Optimal Control of Stochastic Inventory Systems with Multiple Types of Reverse Flows Xiuli Chao University of Michigan Ann Arbor, MI Yunan Lijiang Conference October 31, 2008 Joint work with S. Zhou and Z. Tao 1/44
The Problem Logistics versus reverse logistics system Manufacturing Remanufacturing: Over 7000 remanufacturing firms in US with total sales $53 billion (Lund 1998). Multiple types of returns – Motivating examples 2/44
The Problem Return products R K type-K return J K Manufacturing R m type-m return J m Remanufacturing Demand D type-1 return J 1 Serviceable inventory I R 1 Disposal 3/44
Some related literature Simpson (1978) Inderfurth (1997) Decroix (2006) Decroix and Zipkin (2005) All these papers consider a single type of returned products. 4/44
Other related and review articles Heyman (1977) van der Laan, et al. (1999) van der Laan and Teunter (2005) Fleischmann et al. (1997) Guide and Srivastava (1997) 5/44
Model Details Periodic review system, periods 1 to N . K types of returned products. Disposal may or may not be allowed. Manufacturing and remanufacturing times are equal, and are assumed, without loss of generality, to be 0. The demand for serviceable product over the periods are D 1 , D 2 , . . . , D N . 6/44
Cost Structure Production cost (or ordering cost) for serviceable product, p . Repair cost for type j return is r j , where p ≥ r j , j = 1 , . . . , K . Stocking (holding) cost for type i return is s i , i = 1 , . . . , K . WLOG, assume (1 − α ) r 1 − s 1 ≤ · · · ≤ (1 − α ) r K − s K . 7/44
Cost Structure (Cont’d) There is holding cost for serviceable product. Consider backlog model (lost-sales model can be similarly studied)– shortage cost for backlog. Holding cost for serviceable product and shortage cost for backlog is a general convex function: Expected one-period cost G ( x ) , i.e., G ( x ) = hE [max { x − D, 0 } ] + bE [max { D − x } ] . 8/44
Our Goal Find/characterize the optimal manufacturing (ordering), remanufacturing, and disposal strategy so that the total expected (discounted) cost is minimized. 9/44
Events Timeline How many to How many to dispose? manufacture? How many to All costs remanufacture? incurred n+1 n Demand Returns arrives arrive 10/44
Formulation Type i returns over the periods are R i 1 , R i 2 , . . . , R i N , i = 1 , . . . , K . n , . . . , R K Let R n = ( R 1 n , R 2 n ) . ( D n , R n ) can have arbitrary joint distribution, but ( D 1 , R 1 ) , ( D 2 , R 2 ) , . . . , ( D N , R N ) are assumed to be independent. There is a discount factor α . 11/44
Formulation (Cont’d) I n = inventory level of serviceable product at the beginning of period n ; J i n = inventory level of type i return product at the beginning of period n ; n , . . . , J K J n = ( J 1 n ) ; i n = the inventory level of serviceable product after manufacturing and remanufacturing decisions but before demand is realized in period n ; 12/44
Formulation (Cont’d) j k n = the inventory level of type k returned product after remanufacturing and disposal decisions but before return occurs in period n ; j n = ( j 1 n , . . . , j K n ) ; w k = the remanufacturing quantity of type k return, k = 1 , . . . , K ; w = ( w 1 , . . . , w K ) . 13/44
Formulation (Cont’d) Given ( I n , J n ) , let V n ( I n , J n ) be the minimum total discounted cost from period n to the end of the planning horizon. V n ( I n , J n ) � K � K � � � = min r k w k + p i n − I n − w k w , j n ,i n k =1 k =1 K � � s k ( j k n + E R k + n ) + G ( i n ) + α E V n +1 ( i n − D, j n + R ) k =1 14/44
Constraints This optimization is subject to constraints j k n ≥ 0 , k = 1 , . . . , K 0 ≤ w k ≤ J k n − j k n , k = 1 , . . . , K , � K k =1 w k ≤ i n − I n . As Simpson (1978), let V N +1 ( i, j ) = 0 for any i, j . 15/44
Single type of returns Simpson (1978). Simpson’s result: Strategy for period n is determined by two numbers: ξ 0 ≥ ξ 1 , such that if initial serviceable inventory level is at least ξ 0 , do not manufacture/remanufacture; if initial serviceable inventory level is less than ξ 0 , then try to repair to level ξ 0 ; if after repairing the serviceable product inventory level is less than ξ 1 , then manufacture up to ξ 1 . 16/44
