optimal ordering and transfer policy for an inventory with stock dependent demand

European Journal of Operational Research 196 (2009) 177–185

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European Journal of Operational Research

journal homepage: www.elsevier .com/locate /e jor

Production, Manufacturing and Logistics

Optimal ordering and transfer policy for an inventory with stock dependent demand

Suresh Kumar Goyal a, Chun-Tao Chang b,*

a Department of DS & MIS, John Molson School of Business, Concordia University, Montreal, Quebec, Canada H3G1M8b Department of Statistics, Tamkang University, Tamsui, Taipei 25137, Taiwan, ROC

a r t i c l e i n f o

Article history:Received 28 October 2006Accepted 23 February 2008Available online 8 March 2008

Keywords:InventoryTransferStock dependent demand

0377-2217/$ - see front matter � 2008 Elsevier B.V. Adoi:10.1016/j.ejor.2008.02.029

* Corresponding author.E-mail address: [email protected] (C.-T. Ch

a b s t r a c t

This paper deals with an ordering-transfer inventory model to determine the retailer’s optimal orderquantity and the number of transfers per order from the warehouse to the display area. It is assumed thatthe amount of display space is limited and the demand rate depends on the display stock level. The objec-tive is to maximize the average profit per unit time yielded by the retailer. The proposed models and algo-rithms are developed to find the optimal strategy by retailer. Numerical examples are presented toillustrate the models developed and the sensitivity analysis is also reported.

� 2008 Elsevier B.V. All rights reserved.

1. Introduction

The appropriate amount of inventory has been a major concernfor research. As Levin et al. (1972) observed that ‘‘large piles of con-sumer goods displayed in a supermarket will lead the customer tobuy more. Yet, too much piled up in everyone’s way leaves a neg-ative impression on buyer and employee alike.” Silver and Peterson(1982) also noted that sales at the retail level tend to be propor-tional to the amount of inventory displayed. In order to quantifythis phenomenon, Baker and Urban (1988) established an EOQmodel for a power-form inventory-level-dependent demand pat-tern. Mandal and Phaujdar (1989) then developed a productioninventory model for deteriorating items with uniform rate of pro-duction and linearly stock-dependent demand. Taking another as-pect into consideration, Gerchak and Wang (1994) stated ‘‘someoperations management researchers have incorporated inven-tory-level dependence of demand into various inventory controlmodels” and established periodic-review inventory models withinventory-level dependent demand. Bar-Lev et al. (1994) devel-oped an extension of the inventory-level-dependent demand-typeEOQ model with random yield. Ray and Chaudhuri (1997) ex-tended an EOQ model with stock-dependent demand and short-ages considering the time value of money and different inflationrates for internal and external costs. Wang and Gerchak (2001) sta-ted ‘‘retailers can often affect sales volume of a product by increas-ing the shelf space allocated to it.” Therefore, they studied thesupply chain coordination when demand is shelf-space dependent.Dye and Ouyang (2005) established an EOQ model for deteriorating

ll rights reserved.

ang).

items with stock-dependent selling rate when shortages are par-tially backlogged. Recently, Abbott and Palekar (2008) stated‘‘Retailers have long recognized the relationship between displayspace and product sales” and ‘‘the display space or the facings ofa product makes it more visible and the visibility, in turn, createsadditional demand.” The aforementioned researchers presentedthe retail replenishment models with display-space elastic de-mand. In this paper, we deal with the problem of determiningthe operating policy for an inventory system in which the demandis dependent on the display stock level. Other papers related to thestock-dependent demand are Chang (2004), Chung (2003), Giri andChaudhuri (1998), Pal et al. (1993), Ray et al. (1998), and others.

During the last 30 years, the topics of transfer batching, theintegration of production and inventory model, as well as the pro-curement and shipment of inventory items have been consideredby some researchers. For instance, Goyal (1977) preliminarilydeveloped a single supplier-single retailer integrated inventorymodel. In addition, Banerjee (1986) developed a joint economic-lot-size model and assumed that the supplier followed a lot-for-lot shipment policy with respect to a retailer. Goyal (1988) furtherextended Banerjee’s model and this extended model discussed abatch that consists of a number of equal-sized shipments, but theproduction of the batch had to be finished before the shipmentscould start. Lu (1995) related the assumption implied in Goyal’smodel and explored a model which allowed shipments to takeplace during production. Goyal (1995) proposed a shipment policyin which, during production, a shipment is made as soon as thebuyer is about to run out of stock and all of the manufactured stockmade up to that point is shipped out. Hill (2000) proposed an opti-mal two-stage lot sizing and inventory batching policies. Yang andWee (2003) established an integrated multi-lot-size production

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178 S.K. Goyal, C.-T. Chang / European Journal of Operational Research 196 (2009) 177–185

inventory model for deteriorating items. Nieuwenhuyse andVandaele (2004) developed a model based on minimizing totalcosts in order to determine the optimal number of sub-lots in a sin-gle-product, deterministic flow shop with overlapping operations.Law and Wee (2006) proposed an integrated production-inventorymodel for ameliorating and deteriorating items taking account oftime discounting. Yao et al. (2007) developed a model that exploreshow important supply chain parameters affect the cost savings tobe realized from collaborative initiatives such as vendor-managedinventory. Other papers related to these areas are Hill (1997, 1999),Viswanathan (1998), Goyal and Nebebe (2000), Chiang (2001), Kimand Ha (2003), Siajadi et al. (2006), and others. All these previ-ously-published research are concerned with the integrated ven-dor-buyer inventory model.

