1 / 30

Inventory Management (Deterministic Model): Powers-of-Two Policies

Inventory Management (Deterministic Model): Powers-of-Two Policies. Prof. Dr. Jinxing Xie Department of Mathematical Sciences Tsinghua University, Beijing 100084, China http://faculty.math.tsinghua.edu.cn/~jxie Email: jxie@ math.tsinghua.edu.cn Voice: (86-10)62787812 Fax: (86-10)62785847

penney
Download Presentation

Inventory Management (Deterministic Model): Powers-of-Two Policies

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Inventory Management (Deterministic Model):Powers-of-Two Policies Prof. Dr. Jinxing Xie Department of Mathematical Sciences Tsinghua University, Beijing 100084, China http://faculty.math.tsinghua.edu.cn/~jxie Email: jxie@ math.tsinghua.edu.cn Voice: (86-10)62787812 Fax: (86-10)62785847 Office: Rm. 1202, New Science Building

  2. 经济订货批量（ＥＯＱ）基本模型 基本模型(Single Stage System)： -

  3. 经济订货批量（ＥＯＱ）模型 Ignoring the variable cost (as a fixed constant), Harris (1913): Square-root formula

  4. ＥＯＱ: Sensitivity analysis • If λ is changed to λ’, then In order to lower Q* by half,λ must decrease by 75% • Suppose we use the ‘wrong’ value Q instead of Q* Suppose we overestimate Q* by one-third (Q/Q*=4/3), the cost penalty is only 1/24 (G/G*=25/24). • In this sense, EOQ model is robust.

  5. ＥＯＱ：Powers-of-Two Policy Consider a powers-of-two policy, i.e., T = 2lTL (TL <= T * is the basic planning period, e.g., a shift). What’s the performance of the policy? Find the smallest nonnegative integer value l:

  6. ＥＯＱ：Powers-of-Two Policy It is easy to see Since G(T) is strictly convex, Assuming 2lTL is uniformly distributed over the interval (2-0.5T*, 20.5T*), the average cost

  7. Multi-stage (echelon) system • Serial system （Love 1972, MS): • Assembly system： IN-TREE (1984, MS):

  8. Multi-stage system • Distribution system • General

  9. Serial System Demand 1 J-1 J Graph G=(N(G),A(G))=(N,A) Nested Policy Production does not occur at a stage unless it also occurs at all of the immediate successors. Theorem. (Love 1972; Schwarz 1973) For a serial system it is optimal to follow a nested policy, i.e. Ti = ni Ti+1 where ni is a positive integer.

  10. Two Stage System: Echelon Inventory Sequential stocking points with level demand Two-stage Process ni = 3

  11. Echelon Inventory Echelon Inventory Inventory on hand in this stage and all its successors Echelon Inventory Holding Cost Incremental Inventory Holding Cost hi’is the conventional holding cost rate. In a serial system (i > 1):

  12. Two Stage System: Echelon Inventory Sequential stocking points with level demand

  13. Echelon Inventory vs. On-hand Inventory: 2 Stage Example Theorem. The total carrying cost are the same whether they are calculated using echelon or on-hand inventories.

  14. Optimization Problem Relaxation: Ignoring the constraints (*). What’s the relationship between these two problems? Theorem. The optimal cost (zLB) of the relaxed problem is a lower bound on the cost of any feasible policy.

  15. Relaxed Problem (n-1) constraints Given any subset A= of N\{J}, assume the strict equalities in A= hold, i.e. A= ={i:Ti=Ti+1}. 1 2 3 4 5 J-2 J-1 J NJ NJ-2 N5 Cluster N3 Prev(m) = cluster index just before m; prev(3)=0. Next(m) = cluster index just after m; next(J)=0.

  16. Subproblem for Cluster Nm The optimal solution is given by the EOQ formula This solution may be infeasible for the relaxed problem. feasibility (1) But even it is feasible, it may not be optimal for the relaxed problem.

