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Int. J. Production Economics 97 (2005) 185–195 www.elsevier.com/locate/dsw
A model for shelf space allocation and inventory control considering location and inventory level effects on demand Hark Hwanga,, Bum Choia, Min-Jin Leeb a
Department of Industrial Engineering, Korea Advanced Institute of Science and Technology (KAIST), 373-1 Kusong-Dong Yusong-Gu, Daejeon 305-701, Republic of Korea b Korea Telecom Technology Laboratory, 463-1 Jeonmin-Dong Yusong-Gu, Daejeon 305-811, Republic of Korea Received 5 August 2003; accepted 21 July 2004 Available online 13 September 2004
Abstract Shelf on which products are being displayed is one of the most important resources in retail environment. Retailers cannot only increase their profit but also decrease cost by proper management of shelf space allocation and products display. This paper addresses a problem of retailer who sells various brands of items through displaying on multi-level shelves. It is assumed that the level of shelf on which the product is displayed has a significant effect on sales. We develop an integrated mathematical model for the shelf space allocation problem and inventory-control problem with the objective of maximizing the retailer’s profit. Then, a gradient search heuristic and a genetic algorithm are proposed for the solution to the model. The validity of the model is illustrated with example problems. r 2004 Elsevier B.V. All rights reserved. Keywords: Shelf space allocation; Location effects; Inventory control; Gradient search; Genetic algorithm
1. Introduction Shelf on which products are being displayed is one of the major resources in retail environment. Accordingly, shelf management has been considered as an important decision to retailers. Retailers cannot only increase their profit but also decrease cost by managing shelf well. Increasing sales by Corresponding author. Tel.: +82-42-869-3113, fax:+82-42-
869-3110. E-mail address:
[email protected] (H. Hwang).
attracting the consumer’s attention and encouraging consumers to have additional purchase opportunities can be implemented by proper management of shelf space allocation and products display (location of the product within a display, product adjacencies, aesthetic elements, etc.). There have been many empirical studies to estimate the space elasticity, which is defined as the ratio of relative change in unit sales to relative change in shelf space. Curhan (1972) took a large sample from store experiments, and found the
0925-5273/$ - see front matter r 2004 Elsevier B.V. All rights reserved. doi:10.1016/j.ijpe.2004.07.003
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average value of 0.212 for space elasticity. Also, Heinsbroek (1977) found the average value of 0.15 at the item level, referring to 20 experiments while Corstjens and Doyle (1981) found a low mean value of 0.086. On the other hand, Thurik (1988) found higher average space elasticity close to 0.6 at store level. Desmet and Renaudin (1998) found that space elasticities between categories are different substantially ranging from negative values to levels as high as 0.8. Larson and DeMarais (1990) called the displayed inventory which is carried to stimulate demand ‘‘psychic stock’’ and suggested ‘‘full-shelf merchandising’’ policy; that is, display area is always kept fully stocked, referring the experimental studies which concluded that substantially higher sales could be achieved merely by keeping shelves fully stocked. In addition, Dreze et al. (1994) made a series of field experiments and found that location of the product within a display, especially the level of shelf on which the product is displayed in case of multi-level shelf, has a significant effect on sales. Motivated by those findings, many research articles appeared on the shelf space allocation problem to deal with how to optimally allocate shelf space to each brand of items so as to maximize the total sales volume. These models formulated the demand rate as a function of the shelf space allocated to products. Considering main demand effects, Anderson and Amato (1974) formulated an optimization model as a knapsack problem. Also, Hansen and Heinsbroek (1979) considered the main demand effects in their mathematical programming. Corstjens and Doyle (1981, 1983) developed a generalized geometric programming model with the main effects and cross effects of demand. Also, an optimization model of Bultez and Naert (1988) utilized marginal analysis. Zufryden (1986) solved the model of Corstjens and Doyle (1981) by dynamic programming approach. Borin et al. (1994) considered the main effects and cross effects of substitute items and suggested simulated annealing as a solution methodology. Observing that existing research papers on the shelf space allocation problem ignore the inventory-related decisions, Urban (1998) proposed integrated models of inventory-
control and shelf space allocation problems. However, the above-mentioned models did not consider the location effects Dreze et al. (1994) emphasized. Recently, Yang and Chen (1999) and Yang (2001) proposed a space allocation model, a type of multi-constraint knapsack problem, incorporating the main and cross effects of demand as well as the location effects. Only for simplified versions of the original model, he found an optimal solution. This paper is motivated by the operation of convenience stores in which much of the sales items are displayed on the multi-level shelves in refrigerated show cases. It is assumed that the manager is responsible for the shelf management in addition to the inventory control of the sales items. The reminder of this paper is organized as follows. In Section 2, we develop an integrated mathematical model for shelf space allocation problem and inventory control problem based on ‘full shelf merchandising policy’. Then two heuristic solution procedures, one with gradient search method and the other with genetic algorithm, are developed in Section 3. In Section 4, validity of the model is illustrated through solving example problems. Solutions obtained by the heuristics are compared with those from a total enumeration. Conclusions appear in Section 5.