This paper considers another situation and the objective of thispaper is to determine the ordering and transfer schedule whichmaximizes the average profit per unit time yielded by the retailer.The retailer gets the delivery of the item and some of the itemsare displayed in the shop while the rest of the items are kept inthe backroom/warehouse. The amount of shelf/display space isalways limited. Hence, the inventory cost inside the shop maybe higher as compared to in the backroom. The problem the man-agement faces is how often and how many inventory items theyshould transfer from the back of the shop (i.e., warehouse) tothe display area of the shop. Hence, in this article, we first developan ordering-transfer inventory model with the demand rate thatis influenced by stock level. Next, we provide the fundamentalassumptions for the proposed model and the notations in Section2. In Section 3, a mathematical model is established to manifestthe ordering- transfer model for maximizing profits. The neces-sary conditions for an optimal solution are discussed and itssolution algorithm is developed in Section 4. In Section 5, a specialcase, the demand rate is constant, is considered. The necessaryconditions for an optimal solution and its solution algorithm arealso presented in this Section. Numerical examples are providedto illustrate the proposed model and the sensitivity analysis ofthe optimal solution with respect to parameters of the system iscarried out in Section 6. Finally, we draw the conclusions inSection 7.

2. Assumptions and notation

The following assumptions are adopted:

1. Shortages are not allowed.2. To avoid a negative impression and to indicate the display space

limit, we define the maximum allowable number of displayedstocks in the display area as l.

3. The lead time between the retailer and the supplier is zero.4. The time to transfer items from the warehouse to the display

area is zero.5. The ordering-transfer policy adopted in the paper is as follows:

the retailer orders quantity Q per order from a supplier andstocks these items in the warehouse. The items are transferredfrom the warehouse to the display area in equal lots of q untilthe inventory level in the warehouse falls to zero.

The following notations are used:h1 the unit carrying cost per item in the warehouseh the unit carrying cost per item in the display area, with

h > h1

l the maximum allowable number of displayed stocks in thedisplay area

p the unit selling price of the product per unitc the unit purchasing costS the cost of placing per order

s the fixed cost per transfer from the warehouse to the dis-play area

T the replenishment cycle time in the warehousen the integer number of transfers from the warehouse to the

display area per ordert1 the replenishment cycle time in the display areaQ the order quantity placed on the supplierq the quantity per transfer from the warehouse to the dis-

play area, where 0 6 q 6 lI(t) the inventory level at time t in the display area, which is

always less than or equal to lR the inventory level of the item in the display area regard-

ing the transfer of q items from the warehouse to the dis-play area

D(I(t)) the demand rate at time t. We assume that the demandrate D(I(t)) is a function of the stocks on display I(t). Forsimplicity, we also assume that D(I(t)) = a + bI(t), where aand b is a non-negative constant, respectively.

3. Mathematical models

3.1. The total cost per unit cycle in the warehouse

The retailer orders Q items per order from a supplier and stocksthese items in the warehouse. Next, the quantity q per transfer istransferred from the warehouse to the display area until the inven-tory level in the warehouse falls to zero. Hence, we get Q = nq. Thetotal cost over the period [0, T] in the warehouse consists of (1) thecost of placing orders = S and (2) the cost of stock holding ish1

nðn�1Þ2 q

h it1.

3.2. The total cost per unit cycle in the display area

At time t = 0, the inventory level I(t) reaches the top I ðI 6 lÞ dueto the items are transferred from the warehouse to the displayarea. The inventory level then gradually depletes to R at the endof the cycle. A graphical representation of this inventory systemis depicted in Fig. 1 (see Appendix 1).

The differential equation expressing the inventory level at timet can be written as follows:

dIðtÞ=dt ¼ �DðIðtÞÞ ¼ �½aþ bIðtÞ�; 0 6 t 6 t1; ð1Þ

with the boundary condition I(t1) = R. Accordingly, the solution ofEq. (1) is given

IðtÞ ¼ Rebðt1�tÞ þ ab½ebðt1�tÞ � 1�; 0 6 t 6 t1: ð2Þ

The total cost over the period [0, t1] consists of (1) the cost of plac-ing orders = s and (2) the cost of stock holding which is given

hZ t1

0IðtÞdt ¼ h

1b

Rþ ab

� �½ebt1 � 1� � a

bt1

� �: ð3Þ

The revenue per cycle is

ðp� cÞZ t1

0DðIðtÞÞdt ¼ ðp� cÞ

Z t1

0½aþ bIðtÞ�dt

¼ ðp� cÞat1

þ ðp� cÞb 1b

Rþ ab

� �½ebt1 � 1� � a

bt1

� �: ð4Þ

Using Eq. (2) and I(0) = q + R, we get

q ¼ Rþ ab

� �ðebt1 � 1Þ: ð5Þ

Then, the cost of stock holding in the warehouse is

h1nðn� 1Þ

2q

� �t1 ¼ h1

nðn� 1Þ2

Rþ ab

� �ðebt1 � 1Þ

� �t1: ð6Þ

S.K. Goyal, C.-T. Chang / European Journal of Operational Research 196 (2009) 177–185 179

