optimal single ordering policy with multiple delivery modes and bayesian information updates

Computers & Operations Research 31 (2004) 1965–1984www.elsevier.com/locate/dsw

Optimal single ordering policy with multiple delivery modesand Bayesian information updates

Tsan-Ming Choi∗, Duan Li, Houmin Yan

Department of Systems Engineering and Engineering Management, The Chinese University of Hong Kong,Shatin, N.T., Hong Kong

Abstract

We investigate in this paper a retailer’s optimal single ordering policy with multiple delivery modes. Dueto the existence of di3erent delivery modes, the unit delivery cost (and hence the product cost) is formulatedas a decreasing function of the lead-time. Market information can be collected in the earlier stages and usedto update the demand forecast by using a Bayesian approach. The trade-o3 between ordering earlier or lateris evident. The former enjoys a lower product cost but su3ers a less accurate demand forecast. The latterpays a higher product cost, but bene7ts from a lower uncertainty in the demand forecast. In this paper, amulti-stage dynamic optimization problem is formulated and the optimal ordering policy is derived usingdynamic programming. The characteristics of the ordering policy are investigated and the variance of pro7tassociated with the ordering decision is discussed. Numerical analyses through simulation experiments arecarried out to gain managerial insights. Implementation tips are also proposed.? 2003 Elsevier Ltd. All rights reserved.

Keywords: Inventory management; Optimal stopping problem; Bayesian normal conjugate family; Dynamicprogramming; Computer implementation

1. Introduction

In supply chain inventory management, one of the challenges is how to manage demand uncer-tainty. As an example, consider a problem faced by a fashion apparel retailer: The fashion product’sshelf-life is short, the ordering lead-time is long and market demand is highly volatile. In fact,demand uncertainty is the “root of evil” which accounts for the large supply chain inventory cost. Inorder to reduce demand uncertainty, companies of fashion clothing have proposed the well-known“quick response” policy. Quick response basically refers to the concept of reducing the ordering

∗ Corresponding author. Tel.: +852-31634062; fax: +852-26035505.E-mail addresses: [email protected] (T.-M. Choi), [email protected] (D. Li), [email protected] (H. Yan).

0305-0548/$ - see front matter ? 2003 Elsevier Ltd. All rights reserved.doi:10.1016/S0305-0548(03)00157-6

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1966 T.-M. Choi et al. / Computers & Operations Research 31 (2004) 1965–1984

lead-time and adjusting the stocking decision quickly in response to market changes. In fact, weknow that the closer to the selling season, the more information we have about the uncertain de-mand and hence a smaller demand uncertainty results. Quick response is now widely applied invarious industries all around the world and its impact is substantial. Reducing demand uncertaintyby shortening the lead-time is well-recognized as an important achievement of supply chainmanagement.

In the literature, many articles focus on the issues of postponing the stocking decision and usinginformation updates for forecast improvement. Among them, Iyer and Bergen [1] provide an indepthanalysis of quick response policy and information updating in a manufacturer-retailer channel. They7nd that by postponing the ordering decision under quick response, the retailer is always better-o3.They also suggest methods under which the manufacturer also bene7ts from quick response. A con-sideration of the use of early sales signals from the market for reducing cost and demand uncertaintyunder a quick response type of ordering system is studied in Fisher and Raman [2]. They modeland analyse the ordering decisions under quick response for a single selling season and propose amethod for estimating the demand distribution required in their model. A fashion skiwear 7rm ischosen as the analysis case and the use of information is shown to be very signi7cant. In additionto the work by Fisher and Raman [2], with the consideration of using advance demand informa-tion, Gilbert and Ballou [3], Gallego and Ozer [4] and Tang et al. [5] have formulated inventorymodels under which the early demand signals or commitments are used to improve the forecast ofthe product during the normal selling season. The impacts brought by advance demand informationhave also been discussed. On the other hand, with the idea of including Lexibility into the retailer’sstocking decisions, Lau and Lau [6], Gurnani and Tang [7], Yan et al. [8] and Choi et al. [9] havestudied the ordering model with two ordering opportunities: After placing the 7rst order, the retailercan make market observation and improve the prior forecast. The retailer can then decide how muchto order at the second ordering chance with the updated and more accurate forecast.

Some papers are devoted to modeling the inventory decisions with multiple supply or deliverymodes. For example, in Zhang [10], ordering policies with three supply modes are developed, witha heuristic method proposed to solve the problem. Other related papers for information updates andmultiple production/delivery modes include Murray and Silver [11], Azoury and Miller [12], Azoury[13], Hausmann et al. [14], Brown and Lee [15], Choi et al. [16], Barnes-Schuster et al. [17], Huanget al. [18], Donohue [19], Chiang [20], Chung and Flynn [21], Sethi et al. [22], and Lau and Lau[23].

In this paper, we investigate an inventory stocking problem with multiple delivery modes andinformation updates. A retailer is considering placing an order for a (newsvendor type of) seasonalproduct with an uncertain demand. There are di3erent delivery modes available for him to chooseand the faster delivery mode is also the more expensive one. If he chooses a faster delivery mode,he can delay his stocking decision more and obtain more market observation (e.g. by observing thesales of other related pre-seasonal product). To be precise, the retailer can use market observationto revise and update his current knowledge about the demand distribution in a Bayesian fashion.This results in a better forecast and a lower demand uncertainty. Thus, a tradeo3 between deliverycost and demand uncertainty exists. In this paper, we are interested in studying an optimal singleordering policy. The retailer can only place one order. This situation arises when: (1) The 7xedordering setup cost is high. For instance, if the retailer places one big order, he can make use of afull truckload for delivery; if he makes two orders, he needs to use two which can cost a lot if the

T.-M. Choi et al. / Computers & Operations Research 31 (2004) 1965–1984 1967

truck’s capacity is expensive. (2) A quantity discount scheme o3ered by the supplier is attractiveenough to entice the retailer to make one big order. (3) There is a minimum order quantity imposedby the supplier and the retailer’s demand is not large enough to support him to order more thanonce.

