Assortment Planning and Inventory Management

Assortment Planning and Inventory Management Under Dynamic Stockout-based Substitution Preliminary Draft

Doroth´ee Honhon∗,

Vishal Gaur∗,

Sridhar Seshadri∗

January 2006

Abstract We consider the problem of a monopolist retailer choosing the optimal assortment and inventory levels for a discrete set of products with varying prices and costs in order to maximize expected profit in a single-period setting. Customer preferences are modeled through the definition of consumer types, where a type is a ranking of the products by order of preference. A customer buys the highest ranked product available (if any) in the assortment at the time of his visit to the store (dynamic substitution). We first solve the case where all the customers have the same consumer type, and show that it may be optimal to stock multiple products because of differences in their implied risk-return tradeoffs. The solution is obtained using an efficient dynamic programming algorithm. We then show that this algorithm gives the optimal solution for more general preference structures when customers are partitioned into different types in fixed proportions. When the number of customers of each type is random, we use the algorithm to construct two heuristics and an upper bound, and numerically evaluate the performance of the heuristics.

∗

Department of Information, Operations and Management Science, Leonard N. Stern School of Business, New York University, 8-160, 44 West 4th St., New York, NY 10012. E-mail: [email protected], [email protected], [email protected].

1

Introduction

Assortment planning and inventory management are fundamental decision problems in a retail store. These problems are interrelated due to their dependence on consumer preferences, and the resultant choice and substitution behavior. The substitution behavior is said to be static or assortment-based if consumers choose their preferred products from a given assortment without any knowledge of the availability of products, and do not modify their choice in the event of a stockout. The substitution behavior is said to be dynamic or stockout-based if consumers have knowledge of the availability of products before choosing their preferred products from the given assortment. The assortment planning and inventory management problem under static substitution has been well-studied in the recent past for a variety of consumer choice models, including the multinomial logit model, the locational choice model, and generalizations of these models. However, the problem under dynamic substitution is considerably harder, and thus, relatively little is known about its solution even though dynamic substitution is more representative of real-life applications than static substitution. Indeed, Mahajan and van Ryzin (2001) show that the profit function for this problem is not quasi-concave in inventory levels, and thus, its optimal solution cannot be obtained in general without resorting to enumeration techniques. They propose a sample path gradient based heuristic to solve this problem, which is guaranteed to converge to a local maximum under fluid approximation but is computationally intensive. Smith and Agrawal (2000) propose another heuristic for this problem, which relies on the static-substitution model, and is, thus, computationally simpler. A third heuristic is proposed by Gaur and Honhon (2005) for assortment planning under dynamic substitution when customer preferences are represented by a locational choice model. This heuristic is based on retailer-controlled substitution, wherein the retailer is able to allocate products to customers in order to maximize her profits. Gaur and Honhon evaluate the performance of this heuristic and the static-substitution heuristic, and compare them to an upper bound on the expected profit under dynamic substitution. In general, due to the difficult nature of the dynamic substitution problem, performance bounds on heuristics are rarely available, and little is known about the conditions under which static substitution serves as a good approximation for the dynamic substitution problem. In this paper, we investigate classes of consumer choice models for which the joint assortment planning and inventory management problem under dynamic substitution can be solved to optimality. We represent consumer tastes by ‘customer types’. A type is a list of products that a customer of that type will purchase, arranged in decreasing order of preference. We show that there are types of preference structures for which an optimal solution can be characterized in closed form, and obtained efficiently even though the profit function is not quasi-concave in inventory levels. The simplest preference structure we consider is that of a homogeneous population in which all customers have the same tastes. In this case, there is a single customer type, given by (1, 2, . . . , n), where n is the number of products available. We find the optimal set of products to stock and the corresponding inventory vector for this problem in a single-period (newsvendor) setting using an efficient dynamic program algorithm of complexity O(n2 ). A priori, it appears that the retailer may do well by stocking a single product. Indeed, the optimal static substitution solution consists of stocking the one product that is the most profitable. However, we show that the optimal solution may consist of offering more than one product, and that the value function is non-differentiable in the inventories of products.

1

We then consider a preference model called nested preferences, in which customers can be of the types (1, ..., j) for j = 1, ..., n. Thus, a customer of type (1, ..., j) will prefer the products 1 through j and in that order but will not buy product j + 1 if products 1 to j are not available. We show that the same algorithm can be used to solve the assortment problem to optimality when all customers are of the same type but the retailer does not know a priori which one it is, and assigns probability αj to type (1, ..., j) (this is labelled as a trend-following population). The algorithm also applies to the case where there is a fixed proportion αj of customers of each type (fixed proportion). This solution is then shown to provide an upper bound and a heuristic for the case in which each customer can independently be of each type with probability αj (random proportion). Finally, we consider the case where the set of consumer types can be represented by an outtree, and there is a fixed proportion of customers of each type. We show that the algorithm provides an optimal solution in this case as well. For the random proportion case, we numerically test the performance of the fixed proportion heuristic and that of a modified version. We benchmark the results using two previously known heuristics, one based on static substitution, and the other, the Sample Path Gradient Algorithm of Mahajan and van Ryzin (2001). In our numerical studies, our two heuristics yield average optimality gaps of 0.14% and 0.18%, while the static substitution-based heuristic yields an average optimality gap of 1.57%, and the sample path gradient algorithm yields 0.54%. The optimality gaps of our heuristics decrease as mean demand increases and as the proportion of customers willing to substitute increases. Comparison with the static substitution heuristic reveals that the profit loss caused by ignoring substitution due to stock-outs can be substantial. Indeed, we theoretically show that there are conditions in which a static-substitution based heuristic can be arbitrarily far from the optimal solution, and thus, dynamic substitution plays a significant role in determining profits. The optimality gap of our algorithm is similar to or better than that of the SPGA in almost all problem instances, and further, significantly improves computation time. However, our algorithm is restricted to the specific nature of consumer choice while the SPGA is very general. A significant outcome of our analysis is that even when all consumers have identical preferences, it is optimal for a retailer to offer more than one product in the optimal assortment. This result shows that dissimilar costs and prices of products provide a rationale for variety under dynamic substitution. Traditionally, the drivers of variety studied in the academic literature are heterogeneous tastes of consumers, uncertain or variety-seeking preferences, and competition. With regard to heterogenous tastes, economists have explained the degree of variety in a given product category as the result of the interplay between the demand for variety from consumers and the cost of providing the variety. Lancaster (1998) writes that “[c]onsumers themselves are the ultimate source for the demand for product variety, either because each seeks individual variety or because different individuals prefer different variants.” He further argues that “[i]n the absence of some kind of economies of scale, it would be in the interest of firms [...] to produce goods individually customized to the ideal specification of each customer.” Thus, consumer heterogeneity has been considered as a driver of variety while operational costs have been considered as restricting variety. In the context of a monopolist retailer, van Ryzin and Mahajan (1999) and Gaur and Honhon (2005) show that customer heterogeneity and uncertain preferences increase the variety whereas inventory costs limit the variety when customers choose according to the multinomial logit choice model or a locational choice model, respectively. With respect to competition, Cachon et al. (2005) show that in the context of multi-product oligopolistic competition, lower search costs lead to larger assortments. In our preference model, the traditional reasons for offering variety are assumed away because