From the above result, the total profit TP over the period [0, T] isdenoted

TPðn;R; t1Þ ¼ revenue� ½total cost in the warehouse�� ½total cost in the display area�¼ nðp� cÞat1

þ nðp� cÞb 1b

Rþ ab

� �½ebt1 � 1� � a

bt1

� �

� Sþ h1nðn� 1Þ

2Rþ a

b

� �ðebt1 � 1Þ

� �t1

� �� ns

� nh1b

Rþ ab

� �½ebt1 � 1� � a

bt1

� �¼ �S� nsþ nh

ab

t1

þ Rþ ab

� �ðebt1 � 1Þ nðp� cÞ � h1

nðn� 1Þ2

t1 �nhb

� �:

ð7Þ

Hence, the average profit per unit time is

APðn;R; t1Þ ¼ TP=T ðwhere T ¼ nt1Þ

¼ � Snt1� s

t1þ h

ab

þ Rþ ab

� �ðebt1 � 1Þ p� c

t1� h1ðn� 1Þ

2� h

bt1

� �: ð8Þ

4. Necessary conditions for an optimal solution

According to (8), AP(n, R, t1) is a function of n, R and t1. For fixedR and t1, the effect of n on the average profit per unit time will beexamined. Taking the first and second partial derivative ofAP(n, R, t1) with respect to n, we obtain

oAPðn;R; t1Þon

¼ Sn2t1

þ Rþ ab

� �ðebt1 � 1Þ �h1

2

� �; ð9Þ

and

o2APðn;R; t1Þon2 ¼ � 2

n3

St1< 0: ð10Þ

The results show AP(n, R, t1) is a concave function of n for fixed t1

and R. Consequently, the search for the optimal transfer numbern* is reduced to finding a local optimal solution.

Now, the model is to determine the optimal replenishment cy-cle time in the display area for any given n. Taking the first partialderivative of AP(n, R, t1) with respect to R, we get:

oAPðn;R; t1ÞoR

¼ ðebt1 � 1Þ 1bt1½ðp� cÞb� h� � h1ðn� 1Þ

2

� �: ð11Þ

Based on the size of [(p � c)b � h], three cases are discussed asfollows:

Case 1. [(p � c)b � h] < 0

If [(p � c)b � h] < 0, then oAP(n, R, t1)/oR < 0. That is, AP is a decreas-ing function of R for fixed n. Therefore, the optimal retransfer levelof the item in the display area R* should be zero. ‘‘[(p � c)b � h] < 0”implies that the benefit received from the unit of inventory (p � c)bis smaller than the unit carrying cost h. That is, it is not profitable tobuild up inventory. Substituting R* = 0 into Eq. (8), we obtain AP is afunction of n and t1. The first-order condition for finding the optimalt�1 is oAP(n, t1)/ot1 = 0, which leads to

Snþ sþ a

b2 ½ðp� cÞb� h�½bt1ebt1 � ebt1 þ 1� ¼ h1ðn� 1Þ2

at21ebt1 : ð12Þ

The second-order condition is

o2APðn;t1Þot2

1

��t1¼t�1

¼ 1t21ðat1ebt1 Þ ½ðp� cÞb� h� � h1ðn�1Þ

2 ð2þ bt1Þn o

< 0:ð13Þ

From (13), we know o2APðn; t1Þ=ot21 < 0. Hence, AP(n, t1) is a con-

cave function in t1 for fixed n. There exists a unique value of t1 (de-noted by t#1

1;nÞ such that APðn; t#11;nÞ is the maximum value. t#1

1;n can bedetermined by solving Eq. (12). Substituting t#1

1;n and R* = 0 into (5),the transfer quantity q#1

n can be determined for fixed n.Note: Because q 6 l for all q, q#1

n ¼ l if q#1n P l. Thus, according to

Eq. (5) and q#1n ¼ l; t#1

1;n is resolved as t#11;n ¼ lnðlðb=aÞ þ 1Þ=b.

Case 2. [(p � c)b � h] = 0

If [(p � c)b � h] = 0, then oAP(n, R, t1)/oR < 0 and Eq. (8) can be re-duced to the following function:

APðn;R; t1Þ ¼ �S

nt1� s

t1þ h

ab� h1ðn� 1Þ

2Rþ a

b

� �ðebt1 � 1Þ: ð14Þ

Since oAP(n, R, t1)/oR < 0, AP is a decreasing function of R for fixed n.It is the same as Case 1. Therefore, the optimal retransfer level of theitem in the display area R* should be zero. Substituting R* = 0 into(14), we obtain AP is a function of n and t1. The first-order conditionfor finding the optimal t�1 is o AP(n, t1)/ot1 = 0, which leads to

Snþ s ¼ h1ðn� 1Þ

2at2

1ebt1 : ð15Þ


o2APðn;t1Þot2

1

��t1¼t�1

¼ 1t21� h1ðn�1Þ

2 a

ð2t1ebt1 þ bt21ebt1 Þ < 0:

ð16Þ


cave function in t1 for fixed n. There exists a unique value of t1 (de-noted by t#2

1;nÞ such that APðn; t#21;nÞ is the maximum value. t#2

1;n can bedetermined by solving Eq. (15). Substituting t#2

1;n and R* = 0 (5), thetransfer quantity q#2

n can be determined for fixed n.Note: (1) Since q 6 l for all q; q#2

n ¼ l if q#2n P l. Thus, according

to Eq. (5) and q#2n ¼ l, t#2

1;n is resolved as t#21;n ¼ lnðlðb=aÞ þ 1Þ=b; (2)

the number of transfers from the warehouse to the display areaper order n must be larger than or equal to 2.

