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1 Optimal Inventory-Backorder Tradeoff in an Assemble-to-Order System with Random Leadtimes Yingdong Lu – IBM T.J. Watson Research Center Jing-Sheng Song – University of California, Irvine David Yao – Columbia University
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2 Outline The Assemble-to-Order System Model formulation Properties of optimal solution Solution techniques Numerical results Conclusion
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3 Problem Background Assemble-to-order –Mass customization: Dell, Compaq, Ford –Only keep component inventory –Final product is assembled after an order is realized Optimal tradeoff between inventory and service Service measure –Average number of product backorders E[B] = Average # of customer orders waiting Proportion to average customer waiting time
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4 The Assemble-to-Order System Suppliers Components Products Backorders (Items) (Customer demands) 1 2 m L1L1 L2L2 LmLm Q 12 1 Q 12 2 QK1QK1 QK2QK2 QKmQKm
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5 The Demand Model (multivariate compound Poisson process) m different components Overall demand: Poisson process with rate Type-K demand: requires only the components in K K = any subset of {1,…, m} Q K i = required number of units of component i in K q K = probability a demand is of type-K q K = 1 Aggregate demand of component i: Compound Poisson process with rate i = {i: in K} q K
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6 Other Modeling Assumptions The leadtimes for each component are i.i.d. random variables –L i has distribution G i Base-stock policies (order-up-to policies) –s i = base-stock level for item i FCFS An order is backlogged if it is not yet completely filled. Committed inventory –If we have some items in stock but not others that are requested by an order, we put aside those available items as committed inventory for that order.
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7 The Optimization Problem minimize E[B(s 1, …, s m )] subject to c 1 s 1 +…+c m s m < C where B = total number of customer backorders For any demand type K, let B K = type-K backorders = number of type-K orders not yet completely satisfied Then, B = K B K
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8 Solution Properties Let s i * be the optimal base-stock level for component i. If c i > c j, i E[L i ] < j E[L j ], and i < j, then s i * < s j *. Example: If i and j have the same cost, and are always requested together, then the one with longest leadtime has higher optimal base-stock level.
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9 Solution Techniques Surrogate the objective function by simple lower and upper bounds Both the upper- and lower-bound problems share similar structures, which can be solved by –an exact network flow algorithm (but the number of arcs grows exponentially in the number of components) –faster greedy heuristic algorithms Numerical results show that the heuristic algorithm is effective.
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10 The Supply Subsystem Q 12 1 Q 12 2 QK1QK1 QK2QK2 QKmQKm Arrivals (Replenishment Orders) Suppliers X1X1 X2X2 XmXm
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The Lower Bound X i = outstanding orders of component i in steady state, has a Poisson distribution with mean i E[L i ] B i = number of component i backorders = [X i - s i ] + B K,i = number of type-K backorders that have component i backlogged E[B K,i ] K E[B i ]/ i B K = number of type-K backorders = max i K {B K,i } E[B K ] > max i K {E[B K,i ]} max i K { K E[B i ]/ i } E[B] = average total number of backorders = E[B K ] > max i K { K E[B i ]/ i }
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12 Surrogate Problem I: Lower Bound Using the lower bound to approximate the objective function, we obtain the following surrogate problem After a change of variables, the problem becomes
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13 Solving Surrogate Problem I Hierarchy structure: –There exists a complete ordering of all the order types (the subsets). For any K and (i,j), such that i, j belong to K, but not any set lower than K, we have z i =z j The hierarchy structure enables us to devise –an exact shortest-path algorithm (but still slow for large problems) –a much faster greedy-type heuristic K1K1 K2K2 K3K3
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14 Surrogate Problem II: Upper Bound Applying Lai-Robins inequality, we have the following surrogate problem:
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15 A Personal Computer Example 6 differentiating items 1. built-in zip drive 2. standard hard drive 3. high-profile hard drive 4. DVD-Rom drive 5. standard processor 6. high-profile processor 6 major demand types –{2,5} –{3,5} –{1,2,5} –{1,3,6} –{1,3,4,5} –{1,3,4,6}
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