2. Mathematical model We deal with the problem of a retailer who displays various brands of items within a category to multi-level shelves that have limited space. Orders are initially received into backroom that is placed in the back of the multi-level shelves, and the items are restocked from the backroom into the shelves as the displayed items on shelves are depleted by consumer demand. The retailer’s total profit can be expressed as the gross margin subtracted by the holding cost of the items in the backroom and those on display, display expense, and ordering cost. We want to determine how many items to order for each brand, how much space to allocate to each brand and where each brand to be placed on shelves in order to maximize the retailer’s total profit.
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2.1. Nomenclature Lj li Pi Cpi Chi Csi Coi X min i X max i Ti Xij Xi Qi di bi ð¼ bii Þ
bik aj
ai
the space of shelf j the amount of shelf space required for a unit of item i unit selling price of item i unit purchasing cost of item i inventory holding cost of item i per unit period display cost of item i per unit period per space ordering cost of item i per order minimum number of item i that can be displayed on shelves, X min i 40 maximum number of item i, which can be displayed on shelves replenishment cycle time of item i number of item i displayed on shelf j total number P of item i displayed on shelves ð¼ M j¼1 X ij Þ order quantity of item i the scale parameter of demand function for item i, di40 the space elasticity for item i with respect to a unit of displayed inventory, 0obio1 the cross space elasticity between item i and k the scale parameter that reflects the increase of demand rate with respect to the level of shelf when items are displayed on shelf j weighted average value of aj when item i is displayed on more than one shelve
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(6) ‘‘Full-shelf merchandising’’ policy is adopted, that is, the display area is always kept fully stocked. (7) All N brands of items have to be displayed on shelves. (8) The items are restocked from the backroom into the shelves continuously and the restocking cost is negligible. (9) The selling price and unit purchasing cost of each item are known and constant. 2.3. Model development Suppose that there are M shelves in a particular categorized area of the store and N brands of items within the category are displayed on the shelves. Due to the full-shelf merchandising policy, the display area is always kept fully stocked and so the displayed inventory of item i always equals Xi. In developing the demand rate of the sales items, we focus on the experimental study of Dreze et al. (1994) which concluded that the level of the shelves on which items are displayed has an effect on sales and eye-level position is the best (location effect). Thus the demand rate is assumed to be a function of the displayed inventory level and the display location within the shelves of each item. Integrating the location effects into the model of Corstjens and Doyle (1981), the following demand function is adopted for this study. Di ¼ ðmain space effectÞðcross space effectÞ ðlocation effectÞ " # N Y bi bik X k ai ; ¼ diX i
ð1Þ
kai
2.2. Assumptions (1) The system involves N brands of items within a category and M-level shelves with limited space. (2) Demand rate of the items is a function of the quantity displayed and the location displayed within the shelves. (3) The time horizon of the inventory model is infinite. (4) Replenishments of each item are independent (no joint replenishments) and instantaneous. (5) The retailer order the items with (r, Q) policy.