2

the retailer is a monopolist and the consumer population has homogeneous tastes and well defined immutable preferences. Thus, we show that varying inventory cost economics of products constitute yet another motive for offering variety when customers substitute dynamically. Products with varying inventory cost economics have different risk-return profiles, and thus, variety becomes a mechanism to manage the risk due to demand uncertainty. In addition to this insight, the algorithm we use to solve the assortment planning problem allows us to obtain structural insights into the optimal solution. In particular, we show that the optimal assortment does not necessarily contain the most profitable or the most preferred product. Thus our paper contributes to the literature on assortment planning under dynamic substitution in the following ways. We obtain the first set of optimal results on assortment planning with dynamic substitution. We do so using an efficient algorithm which can also be used to provide a heuristic for a more general preference structure. In this case we also provide an upper bound on the optimal expected profit. Finally we obtain new insights on the product variety question, by showing that the retailer can in some cases manage demand risk better with a larger assortment. van Ryzin and Mahajan (1999) were the first to study assortment planning and inventory decisions under the MNL model for the case of static substitution with exogenous prices. They determine many properties of the optimal solution, the main being that the optimal assortment consists of the most popular products from the finite set of potential products to offer. Aydin and Ryan (2000) apply the MNL model to study the joint assortment planning and pricing problem under static substitution. They find that the optimal solution is such that all products have equal margins (i.e., the difference between price and cost). Cachon et al. (2005) generalize the consumer choice process to incorporate search costs. They show that ignoring consumer search in demand estimation can result in an assortment with less variety and significantly lower expected profits compared to the optimal solution. They further show that search costs can induce a retailer to carry an unprofitable product in its assortment to reduce consumer search. Kok and Fisher (2004) (2004) show how to estimate assortment-based substitution in an MNL model by leveraging data from stores with varying assortments. They present an algorithm to solve the assortment planning problem with one-level stockout-based substitution in the presence of shelf-space constraints. de Groote (1994) and Alptekinoglu & Corbett (2005) integrate product differentiation and inventory costs in the context of locational choice models with deterministic demand and uniform customer preferences. de Groote considers a monopoly firm and analyzes the coordination between the marketing decision of product line breadth and the operations decision of production flexibility. Alptekinoglu & Corbett analyze competitive positioning and pricing for two firms, one offering infinite variety through mass customization and the other offering a finite set of different products. They show the counterintuitive result that the mass producer needs to reduce variety to soften price competition with the mass customizer firm. Chen et al. (1998) study optimal product positioning and pricing, extending Lancaster’s model to incorporate varying prices and quality levels in the attribute space, as well as varying reservation prices of customers. They show that the optimal solution for this model under stochastic demand and static substitution can be constructed using dynamic programming by utilizing a ‘cross-point property’ to determine the demands for individual products. Finally Gaur & Honhon (2005) combine the optimal assortment problem with the inventory management problem and obtain the optimal solution under static substitution for horizontally differentiated products in the context of a one-dimensional locational choice model. They introduce stochasticity of demand and a unimodal distribution of customers on the attribute space. They show that the optimal assortment is such that products should be equally spaced and

3

that it is not necessary optimal to stock the product located at the mode of the distribution. This research is related to our paper in the following ways. First the problem of assortment planning under static substitution predates the research under dynamic substitution. By getting rid of the dependency of demand on inventory levels, it is possible to solve the problem optimally. However the assumption of static substitution has limited applications in real-life settings. Second, this research presents a rich set of consumer choice models that may be considered under dynamic substitution as well. Finally it shows the practical relevance of the assortment planning and inventory management problem. Beside the work previously mentioned on assortment planning with dynamic substitution, some researchers have focused on related questions. Anupindi et al. (1998) consider the problem of estimating demand from sales data while taking into account dynamic substitution. Also Rajaram and Tang (2001) consider the problem of determining optimal inventory levels under dynamic substitution for a fixed assortment of substitutable products. The rest of this paper is organized as follows. Section 2 describes the consumer choice model and profit assumptions. Section 3 presents the solution to the case where all customers have similar preferences. Sections 4 and 5 extends the results to heterogeneous choice models. Numerical results are presented in Section 6. All proofs are in the Appendix, unless otherwise stated.

