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

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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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Outline

• The Assemble-to-Order System• Model formulation• Properties of optimal solution• Solution techniques• Numerical results• Conclusion

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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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The Assemble-to-Order System

Suppliers Components Products Backorders

(Items) (Customer demands)

1

2

m

L1

L2

Lm

Q121

Q122

QK1QK

2

QKm

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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}

QKi = required number of units of component i in K

qK = probability a demand is of type-KqK = 1

• Aggregate demand of component i: Compound Poisson process with rate

i = {i: in K} qK

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Other Modeling Assumptions

• The leadtimes for each component are i.i.d. random variables– Li has distribution Gi

• Base-stock policies (order-up-to policies)– si = 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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The Optimization Problem

minimize E[B(s1, …, sm)]

subject to c1s1+…+cmsm < C

whereB = total number of customer backorders

For any demand type K, let

BK = type-K backorders = number of type-K orders not yet completely satisfied

Then, B = K BK

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Solution Properties

Let si* be the optimal base-stock level for component i.

If ci > cj, i E[Li] < j E[Lj], and i < j ,

then si* < sj*.

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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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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The Supply Subsystem

Q121 Q12

2

QK1

QK2

QKm

Arrivals (Replenishment Orders)

Suppliers

X1 X2Xm

The Lower Bound

Xi = outstanding orders of component i in steady state,

has a Poisson distribution with mean iE[Li]

Bi = number of component i backorders = [Xi - si ]+

BK,i = number of type-K backorders that have component i backlogged

E[BK,i] K E[Bi]/i

BK = number of type-K backorders = maxi K {BK,i}

E[BK] > maxi K {E[BK,i]} maxi K {K E[Bi]/i}

E[B] = average total number of backorders = E[BK]

> maxi K {K E[Bi]/i}

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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

CscsctssBE mmK

iii

K

Ki

.....)]([maxmin 11

min max . . ( )Ki i i i

i KK i

z s t c h z C

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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 zi=zj

• The hierarchy structure enables us to devise

– an exact shortest-path algorithm (but still slow for large problems)

– a much faster greedy-type heuristic

K1 K2 K3

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Surrogate Problem II: Upper Bound

• Applying Lai-Robins inequality, we have the following surrogate problem:

][

1,

11

:

.....]][)([maxmin

ii sX

k

KKi

Ki

mmK

Ki

Kiii

i

K

Ki

YU

CscsctsEUUEsBE

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A Personal Computer Example

• 6 differentiating items1. 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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Coefficients Budget Algorithms s1 s2 s3 s4 s5 s6 Objective Average BackOrder Relative Error(%)

(1,1,1,1,1,1) C=20 Optimal(Simulation) 4 2 3 2 6 3 2.3142 2.3142

Lower Bound 4 2 3 2 6 3 1.1240 2.3142 0

Upper Bound 4 2 3 2 6 3 2.5155 2.3142 0

C=24 Optimal(Simulation) 5 2 4 3 7 3 1.4607 1.4607

Lower Bound 5 2 3 3 7 4 0.7252 1.506 3.10

Upper Bound 5 2 4 3 7 4 1.4406 1.4607 0


Lower Bound 6 3 5 4 9 5 0.1954 0.4941 0

Upper Bound 6 3 5 4 9 5 0.4952 0.4941 0


Lower Bound 3 1 3 2 6 3 1.4448 3.0172 3.76

Upper Bound 4 1 2 2 6 3 3.1152 2.9078 0


Lower Bound 5 2 3 3 7 4 0.7252 1.5127 0

Upper Bound 5 2 3 3 7 4 1.5297 1.5127 0


Lower Bound 6 3 5 3 8 5 0.3157 0.7039 2.27

Upper Bound 6 3 4 4 8 5 0.7616 0.6883 0

Table1: System: Four product: q125=0.3, q136=0.2,q1345=0.4, q1346=0.1Six items: L=(1,1,1,1,2,2)Overall Arrival rate=4

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Lower Bound 6 2 4 3 10 5 2.8021 5.6806 0

Upper Bound 6 2 4 3 10 5 6.7341 5.6806 0


Lower Bound 8 3 5 3 11 6 2.0721 4.0309 0

Upper Bound 7 3 5 4 11 6 4.4166 4.0997 1.71


Lower Bound 9 4 7 5 13 7 0.8695 1.8491 0

Upper Bound 10 3 7 5 13 7 1.9542 1.9371 4.76


Lower Bound 5 2 3 2 9 4 3.9608 7.3644 0

Upper Bound 5 2 3 2 9 4 8.8442 7.3644 0


Lower Bound 6 2 4 3 11 5 2.7003 5.4173 0

Upper Bound 6 2 4 3 11 5 6.4826 5.4173 0


Lower Bound 8 4 5 3 11 5 2.2058 4.1137 4.05

Upper Bound 8 3 5 4 11 5 4.4150 3.9534 0

Table 2: System: Four product: q125=0.3, q136=0.2,q1345=0.4, q1346=0.1Six items: L=(1,1,1,1,2,2)Overall Arrival rate=8

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Lower Bound 3 2 3 2 8 2 0.8675 1.5052 5.17

Upper Bound 3 2 4 1 8 2 1.7808 1.4312 0


Lower Bound 3 2 5 2 9 3 0.4097 0.7918 2.91

Upper Bound 3 2 5 2 10 2 0.8620 0.7694 0


Lower Bound 5 3 6 3 11 4 0.0959 0.2037

Upper Bound 5 3 6 3 12 3 0.2602 0.1857 0


Lower Bound 3 2 3 2 8 2 0.9676 1.5052 0

Upper Bound 3 2 3 2 8 2 2.3844 1.5052 0


Lower Bound 3 2 5 2 9 3 0.4097 0.7918 0

Upper Bound 3 2 5 2 9 3 0.9077 0.7918 0


Lower Bound 4 3 5 3 11 4 0.1671 0.3188 0

Upper Bound 4 3 5 3 11 4 0.3670 0.3188 0

Table 3: System: Six product: q25=0.1,q35=0.4, q125=0.15, q136=0.1,q1345=0.2, q1346=0.05Six items: L=(1,1,1,1,2,2)Overall Arrival rate=4

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Lower Bound 4 2 5 2 13 4 2.1184 3.9141 5.81

Upper Bound 4 2 6 2 14 3 5.2077 3.6992 0


Lower Bound 5 3 6 3 15 4 1.2252 2.2591 4.66

Upper Bound 5 3 7 3 15 3 2.7572 2.1585 0


Lower Bound 6 4 8 4 18 5 0.4027 0.8076 0

Upper Bound 6 4 8 4 18 5 0.8303 0.8076 0


Lower Bound 3 2 4 2 12 3 2.9306 5.1901 0

Upper Bound 3 2 4 2 12 3 5.6483 5.1901 0


Lower Bound 4 2 6 2 14 4 1.8405 3.1963 0

Upper Bound 4 2 6 2 14 4 3.8718 3.1963 0


Lower Bound 5 3 7 3 15 4 1.0385 2.1649 0

Upper Bound 5 3 7 3 15 4 2.7461 2.1649 0

Table 4: System: Six product: q25=0.1,q35=0.4, q125=0.15, q136=0.1,q1345=0.2, q1346=0.05Six items: L=(1,1,1,1,2,2)Overall Arrival rate=8

Documents

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