where Xi ¼
M X
X ij
j¼1
and PM ai ¼
j¼1 X ij aj
Xi
:
The main space effect implies that the demand of item i increases as its displayed inventory Xi increases and the increasing ratio depends on the
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space elasticity bi. If bi equals 1, the increasing ratio is propositional to that of Xi. If bi equals 0, Di is indifferent to Xi. Di is also influenced by the displayed inventory Xk of other items, which is explained by the cross space effect. If item k is substitute to item i, the cross space elasticity bik becomes negative, i.e., Di decreases as Xk increases. If item k is complementary to item i, bik becomes positive, i.e., Di increases as Xk increases. Di is influenced by the locations item i is displayed within the shelves. The constant parameters ajX1, j=1, 2, y, M, are introduced to express the location effect. The shelf in the eye-level position has the largest value of aj, while the shelf in the worst position, i.e., either the bottom or top shelf, has the smallest value of 1. If item i is displayed on more than one shelf, it is assumed that the weighted average value of aj represents the location effect. Now, we consider a deterministic, continuousreview model of an inventory system with the demand rate of Eq. (1). Let Ii(t) be the instantaneous inventory level of item i at time t which includes both the backroom storage and displayed inventories. With this system, item i is continuously restocked from the backroom into the shelves as soon as the items are sold. Thus Ii(t) decreases linearly following the demand rate. Since the shelves are always fully stocked, orders need to be replenished before the backroom inventory reaches zero; that is, at the end of cycle, Ii(Ti) equals the number of units of item i displayed on shelves, Xi. Silver and Peterson (1985) showed that in the case of a deterministic, constant demand rate and instantaneous replenishment, it is optimal to let the inventory level reaches zero before reordering. So, it will be reasonable to replenish orders whenever the backroom inventory reaches zero. Fig. 1 shows the change of the inventory level in the system. The decreasing rate of inventory level of item i at time t is
Xi + Qi Di
(2)
With the initial condition Ii(Ti)=Xi and Ii(0)= Xi+Qi, we have 0 t T i:
(3)
Qi
Xi Ti
t
Fig. 1. The inventory system of item i.
From Eq. (3), the cycle time and average inventory level of item i is Ti ¼
Qi ; Di
(4)
ð2X i þ Qi Þ : (5) 2 From Eqs. (1), (4) and (5), the retailer’s total profit during unit period can be expressed as " N N X Coi Y b TP ¼ d i ðPi Cpi Þ X kik ai Q i i¼1 k¼1 Chi Qi ðChi þ Csi l i ÞX i : ð6Þ 2
Ii ¼
Given the objective function, the model can be formulated as follows: max TP
½P
Qi ;X ij
s.t. N X
l i X ij Lj
(7)
8j;
i¼1
X i X max X min i i Xi ¼
M X
X ij
8i;
8i;
(8) (9)
j¼1
X ij 0
qI i ðtÞ ¼ Di : qt
I i ðtÞ ¼ ðX i þ Qi Þ Di t;
Ii (t)
Qi 40
8i; j; 8i:
(10) (11)
Decision variables are Qi, Xi and Xij, i=1, y, N, j=1, y, M, with the order quantities each of item, the total numbers of each item displayed on the shelves and the shelf space allocated to each of
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item. As a result of Eq. (7), the sum of space allocations does not exceed the space of shelf. Eq. (8) provides operational constraint that may be imposed by the retailer to ensure the desired minimum and maximum shelf space allocated to each item.
differentiating TP with respect to Qi. sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2Coi Di ; Q i ¼ Chi
½P0
s:t:
max TP ¼ X ij
N X
N X M X
C1i aj X ij
i¼1 j¼1
l i X ij Lj
N X
C2i
(12)
i¼1
8j;
i¼1
Xi ¼
M X
X ij
di
N Y
b X kik
8i;
M X
!, X ij aj
X i:
j¼1
k¼1
[P] is a nonlinear programming problem and it is very difficult to find an optimal solution in a closed form. Thus, we develop two heuristic solution procedures. Note that if the values of Xi and Qi for 8i are given, an optimal solution X*ij for 8i,j can be obtained by solving the following linear programming.