2

Model

2.1

Consumer Choice Model

We consider a product category consisting of n potential products to stock, indexed from 1 to n. Each customer is defined by a consumer type, which is a vector of products he would be willing to buy in decreasing order of preference. For example, a customer of type (1, 2, 3) prefers to buy product 1 if it is available, product 2 if product 1 is not available, product 3 if products 1 and 2 are not available and nothing otherwise. In general, a type t is a vector (t1 , ..., tm ) such that m ≤ n, tl ∈ {1, ..., n} for l = 1, ..., m and t1 6= t2 6= ... 6= tm . In theory, of possible consumer types in a product category with n variants can be P the number n! as large as nj=0 (n−j)! . However, some of these types do not make practical sense. For example for a product like yogurt, customers who prefer a non-fat blueberry yogurt are not likely to switch to a creamy vanilla yogurt with granola or chocolate chips as their second choice. Therefore we define a preference structure as a set of restrictions on the possible consumer types. Let T be the set of all consumer types that satisfy those restrictions. We also define a partitioning mechanism of demand as a method to specify how the customers are split between each possible type in T . We refer to the combination of the preference structure and the partitioning mechanism as a consumer choice model. We say that the consumer choice model is a homogeneous population model when there is only one consumer type. We say it is a heterogeneous population model when there is more than one consumer type. In this paper we consider preference structures (ordered from the most to the lest restrictive):

4

• Homogeneous population: there is only one possible consumer type which, without loss of generality, can be defined as (1, ...., n). • Nested preferences: the set of possible consumer types can be defined without loss of generality as: T = {(1), (1, 2), ..., (1, 2, ..., n)} • Outtree-shaped preferences: the set of possible consumer types can be represented by an outtree, that is a directed graph in which nodes represent products, there is a single initial note representing the first choice product for all consumer types and there is a unique directed path from the initial node to any other node. Each such path corresponds to a consumer type (see §5 for an example). We consider the following partitioning mechanisms for the heterogeneous consumer choice models: • Trend-following partitioning or herd-behavior: all customers are of the same consumer type but the retailer does not know a priori which type it is and assigns a probability to each possible type. • Fixed partitioning: customers are split between the different consumer types in fixed proportions. • Random partitioning: the number of customers of each type is a random variable with known distribution. In particular if we assume that each customer has a given probability of being of each possible type, then the number of customers of each type follows a multinomial distribution. Note that the type of a customer can be the result of a utility maximization process, that is, each customer assigns a utility Uj to product j = 1, ..., n and sets the utility of not purchasing any product to U0 . Let U[k] be the kth greatest value of the utility vector (U0 , U1 , ..., Un ), the type of the customer is (t1 , ...tm ) if Utk = U[k] for k = 1, ..., m and U0 = U[m+1] . In practice a retailer does not need to know the utilities in order to use our model. She only needs to estimate the set of possible consumer types for the given product category and associate a probability to each one of them. We assume that customer preferences are not affected by the decisions of the retailer. In particular, the prices of the products are taken to be exogenous variables in our model.

2.2

Demand

Let qj be the inventory level of product j and q = (q1 , ..., qn ) be the inventory vector. Let D be the random variable denoting total number of customers coming to the store with cdf F , pdf f and mean µ. Let x ≤ q be the inventory vector seen by a customer of type t = (t1 , ..., tm ). Under dynamic substitution, each customer buys the highest ranked product available (if any) in the assortment 5

at the time of his visit to the store. So the customer buys product tl ∈ t if and only if xt1 = ... = xtl−1 = 0. We also say the customer “attempts” to buy products t1 to tl in this case. The demand D Pjn for product j is defined as the number of customers who attempt to buy product j. Note that j=1 Dj ≥ D because each customer may be counted multiple times. Let rj be the selling price of product j, cj be its purchasing cost, and sj its salvage value for j = 1, ..., n. We assume that every time a customer attempts to buy product j but does not find it, the retailer incurs a penalty cost of pj . This cost measures the loss of goodwill associated with the customer not buying his first choice and having to substitute to a less preferred product, possibly incurring a search cost. The single-period (newsvendor) expected profit for the product category given by, n X © ª EΠ(q) = rj E[min(qj , Dj )] − cj qj + sj E[qj − Dj ]+ − pj E[Dj − qj ]+ j=1

P Let uj = rj + jk=1 pk − cj and oj = cj − sj respectively denote the underage and overage cost of product j. The one-period (newsvendor) expected profit for the product category given by n X © ª rj E[min(qj , Dj )] − cj qj + sj E[qj − Dj ]+ − pj E[Dj − qj ]+ EΠ(q) = j=1

= =

n X j=1 n X

" (rj +

j X

pk − cj )qj − (rj +

k=1

j X

# +

pk − sj )E[qj − Dj ]

−µ

n X

pj

j=1

k=1

n X £ ¤ uj qj − (uj + oj )E[qj − Dj ]+ − µ pj j=1

j=1

In what follows we assume for simplicity that pj = 0 for j = 1, ...n. The retailer’s objective is to find q∗ such that EΠ(q∗ ) = max EΠ(q) q≥0

3

The Homogeneous Population model

In the case of a homogeneous population model, all customers are of the same consumer type, which without loss of generality, is (1, 2, ..., n). In other words, all customers prefer to buy product 1 if it is available, otherwise they prefer to buy product 2 if it is available, and so on, up to product n. And they do not buy anything if only if products 1 to n are not available. The demand for product j, DjH where H stands for Homogeneous preferences, depends on the inventory levels of product 1 to j − 1, and is given by: D1H DjH (q1 , ..., qj−1 )

= D " =

D−

j−1 X i=1

6

#+ qi

for j = 2, ..., n

The one-period (newsvendor) expected profit for the product category given by EΠ(q) =

n X £ j=1

=

n X

uj qj − (uj + oj )E[qj − DjH ]+

"

¤

à j ! !# à j−1 ! Z Pj X X i=1 qi qi − x f (x)dx uj qj − (uj + oj ) qj F qi + P Ã

j=1

j−1 i=1 qi

i=1

(1)

i=1

We assume that no two products have the same overage and underage costs. If it were the case we will show that the retailer would only stock the most preferred one (if at all).

3.1

Two-product problem

The tradeoffs and computational problems in our model can be illustrated using a two-product problem. We show that a retailer may offer more than one product in her assortment because of differences in the price and cost parameters of products. Thus, even in the absence of competitive pressures and customer heterogeneity, cost economics can be a driver of product variety when customers dynamically substitute in the event of a stock-out. Suppose that n = 2. Expected profit is given by: Z q1 H EΠ (q1 , q2 ) = u1 q1 + (u1 + o1 ) (q1 − x) f (x)dx 0 µ ¶ Z q1 +q2 +u2 q2 − (u2 + o2 ) q2 F (q1 ) + (q1 + q2 − x) f (x)dx

(2)

q1

The first derivative of expected profit with respect to q2 is given by: ∂EΠH (q) ∂q2

= u2 − (u2 + o2 )F (q1 + q2 )

Given that EΠH is concave in q2 , the optimal value of q2 , as a function of q1 is equal to ¤+ £ q2 = F −1 (θ2 ) − q1 where θ2 ≡

(3)

u2 u2 +o2 .