We formulate the ordering policy as an optimal stopping problem (see Bertsekas [24]) and usedynamic programming to derive the optimal policy. Notice that this paper basically extends theordering system in Iyer and Bergen [1] with the incorporation of multiple delivery modes, multipleBayesian information updates and delivery cost di3erence. This enriches the ordering policy andaddresses the possibility of having multiple information updates. The optimal ordering decision ina multi-stage setting is notorious for a growth of complexity with respect to the number of stages.The lack of analytical solutions also negatively a3ects the model’s applicability. Thus, we proposemethods which enhance the implementation of the optimal policy by a computer program. We alsostudy the impact of observation on the variance of pro7t for the stocking problem.

The organization of the rest of this paper is as follows. In Section 2, we de7ne the basic mathemat-ical model and the uncertainty structure of the problem. In Section 3, we determine the optimal orderquantity and derive the optimal ordering policy using dynamic programming. Section 4 is devoted tothe numerical analyses with a real case example. The variance of pro7t associated with the optimalpolicy is discussed in Section 5. We conclude in Section 6 with a discussion of implementation.

2. Mathematical model

We derive in this section the mathematical model for the optimal ordering problem. First, we de7nesome notation: We denote probability density function (pdf) and cumulative distribution function(cdf) of the standard normal by �(·) and �(·), respectively. The inverse of �(·) is represented by�−1(·). We de7ne the standard normal linear loss function, �(·), as follows:

�(a) =∫ ∞

a(x − a) d�(x): (2.1)

Notice that �(·) can be written as

�(a) = �(a) − a[1 − �(a)]: (2.2)

In this paper, we have a multi-stage model in which a retailer can choose one delivery mode amongmany for shipping the ordered products. The faster the delivery mode, the more expensive it is.This is intuitive in real-life. For instance, using air-transport is much faster and requires a muchshorter lead-time than ship-cargo, but it is also much more expensive. Now, suppose there are N +1di3erent delivery modes available for the retailer to choose (N is any non-negative integer). Weorder the delivery modes in a decreasing sequence of their respective shipping times and the fastestdelivery mode is named Stage N and the slowest delivery mode is called Stage 0. As a result, fork = 0; 1; : : : ; N : Stage k’s delivery time is longer than the delivery time of Stage k + 1. (Notice thatwe assume in this paper that the delivery lead-time of each delivery mode be deterministic and theorder placed with a particular delivery mode will arrive on time before the selling season starts. Inreal-life, however, a probable variation of delivery lead time does exist. As a result, the retailer shouldmake sure that the delivery mode is reliable by, for example, signing a supply warranty contractto ensure that the product will be delivered on time (or at least within a small time interval).


This is essentially important because failing to guarantee the reliability of the delivery time, themodeling for the delivery modes becomes inaccurate, which then directly a3ects the optimality of thesolution.) When the retailer chooses to order later (e.g. at Stage 1 instead of Stage 0), he can makeobservations in the market and use this information to revise his forecast estimate. To be speci7c, fork = 0; 1; : : : ; N − 1, market information gathered by the retailer at Stage k can be used to update thedemand forecast at Stage k +1 by using a Bayesian approach. The observations can be visualized asthe market signals obtained from the sales of other related “pre-seasonal” product/product attributein the market. For example, let us consider the seasonal product as a yellow jacket which is goingto be sold in the up-coming selling season. Before the selling season of this yellow jacket starts,observations can be made from another clothing item which is currently selling in the market (e.g. ayellow t-shirt) and the sales of this related “pre-seasonal product” can help to improve the demandforecast for the yellow jacket. In fact, the color yellow is an important product attribute here whichis observed in the observation process. This type of example is well-studied in the literature andmore discussions can be found in Iyer and Bergen [1]. We denote the predicted demand of theseasonal product at Stage k by xk , where k = 0; 1; : : : ; N . Following the basic demand uncertaintystructure as that of Iyer and Bergen [1], we take the distribution of xk to be a normal distributionas follows:

xk ∼ N (k ; �); (2.3)

where k is the mean of xk and is also normally distributed with a mean of �k and a variance ofdk ,

k ∼ N (�k; dk): (2.4)

The variance � in (2.3) is called the seasonal product’s inherent demand uncertainty. It reLects theuncertainty of xk for given k and also captures the uncertainty in using the sales of the related (butnot the same) pre-seasonal products in the market to estimate the seasonal product’s demand (forthe details, please refer to Iyer and Bergen [1, p. 561]).

At Stage k + 1, we have an observation between Stage k and Stage k + 1 and we call it x̂k . ByBayes’ Theorem, with x̂k , we can update the demand distribution with the following parameters’changes:

�k+1 = [dk=(dk + �)]x̂k + [�=(dk + �)]�k; (2.5)

dk+1 = �dk=(dk + �): (2.6)

Before obtaining the observation x̂k , the unconditional distribution for the predicted demand at Stagek + 1; xk+1, is

xk+1 ∼ N (�k+1; dk+1 + �): (2.7)

Using (2.5) and (2.6), we have the unconditional distribution of �k+1,

�k+1 ∼ N (�k; �2k); (2.8)

where �2k = d2

k =(dk + �).