  17. Optimality Optimality requires: If cut Nm into (Nm-, Nm+), where for any prev(m)<j<m, (2) Theorem. The conditions (1) and (2) are necessary and sufficient for T*(m) to be optimal for the relaxed problem. Proof. (Omitted. Can be found in Zipkin 2000).

  18. Algorithm for the relaxed problem • Set Nj ={j}, j=1,2,…,J; • For j=1,2,…,J, do: While prev(j) ≠ 0 and π(Nprev(j)) ≤ π(Nj) : Reset Nj = Nprev(j) U Nj Theorem. The above algorithm determines an optimal partition, and so the corresponding solution solves the relaxed problem in O(J) time. Proof. (Omitted. Can be found in Zipkin 2000).

  19. An Example J=4, λ=1, hj=1, Kj=(4,8,2,2)  gj=1/2 Initialize: Set Nj ={j}, j=1,2,…,4;π(Nj) =(8, 16, 4, 4) j=1: prev(1) = 0, so no change j=2: prev(2)=1≠0 and π(N1)=8 ≤ π(N2)=16 : Reset N2={1,2}, K(N2)=12, G(N2)=1, π(N2)=12 prev(2)=0, so no change j=3: π(N2)=12>π(N3)=4, so no change j=4: prev(4)=3≠0 and π(N3)=4 ≤ π(N4)=4 : Reset N4={3,4}, K(N4)=4, G(N4)=1, π(N4)=4 prev(4)=2 and π(N2)=12 >π(N4)=4 , so no change

  20. An Example (Continued) J=4, λ=1, hj=1, Kj=(4,8,2,2)  gj=1/2 Optimal Partition: P*={N2,N4}, N2={1,2}, N4={3,4} π(N2)=12,π(N4)=4 Optimal Solution to the Relaxed Problem: T1*= T2*= (π(N2))0.5=120.5 = 3.464, T3*= T4*= (π(N4))0.5=40.5 = 2, The cost zLB = 4+2*(12)0.5 =10.928

  21. Powers-of Two Policies Approximation: Use the same partition P*= U Nm Ti+: Resolve ties by rounding down: round(3.5)=3 Theorem.

  22. An Example (Continued) J=4, λ=1, hj=1, Kj=(4,8,2,2)  gj=1/2 Optimal Solution to the Relaxed Problem: T1*= T2*= (π(N2))0.5=120.5 = 3.464, T3*= T4*= (π(N4))0.5=40.5 = 2, The cost zLB = 4+2*(12)0.5 =10.928 Take TL=1: log2 (3.464) = 1.79  2; log2 (2) = 1.0  1 T1+= T2+= 22=4, T3+= T4+= 21 = 2, The cost z+ =11  z+ / zLB = 1.007

  23. Without Base-Period For given TL, denote the cost above by z+(TL). Choose a base period TL to minimize the cost z+(TL). First, only need to consider TL in interval [1,2) : Doubles or halves TL will not change Ti+! Then, what happens to Ti+when TL changes from 1 to 2? • li = li(m) remains unchanged  Ti+is linear in TL. • li drops by 1 at a breakpoint new Ti+is linear in TL.

  24. Without Base-Period For each m, T+(m) is a piecewise function with two pieces.  All T+ is a piecewise function with at most |P|+1 pieces. Within each piece, the cost function can be optimized by an EOQ formula.  Select the best one from them, and denote the cost as z’. Theorem. Proof. See next page. (Can be found in Zipkin 2000)

  25. Break Point T+(m) is a piecewise function with two pieces.  What’s the break point χ(m)? Considerχ(m) >1: Proposition.

  26. Proof of the Theorem z’ = min { z+(TL), 1≤TL<2}  Any weighted average of z+(TL) is an upper bound of z’ f(x)=1/(xln2) is a probability density (i.e. weight) function over [1,2):

  27. One-warehouse N-retailer System

  28. One-warehouse N-retailer System

  29. Review of this lecture: ---- Powers-of-Two Polices • Can be used to Tree System • Single stage • Serial system • Assembly system • Distribution system • How about non-tree system? • General system

More Related