Plugging this value back into (2), we obtain: R q1   (u1 − u2 )q1 − [(u1 + o1 ) − (u2 R+ o2 )] 0¡ (q1 − x) f (x)dx ¢ F −1 (θ2 ) EΠH (q1 ) = + u2 F −1 (θ2R) − (u2 + o2 ) 0 F −1 (q2 ) − x f (x)dx if q1 < F −1 (θ2 )(4)  q u1 q1 − (u1 + o1 ) 0 1 (q1 − x) f (x)dx if q1 ≥ F −1 (θ2 ) Note that EΠH (q1 ) is continuously differentiable in q1 . Taking derivative with respect to q1 , we get: ½ ∂EΠH (q1 ) (u1 − u2 ) − [(u1 + o1 ) − (u2 + o2 )]F (q1 ) if q1 < F −1 (θ2 ) = (5) u1 − (u1 + o1 )F (q1 ) if q1 ≥ F −1 (θ2 ) ∂q1 7

Figure 1: Solution to the 2-product problem u1 – u 2

Case 1

Case 2

Stock both

Stock only product 1

q1 = F −1 (θ12 ), q 2 = F −1 (θ 2 )

q1 = F −1 (θ1 ), q 2 = 0 θ1 - θ2 Case 4

Case 3 Stock only product 2

Stock only product 1 or only product 2

q1 = 0, q 2 = F −1 (θ 2 ) or

q1 = 0, q 2 = F −1 (θ 2 )

q1 = F −1 (θ1 ), q 2 = 0

First, note that EΠH is not concave in q1 if u1 +o1 < u2 +o2 . It is noteworthy that the expected profit function is not necessarily concave even for a simplified problem with homogeneous consumers. Secondly, note that the global maximum of EΠH (q1 ) can occur either at q1 = 0 or at a point where the first order condition is satisfied. The two functions in (5) have unique stationary points −u2 1 . It equal, respectively, to F −1 (θ12 ) and F −1 (θ1 ), where θ12 = (u1 +ou11)−(u and θ1 = u1u+o 1 2 +o2 ) H follows that, because EΠ (q1 ) is differentiable everywhere, it can have at most one interior local maximum, either at F −1 (θ12 ) or at F −1 (θ1 ). The optimal solution depends on the signs of u1 − u2 and θ1 − θ2 . Thus, there are four cases as shown in Figure 1. If u1 ≤ u2 , then EΠH (q1 ) is decreasing at zero, and therefore, q1 = 0 is a local maximum. If θ1 ≥ θ2 , then EΠH (q1 ) reaches an interior local maximum at F −1 (θ1 ). In Case 1, EΠH is concave and reaches a unique local maximum at F −1 (θ12 ). Since this value is less than F −1 (θ2 ), (3) implies that one should stock a positive quantity of both products. This is the only case where the optimal solution is obtained by solving the first order conditions. In Case 2, the only local maximum is F −1 (θ1 ) and since this value is greater than F −1 (θ2 ), one should stock only product 1. In Case 3, EΠH is decreasing in q1 so that it is optimal to stock only product 2. Finally in Case 4, EΠH (q1 ) has two local maxima at q1 = 0 and q1 = F −1 (θ1 ). In that case the optimal solution is not completely defined ¡ −1 by ¢the underage and overage costs, as it depends on the relative H H values of EΠ (0) and EΠ F (θ1 ) , which in turn depend on F . It is interesting to note that the optimal solution may involve stocking both products. The following examples illustrate the impact of optimization on the expected profit by comparing the optimal solution with that obtained from static substitution. Example 1. Let r1 = 18, c1 = 9, s1 = 2, p1 = 0 and r2 = 10, c2 = 2, s2 = 1, p − 2 = 0. Assume that demand is normally distributed with mean and variance 50. This implies that u1 = 9, o1 = 7, θ1 = 0.5 and u2 = 8, o2 = 1, θ2 = 0.9. These parameters fall into Case 1 of Figure 1. If one were to use a static substitution based algorithm to determine inventory levels, then one would stock either product 1 or product 2. When only product 1 is stocked, we obtain q1 = 51.11 and EΠH = 405.41. When only product 2 is stocked, we obtain q2 = 58.63 and EΠ = 387.94. In contrast, the optimal solution is q1 = 42.45 and q2 = 16.18, and gives EΠH = 426.78. Thus, the optimal solution obtained from dynamic substitution provides an increase of 5.00% over the static substitution based solution. To understand why it is optimal to stock more than one product even though customers have the same preferences, we introduce the notion of marginal expected profit of each potential unit 8

Table 1: Example 2 (q , q ) µ = 40 µ = 50 ¡ −11 2 ¢ (θ1 ), 0¢ EΠ =69.09 EΠ =89.09 ¡F −1 0, F (θ2 ) EΠ =68.22 EΠ =93.22

of demand. For a given inventory vector q, let ρ(x) be the product that the xth customer would purchase if D ≥ x, we have: ½ Pj−1 P P j such that qi < x ≤ ji=1 qj if x ≤ ni=1 qi i=1 ρ(x) = 0 otw Now let ξ(x) be the marginal expected profit associated with the xth potential unit of demand: ¡ ¢ ½ P uρ(x) − uρ(x) + oρ(x) P [D ≤ x] if x ≤ ni=1 qi ξ(x) = 0 otw P In other words, the marginal profit associated with the xth unit of demand for x ≤ nj=1 qj is equal to uρ(x) if at least x come to the store and −oρ(x) otherwise. It follows that uρ(x) can be seen as the return of the xth potential unit of demand while uρ(x) + oρ(x) can be seen as its risk. In Example 1, product 1 has a higher risk and a higher return than product 2. For the first units of potential demand, P [D ≤ x] is low so that the marginal expected profit is close to uρ(x) , and it is optimal to have ρ(x) = 1. However P [D ≤ x] increases with x so that beyond a threshold value it is optimal to have ρ(x) = 2. The threshold value is x12 such that u1 − (u1 + o1 )P [D ≤ x12 ] = u2 − (u2 + o2 )P [D ≤ x12 ] P [D ≤ x12 ] = θ12 Example 2. Let r1 = 7, c1 = 5, s1 = 4, p1 = 0 and r2 = 11, c2 = 8.5, s2 = 1, p2 = 0. This implies that u1 = 2, o1 = 1, θ1 = 0.66 and u2 = 2.5, o2 = 7.5, θ2 = 0.25. These parameters fall into Case 4 of Figure 1 therefore it is optimal to stock either only product 1 or only product 2. Let demand be Normally distributed with mean µ and standard deviation 10. Table 1 shows the value of expected profit when stocking only product 1 and only product 2 for two different value of the mean µ. So when µ = 40 the optimal solution is to stock only product 1 whereas when µ = 50 one should stock only product 2.