As a remark, from (2.3) to (2.8), we assume that the uncertain demand is normally distributed.Since demand is non-negative, in order to have a valid model, the coeQcient of variation of xk ,de7ned by

√dk + �=�k , should be suitably small (for example, less than 0.4). 1

Next, we discuss the cost-revenue structure for the seasonal product. We take the seasonal productas a newsvendor type of product. The retailer buys the product with a 7xed unit purchase cost calledpc. The holding cost and salvage price of each unit of product left over at the end of the sellingseason are hc and v, respectively. We represent hc − v by h (the unit cost of the product left over atthe end of the selling season). Observe that h can be positive or negative depending on the valuesof hc and v. The revenue (i.e. selling price) per product sold is called r. We denote the unit deliverycost for using the delivery mode at Stage k as lk . In our multi-stage model, we have the followingstructure for the unit product’s delivery cost:

l06 l1 · · ·6 lN ; (2.9)

Eq. (2.9) is a sound assumption because the faster delivery mode should be more expensive (orat least not cheaper) and the larger stage number implies a faster delivery mode in our modelformulation. To simplify the notation, we de7ne the unit ordering cost at Stage k as the summationof unit purchase cost and the unit delivery cost at Stage k:

ck = pc + lk for k = 0; 1; : : : ; N: (2.10)

Obviously, from (2.9) and (2.10), we have the following:

c06 c16 c2 · · ·6 cN−16 cN : (2.11)

To avoid trivial cases, we have r ¿ck , for k = 0; 1; : : : ; N . Furthermore, we de7ne the service level(also called the “7ll-rate”), which describes the probability of having no stockout during the sellingseason (see Nahmias [25, Chapter 5]), when the retailer orders at Stage k, by

sk = (r − ck)=(r + h) for k = 0; 1; : : : ; N: (2.12)

Fig. 1 shows the basic model structure of the problem. It also shows the trade-o3 between orderingcost and forecast accuracy.

Fig. 1. The basic model structure of the single ordering problem.

1 Thanks are given to an anonymous reviewer for reminding us of this important model validity condition.


3. Optimal ordering policy

We derive in this section the optimal threshold policy for the ordering problem. This policyessentially attempts to address two fundamental questions in inventory management: “When to placethe single order and how much to order”.

Following the approach in solving the classical newsvendor problem, in our multi-stage setting,for k = 0; 1; : : : ; N , if an order is placed at Stage k, the optimal order quantity at Stage k is givenby (see Nahmias [25, Chapter 5])

Q∗k = �k +

√dk + ��−1(sk): (3.1)

We denote the retailer’s expected pro7t with Q∗k by EPk(�k):

EPk(�k) = (r − ck)�k +√

dk + �Bk; (3.2)

where

Bk = {−(ck + h)[�−1(sk)] − (r + h)�[�−1(sk)]}: (3.3a)

Using �(x) = �(x) − x[1 − �(x)], (3.3a) becomes

Bk =−(ck + h)[�−1(sk)] − (r + h)(�[�−1(sk)] − �−1(sk){1 − �[�−1(sk)]})

=−(ck + h)[�−1(sk)] − (r + h)(�[�−1(sk)] − �−1(sk)[1 − sk])

=−(ck + h)[�−1(sk)] − (r + h)(�[�−1(sk)] − ck + h

r + h�−1(sk)

)

=−(r + h)�[�−1(sk)]: (3.3b)

De7ning

�k = (r − ck); (3.3c)

Kk =√

dk + �Bk: (3.3d)

With (3.3c) and (3.3d), (3.2) becomes

EPk(�k) = �k�k + Kk; (3.4)

where �k and Kk are independent of �k .Notice that Kk6 0.With the above, we know that when the retailer decides to place the order at Stage k, the optimal

order quantity is de7ned in (3.1) and the corresponding expected pro7t equals (3.4). Having theoptimal order quantity for Stage k, we use dynamic programming to derive the optimal policy forchoosing the best delivery mode and hence the optimal ordering time. Here, we have N + 1 stagesand our decision is to decide when to place the single optimal order among Stage 0, Stage 1; : : : ;Stage N . We assume in this paper that the retailer must place the order within the N + 1 stages. Itmeans that if the retailer has not placed the order from Stage 0 to Stage N − 1, the order must then


be placed at Stage N . Moreover, for all k = 0; 1; : : : ; N − 1: When the retailer is at Stage k + 1, hecannot regret and place the order back at Stage k (i.e. the ordering decision is not recallable). Thisis similar to the assumption in the classical optimal stopping problems (see Bertsekas [24]). In thefollowing, we follow the dynamic programming algorithm in Bertsekas [24] to obtain the optimaldecision rule.

First of all, we de7ne a termination state T . The system will move into the termination state T assoon as an order has been placed. For k = 0; 1; : : : ; N , we de7ne a state variable zk . zk = T impliesthat the order has been placed at or before Stage k, and zk �= T implies that the order has not beenplaced at or before Stage k. We de7ne vk as a Lag to indicate whether we will order or not at Stagek. vk = v1 means that the order is placed at Stage k and vk = v2 means that the order is not placedat Stage k. The assignment of the state variable zk is thus de7ned as follows:

For k = 0; 1; : : : ; N ,

zk =

{T if zk−1 = T or vk = v1;

0 otherwise:

We set the state variable z−1 = 0. Notice that the purpose of assigning z−1 to be 0 is to help inde7ning z0 because from our de7nition of zk ; z0 is de7ned as

z0 =

{T if z−1 = T or v0 = v1;

0 otherwise:

Moreover, assigning z−1 = 0 is also consistent with our de7nition for the state variable zk , as noorder has been placed before Stage 0. Next, we have the reward function,

E

{gN (�N ) +

∑k=0;1;:::;N−1

gk(�k)

};

where

gN (�N ) =

{EPN (�N ) if zN−1 �= T;

0 otherwise;

and for k = 0; 1; : : : ; N − 1:

gk(�k) =

{EPk(�k) if zk−1 �= T and vk = v1;

0 otherwise:

Based on this formulation, the corresponding dynamic programming algorithm over the state �k isgiven as follows:

For k = N ,

JN (�N ) =

{EPN (�N ) if zN−1 �= T;

0 otherwise:


For k = 0; 1; : : : ; N − 1,

Jk(�k) =

{max{EPk(�k); E[Jk+1(�k+1)]} if zk−1 �= T and vk = v1;

0 otherwise:

Before we make the decision about when the order will be placed, zk will not go into the terminationstate T , i.e. zk �= T . As a result, our optimal policy will be the following:

At Stages 0; 1; : : : ; N − 1, for the case of zk �= T (where k = 0; 1; : : : ; N − 1): We will place anorder when EPk(�k)¿E[Jk+1(�k+1)]; we will not place an order when EPk(�k)¡E[Jk+1(�k+1)].

