effective heuristics for multiproduct partial shipment models

Effective Heuristics for Multiproduct Partial Shipment ModelsAuthor(s): Milind Dawande, Srinagesh Gavirneni and Sridhar TayurSource: Operations Research, Vol. 54, No. 2 (Mar. - Apr., 2006), pp. 337-352Published by: INFORMSStable URL: http://www.jstor.org/stable/25146970 .

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Operations Research infTHnEI. Vol. 54, No. 2, March-April 2006, pp. 337-352

issn 0030-364XI eissn 1526-5463 1061540210337 DQI 10.1287/opre. 1050.0263

?2006 INFORMS

Effective Heuristics for Multiproduct Partial Shipment Models

Milind Dawande School of Management, University of Texas at Dallas, Mail Station SM 30, Richardson, Texas 75083-0688,

[email protected]

Srinagesh Gavirneni Johnson Graduate School of Management, Cornell University, Ithaca, New York, [email protected]

Sridhar Tayur Tepper School of Business, Carnegie Mellon University, Pittsburgh, Pennsylvania, [email protected]

Motivated by real applications, we consider the problem of shipping products to multiple customers from limited inventory. After formulating the optimization problems under different restrictions on partial shipments, we find that commercially available packages, applied directly, are unsatisfactory, as are simple greedy approaches. We develop a scheme of heuristics

that enables the user to select a good balance between computation time and effectiveness. A detailed computational study of one- and two-period industrial-sized problems indicates that these heuristics are computationally practical and generate solutions that are, on average, within 3%-4% of the optimum.

Subject classifications: partial shipments; integer programming models; heuristics.

Area of review: Computing and Decision Technology.

History: Received October 2003; revisions received June 2004, December 2004; accepted January 2005.

1. Introduction

For many purchasing executives, their biggest complaints

with distributors involve elemental services?too many

instances of partial shipments, wrong materials sent, lack of

basic cost management understanding. "Good may not be

good enough," a special report on distribution in Purchasing

(Morgan 1996, p. 38).

In this paper, we consider the problem of partial ship ments and study models that address several important

aspects of the problem. Consider a distributor that supplies, from inventory, multiple products to multiple customers.

For many reasons, including (1) demand uncertainty and

high storage costs (e.g., electronics, fashion goods, etc.),

and (2) restrictions on manufacturing capacity (especially for seasonal items such as lawn mowers, pharmaceuticals,

etc.), available inventory is often less than the total demand.

The distributor has to decide whether to ship complete orders to a few customers or to satisfy partial orders of

more customers. The former approach has the advantage of

providing better service to some of the customers, but may result in the loss of other customers. The latter approach leads to a larger customer base, but may result in customer

dissatisfaction and loss of goodwill. In some cases, real costs are involved with delays or partial shipments as part of the contractual agreements between the supplier and the customer. It has been our experience that, in these situa

tions, the distributor, more often than not, chooses the latter

approach. If the distributor carried enough inventory to sat

isfy all customers all the time, the problem of partial ship ments would not arise at all. To quote from Bolton (1991, p. 42): "Many partials could be eliminated if companies could forecast their requirements accurately, but when deal

ing with maintenance, repair, and operating supplies, this is easier said than done. How many office workers fore cast their requirement for office supplies?" To implement an effective partial shipment strategy, the distributor needs to have a good understanding of the customer perceptions,

transportation costs, and other costs (or revenues) associ

ated with partial shipments. Based on our experience with Procter & Gamble and other companies, we formulate mod

els to help distributors implement effective partial-shipment strategies.

Some customers will not, under any situation, accept

partial shipments, and even enter into a purchasing con

tract that explicitly forbids them. On the other hand, par tial shipments are often accepted because, for example, the customer cannot wait any longer to begin a particu lar job and is willing to accept a shipment big enough to

get the job started. Other reasons may include the promise of filling back orders (i.e., the remaining demand) "soon" or without further partitioning. For an initial partial ship

ment and subsequent back orders, sending a very small

shipment has at least two drawbacks: (1) It does not war rant the additional transportation costs, and (2) it results in additional paperwork and tracking costs. Additional paper work is a big problem, not only for the purchasing depart

337



Dawande, Gavirneni, and Tayur: Effective Heuristics for Multiproduct Partial Shipment Models

338 Operations Research 54(2), pp. 337-352, ?2006 INFORMS

ment, but also for other support functions such as receiving,

accounting, and inventory control (Bolton 1991). To avoid

such situations, customers often stipulate that the quanti

ties in a partial shipment cannot be too small. Distributors

are thus faced with the problem of allocating products to

customers in the presence of these shipping restrictions.

Note that customers could be external entities, as was the

case at Procter & Gamble or organizations under the same

ownership as the distributor, as is usually the case in retail

supply chains (e.g., Shaw's supermarkets and Lowe's home

improvement). As observed in Morgan (1996), there is a

need for a quantitative analysis of these restrictions and

their associated costs.

To the best of our knowledge, quantitative models for

deciding partial shipments in the presence of such shipping restrictions have not been studied in the literature. However, a few articles (see Banerjee et al. 2001) have discussed the

problem in a qualitative manner. One approach discussed in

Dowling (1996) suggests giving an option to the customer

to either receive a partial shipment of the available goods or wait until all the items are available (usually a couple of weeks) to receive one complete shipment. Another sug

gested approach is to have the customer adopt a purchasing

practice aimed at improving the level of communication

and understanding with distributors (Bolton 1991, Morgan

1996). A shipping restriction well studied in the literature

is one of using a discrete set of shipping frequencies to

minimize transportation and inventory costs (Bertazzi et al.

1997, 2000). Our original motivation for solving this problem came

from Procter & Gamble; subsequently, a supplier to a

fast-food chain and a distributor of electrical supplies

approached us with similar problems. We consider the fol

lowing restrictions placed on partial shipments.

(1) If a shipment is made to a customer it should satisfy at least 100)8 percent of the customer's total demand, where

0</3^1.

(2) Supplying a very large fraction, say 90%, of the

customer's demand is also undesirable because this would

necessitate a very small second shipment. Thus, a more

desired condition is that either a customer's demand be sat

isfied fully, or not at all, or that the shipment be between

100/3!% and 100/32% of the customer's demand, where

0 < fix < /32 < 1. We also incorporate the existence of cus

tomers who want either 100% of their demand or none at

all. This is motivated by the observation that some purchas

ing contracts do not allow partial shipments.

(3) Some customers require, in addition, that a partial

shipment includes a minimum fraction of the demand for

each product. This is especially true if the products com

plement each other.

(4) The acceptable minimum fraction (of a customer's

demand) is customer specific. That is, the minimum

requirements could be different for different customers.

We start by incorporating restriction 1, and refer to the

corresponding model as Model 1. Restriction 2 is addressed

by Model 2. Restrictions 3 and 4 will be addressed as

extensions in ?5. As will be shown in Lemma 2.1 below, the problems

resulting from Models 1 and 2 are NP-hard. We model

these problems as mixed-integer programs (MIPs). Solv

ing most of these MIPs to optimality requires significant

computational effort. We make two attempts to solve the

MIPs to optimality. First, we use a commercially available

package (CPLEX, version 8.1) on the MIP formulations.

Second, we reformulate these problems as fixed charge net

work flow (FCNF) problems (Magnanti and Wong 1984, Nemhauser and Wolsey 1988) and attempt to solve them

using MINTO (Nemhauser et al. 1994, Savelsbergh and

Nemhauser 1996) to better exploit the available structure.

While the FCNF approach is better, the required com

putational effort is still large. Obtaining an optimal or

near-optimal solution for these problems (single period or

multiperiod) enables us to provide a myopic solution over

a much longer horizon. Driven by these considerations, we

develop fast, efficient, and easy-to-implement heuristics to

obtain good solutions quickly on industrial-sized problems. The rest of this paper is organized as follows. In ?2,

we present the problem definition and formulate Model 1.

We also present our experience with solving instances of

the model to optimality and applying "greedy" approaches. We propose two heuristics and present a probabilistic anal

ysis to calculate the expected deviations and theoretical

upper bounds on their predictions of the maximum num

ber of customers satisfied. In ?3, we formulate Model 2

and extend the heuristics developed in ?2 to accommo

date additional constraints. In ?4, a detailed computational

study of Models 1 and 2 demonstrates the effectiveness

of the heuristics. Section 5 discusses extensions obtained

from imposing restrictions 3 and 4 above. Section 6 illus trates the performance of the heuristics on a set of hard

instances. Section 7 formulates a two-period problem and

describes how the basic idea behind the one-period heuris

tics can be extended for this problem. We then extend the

computational study to evaluate the effectiveness of the

two-period heuristics. Section 7 ends with a discussion of

some extensions of our two-period model. Conclusions and

future research directions are provided in ?8.

