METRIC

1. METRIC Multi-Echelon Technique for Recoverable Item Control Craig C.Sherbrooke Presented by: Nuriye KAPTANLAR Y.Emre KARAMANOGLU 03.03.2003

2. Contents • General Description • Structure of the Multi-Echelon Problem a.Mathematical Assumptions b.Multi-Echelon Theory • A simple example for METRIC • Conclusion and Interpretations

3. General Description: • METRIC: is a mathematical model of a base-depot supply system in which item demand is compound Poisson with a mean value estimated by Bayesian procedure. • METRIC:is a mathematical model translated into a computer program,capable of determining base and depot stock levels for a group of recoverable items.

4. General Description: [ Cont’d] • The metric theory is the basis used by the several military services because of the existence of recoverable items which have * high cost * low demand * long lead times

5. Recoverable Items being used by TuAF • Spare parts • Range finders • Optical sights and optical detection systems • Infrared Sights for Aircrafts • equipments and night vision devices etc. e.g. Range finder; Product cost = 45000$, LT = 6months, demand for repair = 1/year

6. Recoverable’s Management Process Weapon System Base Stock Serviceable Unserviceable Depot Repair In-house Repair

7. Scenarios With/Without Intermediate Storage Open-loop scenario Warehouse of Serviceable Unserviceable Part(s) Unserviceable Part(s) Repair Process Weapon system Warehouse of Unserviceable Replacement Serviceable Part(s) part(s) Closed-loop scenario Unserviceable Part(s) Repair Weapon system process Serviceable Part(s)

8. Purposes of Metric: • Optimization: • Determining optimal base-depot stock levels for each item • Redistribution: • Allocating the stock between the bases and depot • Evaluation: • Providing an assessment of the performance and investment cost for the system of any allocation between the bases and depot.

9. Mathematical Assumptions... System Objective: Min  E(backorders on all items at all bases pertinent to a specific weapon system) • Fill rate • Service rate • Ready rate • Operational rate

10. Mathematical Assumptions... • Demand for each item~Logarithmic Poisson E.g. Buses arrive at a sporting event in acc. with a Poisson process and numbers of customers in each bus are i.i.d. {X(t), t0}- number of buses who arrived by t Yi - number of customers in ith bus (Ross, 1997)

11. Mathematical Assumptions... Customer’s Arrival ~ Poisson (),Demand ~ Logarithmic (i.i.d)P(x demands in the time period) ~ Negative Binomial (q) q:Demand’s \ (constant over partic. item)

12. Mathematical Assumptions... • Demand is stationary over the prediction period • Demand on where repair is to be accomplished depends on the complexity of the repair only. • Lateral resupply is ignored • System is conservative (no condemnation)

13. Mathematical Assumptions... • Depot repair begins when the reparable base turn-in arrives at the depot. • Metric will accept relative backorder costs or essentialities by base and item. • Initial estimate of is obtained by pooling of the demand from several bases.

14. Base2 Base1 Basej BaseJ Multi-Echelon Theory... The Model: DEPOT • Cus. Arr.~  j(1- rj) • Demand ~  j f(1- rj) • fj = f for  j 1-rj rj ........... ........... fj j

15. Multi-Echelon Theory... Expected # of Backorder(for an item): s : spare stock  : mean customer arrival T : mean repair time p(x/ T): probability of arriving x demand in the period T.

16. Multi-Echelon Theory... B(s) is a CONVEX function: of the base stock level s.

17. Computation of Multi-Echelon Solution… 5 Stages: 1- Average Time that elapses between a base request for a resupply from a depot 2- E(backorders as a function of sj) (each so and each base) 3- Optimal allocation of the stock to the bases to minimize the  E(backorders) 4- For constant so+s select min E(system backorder) 5- Consideration of multi-item problem

18. Computation of Multi-Echelon Solution… Stage 1: (Avg. Response Time) S Avg. Response Time • Oj 0 Oj + Dj Therefore 0 < Delay at depot < D where D: avg. repair time

19. Computation of Multi-Echelon Solution… Stage 1: (Avg. Response Time) x: # of unserviceable parts demanding depot repair If x<s0, no resupply delayed O.w., x - s0 delayed. THEN:

20. Computation of Multi-Echelon Solution… Stage 1: (Avg. Response Time) •  E(system delay over any time period) = E(#of units on which delay is being incurred) *Length of the time period = • Avg. Delay/Demand =

21. Computation of Multi-Echelon Solution… Stage 2: (E(backorders as a function of sj) with the specification s = sj,  = j, T=