Case 3. [(p � c)b � h] > 0

There are three sub-cases in Case 3.

Case 3.1. {[(p � c)b � h]/(bt1)} < h1(n � 1)/2

If {[(p � c)b � h]/(bt1)} < h1(n � 1)/2, then oAP(n, R, t1)/oR < 0. It isthe same as Case 1. The optimal retransfer level of the item in thedisplay area R* = 0 and there exists a unique value of t1 (denotedby t3:1

1;nÞ such that APðn; t3:11;nÞ is the maximum value. t3:1

1;n can be deter-mined by solving Eq. (12). Substituting t3:1

1;n and R* = 0 into (5), thetransfer quantity q3:1

n can be determined for fixed n.Note: (1) If t3:1

1;n 6 2½ðp� cÞb� h�=½bh1ðn� 1Þ�, then t3:11;n is infeasi-

ble. That is, the optimal solution does not exist; (2) because q 6 lfor all q, q3:1

n ¼ l if q3:1n P l. Thus, according to Eq. (5) and

q3:1n ¼ l; t3:1

1;n is resolved as t3:11;n ¼ lnðlðb=aÞ þ 1Þ=b; (3) the number of

transfers from the warehouse to the display area per order n mustbe larger than or equal to 2.

Case 3.2. {[(p � c)b � h]/(bt1)} > h1(n � 1)/2

If {[(p � c)b � h]/(bt1)} > h1(n � 1)/2, then oAP(n, R, t1)/oR > 0.That is, the benefit received from the unit of inventory (p � c)b islarger than the unit carrying cost h and AP is an increasing functionof R for fixed n. Therefore, we should pile up inventory to the max-imum allowable number l. So, I(0) = l. From I(0) = l = q + R and Eq.(5), we know

R ¼ lþ ab

� �e�bt1 � a

b; ð17Þ

which indicates that R is a function of t1.


Substituting (17) into (8), we know that AP is a function of t1

and n. The first-order condition for finding the optimal t�1 isoAP(n, t1)/ot1 = 0, which leads to

Snþ sþ lþ a

b

� �1b½ðp� cÞb� h�½bt1e�bt1 þ e�bt1 � 1�

¼ h1ðn� 1Þ2

t21e�bt1 lþ a

b

� �b: ð18Þ


o2APðn;t1Þot2

1

��t1¼t�1

¼ 1t21

lþ ab

f�bh1ðn� 1Þt1e�bt1 þ t2

1b2e�bt1 ½ðh1ðn� 1Þ=2Þ

�ððp� cÞb� hÞ=ðbt1Þ�g < 0:

ð19Þ


cave function in t1 for fixed n. There exists a unique value of t1 (de-noted by t3:2

1;n) such that APðn; t3:21;nÞ is the maximum value. t3:2

1;n can bedetermined by solving Eq. (18).

Note: The number of transfers from the warehouse to the dis-play area per order n must be larger than or equal to 2.

Case 3.3. {[(p � c)b � h]/(bt1)}=h1(n � 1)/2

If {[(p � c)b � h]/(bt1)} = h1(n � 1)/2, then we obtain

t�1 ¼ t3:31;n ¼

2½ðp� cÞb� h�bh1ðn� 1Þ : ð20Þ

From oAP(n, R, t1)/ot1 = 0 and Eq. (20), the retransfer level of theitem in the display area R� ¼ R3:3

n can be determined. However,

ðo2APðn;R; t1Þ=ot1oRÞ2 � ðo2APðn;R; t1Þ=ot21Þðo

2APðn;R; t1Þ=oR2Þ > 0

at ðt�1;R�Þ ¼ ðt3:3

1;n;R3:3n Þ. Hence, AP(n, R, t1) has a saddle point at

ðt�1;R�Þ ¼ ðt3:3

1;n;R3:3n Þ for fixed n.

Algorithm 1

Step 0: Given S, s, h1, h, p, c, a, b, l.Step 1: If (p � c)b � h < 0, then go to Algorithm 1-a.Step 2: If (p � c)b � h = 0, then go to Algorithm 1-b.Step 3: If (p � c)b � h > 0, then go to Algorithm 1-c.

Algorithm 1-a

Step 1: Set R* = 0 and n = 1.Step 2: t#1

1;1 is determined by Eq. (12) and q#11 is determined by

substituting t#11;1 into Eq. (5).

Step 3: If q#11 < l, then �t#1

1;1 ¼ t#11;1. Otherwise, �t#1

1;1 ¼ lnðlðb=aÞþ 1Þ=b.Step 4: Compute APð1;�t#1

1;1Þ.Step 5: Set n = n + 1.Step 6: t#1

1;n is determined by Eq. (12) and q#1n is determined by

substituting t#11;n into Eq. (5).