If zN−1 �= T (i.e. the order has not been placed at or before Stage N − 1), since Stage N − 1 isthe second last possible ordering time, the ordering must then be made at Stage N .

Using the ordering policy discussed above, we want to 7nd the cutting point (or decision threshold)at each stage for deciding whether the order should be placed. We investigate that stage by stage:

At Stage N (the 7nal stage): For zN−1 �= T , the bene7t-to-go is JN (�N )=�N�N +KN . As it is the7nal stage, the system must stop here and order Q∗

N when zN−1 �= T and we set the cutting point,�∗N , to a very negative value: �∗N = −∞. This is employed in the following algorithm.

At Stage N − 1: For zN−2 �= T , the bene7t-to-go is

JN−1(�N−1) = max{EPN−1(�N−1); E[JN (�N )]}= max{�N−1�N−1 + KN−1; HN−1(�N−1)}; (3.5)

where HN−1(�N−1) is de7ned to be E[JN (�N )].We further de7ne

GN−1(�N−1) = HN−1(�N−1) − EPN−1(�N−1): (3.6)

Thus, solving GN−1(�N−1) = 0 for �N−1 gives us the cutting point �∗N−1:

GN−1(�N−1) =HN−1(�N−1) − EPN−1(�N−1)

= [�N�N−1 + KN ] − [�N−1�N−1 + KN−1];

GN−1(�N−1) = 0

⇒ �∗N−1 =KN − KN−1

�N−1 − �N=

KN − KN−1

cN − cN−1: (3.7)

With this cutting point, at Stage N −1, whenever �N−1¿ �∗N−1, the bene7t-to-go at Stage N −1 willbe larger than or equal to Stage N ’s bene7t-to-go. Thus, we have the following optimal orderingdecision at Stage N − 1: “If �N−1¿ �∗N−1, the optimal decision is to order at Stage N − 1 with theoptimal order quantity Q∗

N−1; if �N−1 ¡�∗N−1, the optimal decision is to wait and place an order atStage N with the optimal order quantity Q∗

N .” Observe that the cutting point �∗N−1 is expressed as afraction with the numerator representing the “value” or “e3ect” of information update with respectto the demand uncertainty, and the denominator representing the di3erence between the unit orderingcosts at Stage N and Stage N − 1.


At Stage k (for k = 0; 1; : : : ; N − 1): Similar to JN−1(�N−1), we de7ne

Hk(�k) = E[Jk+1(�k+1)] for k = 0; 1; : : : ; N − 1 (3.8)

and HN (�N ) = 0.For zk−1 �= T , the bene7t-to-go is

Jk(�k) = max{EPk(�k); E[Jk+1(�k+1)]}= max{�k�k + Kk; Hk(�k)}:

We de7ne

Gk(�k) = Hk(�k) − EPk(�k) for k = 0; 1; : : : ; N − 1: (3.9)

Hence, at Stage k, solving Gk(�k) = 0 for �k gives us the cutting point �∗k . Theorem 3.1 belowsummarizes the structural properties of Gk(�k) and �∗k .

Theorem 3.1. (a) Solution of Gk(�k) = 0 always exists for each k = 0; 1; : : : ; N − 1.(b) Gk(�k) is a non-increasing function.(c) When l0 ¡l1 · · ·¡lN , Gk(�k) is strictly decreasing and a unique cutting point �∗k exists, for

each k = 0; 1; : : : ; N − 1.

Proof of Theorem 3.1. (a) The existence of a solution is proved by showing that Gk(�k) takesopposite signs when �k goes to positive in7nity and negative in7nity, respectively (details are shownin Appendix A.1). (b) and (c) Using mathematical induction, we can prove that dGk(�k)=d�k6 0indicating the non-increasing property of Gk(�k). When l0 ¡l1 · · ·¡lN , we have dGk(�k)=d�k ¡ 0.In this case, a unique �∗k exists for every Stage k. (Details of the proofs for (b) and (c) are shownin Appendix A.2).

From Theorem 3.1, we know that the cutting points can be uniquely determined. The algorithmfor the optimal policy is then as shown below.

Algorithm for the Optimal Policy:Step 1: Initialization: Set k = 0 and �∗N = −∞.Step 2: Determining the decision thresholds, �∗k , for all k = 0; 1; : : : ; N − 1.Step 3: Checking whether �k is larger than or equal to �∗k : If �k¿ �∗k , the optimal action is to

place the order with an order quantity Q∗k and stop; if �k ¡�∗k , the optimal action is: Do not order,

wait, make a market observation to update the demand forecast for the next stage, increment i by 1and repeat Step 3.

Notice that we have de7ned the cutting point at Stage N to approach negative in7nity. As aresult, for all values of �N , the decision at Stage N is to place the order with an order quantityQ∗

N . Moreover, the cutting point �∗k is a constant decision threshold and its value does not a3ect thevalidity of the demand model in the problem formulation. In the following section, we discuss thecase when we have three delivery modes and hence three stages.


4. Numerical analysis with a real case example

We investigate the ordering problems with three di3erent delivery modes in this section. First ofall, following the algorithm developed in Section 3, we derive speci7c expressions for Gi(�i) (fori = 0; 1) which are essential for the calculations of the cutting points for the three-stage problem:

G1(�1) =H1(�1) − [�1�1 + K1]

= [�2�1 + K2] − [�1�1 + K1]: (4.1)

Solving G1(�1) = 0 gives �∗1 = (K1 − K2)=(c1 − c2).For G0(�0), we 7rst derive H0(�0):

H0(�0) =∫ �∗

1

−∞H1(�1)f1(�1) d�1 +

∫ ∞

�∗1

[�1�1 + K1]f1(�1) d�1; (4.2)

where f1(�1) is the normal density function for �1 with a mean �0 and a variance �20. After

simpli7cation, we have

H0(�0) = �2�0 + K2 + (c2 − c1)�0�[(�∗1 − �0)=�0]: (4.3)

Thus,

G0(�0) =H0(�0) − [�0�0 + K0]

= (−c2 + c0)�0 + K2 − K0 + (c2 − c1)�0�[(�∗1 − �0)=�0]: (4.4)

Solving G0(�0) = 0 for �0 gives �∗0 . As an illustration of 7nding the cutting points and the optimalpolicy, consider the following example.