(13)
where Di ¼
3. Solution procedure
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Property 2. Let X ij and Q i be the optimal values for a prescribed value of Xi in [P]. For given value of Xi for all i, X ij and Q* must satisfy Property 1 and Eq. (13), respectively. Thus they can be obtained by the following procedures: qffiffiffiffiffiffiffiffiffiffi idi Step 1. Set Qi ¼ 2Co Chi : If Qi is not an integer value, substitute Qi with the nearest integer. Step 2. With given Xi and Qi, find X ij from Property 1. Step 3. With given Xij and X ij ; find Q i from Eq. (13). If Q i is not an integer, substitute Q i with the nearest integer. If Q i is equal to Qi, stop. Else, set Qi ¼ Q i ; and go step 2.
j¼1
X ij 0
The proof is shown in Appendix B.
8i; j;
where
Note that Property 2 implies that [P] has essentially only one decision variable type, Xi.
n o Q N ðb 1Þ biz i X Xi i C1i ¼ d i ðPi Cpi Þ Co z Q
8i;
C2i ¼ ðChi þ Csi l i ÞX i þ Chi Qi =2
8i:
i
zai
Property 1. For given values of Xi and Qi for all i, assigning items to the shelves in decreasing order of C1i/li produces an optimal solution of [P0 ]. The proof is shown in Appendix A. Thus the items with higher C1i/li have to be placed closer to the eye-level shelf. Note that due to the above property the number of decision variable types is reduced to two, Xi and Qi. Note that for fixed values of Xi and Xij for all i and j, an optimal solution Q*i can be obtained by
3.1. A gradient search In gradient search, an iterative procedure is developed through which the total profit increases. We proceed from an initial value of Xi to the next in such a way that the total profit increases. The partial derivative of the objective function with respect to Xij is " ( ) N X Coy qTP ¼ byi d y ðPy Cpy Þ qX ij Qy yai ! # Y byk ðb 1Þ X k X i yi ay ka1
þ di
Coi ðPi Cpi Þ Qi
Y ka1
! b X kik
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190 ðbi 1Þ
X i
aj ð1 bi ai
ðChi þ Csi l i Þ:
ð14Þ
Eq. (14) represents the ratio of change in the unconstrained profit function to change in Xij. For any given solution, it gives the change in profit incurred by increasing Xi provided that shelf j can accommodate. Since the required shelf space may not be available when an increment in Xi has to be accommodated in shelf j, relocations of some items must be considered. Following Wilson (1976), the gradient of TP* at Xi is expressed as G X i ¼ qTP =qX i ¼ [Rate of change in the total profit with respect to Xij for given j][The cost incurred by the relocation of items, which is resulted from increasing Xi]. TP* is an unconstrained function of Xi. Since constraints (7)–(11) can be satisfied through the following solution procedure, we can proceed from any given value of Xi o the next in the direction of the positive gradient of the unconstrained profit function, TP*. Step 1: Initialization For a starting min X min 2 ; . . . ; X N Þ:
point,
set
X 0 ¼ X min 1 ;
Step 2: Iterative search Step 2.1: Consider the set of display quantities for kth iteration. X k ¼ X k1 ; X k2 ; . . . ; X kN For a given Xk, obtain an optimal solution k X ij ; Qk i from Property 2. Step 2.2: Utilizing Property 1, compute the cost incurred by relocation of items, which is resulted from increasing Xi by one unit. The following additional notations are introduced. Pj the last item allocated to shelf j That is, Pj has the lowest value of C1i/li among all the items allocated to shelf j. Ji the last shelf for which Xij40 Fi the first shelf which has extra empty space when we add a unit of item i. If we increase a unit of item i allocated to shelf j, item Pj would be transferred from shelf j to shelf