2. Model 1

We are faced with demands for a set, M, of products

(indexed by j) from a set, N, of customers (indexed by /). Let

Dj be the demand for product j from customer /. We

are to decide the portion of Dj

that will be satisfied from

inventory / , for product j. We cannot supply more than

what is demanded and we cannot supply more than what

we have on hand. Also, if a shipment is made to a cus

tomer, it must be at least /3 portion of the customer's total

demand accumulated over all the products. Let Sj

be the

amount of product j shipped to customer /. The revenues

received are of two types. We receive a reward Kl if a ship

ment is made to customer /. Kl is a representation of the




Operations Research 54(2), pp. 337-352, ?2006 INFORMS 339

customer appreciation for having received a shipment even

if it is partial. Kl also represents the weight that the distrib

utor attaches to each customer. Furthermore, we receive a

reward of (Sj/Dj)rj

for supplying product j to customer i, where

rj is a measure of the revenue generated when

Dj units of product j are shipped to customer /. The objective function in our models captures the practices at many dis

tributors (see, for example, Cardinal Health Speciality Phar

maceutical Services). It is worth noting that these rewards

(fixed and variable) can be expressed in many forms and

the representation chosen, as long as it is linear, will not

significantly alter the structure of the problem. If wl = 0, we supply nothing to customer /; otherwise w' = 1, and the

customer gets at least a partial shipment. This problem is

formulated as a MIP as follows:

\N\ \N\ \M\ gi Maximize ^W + ̂ ^rj^ 1=1 i=\ j=\ Di

\N\ such that ?$}</;, jeM,

1=1

\M\ AM\ v

?^/3 E^R ieN, (1)

\M\ AM\ x

j=\ V=i /

O^Sj^Dj, ieN, jeM,

u/e{0,1}, ieN.

We have assumed that (3 is not customer specific; this

is reasonable when the customers are of similar size and

are geographically dispersed similarly from the distributor. In ?5, we incorporate customer specific /3s, and show that our heuristics continue to work well. In general, the cor

rect value of (3 is difficult to obtain and depends on the

relationship between shipment economies, loss of good will, customer interaction, and the type of product. In our

experience, a /3 value of 0.6 is typically considered satis

factory. Also, instead of (or in addition to) expressing a

minimum shipment quantity as a percent of a customer's

total demand, it could be a specified minimum quantity.

Lemma 2.1. The problem class arising from Model 1 is NP-hard.

Proof. Consider an instance of a 0-1 knapsack problem:

max{?X, CiXt: ?!Li a^ ^ b, x e {0, 1}"}. This problem is a special case of Model 1 by choosing \N\=n, \M\

= l9 Kl = ch P = 1, r\

= 0, Dj = ai9 and /,

= b. The result follows.

To examine the solvability of this model, we performed a pilot study with 25 problems, each containing 100 cus tomers and 25 products. This was based on our experience that a typical problem has anywhere from 60 to 150 cus

tomers, and anywhere from 10 to 40 products. It is pos sible, however, for some shipping scenarios to result in

instances with as many as 1,000 customers and 100 prod ucts. We examine such large instances in ?6. The problems in the pilot study were generated to simulate various levels

of inventory under similar demand scenarios. The experi

mental setup is described below: \N\ = 100, \M\

= 25. For

all /, j, we chose )8 = 0.6, demand

D) =

U[0,100], Kl =

U[0, 150], and r) =

U[0, 15]. We chose five settings for

the inventory of each product, If Ij =

?/[2,000,4,000] Vy,

Ij =

L7[2,000, 3,000] Vy, Ij =

U{1,000, 3,000] Vy, /, =

U[l,000, 2,500] Vy, and /,. =

U[ 1,000, 2,000] Vy. For

each of these inventory-level settings, we generated five

problems. Using the straightforward integer programming formulation given above, CPLEX (version 8.1, running on a

Pentium IV, 1 GHz, 512 MB RAM machine with Mandrake

Linux, version 9.1, as the operating system) was able to

solve only 11 (out of 25) problems to optimality within the

imposed time limit of 1,800 seconds.

An analysis of when these instances are difficult to solve, and when they can be solved easily, reveals an important

insight: The ratio of the total available inventory to /3 times

the total demand is a good indicator of the difficulty of a problem. For the 25 problems generated above, these

ratios ranged from 1.03 to 0.49. When the ratio is high the

problems are easily solved, and when it reduces, the diffi

culty of these problems increases dramatically. On the one

hand, when this ratio is more than one (or close to one), it is possible to ship at least a partial shipment to most

of the customers and, consequently, the important decision

about which customers to ship to becomes (relatively) easy. On the other hand, when this ratio is much smaller than

one, then in general only a small subset of customers will

receive a shipment, and deciding which customers should

receive a shipment becomes much more difficult. Later, we

use this observation to develop effective heuristics.

2.1. FCNF Formulation

In this section, we formulate our problem as an FCNF prob

lem. The FCNF problem has been shown (Magnanti and

Wong 1984, Nemhauser and Wolsey 1988) to have a spe cial structure and tools such as MINTO (Nemhauser et al.

1994, Savelsbergh and Nemhauser 1996) are available to

solve these problems efficiently. To obtain this formulation, we focus our attention on the unsatisfied demand of prod

uct j to customer /, Uj. Thus, for every (/, j),

S) + U>=D).

Instead of working with the revenue earned, we min

imize the revenue lost, which is equivalent. Thus, for

every customer, we compute the revenue lost by unsatisfied

demand. The approach is illustrated for a three-product, two-customer case in Figure 1. For product j, Ej

= Ij

?

Jli=\ Sj is tne excess inventory (i.e., inventory not shipped

to any customer). Due to the shipment restrictions, the total unsatisfied demand, jjjl] Uj, for customer / is either equal





Figure 1. Fixed-charge network flow formulation.

i/V ; r-r^" tt2^^2 r2 /3 S3 \^2 j DlX(/2/t/3

F2

to his total demand, V = YjfX Dp

or less than (1 ?

/3) x

I^i ?*/ If it is equal to

J2y=x Dj, then we lose the rev

enue Kl. To model this, we use two flows (F( and F2l) for

each customer. We restrict F{ to be equal to zero or equal to Tl with a cost of Kl if the flow is nonzero. Thus, this arc

captures the FCNF structure of the problem. The flow F2l is restricted to be below (1

? j8) x Tl and has no fixed cost

associated with it. In addition, for each of the arcs with

flow Up there is a cost

rj/Dj associated with it, which is

the revenue lost for not meeting the demand. There is con

servation of flow at each node and the dashed arcs have

fixed flows. Recall from ?1 that 0 < )8 ̂ 1. The FCNF for

mulation is as follows:

\N\ \N\ \M\ jji Minimize ]T Kl(1

- ut) + ? L r)

-? i=\ i=\ j=\ Dj

\N\ such that J2sj + Ej

= Ij> JeM>

S) + U^ =

Dl ieN,jeM, \M\

^2UJ = F{ + F2i9 ieN,

\M\

F{ =

(l-wi)Y^Dij, ieN, 7=1

\M\

^(i-j3)?d;, **"> 7=1

S),UlEj^0, ieN, jeM,

F/,F2'>0, ieN,

wle{0,l}, ieN.

The computational results of using MINTO (version 3.0) on this formulation were a significant improvement over

CPLEX (version 8.1). Most (20 out of 25) of the problems were solved to optimality under the imposed time limit

of 1,800 seconds. The reason for this improvement is the

ability of MINTO (Nemhauser and Wolsey 1988, Padberg et al. 1985, Van Roy and Wolsey 1987) to better exploit the

fixed-charge network structure of the problems (Linderoth

2003). However, in addition to the five unsolved problems, the computation times were as high as 1,300 seconds.

A detailed computational study, presented later in ??4-7, reveals that one-period and two-period problems of mod

erate size (25 products, 100 customers) can be solved to

optimality within four to five hours. However, note that

obtaining optimal solutions in a few hours does not provide closure. In practice, solutions to inventory allocation prob lems are often subject to a detailed what-if analysis; models

are typically run a number of times before a final solu

tion is fixed. Moreover, solutions to such problems may be

required on an ad hoc basis following updates to inventory and demand data. There is, therefore, a need for heuristics

that provide near-optimal solutions in real time (i.e., within a few seconds or minutes).

For the rest of the paper, we assume that demands for

all products from all customers are identically and indepen

dently distributed. This means that for any values ix, i2, jx, and j2, demand

Dlj has the same distribution as demand

Dlf. However, as will become clear below, the main idea

behind our heuristics can be easily extended to the case of

nonidentical customers and products.

2.2. Heuristics for Model 1

We first evaluated several simple greedy strategies that rank

customers based on an estimate of their revenue potential.

We mention the following two ranking functions: (1) Kl +

PYjjJx rp which represents the minimum reward received

if the demand for customer i is satisfied, and (2) (Kl +

Y.j=x rj)/Ylj=\ Dp the maximum profit per total demand.