Example 4.1. We consider a three-stage problem with pc = 3, l0 = 1:0, l1 = 1:2, l2 = 1:4, h = 0:01,r = 10, d0 = 3, � = 1. With these parameters, the cutting points are found to be: �∗1 = 2:2890 and�∗0 = 12:7993. The optimal ordering policy follows: At Stage 0: If �0¿ 12:7993, order Q∗

0 and stop;if �0 ¡ 12:7993, do not order, wait and make decision again at Stage 1. At Stage 1: If �1¿ 2:2890,order Q∗

1 and stop; if �1 ¡ 2:2890, do not order at Stage 1 and order at Stage 2 with an order quantityQ∗

2 . Notice that by using (3.1), we have Q∗0 = �0 + 0:504, Q∗

1 = �1 + 0:265 and Q∗2 = �2 + 0:179,

where �0 is known at Stage 0, while �1 and �2 depend on the observations (see (2.5)) and they areknown only after the respective observations are made at Stage 1 and Stage 2.

From Example 4.1, we can see that the optimal policy is indeed simple and intuitive. All weneed to do is to calculate the cutting points. However, an important issue is: “How good is theoptimal policy?” In order to address this issue, we conducted simulation studies with the analysisof the impacts brought by varying di3erent parameters. As a comparison, we de7ne the followingpolicies:

• Policy 0: The policy under which an order of Q∗0 is always placed at Stage 0.




Table 1Simulation results with di3erent d0

d0 �∗0 �∗

1 AP∗ AP0 AP1 AP2 %TAP0 %TAP1 %TAP2

2.0 8.240 1.902 77.617 77.617 65.866 70.649 0.00% 17.84% 9.86%2.5 10.566 2.122 77.827 77.827 67.384 72.083 0.00% 15.50% 7.97%3.0 12.799 2.289 82.680 78.023 68.754 73.383 5.97% 20.25% 12.67%3.5 14.941 2.420 83.930 78.204 70.008 74.577 7.32% 19.89% 12.54%4.0 16.999 2.525 85.091 78.378 71.172 75.687 8.56% 19.56% 12.42%

Table 2Simulation results with di3erent �

� �∗0 �∗


0.5 17.331 1.979 84.235 77.827 74.398 76.739 8.23% 13.22% 9.77%1.0 12.799 2.289 82.680 78.023 68.754 73.383 5.97% 20.25% 12.67%1.5 10.092 2.330 78.204 78.204 64.375 70.796 0.00% 21.48% 10.46%2.0 8.247 2.265 78.378 78.378 60.687 68.638 0.00% 29.15% 14.19%2.5 6.899 2.154 78.544 78.544 57.452 66.769 0.00% 36.71% 17.64%

Table 3Simulation results with di3erent r

r �∗0 �∗


9.0 11.908 2.222 63.882 63.882 56.261 60.188 0.00% 13.55% 6.14%9.5 12.370 2.256 70.925 70.925 62.491 66.769 0.00% 13.50% 6.22%

10.0 12.799 2.289 82.680 78.023 68.754 73.383 5.97% 20.25% 12.67%10.5 13.199 2.321 89.664 85.169 75.042 80.023 5.28% 19.49% 12.05%11.0 13.573 2.351 96.672 92.360 81.354 86.687 4.67% 18.83% 11.52%

With the same parameters as that in Example 4.1 and a prior mean of demand �0 = 12:5, weconducted simulation experiments and the results are shown in Tables 1–5. Notice that each resultis obtained by running the simulation 1000 times, and in the tables, we have the following notation:

• AP∗ = average pro7t under the optimal three-stage ordering policy.• APi = average pro7t under Policy i, i = 0; 1; 2.• %TAPi = 100% × (AP∗ − APi)=APi, i = 0; 1; 2.

Notice that %TAPi represents the percentage of the average pro7t improvement by using the optimalordering policy compared with Policy i.


Table 4Simulation results with di3erent l0

l0 �∗0 �∗


0.50 3.270 2.289 86.237 86.237 68.754 73.383 0.00% 25.43% 17.52%1.00 12.799 2.289 82.680 78.023 68.754 73.383 5.97% 20.25% 12.67%1.05 17.229 2.289 82.680 78.219 68.754 73.383 5.70% 20.25% 12.67%1.10 26.076 2.289 82.680 76.417 68.754 73.383 8.20% 20.25% 12.67%1.19 264.631 2.289 82.680 74.984 68.754 73.383 10.26% 20.25% 12.67%

Table 5Simulation results with di3erent l1

l1 �∗0 �∗


1.01 261.630 1.029 85.258 78.023 71.332 73.383 9.27% 19.52% 16.18%1.10 25.881 1.432 84.034 78.023 70.108 73.383 7.70% 19.86% 14.51%1.20 12.799 2.289 82.680 78.023 68.754 73.383 5.97% 20.25% 12.67%1.30 8.451 4.826 78.023 78.023 67.405 73.383 0.00% 15.75% 6.32%1.39 7.544 50.207 78.023 78.023 66.196 73.383 0.00% 17.87% 6.32%

4.1. Observations and discussion from Tables 1–5

1. In the tables, for all cases with 0% improvement of average pro7t, the optimal single order isactually always placed at Stage 0 (* �0¿ �∗0). Thus, the average pro7t under the optimal policyis the same as the average pro7t under Policy 0.