Step 7: If q#1n < l, then �t#1

1;n ¼ t#11;n. Otherwise, �t#1

1;n ¼ lnðlðb=aÞþ1Þ=b.Step 8: Compute APðn;�t#1

1;nÞ.Step 9: If APðn;�t#1

1;nÞ > APðn� 1;�t#11;n�1Þ then go to Step 5. Other-

wise, go to Step 10.Step 10: The optimal solution ðn�; t�1Þ ¼ ðn� 1;�t#1

1;n�1Þ.

Algorithm 1-b

Step 1: Set R* = 0 and n = 2.Step 2: t#2

1;2 is determined by Eq. (15) and q#21 is determined by

substituting t#21;2 into Eq. (5).

Step 3: If q#21 < l, then �t#2

1;2 ¼ t#21;2. Otherwise,

�t#21;2 ¼ lnðlðb=aÞ þ 1Þ=b.

Step 4: Compute APð2;�t#21;2Þ.

Step 5: Set n = n + 1.Step 6: t#2

1;n is determined by Eq. (15) and q#2n is determined by

substituting t#21;n into Eq. (5).

Step 7: If q#2n < l, then �t#2

1;n ¼ t#21;n. Otherwise,

�t#21;n ¼ lnðlðb=aÞ þ 1Þ=b.

Step 8: Compute APðn;�t#21;nÞ.

Step 9: If APðn;�t#21;nÞ > APðn� 1;�t#2

1;n�1Þ then go to Step 5. Other-wise, go to Step 10.

Step 10: The optimal solution ðn�; t�1Þ ¼ ðn� 1;�t#21;n�1Þ.

Algorithm 1-c

Step 1: Set n = 2.Step 2: t3:1

1;2 is determined by Eq. (12) and q3:12 is determined by

substituting t3:11;2 and R* = 0 into Eq. (5).

Step 3: If q3:12 < l, then �t3:1

1;2 ¼ t3:11;2. Otherwise, �t3:1

1;2 ¼ lnðlðb=aÞþ1Þ=b.

Step 4: If f½ðp� cÞb� h�=ðb�t3:11;2Þg < h1ðn� 1Þ=2, then compute

APð2;�t3:11;2Þ. Otherwise, set APð2;�t3:1

1;2Þ ¼ 0.Step 5: t3:2

1;2 is determined by Eq. (18).Step 6: If f½ðp� cÞb� h�=ðbt3:2

1;2Þg > h1ðn� 1Þ=2, then substitutet3:2

1;2 into Eq. (17) to find R*and compute APð2; t3:21;2Þ. Other-

wise, set APð2; t3:21;2Þ ¼ 0.

Step 7: APð2; t#31;2Þ ¼MaxfAPð2;�t3:1

1;2Þ;APð2; t3:21;2Þg.

Step 8: Set n = n + 1.Step 9: t3:1

1;n is determined by Eq. (12) and q3:1n is determined by

substituting t3:11;n and R* = 0 into Eq. (5).

Step 10: If q3:1n < l, then �t3:1

1;n ¼ t3:11;n. Otherwise, �t3:1

1;n ¼ lnðlðb=aÞþ1Þ=b.

If f½ðp� cÞb� h�=ðb�t3:11;nÞg < h1ðn� 1Þ=2, then compute

APðn;�t3:11;nÞ. Otherwise, set APðn;�t3:1

1;nÞ ¼ 0.
Step 11: t3:2
1;n is determined by Eq. (18).
If f½ðp� cÞb� h�=ðbt3:2
1;nÞg > h1ðn� 1Þ=2, then substitutet3:2

1;n into Eq. (17) to find R* and compute APðn; t3:21;nÞ. Other-

wise, set APðn; t3:21;nÞ ¼ 0.

Step 12: APðn; t#31;nÞ ¼MaxfAPðn;�t3:1

1;nÞ;APðn; t3:21;nÞg.

Step 13: If APðn; t#31;nÞ > APðn� 1; t#3

1;n�1Þ then go to Step 8. Other-wise, go to Step 14.

Step 14: The optimal solution ðn�; t�1Þ ¼ ðn� 1; t#31;n�1Þ.

5. Special case

If b = 0, then the demand rate D(I(t)) = a. That is, the demandrate is a constant. Hence, the inventory level at time t can berewritten as

IðtÞ ¼ aðt1 � tÞ þ R; ð21Þ

and the average profit per unit time AP over the period [0, T] is re-duced to

Aðn;R; t1Þ ¼ ðp� cÞa� Snt1þ ah1ðn� 1Þt1

2

� �

� st1þ h

a2

t1 þ R � �

: ð22Þ

Taking the first partial derivative of AP(n, t1, R) with respect to R, weget:

oAPðn; t1;RÞoR

¼ �h < 0: ð23Þ

It shows that AP is a decreasing function of R. Therefore, the optimalretransfer level of the item in the display area R* should be zero.Substituting R* = 0 into (22), we obtain AP is a function of n and t1

as follows:

APðn; t1Þ ¼ ðp� cÞa� Snt1þ ah1ðn� 1Þt1

2

� �� s

t1þ h

a2

t1

� �: ð24Þ


5.1. Necessary conditions for an optimal solution

From (24), AP(n, t1) is a function of n and t1. For fixed t1, the ef-fect of n on the average profit per unit time will be examined. Tak-ing the first and second partial derivative of AP(n, t1) with respectto n, we obtain

oAPðn; t1Þon

¼ Sn2t1

� h1at1

2; ð25Þ

and

o2APðn; t1Þon2 ¼ � 2S

n3t1< 0: ð26Þ

The results show AP(n, t1) is a concave function of n for fixed t1.Consequently, the search for the optimal transfer number n* is re-duced to finding a local optimal solution.