j+1 by li/lpj, unit. The cost incurred by moving item Pj from shelf j to shelf j+1 by one unit is D0 j ¼ C1pj ½ajþ1 ðX pj ;jþ1 Þ þ aj ðX pj ;j Þ C1pj ½ajþ1 ðX pj ;jþ1 þ 1Þ þ aj ðX pj ;j 1Þ ¼ C1pj ðaj ajþ1 Þ:
ð15Þ
The increased cost per unit space of item Pj, which is relocated, is Dj ¼ D0 j =l pj ¼ ðC1pj =l pj Þðaj ajþ1 Þ:
ð16Þ
The increased cost, CRi, attributable to the relocations is the sum of the costs of transferring an equal space of item between each successive pair of shelves from the last shelf which item i is allocated, Ji to the first shelf which has extra space enough to add a unit of item i, Fi. CRi ¼ l i
FX i 1
Dj :
(17)
j¼J i
Step 2.3: Compute the gradient vector at Xi for I=1,y, N and then select the item using the greatest absolute gradient value. If the gradient value of the selected item is positive and also its displayed inventory equals Ximax, then select another item. If it is negative and its displayed inventory equals Ximax, then select another item. qTP k X Gi ¼ X CRi : (18) qX ij iJ i Step 2.4. Updated the value of X ki for the selected item i from Step 2.3. 8 > X min > i > > > min > if X ki þ STEPX GX > i oX i ; > > < X k þ STEPX G X i i X kþ1 ¼ i min k max > if X X þ STEPX G X ; > i i i Xi > > > max > > Xi > > : if X max oX ki þ STEPX G X i ; i (19) where STEPX ¼
DX max : max G X i i
(20)
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Until a decrease in total profit occurs, the step parameter DXmax is left unchanged in successive iterations. If a decrease in total profit is encountered in any iteration, DXmax is reduced by a factor of one-half.
191
3.2.2. Initialization The next step is the creation of initial populations. In this research, a set of initial populations are randomly generated, i.e., a gene Xi of chromosome is randomly generated within the interval of [Ximin, Ximax].
Step 2.5: Increase k by one unit (k=k+1). Step 2.6: Go to Step 2.1 if neither of the following criteria is satisfied: k ¼ MAX; N X i¼1
l i X ki 4
(21)
M X
Lj ;
(22)
j¼1
DX max ðX Þ;
(23)
where e(X) is substantially small value. MAX is the maximum number of iterations. 3.2. A genetic algorithm Genetic Algorithms (GAs) can be defined as meta-heuristics based on the evolutionary process of natural systems (Holland, 1975). Since their inception, they have been applied to numerous optimization problems with highly acceptable results. Recently Go´mez et al. (2003) used genetic algorithms to resolve layout problems in facilities where there are aisles. Zhou et al. (2003) proposed a genetic algorithm approach to the bi-criteria allocation of customers to warehouses. Balakrishnan et al. (2003) proposed a hybrid genetic algorithm for the dynamic plant layout problem. Given successful application of genetic algorithms to layout problems which are similar to the selfspace allocation problem, the following genetic algorithm is proposed to tackle the self-space allocation problem. 3.2.1. Encoding The first step of developing genetic algorithm is to encode a solution as a finite-length string called chromosome. For the problem considered here, Xi is represented as 0–1 binary digit and a chromosome is encoded as (X1, y, XI).
3.2.3. Crossover and mutation Offspring chromosomes are usually generated from the current populations (i.e., parent chromosomes) using two common genetic operators: crossover and mutation. Crossover plays the role of exchanging information among chromosomes. It usually leads to an effective combination of partial solutions and thus accelerates the search procedure early in the generation. Here is used ‘‘one-cut-point’’ method, which randomly selects one ‘‘cut-point’’ and exchanges the right part of two parents. Probability of crossover is set to 0.25, which means that 25% of chromosomes undergo crossover. Thereafter, a mutation operator is invoked to produce offspring chromosomes. Mutation is to alter one or more genes with probability equal to the mutation rate. In this paper, the probability of mutation is set to 0.01. 