After listing the customers in decreasing order of their rank, a greedy strategy starts at the top of the list, and, working in sequence, meets the minimum demand (i.e., /3 portion of the total demand). In general, the performance of these

and other greedy variants was very poor, with the resulting

solutions being 10%-25% away from the optimum.

A Scheme of Heuristics. We recognize that the prob lem is difficult due to the presence of the minimum ship

ment size requirements, which partition feasible shipments to a customer into two disjoint sets: The shipment size is

either 0, or at least a /3 fraction of the total demand. To

obtain a quick solution, we impose these minimum ship ment restrictions on a subset of the customers and restrict

that the remaining customers do not receive a shipment at

all. We need to make two decisions. First, we decide the

number of customers, F, who will be chosen to be the only ones eligible to receive a shipment. Thus, customers not

chosen will receive no shipment and the binary variable wl

is set to 0 (for these customers). The second decision is

regarding the selection of this subset of the customers.




Operations Research 54(2), pp. 337-352, ? 2006 INFORMS 341

The main idea behind the heuristics is simple: Take an optimal algorithm, remove some "less promising" cus

tomers, and run the algorithm. Defining the set of "less

promising" customers (i.e., how many customers and which

customers) is accomplished in Steps 1 and 2 below. Such a

preselection of customers is similar to the idea of heuristic

concentration proposed by Rosing and ReVelle (1997). As

will be seen in ?4, the problem resulting from the removal

of these customers can be solved quickly and provides a

reasonable solution for the original problem. Step 1. How many customers will be (at least partially)

satisfied? Let us denote this number by F. Indeed, the value of F depends on the total inventory on hand. In general, for

any given inventory position, if F is large, the problem is

hard, and if F is small, the problem is easy but we get an

inefficient solution. Hence, choosing the value of F is a key step. If there is enough inventory to satisfy the ft portion of every customer's demand, then we set F =

\N\. If there is not enough inventory to satisfy all the customers, then

we select F such that, with a prespecified probability, the

inventory will be sufficient to meet the f3 portion of demand of these F customers. The idea is to compare the total

inventory and total demand and then determine the number of customers that the available inventory would satisfy if each shipment is the minimum feasible nonzero quantity. For this purpose, we found a probabilistic approach to be

most appropriate.

Let us assume that Dj

has mean /jl and variance a2 for all values of / and j. Using the central limit theorem (see, e.g., Bain and Engelhardt 1987, p. 235), the demand of F cus

tomers is (approximately) distributed normally with mean

\M\/nF and variance \M\Fcr2. Also, the /3 portion of this total demand has mean /3\M\/jlF and variance fi2\M\Fcr2.

To see how F should be selected, we experimented with different values of / =

?y=l /,. and \M\ = 25, \N\

= 100, ^ = 50, a2 = 833.33, and /3

= 0.6. Let

= Ef=i (, ~

P\M\fi\N\ = / - 75,000

/y|M||7V|o-2 866.03

The value of z provides information about the position (with respect to the mean) of the total inventory level in the normal distribution curve of f3 times the total (aggre gated over all customers and all products) demand. The

plot of the computation times versus the z-values is given in Figure 2. It should be noted that when the z-values were close to 0 or positive, the problems were easily solved, while for problems in which the z-values are far below

zero, the problems were, in general, harder. We now use

this observation to select F so that the total inventory posi tion exceeds the /3 fraction of the total demand of F cus tomers with a very high or a very low probability. Suppose, for 0 ^ 8 ^ 1, we impose that /3(total demand of F cus

tomers) is less than the total inventory 8 x 100% of the time. That is,

Prob[(/3 demand of F customers) ^I] = 8.

Figure 2. Plot of CPU time vs. z-value.

1,800

Solved

Unsolved

1,440

a o a 8 1,080

?

.S #

e ~ 720 ?

P Ph

u

360-- #

?P-*-'-f-1?h?^-h~M -45 -37 -29 -21 -13 -5 3

z-Value

Choosing a value of 5 close to 1 would result in a choice of F such that the total inventory position is on the right side of the demand distribution. Similarly, by choosing 8 close to 0, we choose F such that the total inventory posi tion is on the left side of the demand distribution. The idea, then, is to (1) choose a value of 8 to obtain a value of F; and (2) to restrict that only F customers are eligible to

receive a shipment. The resulting problem (which will still be a MIP with \N\

? F integer variables set to 0) will be

computationally easier to solve and hopefully provide good feasible solutions.

For a fixed value of 8, we have

\M\?F + zsV\W^2 = ^j^,

(2)

where Prob[Z < z$] = 8 for a standard normal random vari

able Z. In (2), the value of F can be computed because all the other quantities are known. Note that because z? ^ 0

for 8 ^ 0.5, the value of F from (2) will increase as 6

approaches 0. For any fixed value of 8, we let H(8) denote the resulting heuristic. For illustrative purposes, we choose two extreme values of 8: (1) 8 = 0.95 (we call the result

ing heuristic H(8 =

0.95)), and (2) 8 = 0.01 (we refer to this heuristic as H(8

= 0.01)). H(<5

= 0.95) is a more

conservative heuristic compared to H(8 =

0.01): For the same aggregated inventory level, the value of F predicted by H(c5

= 0.01) will be higher than the one predicted by

H(8 =

0.95). Thus, in general, the MIP problem result

ing from H(5 =

0.01) will be computationally harder to

solve, but will produce a better solution than that given by H(8

= 0.95).

In summary, while the H(8 =

0.95) heuristic is expected to be quicker, we may be sacrificing some customers that can be satisfied from the inventory available. This is cor rected in the H(8

= 0.01) heuristic by attempting to sat

isfy as many customers as possible. (Note that choosing



Dawande, Gavirneni, and Tayur: Effective Heuristics for Multiproduct Partial Shipment Models 342 Operations Research 54(2), pp. 337-352, ? 2006 INFORMS

8 = 0.90 will result is a larger F as compared to that with 5 = 0.95; however, the resulting MIP will take longer to

solve. As we see in ?4, H(8 =

0.95) solves all instances within 40 seconds.) In ?4.3, we also present aggregate results for 8 values of 0.01, 0.25, 0.50, 0.75, and 0.95.

Step 2. Which F customers? After having decided on the value of F, we need to decide the F customers on whom the minimum shipment restrictions are imposed. We do that

by sorting the customers in the decreasing order of (Kl +

PY}j=i rj)/03?^i Dj) and selecting the first F customers.

The intuition here is to rank customers by the profit per unit realized if the quantity shipped is a j8 fraction of the total

demand. Other sorting functions such as (1) the total profit Kl + PYljJi rp and (2) maximum profit per total demand,

(Kl + Jjj=\ rj)/ JljJi Dlj, did not improve the performance.

It is possible to adapt this idea to accommodate more

complex situations such as nonidentical customers and

products. In some cases, it may be necessary to rank the

customers first and then determine a subset to be consid

ered for shipment. It may also be necessary to change the

way the statistical/aggregate properties of the demands are

used. Actual demands (instead of a demand distribution) could be used in both the steps, possibly losing the ability to adjust the effectiveness of the solution.

2.3. Analysis of H(a =

0.95) and H(5 = 0.01)

The success of H(8 =

0.95) and H(8 =

0.01) crucially

depends on predicting the number of customers on which

the minimum shipment restrictions are imposed. In this sec

tion, we offer an analysis to show that the predictions by both these heuristics are satisfactory.

We know that the /3 portion of the total demand

of JA customers is (approximately) distributed normally with mean Wj = /3/ul\M\Ja and standard deviation Ay

=

P(Ty/\M\JA. Let Wj (.) be the cumulative distribution

function (c.d.f.) of this distribution. Given any inventory level, /, let N(I) be the maximum number of customers

that can be satisfied by this inventory over all feasible solu

tions. Wj (I) gives the probability that N(I) is greater than or equal to JA. Similarly, %a_x(I) is the probability that

N(I) ^ JA ? 1. Using these two observations, we get

PjA=?iob{N(I) =

JA}

= %^(I)-%a(I)

= <l>(zjA-i)-<i>(zjA),

where <?(.) (respectively, </>(.)) is the c.d.f. (respectively,

probability density function (p.d.f.)) of the standard nor

mal variable and zt = (I

- W^/A,-. For any given inventory

level, these probabilities {PJa; Ja =

1,2...N} are easily estimated.