3.2

Dynamic programming formulation

In the n-product problem, if the quantities q1 , ..., qn−2 are fixed, then the problem reduces to a twoproduct problem. However, because of Case 4 in Figure 1 we cannot associate a unique fractile of the distribution F with the optimal quantity of product n−1 and solve backward using substitution. The preference ordering in the homogeneous population model suggests that the n-product assortment planning and inventory management problem can be formulated as a dynamic program. Let hj (qj , Q) be the expected profit from product j when the inventory of product j is qj , and the

9

total inventory of products 1, ..., j − 1 is Q ≡

Pj−1

k=1 qk .

We have Z

hj (qj , Q) = uj qj − (uj + oj )qj F (Q) − (uj + oj )

Q+qj

Q

(Q + qj − x) f (x)dx

= hj (qj + Q, 0) − hj (Q, 0)

(6)

Also let hj (q) ≡ hj (q, 0) be the expected newsvendor profit when q of product j are stocked in a one-product problem. We have Z q hj (q) = uj q − (uj + oj ) (q − x) f (x)dx 0

and h0j (q) ≡ h00j (q) ≡

∂hj (q) = uj − (uj + oj )F (q) ∂q ∂ 2 hj (q) = −(uj + oj )f (q) < 0 ∂q 2

(7) (8)

hj (q) is a strictly concave function of q with a unique maximum at F −1 (θj ) where θj = j = 1, ..., n.

uj uj +oj ,

for

We are also interested in the following quantity: for j, k ∈ {1, ..., n} and j 6= k, Z q hj (q) − hk (q) = (uj − uk )q − [(uj + oj ) − (uk + ok )] (q − x) f (x)dx 0

and h0j (q) − h0k (q) ≡ h00j (q) − h00k (q) ≡

∂[hj (q) − hk (q)] = (uj − uk ) − [(uj + oj ) − (uk + ok )]F (q) ∂q ∂ 2 [hj (q) − hk (q)] = −[(uj + oj ) − (uk + ok )]f (q) ∂q 2

(9) (10)

hj (q) − hk (q) is a concave function if uj + oj ≥ uk + ok and convex otherwise. If uj + oj 6= uk + ok , let θjk ≡

uj − uk (uj + oj ) − (uk + ok )

If θjk ≥ 0, then hj (q) − hk (q) achieves a unique stationary point at F −1 (θjk ). This implies that h0j and h0k intersect at most once. We refer to this as the at most-one-time-crossing property. Using (6), we rewrite (1) as EΠH (q) = h1 (q1 , 0) +

n X

à hj

j=2

= h1 (q1 ) +

n X

"

hj

j=2

Ã

qj ,

! qk

k=1 j X k=1

10

j−1 X

!

qk

− hj

à j−1 X k=1

!# qk

(11)

We now formulate the dynamic program to determine inventory levels as follows. Let Vj (Q) be the maximum expected profit that can be obtained from products j, ..., n given that total inventory for products 1, ..., j − 1 is Q. We have: Vj (Q) = max {hj (qj , Q) + Vj+1 (Q + qj )} qj ≥0

= max {hj (qj + Q) − hj (Q) + Vj+1 (Q + qj )} qj ≥0

= max {hj (q) − hj (Q) + Vj+1 (q)} q≥Q

for j = 1, . . . , n, and Vn+1 (Q) = 0 for all Q. Let Gj (q) = hj (q) + Vj+1 (q),

(12)

and define Gj (Q) = max Gj (q), q≥Q

so that Vj (Q) = max Gj (q) − hj (Q)

(13)

q≥Q

= Gj (Q) − hj (Q). Lemma 1. EΠ(q∗ ) = V1 (0). Proof. (omitted) Solving the dynamic program (13) is not straightforward, because the function Gj (q) is generally not well-behaved. In the two-product problem of §3.1, (3) implies that: ³£ ¤+ ´ V2 (q) = h2 F −1 (θ2 ) − q which is a convex function of q. As a result, G1 (q1 ) = h1 (q1 ) + V2 (q1 ), which is equal to EΠH (q1 ) in (4), is the sum of a concave and a convex function. In Case 4 of Figure 1, this function has multiple local optima. Even so , we are able to solve this dynamic program efficiently by proving some properties of Vj (Q). Proposition 1 shows that the value function has j = 1, ..., n:  j  A1 − hij (Q)   1  j    A2 − hij2 (Q) Vj (Q) = ..   j    AK j − hiK j (Q)   0

the following piecewise structure for each 0 ≤ Q ≤ aj1 aj1 < Q ≤ aj2 (14) ajK j −1 Q>

Gj (Q1 )

From continuity of Gj and above, we have Gj (Q− 1 ) = Gj (Q1 ) and therefore Gj (Q1 ) > Gj (Q+ 1) + however Gj (Q+ 1 ) ≥ Gj (Q1 ) by definition of G, which gives a contradiction because Gj is continuous. Therefore Gj (Q) is continuous in Q and so is Vj (Q) = Gj (Q) − hj (Q).

We see that in going from Vj+1 to Vj , the index of the h function (i.e, ik for some k) at a certain Q can either be replaced with j or remain at its previous value, that is greater than j. We refer to this as the “non-increasing index” property: as we perform the backward induction of the dynamic program, the index of the product that determines the slope of the value function at a given Q can only decrease. 1

These limits exist because the function is monotone.