What Happens if Multiple Types of Returns? One might want to expect that Simpson’s result extends to multiple-type of returns. This is not true. Under some conditions the control parameters of the optimal strategy is state-independent, but in general, they are not. 17/44
System without Disposal w k = J k − j k . Change of variable and let x = ( x 0 , . . . , x K ) : x 0 = I, k � x k J ℓ , = I + k = 1 , . . . , K, ℓ =1 y 0 = i, k � y k j ℓ , = i + k = 1 , . . . , K 18/44 ℓ =1
Modified Formulation Given x , let V n ( x ) be the value function. K − 1 y { H n ( y ) } − r 1 x 0 + � ( r k − r k +1 ) x k V n ( x ) = min k =1 +( r K − p ) x K x 0 ≤ y 0 ≤ y 1 ≤ · · · ≤ y K , s.t. x K ≤ y K , y k +1 − y k ≤ x k +1 − x k , k = 0 , . . . , K − 1 . 19/44
H n ( y ) H n is given by H n ( y ) ( r 1 − s 1 ) y 0 + G ( y 0 ) = K − 1 � ( r k +1 − r k + s k − s k +1 ) y k + k =1 +( p − r K + s K ) y K + α E [ V n +1 ( y 0 − D, y 1 + R 1 − D, y 2 + R 1 + R 2 − D, . . . , y K + Re T − D )] . 20/44
A Technical Result Lemma: If system parameters satisfy (1) r 1 − s 1 ≤ r 2 − s 2 ≤ · · · ≤ r K − s K , then V n ( x ) can be decomposed as K � Q k n ( x k ) , V n ( x ) = k =0 in which Q k n ( · ) is a univariate convex function for each k . 21/44
Theorem Under condition (1), the optimal manufacturing/remanufacturing strategy is determined by K + 1 parameters ξ 0 > ξ 1 > · · · > ξ K , such that, when ξ ℓ ≤ x 0 < ξ ℓ − 1 , then do not use returned product of type ℓ + 1 , . . . , K + 1 ξ K +1 = −∞ , ξ − 1 = ∞ and K + 1 is new product (manufacturing or ordering). 22/44
Theorem (Cont’d) Repair type 1 to bring inventory level to ξ 0 , otherwise, repair type 2 to ξ 1 , ..., and the process continues, until, repair (or manufacture) type ℓ + 1 to ξ ℓ . Illustrate the case ℓ = 0 , K + 1 . Example K = 2 . 23/44
� � � Illustration I 2 2 2 2 1 1 1 0 1 n 1 n 2 n 24/44
� � � Illustration II 0 n 2 2 2 1 n 2 1 2 1 2 n 1 1 25/44
� � � Illustration III 0 n 1 n 2 n 2 2 2 2 1 1 1 1 x 2 x 1 x 0 26/44
What happens if ... What happens if r 1 − s 1 ≤ r 2 − s 2 is not satisfied? The optimal policy will no longer be determined by simple thresholds. Example 27/44
Example K = 2 , r 1 = 4 , r 2 = 2 , s 1 = 2 , s 2 = 1 , p = 5 , α = 1 , h = 3 , b = 5 , N = 2 . Poisson demand rates 3, and 4. ( y 0 ∗ , y 1 ∗ , y 2 ∗ ) ( x 0 , x 1 , x 2 ) (4,14,17) (12,14,17) (4,15,16) (13,15,16) (4,15,17) (13,15,17) (4,15,18) (12,15,18) (4,15,19) (12,15,19) 28/44
What is optimal, then? Suppose r 1 − s 1 > r 2 − s 2 . H n ( x ) is no longer decomposable. We can characterize the optimal policy, which is complicated with state-dependent control parameters. We also develop simple heuristic policies. 29/44
Systems with Disposals Suppose there exists an M , for k ≥ M , type k returns can be disposed. Under stronger condition s 1 ≤ · · · ≤ s K , the optimal policy is determined by a set of control parameters. Otherwise the optimal policy can be characterized, and it is complicated with state-dependent control parameters. 30/44
Theorem Under condition (2), the optimal remanufacturing/manufacturing and disposal policy for period n , is determined by two sets of parameters { ξ k , k = 0 , . . . , K } and { η k , k = M, . . . , K } , satisfying ξ K ≤ · · · ≤ ξ 1 ≤ ξ 0 , and η K ≤ · · · ≤ η M , and ξ k ≤ η k +1 , k = M − 1 , . . . , K − 1 . 31/44
� � � � � Illustration IV 1 n 2 2 n 2 2 2 1 1 0 n 1 1 1 n 2 n 32/44
� � � � � Illustration V 2 1 2 n 2 1 n 1 1 1 0 n 1 n 2 n 33/44
Then What? Thus, only under conditions (1) and (2) the optimal policy has a simple form. If these conditions are not satisfied, optimal policy is complicated and state-dependent. We develop simple heuristic policies with state-independent control parameters. 34/44
Heuristic I Illustrate the heuristic solution for K = 2 . Suppose the data is stationary. � (1 − α ) r 1 − s 1 + h � ξ 0 = F − 1 , D h + b � (1 − α ) r 2 − s 2 + h � ξ 1 = F − 1 , D h + b � (1 − α ) p + h � ξ 2 = F − 1 . D h + b 35/44
Heuristic I (Cont’d) � P ( η 1 − D + R 1 ≤ ξ 1 ) s 1 + α ( r 1 − r 2 ) � �� D − R 1 η 1 − ξ 1 1 ( ξ 1 < η 1 − D + R 1 < η 1 ) + E � P ( η 2 − D + R 1 + R 2 ≤ ξ 2 ) + α ( r 2 − p ) � η 2 − η 1 + D − R 1 + R 2 1 ( ξ 2 < η 1 − D + R 1 + R 2 < η 2 ) + E η 1 − ξ 2 = 0 36/44
Heuristic I (Cont’d) s 2 + α ( r 2 − p ) P ( η 2 − D + R 1 + R 2 ≤ ξ 2 ) D − R 1 + R 2 � 1 ( ξ 2 < + α ( r 2 − p ) E η 2 − ξ 2 � η 2 − D + R 1 + R 2 < η 2 ) = 0 . 37/44
Recommend
More recommend