2. When l0 is close to l1, �∗0 becomes very large. In fact, when l0 = l1, �∗0 approaches in7nity,which implies that no order will be placed at Stage 0. This result makes sense as it is alwaysbene7cial to order after observation if the unit delivery cost (and hence the unit ordering cost)at Stage 1 is the same as (or very close to) the unit delivery cost at Stage 0. Similarly, when l1

is close to l2, we have a very large value for �∗1 .3. In our model, an observation can reduce d0 but not �. Thus, in Table 1, the larger the value of

d0, the larger the value of %TAP0. On the contrary, � is the inherent variance which cannot bereduced via observation. As a result, in Table 2, the larger the value of �, the smaller the valueof %TAP0.

4. When we compare the performance of the optimal policy with Policies 1 and 2, the optimalpolicy always out-performs Policies 1 and 2 in generating a larger average pro7t. Notice thatthe optimal policy is a dynamic one. Depending on di3erent observations, the optimal orderingdecision varies. As a result, if the optimal order is not placed at Stage 0, it will be placed at Stage1 or Stage 2, depending on di3erent observed values. Please also observe that the impacts broughtby varying parameters are not monotone when we compare the performance of the optimal policywith Policies 1 and 2.

5. The e3ect of the unit selling price r is rather surprising. Whenever �0 ¡�∗0 , which implies thatthe optimal order is not placed at Stage 0: The larger the unit selling price r, the larger the pro7t


margin and the optimal policy brings a less improvement (%TAP0) compared to Policy 0 asshown in Table 3. The reason can be explained intuitively because the larger the pro7t margin,the larger the potential pro7t and the precision of the ordering quantity is less important in thesense that excess stock does not cost much relatively. However, if the pro7t margin is small, thenthe accuracy of the stocking decision becomes more important as the ordering and delivery costsbecome relatively more signi7cant in the pro7t function.

6. For the unit delivery costs: In Table 4, when l0¿ 1:0, the optimal order is never placed at Stage0 because �0 ¡�∗0 . As a result, the average pro7t for the optimal policy remains the same whenl0 changes. Moreover, when l0, l1 and l2 increase, the average pro7t of ordering at the respectivestage decreases. On the other hand, when li changes, APj is not changed at all for any i �= jbecause the change of the unit delivery cost for delivery mode i does not a3ect the average pro7tgenerated by another delivery mode under Policies 0, 1, and 2. Notice that we 7x the value ofl2 as a reference value (i.e. the upper bound) for l0 and l1, and so we do not vary its value.

7. From Tables 1–5, we can see that the percentage increase of average pro7t by using the optimalpolicy can be very substantial. The e3ects of the individual parameters have been mentionedabove. It helps us to understand the conditions under which the optimal policy is especiallye3ective.

As a further illustration of the applicability of the optimal multi-stage ordering policy, let usconsider the following real case example.

4.2. A real case example

OR Ltd. (the company described here is a real company but the name is 7ctitious) is a China basedconsumer electronics manufacturer which acts as both an OEM and an ODM. OR Ltd. producesmany audio-visual and security electronics products in its factories in mainland China. This realcase example talks about a production problem of a seasonal consumer electronics product that isproduced by OR Ltd.

A seasonal consumer electronics product is to be manufactured by OR Ltd. and be sold in theup-coming selling season. In order to produce this electronics product, OR Ltd. needs to order acritical component. There are two di3erent but equally good components available in the market andany one of them can be the right candidate to produce the electronics product. For simplicity, we callthese components Chip A and Chip B, respectively. Chip A is much cheaper than Chip B. However,Chip A will take a longer time to process before it can be used for production and assembly of theelectronics product. Speci7cally, it takes about 6 days more to process the whole batch of Chip Athan the whole batch of Chip B before assembling the chip into the electronics product. In additionto the processing time, Chip A, which is supplied by a distributor in South Korea, requires a deliverylead-time of 15 days if it is shipped by ship-cargo and the delivery lead-time is shortened to be 8days if it is shipped by air-transport. Chip B can be purchased from a distributor in China and thelead-time is 2 days only. Taking the other production and operations processes into the consideration,it is estimated that the total production time (including the processing, delivery, 7nal assembly andQC process) for the seasonal electronics product with Chip A (under two delivery modes) and ChipB can be ranked as follows: “Chip A with ship-cargo (slowest), Chip A with air-transport, ChipB (fastest)”. In the market, the sales of an earlier model of this seasonal electronics product can


help OR Ltd. to improve the forecast for the seasonal electronics product. Notice that the suppliersof Chips A and B have imposed certain minimum purchasing quantities with which the productionquantity of OR Ltd. cannot support itself to order more than once. Thus, OR Ltd. can only make useof a single order (of Chip A via ship-cargo, Chip A via air-transport, or Chip B) for the productionof the seasonal electronics product. We have collected from OR Ltd. some real data related to thedemand forecasts and costs-revenue parameters for this seasonal electronics product and we modelthe ordering problem faced by OR Ltd. as follows (notice that the units and scales of the followingparameters are adjusted for a con7dentiality reason). First, we treat the ordering decision time pointfor Chip A with ship-cargo, Chip A with air-transport, and Chip B as Stages 0, 1, and 2, respectively.The overall unit production cost of the seasonal product with Chip A is 8.45, with Chip B is 10.27.The unit shipping cost for Chip A via ship-cargo is 0.2 while the unit shipping cost for Chip A viaair-transport is 1.6. The unit revenue is 21.60 and the unit holding cost is 0.024. The demand of theseasonal electronics product at Stage 0 (x0) is estimated to be normally distributed with parameters:d0 =3:701, �=0:787 and �0 =5:33 (the coeQcient of variation of the unconditional distribution of x0

is 0.397). With these details, we have the following values for cost-revenue parameters: r = 21:60,h=0:024, c0 =8:45+0:2=8:65, c1 =8:45+1:6=10:05, c2 =10:27. After solving for the cutting points,we have: �∗1 = 4:972 and �∗0 = 5:369. Since (�0 = 5:33)¡ (�∗0 = 5:369), OR Ltd. should not order atStage 0 (i.e. should not order Chip A via ship-cargo) and the optimal decision should be made afterobservation. As a result, after observation at Stage 1: If �1¿ 4:972, order Q∗

1 (=�1 + 0:103) units ofChip A via air-transport and stop; if �1 ¡ 4:972, do not order at Stage 1 and order Q∗

2 (=�2 +0:064)units of Chip B at Stage 2.