3.2.4. Handling constraints The central problem for applying genetic algorithms to the constrained optimization is how to handle a set of constraints because genetic operators used to manipulate the chromosomes often yields infeasible offspring. In this study, we use the penalizing strategy in which penalties are imposed on the chromosomes according to the degree of violating constraints. 3.2.5. Fitness function and scaling mechanism In the genetic algorithm, fitness is computed for each chromosome in the population and the objective is to find the chromosome with the maximum fitness. In this paper, we use the total profit as the fitness value. Since performance of the genetic algorithm is highly sensitive to the fitness, it is very important to scale the fitness not only to avoid a premature convergence but also to diversify the population. In this study, the fitness value pk of
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0.008 0.014 0.014 0.29 0.007 0.011 0.23 0.008 0.014 0.44 0.009 0.011 0.34 0.014 0.005 0.008 1 2 3 4 0.008 0.01 0.012 0.29 0.007 0.015 0.23 0.011 0.014 0.44 0.011 0.01 0.34 0.007 0.014 0.007 1 2 3 4 0.008 0.005 0.011 0.29 0.013 0.01 0.23 0.014 0.01 0.44 0.014 0.006 0.34 0.006 0.006 0.012 1 2 3 4 1.0 1.0 1.0 1.0 60.04 68.55 74.32 63.6 5.0 5.0 5.0 5.0 0.5 0.5 0.5 0.5 3.58 2.57 2.1 3.1 Data set 3 1 6.44 2 4.63 3 3.78 4 5.58
0.89 0.64 0.52 0.78
0.014 0.014 0.007 0.43 0.008 0.008 0.24 0.009 0.014 0.27 0.006 0.011 0.48 0.012 0.005 0.011 1 2 3 4 0.011 0.007 0.013 0.43 0.01 0.01 0.24 0.008 0.011 0.27 0.01 0.01 0.48 0.007 0.006 0.007 1 2 3 4 0.01 0.006 0.011 0.43 0.008 0.005 0.24 0.015 0.007 0.27 0.005 0.01 0.48 0.012 0.015 0.013 1 2 3 4 1.0 1.0 1.0 1.0 59.71 69.31 75.19 58.5 5.0 5.0 5.0 5.0 0.5 0.5 0.5 0.5 3.63 2.5 2.04 3.82 Data set 2 1 6.53 2 4.5 3 3.67 4 6.88
0.91 0.62 0.51 0.96
0.008 0.013 0.4 0.012 0.011 0.25 0.007 0.015 0.5 0.008 0.009 0.009 1 2 3 4 0.01 0.014 0.011 0.34 0.014 0.009 0.4 0.015 0.011 0.25 0.006 0.009 0.5 0.01 0.005 0.008 1 2 3 4 0.013 0.012 0.007 0.34 0.012 0.013 0.4 0.005 0.015 0.25 0.013 0.011 0.5 0.009 0.006 0.013 1 2 3 4 1.0 1.0 1.0 1.0 62.26 70.11 66.92 61.22 5.0 5.0 5.0 5.0 0.5 0.5 0.5 0.5 0.82 0.61 0.68 0.85 3.27 2.43 2.73 3.41 Data set 1 1 5.89 2 4.37 3 4.91 4 6.14
3 1
2
3
4
1 bik bik
li (Facings) di (Units/day) Coi ($) Csi ($) Chi ($) Cpi ($) Pi ($)
Table 1 Three sets of data for four brands of items
The proposed solution procedures are illustrated with example problems. And solutions are compared with those obtained from a total enumeration. We considered 18 cases, i.e., three different set of bik values (all negative, some negative, and all positive), two different brands of items (3, 4), and three different levels of shelves (4, 5, and 6). Note that negative bik implies item i and k are substitutive each other while positive bik complementary each other. For each case, 3 different data sets were generated, which result in 54 problem instances. The data sets were shown in Table 1. The problem instances were solved both by the proposed solution procedures and a total enumeration. The item parameters of each data set in Table 1 were obtained from: Cpi=U(2, 4), Pi=1.8 Cpi, Chi=0.25 Cpi, Csi=0.5, Coi=5.0, di=100/Cpi0.4, li=1.0, bI= bii=U(0.2, 0.5), |bik|=U(0.005, 0.015), I=1,y, N, k6¼i. Ximin=1, Ximax=1.5 M, i=1,y, N. The shelf parameters were fixed as; Lj=3, aj=1+(Mj) 0.1, j=1,y, M. Furthermore, the search parameters used in the gradient search were: DXmax=0.1, MAX=100, e(X)=0.0001.