Let JH be the number of customers predicted by our heuristic. Using these probabilities we calculate the

expected deviation (ED) of the heuristic from the maxi

mum that can be satisfied as ED = J2ja>jh{Ja

~ ^h)^ja

=

T,ja>jh{Ja -

4/)[*(z/A-i) -

*(z/A)l- These deviations are

easily calculated, and for the inventory levels we were

studying, they were found to be very small, the highest being 0.01 for H(8

= 0.01) and around 3.5 for H(8

=

0.95). Instead of obtaining ED computationally, we can

obtain an analytical formula for an upper bound on ED for

H(S =

0.01). Recalling that the number of customers, JH, chosen by H(S

= 0.01) satisfies Zj ^ ?1, we can obtain

an upper bound on the expected deviation as follows:

ED= E(A-^)[1>(^-.)-*(^)] JA>JH

< E('A-^)<?Z/A-i)[ZyA-.-z,J JA>JH

< E(^-4)i^-l/2[vi-^ Ja>Jh v27T

Let us denote et =

zt -

zi+x, & =

z2i+x -

zf, and vt

= e~8i.

Observe that zt is decreasing and convex in /, while z] is

increasing and convex in /. This implies that et is decreas

ing in / while gt is increasing in /, which in turn implies that v( is decreasing in /. Incorporating these observations

into the right-hand side of the inequality above results in

^7fe[Zy?"Zy^^ < _e p~zJHll_

^^6j-e (\-vjHr

Because ij < ?1, we know that JH _> I/({}\M\ia). In

the equation above, we can replace Vj by v? where t =

I/(/3\M\n,). We know that z, = 0 and z1, = 0. Estimating

zt+x as we do below, we get an upper bound for vJh:

-\M\P Z'+[

PcrJl/p? + \M\'

2 |M|V Z,+1

fP<ri(l/f3v + \M\)'

implying

V < ^"(lMlV2)/(i82o-2(///3/x+|M|)) JH ̂

Incorporating this into the upper bound above results in the

inequality

1 1/2 \M\p ED < ??p.'1'2- ' '^

V2^ Po-Jl/Pp+\M\ 1

(1 _

e-(\M\2V>2)/(P2<rHl/fa+m))y

'

From the expression of the upper bound we can draw two

conclusions. As a increases, the upper bound increases,





thus implying that our prediction is less reliable. As

p increases, the upper bound decreases. This is due to the

fact that it takes more inventory to satisfy a new customer, and thus we can expect our predicted value to be more

reliable. We calculated these upper bounds on the expected deviations for H(8

? 0.01) for each of the 25 inventory

levels in our pilot study (?2). It is interesting to note that

this upper bound was never greater than 0.75. Thus, for the

25 inventory levels in our pilot study, the number of cus

tomers predicted by the H(5 =

0.01) heuristic is at most

one customer away from the maximum number of cus

tomers that can be satisfied.

3. Model 2

We consider two extensions to Model 1. First, we include

the presence of some customers who want all of their

demand or none at all: This is motivated by the observation

that some purchasing contracts prohibit any partial ship ments. Let Nx represent this subset of customers. The sec

ond extension is motivated by the observation that a very

large first shipment leads to a small second shipment. Thus, we add the restriction that the customer is supplied with

either his full demand, or between 100/3!% and 100/32% of

his demand, where 0 < y8x < /32 ̂ 1, or none at all. Let K\ be the reward received when customer / is supplied with

the full demand and K2 be the reward received when cus

tomer / is supplied only part of the demand.

Note that Model 1 is a special case of Model 2 cor

responding to /32 = 1 and Nx

= 0. Therefore, it is trivial

to generate instances of Model 2 that are at least as hard

as Model 1. The need to develop heuristics for Model 2

is, therefore, apparent. In the next subsection, we extend

H(5 = 0.95) and H(<5

= 0.01) from ?2.

3.1. Heuristics for Model 2

As before, our first task is to decide, based on the inven

tory level, the number of customers to whom we consider

shipping at least a partial shipment. To this end, we start

by estimating the number of customers that can be supplied

100/3!% of their demand from the given inventory. 100/3}% of the demand of J customers has a normal distribution with mean fixp\M\J and variance /32cr2\M\J. Choose J such that

I-Ptli\M\J ^

Case 1. If J < iNy^l, choose J customers from N\NX. For these J customers, we enable the possibility of receiv

ing a partial (between /3j and /32) shipment. All of the

remaining customers will receive no shipment.

Case 2. If J > \N\NX |, then we know that the inventory is probably enough to satisfy the px portion of the N\NX customers and, therefore, we may be able to supply some

customers in Nx with their full demand. Recall that cus

tomers in Nx require their full demand or none at all.

We know that the fix portion of N\NX customers and the

full demand of Jx customers has mean (3x/jl\M$\N\Nx$ +

tx\M\Jx and variance cr2((\N\Nx\)(l2\M\+JX\M\). There

fore, select Jx such that

I-MM$\N\Nx$-tL\M\Jx=i7

<7V'((|;vvvl|)tf|M| + ./1|M|) Zs'

Case 2(a). If Jx < \NX\, then choose Jx customers out

of Nx. For these Jx customers, we enable the possibility of receiving all or none of their demand. Other customers

in Nx will not receive any shipment. To summarize, we

enable the possibility of (1) all customers in N\NX receiv

ing a partial (between /3X and /32) shipment, and (2) Jx customers from A^ receiving all or none of their demand.

Case 2(b). If Jx > \NX\, then we probably have enough

inventory to satisfy the full demand of more than Nx cus

tomers. Therefore, we choose J2 customers from N\NX to

whom we enable the possibility of giving the full demand.

We know that the full demand of \NX\ + J2 customers

and the (3X portion of the demand of |A^\A^i| ?

J2 has

a normal distribution with mean (1xijl\M\(\N\Nx\ ?

J2) +

fi\M\{\Nx\ + J2) and variance a2(p2(\N\Nx\ -

J2)\M\ +

IMKINJ -\-J2))9 so we select J2 such that

I-^\M\(\N\Nx\-J2)-ii\M\(\Nx\^J2) ^^ a^(p2(\N\Nx\-J2)\M\ + \M\(\Nx\+J2))

Z8'

(If J2 > \N\NX\, we set J2 =

\N\NX\.) For these J2 cus

tomers, we enable the possibility of receiving either a

partial or full shipment. To summarize, we enable the pos

sibility of (1) all customers in Nx receiving all or none

of their demand, (2) J2 customers from N\NX receiving either a partial (between (3X and /32) or a full shipment, and (3) the remaining iwyvj

? J2 customers from N\NX

receiving only a partial shipment. The process of selecting which customers are in the

subsets is done using the sorting procedure as before. By extending the probabilistic analysis presented in ?2.3, we

estimated the upper bounds for Model 2 as well and noted that the expected difference between the heuristic and opti

mal solutions is low. Because the probabilistic analysis is

similar, we skip the details.

Remark. Alternatively, we can start by looking at the Nx customers first; this will imply that after these "all-or

nothing" customers are taken care of, we will satisfy others.

If we follow this approach, we can divide the problem into two subproblems: first, consider the all-or-nothing group; second, we have Model 2 without any customers from

the all-or-nothing group. A similar approach to the one

described above can be used for this case as well.

4. Computational Experience for Models 1 and 2

We summarize the performance of the heuristics for both models. The experimental set-up used to generate the




344 Operations Research 54(2), pp. 337-352, ? 2006 INFORMS

Table 1. Minimum, average, and maximum computation times for solving instances

of Models 1 and 2 to optimality.

Model 1 Model 2

Inventory level Minimum Average Maximum Minimum Average Maximum

/,-= tf[2,000,4,000] 0.16 23.47 350.71 0.21 0.41 0.54

/; = ?/[2,000,3,000] 0.26 221.96 2,374.42 0.87 12.03 50.29

Ij = U[l,000,3,000] 6.29 180.74 1,192.60 2.54 68.92 420.35

Ij = U[l, 000,2,500] 8.32 244.27 1,830.19 5.62 68.76 256.39

Ij = U[l,000,2,000] 25.74 368.85 3,506.25 6.15 139.66 1,550.77

instances for all the models is the same as that described

in ?2. All the computations were carried out using CPLEX

(version 8.1), running on a Pentium IV machine (1 GHz, 512 MB RAM) with Mandrake Linux (version 9.1) as the

operating system. For each inventory level, we generated 50 problems and solved them using the FCNF formula

tions as well as the H(S =

0.01) and H(S =

0.95) heuristic

procedures. For each problem instance, we computed the

percentage gap between the optimal and heuristic solutions

as follows:

optimal solution ? heuristic solution % gap

= ?-:-x 100.

optimal solution

For each combination of the model and the heuristic, we

provide a table containing the inventory level, the percent

age gap of the heuristic solution (minimum, average, and

maximum over the 50 instances), and the computation time

in seconds (minimum, average, and maximum over the

50 instances).

4.1. Model 1

Table 1 indicates the minimum, average, and maximum

computation time to solve instances of Model 1 to optimal

ity. Table 2 presents the results of H(8 =

0.95) for Model 1.