13

3.3

Convexity of the value function

With the example of a two-product problem, we established that the value function in not concave. In this section, we prove that it is actually convex and decreasing in Q. First we show that there are two types of breakpoints in the value function. We say that ak is a differentiable breakpoint (DBP) of Vj if Vj0 is continuous at ak otherwise we say it is a non-differentiable breakpoint (NDBP). The following Lemma show that for the right derivative at NDBP’s is strictly greater to its left derivative. This lemma is useful in solving the dynamic program because it implies that the NDBP’s of Vj+1 cannot constitute local maxima of Gj and therefore interior local maxima can only be found at stationary points of Gj . Lemma 2. For each ak in (14), 0 + Vj0 (a− k ) ≤ Vj (ak ).

(18)

Proposition 2. Vj (Q) is convex and decreasing in Q.

3.4

Algorithm

From Proposition 1, it follows that at each stage of the dynamic program, we need to keep only the following quantities: the vector of product indices (i1 , ..., iK ), the vector of breakpoints (a1 , ..., aK ), and the first constant A1 because by continuity of Vj , we can derive A2 , ..., AK using the following recursion: Ak = Ak−1 − hik−1 (ak−1 ) + hik (ak−1 )

k = 2, ..., K

Thus, given (i1 , ..., iK ), (a1 , ..., aK ) and A1 , it is possible to compute Vj (Q) for every value of Q. Since by Lemma , the NDBP’s of Vj+1 cannot constitute local maxima of Gj , one needs to consider only stationary points of Gj when looking for local maxima. Because of the piecewise structure of Vj established in Proposition 1, and the fact that hj is a concave function for j = 1, ..., n, there is at most one stationary point between every pair of breakpoints of Vj+1 . Therefore we propose the following algorithm to solve the dynamic program. For use in the algorithm, let iK+1 be a dummy product with overage cost and underage cost equal to zero so that θj,iK+1 = θj . Algorithm 1 For j = 1, ..., n, repeat the following steps, given at step j, (i1 , ..., iK ), (a1 , ..., aK ) and A1 corresponding to Vj+1 , Step 1: For k = 1, ..., K + 1 such that uj + oj > uik + oik , check if ak−1 < F −1 (θj,ik ) ≤ ak . Let y1 ≤ ... ≤ yS be the set of values F −1 (θj,ik ) such that this condition is satisfied. Step 2: Let x1 = arg maxys {Gj (ys ), s = 1, ..., S}. For r = 2, ..., m, let xr = arg maxys {Gj (ys ) : ys > xr−1 } such that xm = yS . Step 3: If Gj (x1 ) < A1 then this is Case 1 of Figure 2. Otherwise it is Case 2. In Case 1, find b1 < x1 such that Gj (b1 ) = Gj (x1 ). In both cases, find xr−1 < br < xr and Gj (br ) = Gj (xr ), for r = 2, ..., m. 14

Step 4: For r = 1, ..., m find cr such that acr < br < acr +1 and kr such that akr −1 < xr < akr . The new vector of breakpoints is : (a1 , ..., ac1 , b1 , x1 , ak1 , ..., ac2 , b2 , ...., xm , akm , ..., aK )

in Case 1, if xm < aK

(a1 , ..., ac1 , b1 , x1 , ak1 , ..., ac2 , b2 , ...., xm )

in Case 1, if xm ≥ aK

(x1 , ak1 , ..., ac2 , b2 , ...., xm , akm , ..., aK )

in Case 2, if xm < aK

(x1 , ak1 , ..., ac2 , b2 , ...., xm )

in Case 2, if xm ≥ aK

The new vector of product indices is: (i1 , ..., ic1 , ic1 +1 , j, ik1 , ..., ic2 , ic2 +1 , ...., j, ikm , ..., iK )

in Case 1, if xm < aK

(i1 , ..., ic1 , ic1 +1 , j, ik1 , ..., ic2 , ic2 +1 , ...., j)

in Case 1, if xm ≥ aK

(j, ik1 , ..., ic2 , ic2 +1 , ...., j, ikm , ..., iK )

in Case 2, if xm < aK

(j, ik1 , ..., ic2 , ic2 +1 , ...., j)

in Case 2, if xm ≥ aK

In Case 1, the value of A1 remains unchanged. In Case 2, the value of A1 is replaced by Gj (x1 ). Proof. Steps 2-4 follow directly from Proposition 1. In what follows we explain why in Step 1, it is enough to look at F −1 (θj,ik ) for k = 1, ..., K + 1 when searching for the local maxima x1 , ..., xm of Gj as defined in (16). As shown in the proof of Lemma 3.4, if x is a local Proposition 1, we have  A1 − hi1 (q) + hj (q)      A2 − hi2 (q) + hj (q) ... Gj (q) =   A − hiK (q) + hj (q)    K hj (q)

maximum of Gj then G0j (x) = 0. By 0 ≤ q ≤ a1 a1 < q ≤ a2 aK−1 < q ≤ aK q > aK

It follows that at any q where Gj is differentiable, G0j (q) = h0j (q) − h0ik (q) for some k. And h0j (q) − h0ik (q) = 0 implies that q = F −1 (θj,ik ). Finally this point is a local maximum if h00j (q) − h00ik (q) ≥ 0, that is if uj + oj > uik + oik . This algorithm yields the value of the optimal expected profit. Now, the optimal stocking quantities qj∗ for j = 1, ..., n can be recovered from (ij1 , ..., ijK j ), (aj1 , ..., ajK j ) for j = 1, ..., n, in the following way: Algorithm 2 • Set Q = 0, j = 1 • while Q ≤ ajK j , – find k such that ajk−1 < Q ≤ ajk 15

– if ijk = j then qj∗ = ajk − Q and Q → Q + qj∗ . Otherwise qj∗ = 0. – j →j+1 • if j ≤ n, qj∗ = ... = qn∗ = 0. ∗ Proof. From (13) we get that the optimal quantity of product j, as function of Q = q1∗ + ... + qj−1 is given by:

qj∗ = arg max Gj (q) − Q q≥Q

(19)