5. Variance of pro%t

How does the demand uncertainty reduction (when we use a faster delivery mode and order witha shorter lead-time) a3ect the variation of the pro7t with the optimal order quantity? Since thepro7t generated by selling the product is a random variable before the end of the selling season,the variance of pro7t reLects the likelihood of achieving the corresponding expected pro7t. In thefollowing, we can 7nd that the later the order is placed, the smaller the variance of pro7t (with thecorresponding optimal order quantity). In our model, at Stage k, for k = 0; 1; : : : ; N , the predicteddemand xk distributes as a normal distribution with a mean of �k and a variance of dk + �. Fromthe derivation of variance of pro7t in Choi et al. [26], for a given order quantity Qk , the varianceof pro7t at Stage k, called VPk(Qk), is found to be

VPk(Qk) = (r + h)2(dk + �)2*(ak); (5.1)

where

*(ak) = [ak�(ak) + �(ak) + a2k�(ak) − (�(ak) + ak�(ak))2]; (5.2)

and ak = (Qk − �k)=√dk + �.

Since the optimal order quantity for ordering at Stage k, Q∗k , is given by (3.1): Q∗

k = �k +√dk + ��−1(sk), the variance of pro7t with an order quantity Q∗

k becomes,

VPk(Q∗k ) = (r + h)2(dk + �)2[*(�−1(sk))]: (5.3)

With (5.3), we have the following lemma.


Lemma 5.1. For all Stages k = 0; 1; : : : ; N − 1; VPk(Q∗k )¿VPk+1(Q∗

k+1).

Proof of Lemma 5.1. First, by the de7nition of dk+1, we have dk+1 =dk�=(dk +�). Simple algebraicmanipulation gives: dk − dk+1 = dkdk+1=�¿ 0 which implies

dk ¿dk+1: (5.4)

Next, since sk = (r − ck)=(r + h) and ck+1¿ ck , we have sk+16 sk . As �−1(sk) is increasing in sk ,sk+16 sk implies �−1(sk)¿�−1(sk+1). Moreover, taking the derivative of *(ak) with respect to akgives

d*(ak)=dak = 2[�(ak) + ak�(ak)] − 2�(ak)[�(ak) + ak�(ak)]¿ 0: (5.5)

Eq. (5.5) tells us that *(ak) is an increasing function and it gives

*[�−1(sk)]¿ *[�−1(sk+1)]: (5.6)

Eqs. (5.4) and (5.6) together complete the proof of the lemma.

Lemma 5.1 reinforces the idea of making market observation because we are guaranteed that thevariance of pro7t (with the optimal order quantity) with more observation is always smaller thanthe variance of pro7t (with the optimal order quantity) with less observation. A smaller varianceof pro7t means that the uncertainty towards achieving the corresponding expected pro7t is smaller.In other words, a pro7t which is near to the optimal expected pro7t is more likely to be achievedwhen the variance of pro7t is smaller.

6. Implementation remarks and conclusion

In Section 3, we have shown the optimality of the ordering policy and its structural properties. Oneof the major concerns for multi-stage inventory problems is its potential diQculty in implementation.In the following, we outline the methods for implementing the optimal ordering policy derived inthis paper by a computer program. First, observe that the demand distribution used in this paperis the normal distribution. In order to solve for the cutting points at all stages, we may need tosolve equations with the standard normal pdf, cdf and its inverse. Since there are no closed formexpressions for the standard normal cdf and its inverse, the implementation of the algorithm forsolving all the equations Gk(�k) = 0, ∀k = 0; 1; : : : ; N − 1, seems to be diQcult. Moreover, in themulti-stage setting, for some k, Gk(�k) may contain the integration of standard normal cdf and manyother complicated expressions, giving another implementation diQculty. Fortunately, we can solvethese problems using the methods proposed in Choi [27]. To be speci7c, there are program modulesfor implementing the standard normal cdf and its inverse, details of which can be found in Choi [27].For the implementation challenge with respect to the integration of standard normal cdf and othercomplicated expressions, we can use numerical integration (Choi [27]) or Monte Carlo simulation(see Law and Kelton [28, p. 113]) for calculating the values of functions which involve problemswith multiple-integrals.

In conclusion, this paper derives an optimal single ordering policy with multiple delivery modes.The importance of the multiple delivery modes and information update should not be neglected. In


fact, from the literature 7ndings and our research, we can see that making better forecasts of productdemand can be very important. Multiple delivery modes can help us to include some Lexibility interms of the stocking decision. In this paper, information update is done by a Bayesian approachin which information from the market is used to update the product’s prior demand forecast. Thisinformation collection and updating process has been incorporated into the derivation of the optimalordering policy throughout this paper. We believe that the optimal ordering policy developed inthis paper is applicable in real-life, especially with a market full of di3erent delivery and logisticsservices. From the computer simulation experiments, we can see that the optimal ordering policy canbring a 7nancial bene7t, which is especially pronounced when the prior demand variance is large,the inherent demand variance is relatively small, and/or the pro7t margin is low.

Acknowledgements

We sincerely thank the constructive comments from the anonymous reviewers which led to asubstantial improvement of this paper. We also thank the comments we received from the conferenceparticipants at INFORMS Meeting 2000, Salt Lake City, Utah, USA.