The mean percentage gap, across the 250 instances,

is 1.02% (with a worst case of 3.39%). Moreover, H(8 =

0.95) is very quick. As seen from Table 2, each of the

250 problems was solved within 38 seconds and the aver

age computation time was less than one second. The

results for H(8 =

0.01) on Model 1 are given in Table 3.

As expected, there is an improvement in the quality of the

solutions. Over all 250 problems, the mean percentage gap is 0.21% (with a worst case of 0.87%). Using R(8

= 0.01)

may not be affordable in some cases because the mean time

to solve these problems is 24 seconds (with a worst case

of 622 seconds).

4.2. Model 2

We set fr = 0.5, j32

= 0.7, \NX\ = 10, and K\

= K\

=

U[0, 100]. All the other parameters are the same as before.

Table 1 indicates the minimum, average, and maximum

computation time to solve instances of Model 2 to optimal

ity. For each inventory level, the computation time required to solve the instances of Model 2 in our test bed is notice

ably smaller than that for instances of Model 1. While

much of the structure of Model 1 is preserved in Model 2, the added restriction that any partial shipment to a cus

tomer be at most 100/32% of demand typically leads to a

much smaller percentage integrality gap (ratio of the dif

ference between the optimal value of the LP relaxation and

the optimal integer solution to the value of the optimal inte

ger solution) for the chosen values of f3x and /32. Note,

however, that Model 1 is a special case of Model 2 corre

sponding to /32 = 1 and Nx=0. The difficulty of Model 2

instances can, therefore, be increased easily.

As was the case for Model 1, H(8 = 0.01) provides bet

ter solutions than H(8 =

0.95), but requires more computa tion time. In Table 4, we report the results of H(8

= 0.95)

on these problems. The mean percentage gap, over the

250 instances, of the solutions found is 1.94% (with a worst

Table 2. Minimum, average, and maximum deviations and computation times for the

H(8 =

0.95) heuristic when applied to Model 1.

Percentage gap Computational time


Ij = U[290OO, 4,000] 0.00 0.17 2.94 0.05 0.96 14.54

/,- = tf[2,000,3,000] 0.07 0.58 1.99 0.05 1.65 37.78

Ij = U[l,000,3,000] 0.09 1.15 2.82 0.03 0.51 5.46

Ij = U[l,000,2,500] 0.19 1.50 3.36 0.03 0.25 4.62

Ij: =

U[1,000,2,000] 0.07 1.72 3.39 0.03 0.11 3.12






H(8 =




/, = ?/[2,000,4,000] 0.00 0.00 0.01 0.05 3.84 33.67

/; = L/[2,000,3,000] 0.00 0.11 0.36 4.19 30.75 134.99

/7 = L/[1,000,3,000] 0.00 0.30 0.68 0.05 23.01 240.73

Ij = U[l, 000,2,500] 0.00 0.30 0.77 0.05 41.66 621.49

/, = c/[l, 000,2,000] 0.00 0.36 0.87 0.04 20.56 139.94

case of 5.35%), and the mean time to solve the problems was 0.10 seconds (with a worst case of 2.85 seconds). The H(5

= 0.01) heuristic (Table 5) was able to obtain

solutions with a mean percentage gap of 0.21% (with a

worst case of 1.02%). The mean time to solve these prob lems was 7.28 seconds (with a worst case of 72.90 sec

onds). A probabilistic analysis, similar to that presented for

Model 1 in ?2.3, explains the success of the heuristics. As

with Model 1, the drawback of H(S = 0.95) for Model 2 is

that it can be too conservative in predicting the number of

customers that can be satisfied with the available inventory. Of course, one can choose a smaller value of 8 to make the

heuristic predict with more accuracy. However, a trade-off

is that the computation time will increase. One characteris

tic of these two heuristics is the trade-off between solution

quality and computation time. While the solution quality of H(5 =

0.01) is quite good, there are instances where

the time required may be more than what is acceptable. In

general, other values of 8 can be chosen to customize the

heuristics for the specific requirements at hand. To illus trate this point, in the next subsection we provide aggregate results of the heuristic performance for various values of 8.

4.3. Heuristic Behavior for Various 8 Values

In this section, we present the aggregate results (of the

percentage gap and the computation time) for Models 1

and 2 for various values of 8. To understand how the 8

value affects the performance of the heuristics, we ran

10 instances for each inventory level for 8 values of 0.01, 0.25, 0.50, 0.75, and 0.95. The results are shown in Table 6.

Observe that for both models, as 8 increases, the per

centage gap increases and the computation time decreases.

Note that the percentage gap increases consistently as

8 increases, whereas the computation times are much

higher at lower values of 8. Depending on the requirements of a particular situation, the analyst can choose a value of 8 to achieve a good balance between computation time and

effectiveness of the heuristic solution. The number of cus

tomers set to receive no shipment measures the reduction in

the problem size. For a given inventory level, this reduction increases with 8.

5. Some Extensions of the Basic Models

In this section, we study two extensions for the partial

shipment models detailed above. The first one is with

respect to minimum shipment restrictions being applied to

each of the products, and the second one is with respect to customer specific minimum requirements. For reporting computational results, we incorporate these extensions into the constraint set of Model 1.

5.1. Minimum Restriction on Each Product

In Models 1 and 2, although a customer receives a ship ment, he may not receive anything (or may receive a very

small portion) of a particular product. In some cases, what is desired is that if a customer receives a shipment, then he receives at least some minimum portion of each prod uct. Thus, we require that in addition to the constraints in

Model 1, a minimum required fraction of each product, represented by f33, be shipped to a customer to whom a


H(5 =




Ij =

U[2,000,4,000] 0.00 0.01 0.15 0.13 0.17 0.32

Ij =

U[2,000,3,000] 0.24 0.92 2.05 0.10 0.22 2.85

Ij = U[l,000,3,000] 0.90 2.24 4.02 0.03 0.06 0.15

Ij =

U[l,000,2,500] 0.97 2.88 4.74 0.03 0.04 0.06

Ij = U[l,000,2,000] 2.04 3.66 5.35 0.03 0.03 0.05



Dawande, Gavirneni, and Tayur: Effective Heuristics for Multiproduct Partial Shipment Models 346 Operations Research 54(2), pp. 337-352, ?2006 INFORMS


H(6 =




/,. =

C/[2,000,4,000] 0.00 0.00 0.01 0.16 0.24 0.43

/,- =

l/[2,000,3,000] 0.00 0.02 0.16 0.57 5.00 19.42

Ij = U[l,000,3,000] 0.00 0.32 0.94 0.09 12.15 65.87

Ij = U[l,000,2,500] 0.00 0.32 0.83 0.05 10.43 72.90

Ij = U[l,000,2,000] 0.00 0.41 1.02 0.04 8.60 70.68

shipment is made. We formulated the corresponding inte

ger program and used MINTO to solve it to optimality. For the H(8

= 0.01) and U(8

= 0.95) heuristics for this model,

determining the subset of customers on which to impose the constraints is done as before, using Steps 1 and 2 of ?2.2. The experimental set-up to generate instances of this model is the same as that described in ?2. The results, averaged over 25 problems (5 problems for each inventory level), are

summarized below.

Case 1. /33 =

/3. For this case the optimal solution com

putations required, on the mean, 518.32 seconds of com

putation time, while the mean computation time for the

heuristics, H(8 =

0.01) and H(S =

0.95), was 7.52 sec

onds and 0.09 seconds, respectively. The heuristic solutions

were, on the mean, 1.47% and 4.08% below the optimal for H(S

= 0.01) and H(S

= 0.95), respectively.

Case 2. 0 < /33 < /3. As /33 decreases from /3 to 0, the

problem looses its fixed-charge network flow structure and

the problems are generally harder. For the case /33 = 0.25

and /3 = 0.6, the mean time required to solve the problems to optimality was 837.45 seconds. For H(8

= 0.01), the

mean time required was 9.72 seconds, while the mean per

centage gap was 0.45%. For H(8 =

0.95), the mean time

and the mean percentage gap were 0.53 seconds and 2.18%,

respectively.

Thus, for both cases in this extension, the heuristics

resulted in excellent solutions with reasonable computation

times.

5.2. Customer-Specific Minimum Requirements

All the models considered previously in this paper assume

that the acceptable minimum fraction of a customer's total

demand is the same for all customers. That is, j8 is the same

for all customers in Model 1, and (3X and j82 are the same

for all customers in Model 2. In this section, we consider the case when these minimum requirements are different

for different customers. Both H(8 =

0.01) and H(8 = 0.95)

can be easily extended to include this case: Both heuristics

predict the number, F, of customers to be considered using the average of the minimum requirements for the different customers.