Now, the model is to determine the optimal replenishmentcycle time in the display area for any given n. Taking the firstand second partial derivatives of AP(n, t1) with respect to t1, weget:

oAPðn; t1Þot1

¼ Snt2

1

� h1aðn� 1Þ2

þ sn� ha

2; ð27Þ

and

o2APðn; t1Þ

ot21

¼ � 2Snt3

1

� st3

1

< 0: ð28Þ


cave function of t1 for fixed n. Therefore, there exists a unique valueof t1 (denoted by t1, n) such that AP(n, t1,n) is the maximum value.t1,n can be determined by oAP(n, t1)/ot1 = 0 in Eq. (27), and is givenby

t1;n ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2½ðS=nÞ þ s�a½h1ðn� 1Þ þ h�

s: ð29Þ

Hence, the optimal quantity per transfer from the warehouse to thedisplay area for fixed n is

qn ¼ at1;n ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2a½ðS=nÞ þ s�½h1ðn� 1Þ þ h�

s: ð30Þ

To ensure qn 6 l, we substitute (30) into inequality qn 6 l, and obtain

2aSnþ s

� �6 l2½h1ðn� 1Þ þ h� if and only if qn 6 l: ð31Þ

By combining (29)–(31), we have the following theorem.

Theorem 1. For fixed the number of transfers from the warehouse tothe display are per order n, we can obtain the following results.

(1) If 2a Snþ s�

< l2½h1ðn� 1Þ þ h�, then q�n ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2a½ðS=nÞþs�½h1ðn�1Þþh�

qand

t�1;n ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2½ðS=nÞ þ s�a½h1ðn� 1Þ þ h�

s:

(2) If 2a Snþ s�

P l2½h1ðn� 1Þ þ h�, then q�n ¼ l and t�1;n ¼ la.

Algorithm 2. Let D1;n ¼ 2a Snþ s�

and D2,n = l2[h1(n � 1) + h].

Step 0: Given S, s, h1, h, p, c, a, l.Step 1: Set n = 1 and calculate D1,1 and D2,1.Step 2: If D1,1 > D2,1, then ~q1 ¼ l and ~t1;1 ¼ l=a.ffiffiffiffiffiffiffiffiffiffiffiq ffiffiffiffiffiffiffiffiffiq

Otherwise, ~q1 ¼ q1 ¼ 2a½Sþs�h and ~t1;1 ¼ t1;1 ¼ 2½Sþs�

ah .

Step 3: Compute APð1;~t1;1Þ.Step 4: Set n = n + 1 and calculate D1, n and D2,n.Step 5: If D1,n > D2,n, then ~qn ¼ l and ~t1;n ¼ l=a.ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiq ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiq

Otherwise, ~qn ¼ qn ¼ 2a½ðS=nÞþs�½h1ðn�1Þþh� and ~t1;n ¼ t1;n ¼ 2½ðS=nÞþs�

a½h1ðn�1Þþh�.

Step 6: Compute APðn;~t1;nÞ.Step 7: If APðn;~t1;nÞ > APðn� 1;~t1;n�1Þ, then go to Step 4.

Otherwise, go to Step 8.
Step 8: The optimal solution ðn�; t�1Þ ¼ ðn� 1;~t1;n�1Þ.
6. Numerical examples

Example 1. Given a = 1000 units/unit time, b = 0, h = $0.6/unit/unittime, h1 = $0.3/unit/unit time, c = $1.0 per unit and p = $3.0 perunit. Since b = 0, the demand rate is constant. Using the Algorithm2, we have obtained the optimal solution. We then performedsensitivity analyses, and obtained the numerical results as shownin Table 1 (see Appendix 2).

Based on the computational results as shown in Table 1, weobtain the following managerial insights:

1. A higher value of l causes higher values of t�1, q* and AP*. In addi-tion, the influence of l on n* is non-positive. These results implythat the retailer should increase the quantity per transfer fromthe warehouse to the display area and the replenishment cycletime in the display area. This leads to the increase of averageprofit per unit time, but at this point the retailer may decreasethe number of transfer from the warehouse to the display areaper order, when the maximum allowable number of displayedstocks in the display area increases.

2. A higher value of S causes higher values of T* and Q*, but a lowervalue of AP*. According to these results, the retailer shouldincrease the order quantity and the replenishment cycle timein the warehouse as well as may increase the number of trans-fer from the warehouse to the display area per order, when thecost of placing per order increases.

3. A higher value of s causes a lower value of AP*. The value of n* isnot influenced by the change of s. T*, q* and Q* have low sensi-tivity to changes in s. That is, the influences of s on the decisionsof T*, q* and Q* are little.

Example 2. Given a = 1000 units/unit time, b = 0.2, h = $0.6/unit/unit time, h1 = $0.3/unit/unit time, c = $1.0 per unit and p = $3.0per unit. Since (p � c)b � h < 0, we apply Algorithm 1-a. The com-putational results for different maximum allowable numbers(l = 150, 200, 250 or 300), ordering costs (S = $10, 30, 50, 70 or90) and transferring costs (s = $10, 30, 50, 70 or 90) are shown inTable 2 (see Appendix 3).