Let ijk be such that Vj (Q) = Aij − hij (Q) for k ∈ {1, ..., K j }. k

k

From (16), we know that maxq≥Q Gj (q) can be either equal to Gj (Q) or Gj (xr ) where xr > Q is a local maximum of Gj . In the first case we have Vj (Q) = Vj+1 (Q), so that ijk 6= j and (19) implies that qj∗ = 0. In the second case we have Vj (Q) = Gj (xr ) − hj (Q), so that ijk = j and (19) implies that qj∗ = xr − Q. Finally by Proposition 1, we have that ajk = xr . Lemma 3. Let J = {j1 , ..., jm } ⊆ {1, ..., n} be the set such that qj∗ > 0 if j ∈ J and qj∗ = 0 otherwise. Assume that j1 < j2 < ... < jm . Then ( ¢ P ¡ q ∗ l = 1, ..., m − 1 F −1 θjl ,jl+1 − l−1 ∗ Pm−1k=1∗ jk qjl = −1 F (θjm ) − k=1 qjk l=m Also for l = 1, ..., m − 1, ujl + ojl

> ujl+1 + ojl+1

ujl

> ujl+1

θjl

< θjl+1

Proof. (omitted) We say that (θj1 ,j2 , ..., θjm−1,m , θjm ) is the set of optimal critical fractiles corresponding to optimal solution q∗ . Recall from §3.1 that ξ(x) denotes the marginal expected profit of the x-th potential unit of demand. Lemma 3 implies that ξ(x) is a continuous decreasing and piecewise convex function, given by:  uj1 − (uj1 + oj1 )F (x) x ≤ qj1     uj2 − (uj2 + oj2 )F (x) qj1 < x ≤ qj1 + qj2  ... ξ(x) = Pm−1 P   ujm − (ujm + ojm )F (x) qj < x ≤ m  j=1 j=1 qj  P  0 x> m q j=1 j

16

P1 6.67 15.16

u o

Table 2: Example 3 P2 P3 P4 0.46 16.06 16.65 0.47 22.68 30.19

P5 14.00 15.08

Figure 3: Example 3: V2 (q1 ) 500 V2 450

400 stock 4&5 350 stock 3&5

300

stock 5

250

stock 2

NDBP

200

150

100

50 NDBP 0

0

5

10

15

20 a1

25 q1

a2

30

35

40 a3

45 a4

Lemma 3 also implies that in the optimal assortment, the preference order matches the risk and return order, i.e., the most preferred product of all customers has the highest values of risk and return, their second choice has the second highest values of risk and return, etc. The following example illustrates the properties of the value function and provides some insights on the optimal solution. Example 3. The following table gives the cost parameters for products 1 to 5: Demand is normally distributed with mean and variance equal to 30 and that pj = 0 for j = 1, ..., 5. The optimal assortment is to stock q1 = q2 = q3 = 0 , q4 = 24.89 and q5 = 13.13 for a total expected profit of 466.85. One might thing that the retailer would stock either the most popular or the most profitable product in the assortment. However in this example, the optimal assortment does not contain P1, which is the most preferred product of all customers. Nor does it contain P3, which is the product which, when stocked exclusively, yields the highest expected profit (439.28 as opposed to 167.48, 13.94, 434.08 408.73 respectively for P1, P2, P4 and P5). The following graph plots the value of V2 as a function of q1 . The value function has four breakpoints, two of which are NDBP’s. Between each pair of breakpoints, the products included in the optimal solution vary. For example, if we decide to stock P1 such that q1 ≤ a1 = 17.35, then it is optimal to stock P4 and P5 along with that quantity. If 17.75 < q1 ≤ a2 = 26.67 then it is optimal to stock P3 and P5 along 17

with that quantity, etc. The optimal quantity of q1 is zero therefore we should stock only products P4 and P5.

3.5

Complexity of the Algorithm

In order to obtain the complexity of Algorithm 1 we need to to establish a bound on K j , the number of breakpoints in the value function. Lemma 4. K j ≤ 2(n − j) + 1 where K j is defined in (14). The intuition behind this result is the following. The indices ij1 , ..., ijK j of the h function in each piece of the value function in (14) belong to the set {j, ..., n}. However two indices can correspond to the same product, that is we can have ijk = ijz for 1 ≤ k < z ≤ K j . In order to obtain a bound on K j , we establish that between two repetitions of the same product in the sequence of indices, there should be at least one product that did not anywhere before in the sequence. This creates a limit on the number of repetitions of one product since there could be at most n − j + 1 different products in the sequence. In the following we assume that a line search is assumed to be O(1). Proposition 3. The complexity of the algorithm is O(n2 ) Proof. Step 1 of Algorithm 1 has to be repeated at most K j number of times. In order to perform Step 2, we suggest the following steps: • Set m = 1, xm = yS and max = Gj (xm ). • For s = S → 1 – if Gj (ys ) > max then m = m + 1, xm = ys and max = Gj (xm ) Finally in Step 3, we assume that finding br takes O(1) time. This implies that all steps can be done in O(n) time. Since all the steps should be repeated n times, we obtain that the complexity is O(n2 ).

3.6

Static versus Dynamic substitution

Because, in the case of more general preference structures, the assumption of dynamic substitution makes the problem intractable, many researchers have assumed static substitution instead, i.e. customers do not substitute in the event of a stockout of their most preferred product. In this section, we compare the expected profit obtained under the assumption of dynamic substitution with that obtained under the assumption of static substitution in the homogeneous population setting. In particular, we measure the percentage increase in expected profit by considering dynamic substitution: g =

EΠH (q∗H ) − EΠSS (q∗SS ) EΠSS (q∗SS ) 18

where ΠSS and q∗SS respectively denote the profit and optimal inventory vector under static substitution. Because all customers have the same preferences and do not substitute in the store, the optimal assortment under static substitution, contains only one product, which is the ¡ one with ¢ the largest expected profit. Let j ∗ be the lowest product index such that j = arg maxj hj F −1 (θj ) , ½ qj∗SS =