Appendix A

A.1. Proof of the existence of a solution for Gk(�k) = 0

By de7nition,

Gk(�k) = Hk(�k) − (�k�k + Kk) and

Hk(�k) =∫ �∗

k+1

−∞Hk+1(�k+1)fk+1(�k+1) d�k+1 +

∫ ∞

�∗k+1

[�k+1�k+1 + Kk+1]fk+1(�k+1) d�k+1

=∫ (�∗

k+1−�k)=�k

−∞Hk+1(�kxk + �k)�(xk) dxk

+∫ ∞

(�∗k+1−�k)=�k

[�k+1(�kxk + �k) + Kk+1]�(xk) dxk : (A.1)

Notice that (A.1) is obtained after the standardization of the normal density. Observing from (A.1),we have

lim�k→∞Gk(�k) = lim

�k→∞ [Hk(�k) − (�k�k + Kk)]

= lim�k→∞

[∫ ∞

(�∗k+1−�k)=�k

[�k+1(�kxk + �k) + Kk+1]�(xk) dxk − (�k�k + Kk)

]

= lim�k→∞ [�k+1�k − �k�k + Kk+1 − Kk]6 0: (A.2)


To prove lim�k→−∞Gk(�k)¿ 0, we make use of mathematical induction. First,

lim�N−1→−∞GN−1(�N−1) = lim

�N−1→−∞HN−1(�N−1) − (�N−1�N−1 + KN−1)

= lim�N−1→−∞ (�N − �N−1)�N−1 + KN − KN−1¿ 0:

Assuming lim�k+1→−∞Gk+1(�k+1)¿ 0, then we have

lim�k+1→−∞Hk+1(�k+1) − (�k+1�k+1 + Kk+1)¿ 0;

lim�k+1→−∞Hk+1(�k+1)¿ lim

�k+1→−∞ (�k+1�k+1 + Kk+1): (A.3)

Now,

lim�k→−∞Gk(�k) = lim

�k→−∞

[∫ (�∗k+1−�k)=�k

−∞Hk+1(�kxk + �k)�(xk) dxk − (�k�k + Kk)

]:

With (A.3), we have

lim�k→−∞Gk(�k)¿ lim

�k→−∞

[∫ (�∗k+1−�k)=�k

−∞(�k+1�k + Kk+1)�(xk) dxk − (�k�k + Kk)

];

lim�k→−∞Gk(�k)¿ lim

�k→−∞ [(�k+1 − �k)�k + Kk+1 − Kk]¿ 0: (A.4)

(A.2) and (A.4) together complete the proof.

A.2. Proof of the non-increasing property of Gk(�k)

We prove by mathematical induction that dGk(�k)=d�k6 0; ∀k = N − 1; : : : ; 1; 0 is true. Whenk = N − 1: We have dGN−1(�N−1)=d�N−1 = �N − �N−16 0. Thus, the proposition is true for StageN − 1. Assuming the proposition is true for Stage k + 1, then we have

dGk+1(�k+1)=d�k+16 0: (A.5)

Eq. (A.5) implies that

d[Hk+1(�k+1) − (�k+1�k+1 + Kk+1)]d�k+1

6 0

⇒ dHk+1(�k+1)d�k+1

6 �k+1: (A.6)

Now, at Stage k:

dGk(�k)d�k

=d[Hk(�k) − (�k�k + Kk)]

d�k

=dHk(�k)

d�k− �k : (A.7)


dHk(�k)d�k

=d

d�k

( ∫ (�∗k+1−�k)=�k

−∞Hk+1(�kxk + �k)�(xk) dxk

+∫ ∞

(�∗k+1−�k)=�k

[�k+1(�kxk + �k) + Kk+1]�(xk) dxk

)

=− 1�k

Hk+1

[�k

(�∗k+1 − �k

�k

)+ �k

]�(�∗k+1 − �k

�k

)

+∫ (�∗

k+1−�k)=�k

−∞d

d�kHk+1(�kxk + �k)�(xk) dxk

−(−1

�k

)[�k+1

(�k

(�∗k+1 − �k

�k

)+ �k

)+ Kk+1

]�(�∗k+1 − �k

�k

)

+∫ ∞

(�∗k+1−�k)=�k

�k+1�(xk) dxk

=− 1�k

Hk+1[�∗k+1]�(�∗k+1 − �k

�k

)

+∫ (�∗

k+1−�k)=�k

−∞d


+1�k

[�k+1�∗k+1 + Kk+1]�(�∗k+1 − �k

�k

)

+∫ ∞

−∞�k+1�(xk) dxk −

∫ (�∗k+1−�k)=�k

−∞�k+1�(xk) dxk : (A.8)

By de7nition, �∗k+1 satis7es Gk+1(�k+1) = 0. Thus,

Gk+1(�∗k+1) = Hk+1(�∗k+1) − (�k+1�∗k+1 + Kk+1) = 0: (A.9)

Putting (A.9) into (A.8), we have

dHk(�k)d�k

=∫ (�∗

k+1−�k)=�k

−∞d


+∫ ∞

−∞�k+1�(xk) dxk −

∫ (�∗k+1−�k)=�k

−∞�k+1�(xk) dxk

=∫ (�∗

k+1−�k)=�k

−∞

[dHk+1(�kxk + �k)

d�k− �k+1

]�(xk) dxk + �k+1: (A.10)


By chain rule, we havedHk+1(�kxk + �k)

d�k=

dHk+1(�kxk + �k)d(�kxk + �k)

d(�kxk + �k)d�k

=dHk+1(�kxk + �k)

d(�kxk + �k): (A.11)

By induction assumption in (A.6), (A.11) implies thatdHk+1(�kxk + �k)

d�k=

dHk+1(�kxk + �k)d(�kxk + �k)

=dHk+1(�k+1)

d(�k+1)6 �k+1: (A.12)

Using (A.12), we have the following relationship for (A.10):dHk(�k)

d�k6 �k+1: (A.13)

Thus,dGk(�k)

d�k=

dHk(�k)d�k

− �k6 �k+1 − �k6 0: (A.14)

When l0 ¡l1 · · ·¡lN is true, we have �0 ¿�1 · · ·¿�N . Hence, dGk (�k)d �k

¡ 0.

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