From Table 2, we obtain the following managerial insights:

1. A higher value of l causes higher values of t�1, q* and AP*. In addi-tion, the influence of l on n* is non-positive. These results indi-cate that the retailer should increase t�1 and q*, but may decreasen*, when l increases.

2. A higher value of S causes a lower value of AP*.3. The influences of S on n*, T* and Q* are nonnegative. These

results show that the retailer may increase the number of trans-fer from the warehouse to the display area per order, the orderquantity place on the supplier and the replenishment cycle timein the warehouse, when the cost of placing per order Sincreases.

4. The values of n*, t�1, T*, q* and Q* are not impacted by the changeof s. It means that the decisions on n*, t�1, T*, q* and Q* by retailerare not affected by the value of s. However, a higher value ofscauses a lower value of AP*.


Example 3. Given a = 1000 units/unit time, b = 0.3, h = $0.6/unit/unit time, h1 = $0.3/unit/unit time, c = $1.0 per unit, p = $3.0 perunit and l = 300 units/unit. Since (p � c)b � h = 0, we apply Algo-rithm 1-b. The computational results for different maximumallowable numbers (l = 150, 200, 250 or 300), ordering costs(S = $10, 30, 50, 70 or 90) and transferring costs (s = $10, 30, 50,70 or 90) are shown in Table 3 (see Appendix 4).

The following managerial insights are obtained from Table 3:

1. A higher value of l causes higher values of t�1, q* and AP*. In addi-tion, the influence of l on n* is non-positive. These results indi-cate that the retailer should increase t�1 and q*, but may decreasen*, when l increases.

2. The values of t�1 and q* are not impacted by the size of S.3. The influences of S on n*, T* and Q* are nonnegative. But, a higher

value of S results in a lower value of AP*.

Inventory Level I(t)Inventory of the item

q

R

1tn 1t

q

nq

n 1t

Inventory of the ite

Fig. 1. Combined inventory diagram from the it

4. The values of n*, t�1, T*, q* and Q* are non-sensitive to the changeof s. However, a higher value of s causes a lower value of AP*.

Example 4. Given a = 1000 units/unit time, b = 0.9, h = $4/unit/unittime, h1 = $3/unit/unit time, c = $5.0 per unit, p = $10 per unit andl = 300 units/unit. Since (p � c)b � h > 0, we apply Algorithm 1-c.The computational results for different maximum allowable num-bers (l = 150, 200, 250 or 300), ordering costs (S = $300, 400, 500,600 or 700) and transferring costs (s = $50, 60, 70, 80, 90, or 100)are shown in Table 4 (see Appendix 5).

Table 4 indicates the following results:

1. The influence of l on t�1, q* and AP* is nonnegative as well as theinfluence of l on n* is non-positive. These results indicate thatthe retailer may increase t�1 and q*, but may decreasen*, whenl increases.

in the display area

m in the warehouse

ems in the warehouse and the display area.

Table 1Sensitivity analyses

l S s n* t�1 T� ¼ n�t�1 q* Q* = n*q* AP*

150 90 10 5 0.150000 0.750000 150.0000 750.0000 1678.8333200 4 0.200000 0.800000 200.0000 800.0000 1687.5000250 3 0.250000 0.750000 250.0000 750.0000 1690.0000300 3 0.258199 0.774597 258.1989 774.5967 1690.1613

300 10 10 1 0.258199 0.258199 258.1989 258.1989 1845.080730 2 0.235702 0.471404 235.7023 471.4046 1787.868050 2 0.278887 0.557774 278.8867 557.7734 1749.002070 3 0.235702 0.707106 235.7023 707.1069 1717.157390 3 0.258199 0.774597 258.1989 774.5967 1690.1613

300 90 10 3 0.258199 0.774597 258.1989 774.5967 1690.161330 3 0.300000 0.900000 300.0000 900.0000 1620.000050 3 0.300000 0.900000 300.0000 900.0000 1553.333370 3 0.300000 0.900000 300.0000 900.0000 1486.666790 3 0.300000 0.900000 300.0000 900.0000 1420.0000

Table 2Optimal solutions for (p � c)b � h < 0

l S s n* t�1 T� ¼ n�t�1 q* Q* = n*q* AP*

150 90 10 5 0.147794 0.738970 150.0000 750.0000 1705.6210200 4 0.196104 0.784416 200.0000 800.0000 1724.4020250 3 0.243950 0.731850 250.0000 750.0000 1736.2358300 3 0.291345 0.874035 300.0000 900.0000 1742.9968

300 10 10 1 0.291345 0.291345 300.0000 300.0000 1901.644130 2 0.291345 0.582690 300.0000 600.0000 1839.482350 2 0.291345 0.582690 300.0000 600.0000 1805.158770 3 0.281318 0.843954 289.3830 868.1490 1766.028590 3 0.291345 0.874035 300.0000 900.0000 1742.9968

300 90 10 3 0.291345 0.874035 300.0000 900.0000 1742.996830 3 0.291345 0.874035 300.0000 900.0000 1674.349650 3 0.291345 0.874035 300.0000 900.0000 1605.702370 3 0.291345 0.874035 300.0000 900.0000 1537.055190 3 0.291345 0.874035 300.0000 900.0000 1468.4079


2. The values of n* and R* are not influenced by the value of S (or s).3. A higher value of S (or s) causes higher values of t�1, T*, q* and Q*,

but a lower value of AP*. It indicates that the retailer shouldincrease t�1, T*, q* and Q* when S (or s) increases.