For this extension, we report the results of H(S =

0.01) and H(5

= 0.95) on 10 problems (2 problems for each

inventory level). For each customer, the minimum require

ment was randomly chosen in the interval [0.4, 0.8]. For

H(5 =

0.01), the mean percentage gap of the solutions

found was 1.00% and the mean time to solve the prob lems was 15.46 seconds, while for H(8

= 0.95), the mean

percentage gap was 4.32% and the mean time was 0.61 sec

onds. The mean time required to solve the problems to

optimality (using MINTO) was 621.34 seconds.

5.3. Other Extensions

We discuss three other extensions that may be handled in a

similar manner. The first extension considers the possibility of (simple) substitution of a product by another product. This substitution is possible in a multiproduct environment,

mainly due to two reasons: (1) a product j is identical to

another product / except for the fact that it is of higher

quality, and (2) a product j is identical to another prod uct / except that it is of different packaging (for example, a 12-pack versus a 6-pack). The second extension incor

porates a trucking restriction into the models. That is, the

Table 6. Behavior of the heuristic scheme H(<5) for various values of 8.

Model 1 Model 2

Percentage gap CPU time Percentage gap CPU time

Value of 8 Min. Avg. Max. Min. Avg. Max. Min. Avg. Max. Min. Avg. Max.

0.01 0.00 0.24 0.81 0.04 19.39 152.50 0.00 0.24 0.94 0.05 6.60 34.02 0.25 0.00 0.68 3.19 0.03 3.13 37.51 0.00 0.55 2.40 0.04 1.58 14.21 0.50 0.00 0.90 3.19 0.03 2.02 37.27 0.00 0.71 2.85 0.04 0.58 4.96 0.75 0.00 1.26 4.49 0.03 0.67 10.85 0.00 0.98 3.69 0.04 0.28 4.98 0.95 0.00 2.48 5.72 0.02 0.26 5.21 0.00 2.01 5.55 0.03 0.10 0.32





total physical volume of the shipment to a customer should

be large enough to warrant the use of a truck. (This may

replace or add on to the (3X fraction constraint.) As a third

extension, consider a situation in which the customers are

already prioritized (based on some other corporate criteria). In this case, we need not use the sorting procedure that uti

lizes the actual demands. Hence, using the number of cus

tomers predicted by our heuristics and the prioritized list, it will be possible to inform the customer of the possibility of receiving a shipment at the time the order is placed (an "on-line quote" versus an end-of-the-day decision).

6. A Set of Hard Instances

The problem instances used to obtain the results in

Tables 1-6 were of moderate size (25 products, 100 cus

tomers), and were mainly intended to assess the solution

quality of the scheme of heuristics. To this end, it was ben

eficial that most of the instances were solved to optimal

ity within a reasonable amount of time. However, it is not

uncommon for retail distribution centers to face inventory allocation problems involving up to 100 products and up to

1,000 customers. We now illustrate the performance of the

heuristics on such instances. Our main purpose here is to

demonstrate that, as problem size increases, a practitioner

has to rely on the heuristics to obtain quality solutions

quickly. Table 7 illustrates our computational experience with

Model 1 on four problem sizes: (1) \N\ = 300, \M\

= 75;

(2) \N\ = 500, \M\ = 50; (3) \N\ = 600, \M\ = 50; and

(4) \N\ = 1,000, \M\

= 100. Five instances were gener ated for each problem size. Using the well-known intuition

from generating computationally hard instances of pack

ing problems (e.g., the knapsack problem (Chvatal 1980, Martello and Toth 1990)), the inventory for each prod uct is chosen to make a minimum shipment possible to

approximately half the total number of customers. The

other details of the experimental set-up are as described

in ?2. MINTO was unable to solve any of the 20 prob lems within an imposed limit of 24 hours of CPU time for

each problem. The percentage gaps for the feasible solu

tions were, therefore, computed using the best upper bound

available from MINTO when the problems were termi

nated. Columns 3 and 4 (respectively, 5 and 6) denote the

percentage gap and CPU time, respectively, when MINTO

obtains the first (respectively, best) solution. MINTO was

unable to find any feasible solution for six instances. For

the remaining instances, the time for the first (respec

tively, best) solution ranges from about 25 minutes to about

eight hours (respectively, from about 30 minutes to about

Table 7. Performance of MINTO and the heuristics on hard instances of Model 1.

Heuristic performance Performance of MINTO -

H(S = 0.95) H(5 = 0.01) First solution Best solution solution solution

% Gap++ Time % Gap++ Time Time for % Gap++ Time % Gap++ Time Problem parameters Instance (%) (sec.) (%) (sec.) optimality (%) (sec.) (%) (sec.)

N = 300 1 9.50 1,445.71 9.49 2,394.49 **** 10.91 0.43 9.89 0.62

M = 75 2 0.09 1,338.35 0.08 35,063.74 **** 0.95 0.45 0.25 65.91

Ij = U[4,000,5,000] 3 0.19 28,306.62 0.14 60,475.96

**** 1.26 0.46 0.58 48.43 4 11.92 2,244.09 11.92 2,244.09

**** 12.97 0.40 12.10 212.33

5 9.53 7,390.96 9.53 7,390.96 **** 10.91 0.39 9.95 0.56

_V = 500 1 5.76 2,563.88 5.76 2,563.88 ****

7.32 0.44 5.94 44.37

M = 50 2 0.08 27,070.60 0.01 57,213.14 **** 1.85 0.43 0.35 42.00

Ij = U[7,000,8,000] 3 8.18 5,092.94 8.18 5,092.94

**** 10.20 0.45 8.67 0.66 4 8.36 1,866.50 8.36 1,866.50

**** 10.22 0.39 8.62 0.74

5 10.31 8,417.80 10.31 8,417.80 ****

11.77 0.45 10.54 43.88

TV = 600 1 10.18 3,506.65 10.18 3,506.65 **** 11.83 0.59 10.86 0.84

M = 50 2 0.18 2,033.70 0.15 4,514.07 **** 1.95 0.62 1.04 0.84

Ij = U[S,500,9,500] 3 8.87 1,581.69 8.85 3,998.76

**** 10.34 0.67 9.37 0.86 4 xxxx xxxx xxxx xxxx

**** 1.51 0.62 0.64 0.78

5 7.82 3,642.64 7.82 3,642.64 ****

9.23 0.69 8.26 0.95

N = 1,000 1 xxxx xxxx xxxx xxxx ****

8.05 3.07 6.78 4.87

M=100 2 xxxx xxxx xxxx xxxx ****

7.95 2.72 6.17 7.94

Ij =

U[14,500, 15,500] 3 xxxx xxxx xxxx xxxx **** 4.34 2.80 3.50 6.05

4 xxxx xxxx xxxx xxxx ****

8.03 2.53 6.76 7.32

5 xxxx xxxx xxxx xxxx ****

5.98 2.30 4.63 6.06

****No optimal solution within a CPU time limit of 24 hours.

++Percentage gap computed with respect to the best available upper bound.

xxxx No feasible solution found within a CPU time limit of 24 hours.



Dawande, Gavirneni, and Tayur: Effective Heuristics for Multiproduct Partial Shipment Models 348 Operations Research 54(2), pp. 337-352, ?2006 INFORMS

16 hours). The heuristics, H(8 =

0.95) and H(8 =

0.01), provide comparable solutions quickly. Columns 8 and 9

(respectively, 10 and 11) denote the percentage gap and

CPU time, respectively, of the solution from H(8 =

0.95)

(respectively, H(S =

0.01)). The time for the solutions from H(8

= 0.95) (respectively, H(8

= 0.01)) ranges from

0.39 seconds to 3.07 seconds (respectively, from 0.56 sec

onds to 212.33 seconds). As mentioned earlier, H(8 =

0.95) provides a quicker solution at the expense of solution qual

ity. The success of the heuristics on these instances is

easy to explain: Irrespective of the total number of cus

tomers, the number of customers considered by the inte

ger programs generated by the heuristics has approximately half the total number of customers. For example, for

instances with 1,000 customers, the integer programs for the heuristics consider around 500 attractive customers.

Apart from the reduction in problem size, because the avail

able inventory is sufficient to make a minimum shipment to

most of these chosen customers, the resulting optimization

problems are easy to solve.

7. A Two-Period Problem

Our aim in this section is to illustrate the effectiveness of

the main ideas behind the single-period heuristics (?2.2) for a multiperiod setting. While there are many ways in

which a multiperiod system can be managed, we consider a representative, albeit simplified, two-period optimization

problem. We first present a problem formulation followed

by two heuristic procedures. The system dynamics for the two-period problem can be

described as follows.

