uncertainty – supplier output in a given time period is random, independent of quantity ordered Lead time uncertainty – can result from supplier shortages or unforeseen production/transportation delays © 2014 Lawrence V. Snyder

4. EOQ with Disruptions © 2014 Lawrence V. Snyder 4 As in the basic EOQ model, we have a constant demand rate of d , a fixed order cost K , and a holding cost of h per unit per year (assume for convenience zero lead time between the retailer and supplier) The supplier goes through cycles of being up and down according to a continuous time Markov process

5. EOQ with Disruptions © 2014 Lawrence V. Snyder 5 If the retailer stocks out when the supplier is down, demand during the down period is lost at a cost of p per unit Uptime and downtime intervals are random The retailer always orders in batches of size Q Let X and Y denote the duration of the up and down interval, respectively X and Y are exponentially distributed with rates  and  , respectively

6. EOQ with Disruptions © 2014 Lawrence V. Snyder 6 The length of a cycle is now random: If the supplier is up when the retailer’s inventory hits zero, then the cycle length T = Q / d ; otherwise T > Q / d For tractability reasons, we ignore the variable purchase cost c Let  denote the probability the supplier is down when the retailer’s inventory hits zero It is possible to show that  

7. EOQ with Disruptions © 2014 Lawrence V. Snyder 7 Let f ( t ) be the pdf of the cycle length T Each cycle lasts at least Q / d ; after this, with probability  it lasts and additional 1/  time units E[T] = Q / d +  /   

8. EOQ with Disruptions © 2014 Lawrence V. Snyder 8 We can calculate the expected cost per cycle and expected cycle length Then, by the renewal-reward theorem, we can calculate expected cost per year, g ( Q ) using Expected inventory cost in a cycle is hQ 2 /2 d , while expected shortage cost is pd  /   

9. EOQ with Disruptions © 2014 Lawrence V. Snyder 9 As a result, we can characterize g ( Q ) as Because  depends on Q in a messy way, we cannot simply take a derivative and set it to zero It turns out that g ( Q ) is quasiconvex , which implies it has only one local minimum, so that we can use a bisection search to find the value of Q that minimizes g ( Q )  

10. EOQ with Disruptions © 2014 Lawrence V. Snyder 10 For any Q we can calculate  and g ( Q ) using and Because we know that g ( Q ) has only one local minimum (and this must be a global minimum), we can quickly do a numerical search to find Q * One simple way is to just keep increasing Q from zero until we see it increase, and stop, although bisection search is more efficient  

11. Example 9.1 © 2014 Lawrence V. Snyder 11 This example uses the Joe’s Corner Store setting from Example 3.1 Joe’s supplier has up and down availability periods with exponential rates of 1.5 and 14 This means that, on average, the supplier is disrupted 1/1.5 = 2/3 years after a past disruption ended, and the disruption lasts, on average, 1/14 years, or 0.8571 months Joe’s demand rate is d = 1300, fixed order cost K = 8, and h = 0.225, with a stockout cost of p = 5 per lost sale

12. Example 9.1 © 2014 Lawrence V. Snyder 12 Using Excel we can plot g ( Q ) as shown below (the minimum occurs around Q = 773, which g ( Q ) = 174

13. Example 9.1 © 2014 Lawrence V. Snyder 13 In Example 3.1, when there were no disruptions, we had Q * = 304.1 and g ( Q * ) = 68.4 Now we have Q * = 773 and g ( Q * ) = 174! As you would expect, we order in much larger batches, which covers our supply over a longer time frame in case the supplier becomes disrupted

14. Newsvendor Problem with Disruptions © 2014 Lawrence V. Snyder 14 On-hand inventory and backorders incur costs of h and p per period Demand is deterministic and equals d units per period Let  = P(supplier down next period|up this period)  = P(supplier up next period|down this period)  is disruption probability;  is recovery probability

15. Newsvendor Problem with Disruptions © 2014 Lawrence V. Snyder 15 Let  u and  d denote the probability of being up and down, respectively, in steady state Then Let  n denote steady-state probability that the supplier is in a disruption state that has lasted n periods (  is the probability of no disruption)  