22. Computation of Multi-Echelon Solution… Stage 3: (Optimal allocation of stock) Simple Marginal Allocation: Each unit of stock is added to the base where it will cause LARGEST decrease in EBO *EBO is a convex function of s * Bayesian Procedure (will be described later)

23. Computation of Multi-Echelon Solution… Stage 4:(Min EBO for each level of s0+s) From the table showing EBO versus s0+s * select MIN EBO * record the actual allocation of stock btw bases and depot (will be used in stage 5)

24. Computation of Multi-Echelon Solution… Stage 5: (Multi-Item problem) Using the B(s) functions for each item allocate the next investment to the item which produces the MAX decrease in EBO/unit cost !!! Item backorder functions are not necessarily CONVEX. After each allocation compute the system investment & backorder

25. Linear Program of METRIC… Min System Cost s.t. EBO  bj for j=1,..,J Where is EBO for item I at base j when the depot stock level is sio and the stock at base j is sij

26. Lagrange Multipliers Introduced… where ij are ‘Lagrange Multipliers’ For all ij identical ; we can restrict attention to a single item.

27. Lagrange Multipliers Introduced… Optimal allocation of m item i units Necessary conditions: & Min Min

28. Lagrange Multipliers Introduced… Sufficient condition for Optimality: should be a convex function of m. !!! Not necessarily convex. is defined which lies on the boundary of the convex hull of

29. Lagrange Multipliers Introduced… To ‘m’ be a solution:

30. Limits on Employing Metric’s Objective.. • Be able to specify targets at each base for EBO on all units of all items have Metric determine the set of ıj and corresponding stock levels.

31. Bayesian Procedure… E.g. Case 1 Demand ~ Poisson s = 2 Case 2 w.p. 0.5 Low demand, 0.5 w.p. 0.5 High demand, 1.5 SO: we will understate backorders by using a point estimate.

32. Numerical Example for METRIC • The variable notations different from the paper: mj = avg. annual demand at base j μj = avg. pipeline at base j

33. Numerical Example… 5 identical bases; mj= 23.2 demands/year Tj = .01 years rj = .02 Oj = .01 years T0 = .02531 years  j = 1,2,…,J

34. Numerical Example… [Cont’d] Start with a depot stock level of 0 and compute μj for any base ; j=23.2 { (.02) (.01) + (.8) [.01+ 2.349/92.8] } = .7017

35. Numerical Example…[Cont’d] • Table 3.1. Expected Backorders at any base (DSL=0) • Table 3.2. Expected Backorders • Table 3.3. Expected Backorders • Table 3.4. Expected Backorders

36. Numerical Example…[Cont’d] • Table 3.1 s EBO(s) EBO (s-1)-EBO(s) 0 .07017 ----- 1 .1975 .5042 2 .0411 .1564

37. Numerical Example…[Cont’d] Table 3.2.Expected Backorders: DSL : Total stock levels at bases,optimally allocated: 0 1 2 3 4 5 6 7 8 0 3.5087 3.0044 2.5002 1.9959 1.4916 .9873 .8309 .6745 .5181 Optimal Base to 1 2 3 4 5 1 2 3 Allocate

38. Numerical Example… [cont’d] Table 3.3.Expected Backorders: DSL Total stock at bases,optimally Allocated: 0 1 2 3 4 5 6 7 8 0 3.5087 3.0044 2.5002 1.9959 1.4916 .9873 .8309 .6745 .5181 1 2.6043 2.1983 1.79231.3863 .9803 .5743 .4777 .3811 2 1.9240 1.6046 1.2852 .9658 .6464 .3269 .2694 3 1.5072 1.2469 .9867 .7264 .4662 .2060 4 1.2965 1.0681 .8397 .6113 .3829 5 1.2070 .9925 .7780 .5636 6 1.1743 .9650 .7557 7 1.1639 .9562 • 1.1610

39. Conclusion… • Metric can be utilized by managers in the cases: * Optimization for new procurement * Evaluation of the existing distribution of stock * Redistribution of system stock between the bases and depot.

40. Numerical Example… [cont’d] Table 3.4.Expected Backorders: STOCK TOTAL BACKORDER TOTAL DEPOT BASES BACKORDERS : REDUCTION: 0 0 0 3.5087 1 1 0 2.6043 0.9044 2 2 0 1.9240 0.6803 3 3 0 1.5072 0.4168 4 3 1 1.2469 0.2602 5 2 3 0.9658 0.2811 6 1 5 0.5743 0.3915 7 2 5 0.3269 0.2474 8 3 5 0.2060 0.1209

42. Questions & Answers