Mixed items
4. Numerical examples
All substitutive items
2
3
4
3.2.6. Population update In this step, all of the existing populations are removed and new populations are selected from the off-springs yielded by crossover and mutation. The roulette wheel technique is adopted for the selection where the probability of selection of a certain offspring chromosome is proportional to its scaled fitness value.
2
where spk is the scaled fitness value of chromosome k and g is the predetermined value for scaling purpose.
1
k
bik
All complementary items
(24)
Item (i)
pk g ; spk ¼ P ðpk gÞ
4
chromosome k is scaled as
0.008 0.014 0.009 0.34
H. Hwang et al. / Int. J. Production Economics 97 (2005) 185–195
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Table 2 Comparison of the proposed solution procedures with a total enumeration Number of items
Level of shelf
Gradient Search Av. % deviation
(1) All substitutive items 3
4
(2) 3
4
(3) 3
4
4 5 6 4 5 6 Mixed items (some substitutive and some complementary) 4 5 6 4 5 6 All complementary items 4 5 6 4 5 6
Genetic Algorithm a
Av. % deviationa
1.374 1.836 1.506 3.572 2.675 2.635
0.098 0.079 0.234 0.149 0.512 0.581
2.136 2.239 2.668 3.825 3.150 2.447
0.177 0.132 0.181 0.673 0.584 0.316
1.827 1.999 2.102 3.519 2.619 3.288
0.179 0.181 0.078 0.508 0.143 0.328
jTE Aj 100 ð%Þ; TE where TE is the total profit from a total enumeration and A the total profit from the corresponding solution procedure. a Av. % deviation=the average of % deviations from the three data sets.
% deviation ¼
Experimental tests were done on a personal computer with a Pentium processor (2.4 GHz) and the codes written in the Visual Basic language. Table 2 shows the computational results. Compared to the gradient search, the genetic algorithm performs better and generates near optimal solution. The minimum and maximum percent deviations of the genetic algorithm are 0.027 and 0.813 with the average percent deviation of 0.285. The minimum and maximum percent deviations of the gradient search are 0.503 and 5.490. And the average deviation is 2.523 which are substantially larger than those of the genetic algorithm. It is observed that the percent deviations of the proposed heuristics tend to increase as the number of items increases. The required computation time is shown in Table 3 where the gradient search needs almost no time.
5. Conclusions In this paper, we dealt with the shelf space allocation problem with location and inventory level effects on demand. The demand rate was assumed as a function of the displayed inventory level and the location within shelves of each brand of items. Then a mathematical model was formulated with the objective of maximizing the retailer’s profit. Investigation of the characteristics of the model led to two properties an optimal solution has to satisfy. Consequently, the number of decision variable types was found to be one, which contributed a drastic reduction of the computation time. To solve the model, two different solution procedures were presented, the gradient search and genetic algorithm. To examine the validity of the proposed procedures, 54
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194 Table 3 Comparison of running times Number of items
3
Level of shelf
4 5 6 4 5 6
4
Average running time (s) Gradient search
Genetic algorithm
Total enumeration
50.001 50.001 50.001 50.001 50.001 50.001
0.078 0.080 0.080 0.129 0.125 0.137
0.156 0.314 0.718 0.936 1.871 7.029
example problems were generated by varying the number of item brands, level of the shelves, and item characteristics in terms of bik values with three sets of data. The solutions of the example problems were compared with those obtained from a total enumeration. The performance of the two procedures turns out to be satisfactory with the total average percent deviation of 0.027 and 0.813, respectively. The followings can be suggested for further studies: (1) Product assortment problem with the objective of maximizing the retailer’s profit. (2) Shelf space allocation problem with perishable items.
Appendix A. The proof of Property 1 We will show that violation of CLI rule never raises the total profit. Let item i be assigned to shelf k and item j to shelf u, where CLIiXCLIj and 0 ak4au. From [P ], the total profit becomes TP ¼
N X M X i¼1 j¼1
C1i aj X ij
N X
C2i :
i¼1
If we exchange each unit space of item i and j between shelf k and u, then the change of the total profit with Qi fixed can be expressed as DTP ¼ ½total profit after exchange ½total profit before exchange ¼ C1j ak =l j þC1i au =l i C1i ak =l i þC1j au =l j
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