(1) At the beginning of the first period, each product has some initial inventory and a known replenishment quan

tity that is to be received at the beginning of the second

period,

(2) The first period demands are realized and shipping decisions are made adhering to all necessary shipping restrictions,

(3) Demands not satisfied in the first period are back

logged and penalty costs are levied on them,

(4) Holding costs are accrued on inventory carried over

into the second period,

(5) Replenishment quantities for each product arrive at

the beginning of the second period,

(6) All or part of the backlogged first-period demand

are satisfied (at a lower per-unit revenue) from the new

inventory,

(7) That portion of the backlogged first-period demand

not satisfied at the beginning of the second period is

assumed to be lost,

(8) The second-period demands are realized and ship

ping decisions are made, and

(9) At the end of the second period, holding costs are

accrued on the leftover inventory, and penalty costs are

levied on the backlogged demands.

Two assumptions warrant further discussion. The first

pertains to the decision-maker's ability to (ultimately) let some demand go unsatisfied. In the absence of this option, all the demand for the first period must be met, which in turn results in the problem getting decoupled. In this case, the two-period problem can be essentially solved as two

independent one-period problems. The second assumption relates to the revenue generated from satisfying backlogged demand. We distinguish these backlogged shipments from the regular shipments that satisfy nonbacklogged demand.

We assume that the backlogged shipments need not meet the shipping restrictions imposed on the regular shipments. In addition, the revenue generated by a backlogged ship

ment is assumed to be lower than that generated by a reg ular shipment?this reduction could result from a variety of factors including additional paperwork, loss of good will, and alternate shipping strategies that might need to

be resorted to. For ease of exposition, we further assume

that backlogged shipments for a product to all customers

generate the same per-unit revenue, as doing so allows us

to keep track of the backlogged shipments for a product in aggregate, rather than as individual shipments to each customer.

7.1. Problem Formulation

As before, we are faced with demands for a set, M, of

products from a set, N, of customers. Let Dj (respec

tively, Elj) be the demands for product j from customer / in

Period 1 (respectively, 2) and let Sj (respectively, Tj)

be the

shipments of product j to customer i in Period 1 (respec tively, 2). Let

Ij (respectively, ij) denote the inventory of

product j at the beginning of Period 1 (respectively, 2). For each product j e M, we are to decide the portions of

the demands Dlj and Ej. that will be satisfied from inven

tory. We cannot supply more than what is demanded and we cannot supply more than the available inventory. We

consider the shipping restriction used in Model 1 (?2): If a

regular shipment is made to a customer, it must be at least

the /3 portion of the customer's total demand. As in the

one-period models, the revenues received are of two types:

a fixed reward Kl if a regular shipment is made to cus

tomer / and a shipment-quantity-based reward of (Sj/Dprj

(respectively, (Tj/Efirfi for supplying product j to cus

tomer / in Period 1 (respectively, 2). We use binary vari

ables w\, t = 1, 2; / N, to impose the shipping restrictions.

If w\ = 0, customer / receives no shipment in period t\

otherwise, utt =

1, and the customer gets at least a partial

shipment. Let Rj, j e M, be the replenishment of prod uct j received in Period 2. Let Q\ (respectively, Uj)

be

the ending inventory (respectively, backlogged demand) of

product j in Period 1. The holding cost and the backlog cost per unit per period are assumed to be h and b, respec

tively. For product j, let Vj be the revenue per unit of sat

isfied backlogged demand (recall that we have assumed it

to be independent of the customers) and let L; be the lost

(backlogged, and ultimately unsatisfied) demand. Finally,





let Q2 (respectively, U2) be the ending inventory (respec

tively, backlogged demand) of product j in Period 2. A MIP

formulation of the problem follows:

2 \N\ \N\ \M\ $i \M\ Maximize ZZK'^ +

IlIl'-jji + Y,?j(u} ~Lj)

r=i /=i 1=1 j=\ uj j=\

\N\ \M\ ji 2 \M\

/=1 j=\ ?*j t=x j=x

\N\ such that /]-Esj

= G). JeM> (3)

\M\ /\M\ \

|M| /\M\ \

E^(E^K> ieN, (5)

\N\ \N\

ZD)-ZS) =

u;, jeM, (6)

\N\

Ij-J2Dij + Rj

= I2-Lj, jeM, (1)

i=\

Lj^Uj, jeM, (8)

\N\

1]-YJ) =

&p 7'eM, (9) 1=1

\M\ (\M\ \

E^ME^K, ieN, (10)

\M\ /\M\ \

E^<(E?;K, * tf, (ii)

E?|-E^ =

^> ;eM, (12) i=l i=l

0^5J<D), ieN,jeM, (13)

O^r/^^;, ieN, jeM, (14)

wj,w4e{0,1}, /GiV,

G], tf/, /?, Ly> fi2, tf,2 nonnegative, y e M.

The objective maximizes, over the two periods, the revenue for shipments (both regular and satisfied back

logged shipments) minus the costs associated with holding

inventory and backlogged demand. The regular ship ments in the two periods have both shipment rewards

(E2=iE!=U^;) and per-unit revenues (?l=i ?%\ r)S)/ D) and El=l E%\ r>Tj/E<). The term ?^| Vj(U}

- Lj)

computes the revenue from backlogged shipments. The

holding and penalty costs are computed using the term

EltETJ^ffj + bUj]. Constraints (3) and (9) ensure that shipments are con

strained by the available inventories. Constraints (4) and (5)

(respectively, (10) and (11)) impose the shipping restric

tions on the regular shipments in Period 1 (respec

tively, Period 2). The excess inventory and backlogged demand are computed using constraints (3) and (6) (respec

tively, constraints (9) and (12)) in Period 1 (respectively, Period 2). Constraint (7) ensures that the first-period demands are either satisfied (from initial inventory or

replenishment) or backlogged and ultimately lost. This con

straint also determines the inventory available to satisfy demands in the second period. Constraint (8) ensures that

the lost demand is not greater than the backlogged demand.

Constraints (13) and (14) impose the lower and upper limits

on the shipments in the two periods.

7.2. Heuristic Procedures

We consider two extensions of the single-period heuristics

proposed previously. The first is a myopic, first-come-first

served approach; the first-period demands are satisfied to

the extent possible before the second-period demands are

considered. The second approach looks ahead into the sec

ond period, and takes the first-period shipment decisions

after incorporating the impact of the demands in the second

period.

7.2.1. Myopic Heuristic. The myopic heuristic takes

decisions using just the information available in the current

period. In the first period, the number of customers, F1, on whom shipping restrictions are imposed, is computed as

before, based on the cumulative inventory level as

r m -\

Prob (/3 demand of Fl customers) ^ ?Vj = ^'

The idea, then, is to (1) choose a value of 8 to obtain F1,

(2) impose that a set of (\N\ ?

F1) customers receive no shipment, and (3) optimize the shipping decisions for

the remaining F1 customers. Once the first-period ship

ping decisions have been made, the resulting backlogs and

excess inventories can be determined in a straightforward manner. The replenishment quantities are received at the

beginning of the second period. From the new inventory levels, backlogged demands are satisfied to the extent possi

ble and the rest are assumed lost. After satisfying the back

logged demand, the resulting inventory level ij

of product j in Period 2 is

I) =

[Ij + Rj -

??, D/]+. Next, the num

ber of customers F2 to be included in the second-period

analysis is computed as follows:

r M "i Prob (/3 demand of F2 customers) ^ ]T I2 = 8.

L j=\ J

Again, we decide on a set of (\N\ ?

F2) customers to

receive no shipment in the second period and optimize the

shipping decisions for the remaining F2 customers. As in

the one-period models, the idea is to choose appropriate values of F1 and F2 (based on the starting inventory lev

els, replenishment quantities, and the chosen 8 value) so

that the two optimization problems (one each for Periods 1 and 2) are easily solvable.





7.2.2. Look-Ahead Heuristic. The look-ahead heuris

tic chooses first-period shipments and late-fulfillment

quantities by considering demand information from both

periods, along with the replenishment quantities to be

received at the beginning of the second period. We compute two numbers Fl and F(1+2) using the relations

r m -\ Prob (j8 demand of Fx customers) < E ^/

= ^' L j=\ J

r m -i

Prob (/3-demand of F(1+2) customers)<^[/]+^] \=8. ;=1

As before, F1 is an estimation of the number of customers

that could be satisfied in the first period. F(1+2) is an estima

tion of the total (over Periods 1 and 2) number of customers

that could be sent a regular shipment at the current levels of

inventory and replenishment. We allow for the possibility that some inventory could be carried over (while backlog

ging and losing demand at the same time) from one period to the next to increase the total number of regular ship

ments. Based on the values of F1 and F(1+2), two scenarios

could exist?if Fl ^ |F(1+2)A|, we set F1 = [F(1+2)/2j

and F2 = (F(1+2)

? F1); otherwise, we leave F1 unchanged

and set F^F^-F1.

The main idea behind this calculation is easy to explain: The inventory left over from the first-period regular ship ments and the replenishment quantities need not be imme

diately used to satisfy the backlogged demand. Instead, we

let the optimization problem (resulting from the relation

ship for F2) decide the division of that inventory between

backlogged first-period shipments and regular shipments in

the second period.