16. Newsvendor Problem with Disruptions © 2014 Lawrence V. Snyder 16 Let F ( n ) corresponds to steady-state probability that the supplier is in a disruption that has lasted n or fewer periods (including no disruption) Lemma 9.2: , n  1 , n   

17. Newsvendor Problem with Disruptions © 2014 Lawrence V. Snyder 17 Theorem 9.3 states that a base-stock policy is optimal (although we can only order up to the base stock level if the supplier is up) We wish to minimize g ( S ), where g ( S ) is piecwise linear with breakpoints at integer multiples of d ; implies an optimal solution exists at an integer multiple of d  

18. Newsvendor Problem with Disruptions © 2014 Lawrence V. Snyder 18 Theorem 9.5 shows that the optimal base stock level is given by the formula Here F (.) is not a demand distribution, however; it is the supply disruption distribution  

19. Example 9.3 © 2014 Lawrence V. Snyder 19 G&P has an unreliable supplier for a key ingredient for toothpaste. G&P makes toothpaste at a rate of 2000 cases/day. A case has a holding cost of 0.25, while the stockout cost is $3/day. The supplier’s disruption probability is  = 0.04, and its recovery probability is 0.25 (this means a disruption occurs, on average, every 25 days, and is restored after 4 days). What is their optimal base stock level for the ingredient?

20. Example 9.3 © 2014 Lawrence V. Snyder 20 We have d = 2000, h = 0.25, p = 3 We need to compute F ( n ) for n integer and nonnegative Using , n  0, we can compute:   n F(n) 0.8621 1 0.8966 2 0.9224 3 0.9418 4 0.9564 5 0.9673 6 0.9755

21. Example 9.3 © 2014 Lawrence V. Snyder 21 Note that p /( p + h ) = 3/3.25 = 0.9231 We want the smallest n such that F( n )  p /( p + h ) This gives n = 3 (see table below) We therefore have = 2000 + 2000(3) = 8000   n F(n) 0.8621 1 0.8966 2 0.9224 3 0.9418 4 0.9564 5 0.9673 6 0.9755

22. 9.6.1 Introduction 9.6 A Facility Location Model with Disruptions 22 © 2014 Lawrence V. Snyder

23. Introduction 23 UFLP assumes facilities always operate as planned But disruptions may occur Weather, strikes, natural disasters, etc. Customers must be re-assigned to other facilities Model in this section minimizes expected cost, accounting for disruptions Tries to find facility locations that are inexpensive and reliable © 2014 Lawrence V. Snyder

24. 24 49-Node Data Set © 2014 Lawrence V. Snyder

25. 25 Optimal UFLP Solution Transportation cost = $509,000 © 2014 Lawrence V. Snyder

26. 26 Transportation cost increases to $1,081,000 If Sacramento (CA) Facility is Disrupted © 2014 Lawrence V. Snyder

27. 27 If Harrisburg (PA) is Disrupted Transportation cost = $917,000 © 2014 Lawrence V. Snyder

28. 28 If Springfield (IL) is Disrupted Transportation cost = $697,000 © 2014 Lawrence V. Snyder

29. 29 Disruption Costs and Reliability Location % Demand Served Disruption Cost % Increase Sacramento, CA 19% 1,081,229 112% Harrisburg, PA 33% 917,332 80% Springfield, IL 22% 696,947 37% Montgomery, AL 16% 639,631 26% Austin, TX 10% 636,858 25% No disruption 508,858 0% Reliability depends on location and demand served © 2014 Lawrence V. Snyder

30. 30 A More Reliable Solution Max transportation cost = $697,000 (but more facilities) © 2014 Lawrence V. Snyder

31. Cost vs. Reliability 31 Is it worth opening more facilities to gain reliability? Often, yes Large improvements in reliability with small increases in UFLP cost The reliable fixed-charge location problem (RFLP) Snyder and Daskin (2005) Similar model by Berman et al. (2007) Review: Snyder et al. (2006) © 2014 Lawrence V. Snyder