7. Conclusions

In this paper, we formulated an ordering-transfer inventorymodel when the amount of display space is limited and the de-mand rate depends on display stock level. This objective is tosimultaneously determine the retailer’s optimal ordering quantityand the number of transfer per order from the warehouse to thedisplay area for maximizing the average profit per unit yieldedby the retailer. Some algorithms are developed to find the optimalsolution. Furthermore, we establish Theorem 1, which provides usa simple way to obtain the optimal solution by examining the ex-plicit condition for a special case which the demand rate is con-stant. Numerical examples are provided to demonstrate theapplicability of the proposed models. The sensitivity analysis ofthe optimal solution with respect to the parameters is also in-cluded. The computation results show some phenomena as fol-lows: (1) A higher value of S (or s) causes a lower value of AP*

for every situation. (2) If (p � c)b � h 6 0, then the following re-sults are obtained: (a) A higher value of l causes higher values oft�1, q* and AP*, but a lower value of n*. (b) The values of n*, t�1, T*,q* and Q* are non-sensitive to the change of s. That is, the retailershould increase the quantity per transfer from the warehouse tothe display area and the replenishment cycle time in the displayarea. This leads to the increase of average profit per unit time,

but the retailer may decrease the number of transfer from thewarehouse to the display area per order, when the maximumallowable number of displayed stocks in the display area increasesunder the condition that the demand rate is constant or buildingup inventory is not profitable. (3) If (p � c)b � h > 0, then a highervalue of S (or s) causes higher values of t�1, T*, q* and Q*. That is, theretailer should increase the quantity per transfer from the ware-house to the display area, the replenishment cycle time in thedisplay area, the order quantity place on the supplier and thereplenishment cycle time in the warehouse when the cost of plac-ing per order (or the transfer cost from the warehouse to thedisplay area) increases under the condition that building up inven-tory is profitable.

The proposed model can be extended in several ways. For in-stance, we may extend the model to allow for a constant deterio-ration rate or a two-parameter Weibull distribution. In addition,we can consider the demand as a function of price, quality as wellas time varying. Finally we can generalize the model to allowshortages, quantity discounts, discount and inflation rates, andothers.

Acknowledgement

The authors are grateful to these referees for their encourage-ment and constructive comments.

Appendix 1

See Fig. 1.

Table 3Optimal solutions for (p � c)b � h = 0

l S s n* t�1 T� ¼ n�t�1 q* Q* = n* q* AP*

150 90 10 5 0.146730 0.733650 150.0000 750.0000 1719.1641200 4 0.194230 0.776920 200.0000 800.0000 1742.6723250 3 0.241069 0.723207 250.0000 750.0000 1759.0723300 3 0.287259 0.861777 300.0000 900.0000 1770.7528

300 10 10 2 0.287259 0.574518 300.0000 600.0000 1902.782330 2 0.287259 0.574518 300.0000 600.0000 1867.970550 2 0.287259 0.574518 300.0000 600.0000 1833.158770 2 0.287259 0.574518 300.0000 600.0000 1798.346990 3 0.287259 0.861777 300.0000 900.0000 1770.7528

300 90 10 3 0.287259 0.861777 300.0000 900.0000 1770.752830 3 0.287259 0.861777 300.0000 900.0000 1701.129350 3 0.287259 0.861777 300.0000 900.0000 1631.505770 3 0.287259 0.861777 300.0000 900.0000 1561.882190 3 0.287259 0.861777 300.0000 900.0000 1492.2585

Table 4Optimal solutions for (p � c)b � h > 0

l S s n* t�1 T� ¼ n�t�1 q* Q* = n*q* R* AP*

150 300 50 4 0.140703 0.562812 150.0000 600.0000 0 3473.3123200 4 0.159411 0.637644 171.4131 685.6524 0 3486.3321250 3 0.211741 0.635223 233.2615 699.7845 0 3648.2673300 3 0.211741 0.635223 233.2615 699.7845 0 3648.2673

300 300 50 3 0.211741 0.635223 233.2615 699.7845 0 3648.2673400 3 0.231915 0.695745 257.8943 773.6829 0 3498.0318500 3 0.250018 0.750054 280.3813 841.1439 0 3359.7211600 3 0.265574 0.796722 300.0000 900.0000 0 3230.6591700 3 0.265574 0.796722 300.0000 900.0000 0 3105.1450

300 300 50 3 0.211741 0.635223 233.2615 699.7845 0 3648.267360 3 0.218048 0.654144 240.9148 722.7444 0 3601.733970 3 0.224128 0.672384 248.3332 744.9996 0 3556.503980 3 0.230001 0.690003 255.5374 766.6122 0 3512.464490 3 0.235684 0.707052 262.5449 787.6347 0 3469.5175100 3 0.241192 0.723576 269.3711 808.1133 0 3427.5784


Appendix 2

See Table 1.

Appendix 3

See Table 2.

Appendix 4

See Table 3.

Appendix 5

See Table 4.

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