7.3. Computational Results

The setup for testing the effectiveness of the heuristics is

similar to that used for the one-period problems: \N\ = 100,

\M\ = 25, j8

= 0.6. For all i, j, D) =

U[09 100], Kl =

U[09150], and r)

= U[09 15]. The per-unit revenue of the

satisfied backlogged demand Vj was assumed to be 60% of

the average per-unit revenue across all customers. We con

sider four combinations of the starting inventory levels in

the first period and the replenishment quantities arriving in

the second period.

Case Starting inventory Replenishment quantity number for each product for each product

1 tf[2,000,4,000] L7[4,500,5,500] 2 ?/[2,000,4,000] t/[2,000,3,000]

3 c7[l,000,3,000] c7[4,500,5,500] 4 U[l,000,3,000] L7[2,000,3,000]

Recall that the average total demand in a period is

5,000 units. These cases are designed to capture high and

low starting inventories as well as high and low replen ishment levels. For each case we generated 50 problems.

Table 8 contains the average, minimum, and maximum per

centage gap of the heuristic solution (with 8 = 0.01) from

the optimal solution.

As expected, when the replenishment is close to the aver

age total demand and the starting inventory level is high, the myopic heuristic performed reasonably well. However, its performance deteriorated rapidly as the replenishment

quantity or the starting inventory level decreased. Table 9

contains the average, minimum, and maximum computa

tion times for each of the four cases. Over all the problems, the average computation time for both heuristics was less

than 10 seconds. The maximum time for an instance was

80 seconds for the myopic heuristic and 122 seconds for

the look-ahead heuristic.

It is reasonable to conclude that the look-ahead heuris

tic, with its low computation time and high-quality solu

tions is an effective method for solving the two-period

problem. The myopic heuristic is useful when the start

ing inventory levels and replenishment quantities are suf

ficiently large, but its performance deteriorates sharply as

either one decreases.

7.4. Extensions

We envision the main ideas presented in this paper to be

easily applicable in developing heuristics for a wide variety of inventory-constrained multiproduct, multicustomer, mul

tiperiod distribution problems. These problems are usually characterized by two decisions that must be made simulta

neously: (1) determination of a set of customers that could

receive a shipment, and (2) efficient allocation of scarce

inventory to the chosen customers subject to various ship

ping restrictions. Making both these decisions simultane

ously to obtain a globaloptimum becomes computationally

Table 8. Average, minimum, and maximum deviations of the myopic and look-ahead

heuristic solutions from the optimal solution.

Myopic heuristic Look-ahead heuristic

Case number Minimum Average Maximum Minimum Average Maximum

1 0.00 0.40 4.02 0.00 0.13 1.03

2 29.23 37.11 47.70 0.41 2.21 4.15

3 14.04 19.98 36.24 2.26 3.69 5.48 4 56.73 67.17 77.72 4.96 7.13 9.92





Table 9. Average, minimum, and maximum computation times for the myopic and look-ahead

heuristic procedures and the optimum solutions.

Myopic heuristic Look-ahead heuristic Optimum solution

Case number Min. Avg. Max. Min. Avg. Max. Min. Avg. Max.

1 0.21 6.17 79.84 0.21 6.15 79.95 0.22 15.15 401.75 2 0.22 4.47 31.05 0.87 7.89 122.39 5.76 551.18 8,329.90 3 0.29 5.17 44.90 0.37 6.26 48.42 8.96 177.62 1,228.50 4 0.20 2.92 17.85 0.56 8.89 76.94 1,097.33 3,481.18 16,076.05

intractable, especially for multiperiod problems with a large number of customers and products.

The basic intuition behind the heuristics is the ability to

construct, for each period, a decision problem in which the

available inventory is sufficient to satisfy most of the cus

tomers considered. For multiperiod problems, the choice of

the customers to be considered (for shipment) in individ

ual periods will typically be based on global considera tions and involve predictions based on the available demand

and supply information for the entire planning horizon. In

developing such a scheme for problems with a large num

ber of customers and products, two specific properties can

be exploited: first, the law of large numbers, which leads to

the Central Limit Theorem (see, e.g., Bain and Engelhardt 1987) and approximate normal distributions for sums of

individual observations; and second, the inherent dispersion in the individual customer demands for the products.

A notable advantage of this approach is that the size

(determined by the number of customers considered) of the

decision problems to be solved is driven solely by available

inventory. Thus, the total number of customers, which may sometimes be much larger than the number of customers

that can be sent a partial shipment, does not dominate the

complexity of the problems solved by the heuristics. We believe that similar heuristics can be designed and

shown to be effective for several extensions to the mod els considered here. Below we discuss three of the most salient ones.

Supplier Lead Times. We have assumed that the distrib utor is aware of the replenishment expected to arrive in the next period. In some cases, the replenishment itself may be a decision variable at the distributor and must be decided based on the lead times and the shipping strategy of the distributor's supplier(s). If the supplier has long lead times,

say A periods, then the reorder decision made in a period impacts the costs A periods hence. It then becomes neces

sary to incorporate information about the demands in those

intervening A periods while determining the reorder quan

tity. If the demand forecasts into the future have high vari

ability, it may be necessary to use stochastic programming models (Birge 1994, Kail and Wallace 1995).

Minimum Order Quantities. When dealing with high demand customers (commonly referred to as A-type cus

tomers), a restriction on the percentage of the total demand

typically leads to reasonably large shipments. This may not be the case when the demands from some customers

are relatively small. In the presence of such customers, the

formulation of the shipping restrictions in the form done

here may still result in shipments that are small and cause

major inefficiencies. If such customers can be identified ahead of time, then their /3 value can be changed and the

problem solved along the lines of customer-specific j8 val ues detailed in ?5.2. Otherwise, it may be necessary to

explicitly formulate the restrictions on minimum shipment quantities.

Ordering in Sets. Another extension that may have broad

applicability is the one that deals with orders for sets of

products. In this case, customers want to receive a mini

mum portion of the demand for every product. This exten

sion for the single-period problem was discussed in ?5.1 and similar modifications can be made to the multiperiod problem. Our heuristics can be easily extended to address this situation.

8. Conclusions

We addressed an important and widespread problem faced

by distributors?efficient allocation of scarce inventory to fulfill orders from customers, who impose stringent restric tions on the size of acceptable partial shipments. For customers ordering for sets of products, we considered mul

tiproduct partial-shipment models that incorporate a variety of practical shipping restrictions. Typically, these allocation

problems become computationally intractable with (1) an

increase in the number of products and the number of cus

tomers, and (2) a decrease in the ratio of the available

inventory to the demand. Given a need for real-time solu

tions, we provided a scheme of heuristics that allows users to customize the specific requirements of the quality of the solutions and the speed with which they are desired.

A detailed computational study for two extreme cases of this scheme established its versatility and practicality for a variety of industrial settings. For single-period prob lems of moderate size (25 products, 100 customers), one extreme offered solutions with a mean gap of 1.48% in a few seconds, while the other extreme offered solutions with a mean gap of 0.21% in a few minutes. An important characteristic of the heuristics is that they scale well with

problem size: Large problems with up to 100 products and

1,000 customers were solved in a few minutes, and with a mean gap of 6.24% from the best available upper bound. The effectiveness of the heuristics was also demonstrated for two-period problems.





The difficulty of the partial-shipment models we consider

stems from the fact that they are required to simultaneously decide on the subset of customers receiving a shipment and

the allocation of inventory to these customers. By incorpo

rating demand information for the entire planning horizon, the heuristics construct problems for individual periods that

balance the available inventory and the choice of customers.

For large, inventory-constrained distribution problems such a construction should typically be applicable in the pres ence of a reasonable amount of dispersion in individual

customer demands for the products.

Typically, in practice, inventory policies (min-max,

safety-stock rules, order sizes, etc.) are set arbitrarily (or

using some aggregate analysis) across many SKUs. This

leads to high inventories and low service levels at the same

time (i.e., there is a mismatch in the product mix) due to the

inefficient accommodation of orders that come in sets. The

goal of our analysis is to provide a bottom-up method to set

inventory rules. Because these inventory rules are generally

set for a longer time horizon, it may be necessary to ana

lyze a detailed multiperiod model to set them. The analysis

presented in this paper is a first step in that direction and

a comprehensive multiperiod model is expected to be the

subject of future work if there is sufficient industry demand.

Multiproduct problems are considerably more complex in

the multiperiod setting when the demands come for sets

of products and are known a priori only through a proba

bility distribution with actual values revealed on a period

by-period basis?in fact, the structure of optimal solutions

is not known even for simple cases (for example, without

shipping restrictions). Thus, good heuristics are even more

vital for these types of problems.

Acknowledgments The authors thank Martin Savelsbergh and Jeff Linderoth

for providing the latest version of MINTO for the

computations.

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