32. 9.6.2 Notation 9.6 A Facility Location Model with Disruptions 32 © 2014 Lawrence V. Snyder

33. Modeling Disruptions 33 Assume each facility is disrupted following 2-state Markov process q = probability that a given facility is disrupted Same for every facility Harder to allow q to depend on j : Berman et al. (2007), Shen et al. (2010), Cui et al. (2010), etc. Estimate q from historical data, forecasting techniques, subjective methods, etc. © 2014 Lawrence V. Snyder

34. Penalty and Emergency Facility 34 q i = cost of not serving customer i , per unit of demand Lost-sales cost Cost to purchase product from competitor, etc. Modeling trick: Add emergency facility u to J f u = (⇒ x u = 1 ) c iu = q i for all i q i q i © 2014 Lawrence V. Snyder

35. Customer Assignments 35 Each customer gets a primary facility and a set of backup facilities Level- r assignment: there are r closer facilities that are open r = 0 : primary assignment r = 1 : first backup assignment, etc. Each customer needs level- r assignment for r = 0 , …, | J | – 1, unless assigned to u at level s < r © 2014 Lawrence V. Snyder

36. 9.6.3 Formulation 9.6 A Facility Location Model with Disruptions 36 © 2014 Lawrence V. Snyder

37. Objective Function 37 Let Then objective function = fixed cost e xpected transportation cost: i is served by level- r facility w.p . q r (1 – q ) e xpected penalty: i is served by u at level r w.p . q r © 2014 Lawrence V. Snyder

38. IP Formulation 38 ( R FLP) Assign every customer at every level, unless already assigned to u C an’t assign customer to same facility at >1 level Integrality Linking constraints Open u © 2014 Lawrence V. Snyder

39. Assignment Ordering 39 Theorem 9.10. In any optimal solution, if then c ij ≤ c ik (Always assign in order of distance) Therefore, don’t need constraints requiring this Proof. Omitted. © 2014 Lawrence V. Snyder

40. 9.6.4 Lagrangian Relaxation 9.6 A Facility Location Model with Disruptions 40 © 2014 Lawrence V. Snyder

41. Lagrangian Relaxation 41 Relax assignment constraints: ( R FLP- LR l ) © 2014 Lawrence V. Snyder

42. Objective Function 42 where … (some algebra) … © 2014 Lawrence V. Snyder

43. Subproblem for j 43 i can only be assigned to j at one level r Choose the one that minimizes Therefore, © 2014 Lawrence V. Snyder

44. Solving ( RFLP- LR l ) 44 Set x j = 1 if b j + f j < , or if j = u Set y ijr = 1 if x j = 1 < 0 minimizes © 2014 Lawrence V. Snyder

45. Other Aspects of the Algorithm 45 Upper bound: Open facilities that are open in subproblem Assign each customer to level- r facilities in increasing order of distance Improvement heuristics Subgradient optimization, branch and bound—like UFLP © 2014 Lawrence V. Snyder

46. 9.6.5 Tradeoff Curves 9.6 A Facility Location Model with Disruptions 46 © 2014 Lawrence V. Snyder

47. Multi-Objective Model 47 Want to minimize: Classical UFLP cost Expected transportation cost And express preference between the two objectives New objective: where © 2014 Lawrence V. Snyder

48. Tradeoff Curve 48 ≤ a ≤ 1 is a constant Large a ⇒ more emphasis on w 1 Small a ⇒ more emphasis on w 2 Typically, solve problem for multiple values of a Generate tradeoff curve Plots both objectives for each solution found aka Pareto curve , efficient frontier Generate by varying a systematically— weighting method ( Cohon 1978) © 2014 Lawrence V. Snyder

49. Tradeoff Curve for RFLP 49 UFLP solution 13.4% better disruption cost 3.1% worse UFLP cost 26.5% better disruption cost 7.3% worse UFLP cost © 2014 Lawrence V. Snyder