F −1 (θj ) 0

if j = j ∗ otw

Lemma 5. When n = 2, g can be made arbitrarily close to 1. Proof. Let n = 2. We construct an example where g → 1. Fix u1 , o1 , u2 such that u1 > u2 then let o2 be such that ¡ ¢ ¡ ¢ h1 F −1 (θ1 ) = h2 F −1 (θ2 ) (20) This implies that θ1 < θ2 and the underage and overage costs fall into Case 1 of Figure 1. Case 1 being the case where it is optimal to stock both products, we get ¡ ¢ ¡ ¢ h1 F −1 (θ12 ) − h2 F −1 (θ12 ) g = ≥0 h1 (F −1 (θ1 )) Now if we let o1 → ∞ and u2 , o2 → 0 such that θ1 → 0 and θ2 → 1 while making sure that condition (20) holds. This implies that: ¡ ¢ ¡ ¢ h1 F −1 (θ12 ) − h2 F −1 (θ12 )

Z =

(u1 − u2 )F

−1

(θ12 ) − [(u1 + o1 ) − (u2 + o2 )] Z

−1

→ u1 F (θ1 ) − (u1 + o1 ) ¡ ¢ = h1 F −1 (θ1 )

F −1 (θ12 )

(q − x) f (x)dx 0

F −1 (θ1 )

(q − x) f (x) 0

So that g → 1. With n products the percentage increase g can be larger than 100%. This shows that assuming static substitution can cause a substantial drop in expected profit because the retailer is not able to take advantage of the differences in return and risk between products.

4

The Nested Preferences model

In the nested preference model, customers can be of the following types: (1), (1, 2), or (1, 2, ..., n).

4.1

Trend-following partitioning

We assume that with probability α1 all customers are of type (1), with probability P α2 they are of type (1, 2),... and with probability αn they are of type (1, 2, ..., n). Assume that nj=1 αj = 1. Let 19

P βj = nk=j αk be the probability that a customer is willing to buy product j, we have therefore β1 = 1 and 1 ≥ β2 ≥ ... ≥ βn . The demand for product j is denoted DjN T , where N T stands for Nested preferences with Trend-following partitioning, and is given by: D1N T DjN T

= D ( h i+ P q D − j−1 i i=1 = 0

w.p. βj

j = 2, ..., n

w.p. (1 − βj )

Let ΠN T (q, u, o) denote profit in this setting under the continuous approximation of demand, for inventory vector q and underage and overage cost vectors u and o respectively. We have: ˜ , o) where u˜j = βj uj − (1 − βj )oj for j = 1, .., n Lemma 6. EΠN T (q, u, o) = EΠH (q, u Proof. EΠN T (q, u, o) =

n X £ ¤ uj qj − (uj + oj )E[qj − Dj ]+ j=1

=

n X

"

" uj qj − (uj + oj )qj

j=1

Z −(uj + oj )βj

à j X i=1 qi

Pj−1

=

qi

i=1

!

Pj

i=1

n X

à j−1 !# X (1 − βj ) + βj F qi #

qi − x f (x)dx

i=1

[(βj uj − (1 − βj )oj )qj

j=1

Ã

à j−1 ! Z Pj à j ! !# X X i=1 qi −βj (uj + oj ) qj F qi + P qi − x f (x)dx j−1 i=1 qi

i=1

i=1

If we set u ˜j = βj uj − (1 − βj )oj , we get " à à j−1 ! Z Pj à j ! !# n X X X i=1 qi EΠN T (q, u, o) = u ˜j qj − (˜ uj + oj ) qj F qi + P qi − x f (x)dx j=1

i=1

j−1 i=1 qi

i=1

H

˜ , o) = EΠ (q, u

4.2

Fixed partitioning

In this case we assume that a fixed proportion α1 of customersPare of type (1), α2 are of type Pn (1, 2),... n and αn are of type (1, 2, ..., n). Again we assume that that j=1 αj = 1 and let βj = k=j αk be the probability that a customer is willing to buy product j. Also let γj = βj /βj−1 for j = 2, ...n be the proportion of customers who are willing to buy product j − 1, who are also willing to buy 20

product j. Note that the preferences can be fully characterized by either one of the three following vectors: (α1 , ..., αn ), (β1 , ..., βn ) or (γ1 , ..., γn ). The demand for product j is denoted DjN F , where N F stands for Nested preferences with Fixed partitioning, and is given by: D1N F

= D

DjN F

= [Dj−1 − qj−1 ]+ γj " #+ j−1 X qi = D− βj βi

j = 2, ..., n

i=1

Let ΠN F (q, u, o) denote profit in this setting under the continuous approximation of demand, for inventory vector q and underage and overage cost vectors u and o respectively. We have: ˜, o ˜ ) where, for j = 1, ..., n Lemma 7. EΠN F (q, u, o) = EΠH (˜ q, u q˜j u ˜j o˜j

q

= βjj = uj βj = oj βj

(21)

Proof. n X £

EΠN F (q, u, o) =

j=1

uj qj − (uj + oj )E[qj − Dj ]+

 !+ + j−1 X q i uj qj − (uj + oj ) qj − x − = βj  f (x)dx β i 0 j=1 i=1 " à ! j−1 n X X qi uj qj − (uj + oj )qj F = βi j=1 i=1  " à ! #+ Z ∞ j−1 X qi −(uj + oj ) P q qj − x − βj f (x)dx j−1 i β i i=1 βi i=1 " à ! j−1 n X X qi = uj qj − (uj + oj )qj F β j=1 i=1 i à ! # # Z Pj qi " j−1 X i=1 βi qi −(uj + oj ) P qj − x − βj f (x)dx j−1 qi β i i=1 n X



Z



¤

∞

Ã

i=1

βi

We do the following transformation of variables in (21), we get: " à à j−1 ! Z Pj à j ! !# n ˜i X X X i=1 q EΠN F (q, u, o) = u ˜j q˜j − (˜ uj + o˜j ) q˜j F q˜i + P q˜i − x f (x)dx j=1

i=1 H

˜, o ˜) = EΠ (˜ q, u

21

j−1 ˜i i=1 q

i=1

4.3

Random partitioning

In this case, the number of customers of each type is a random variable. Let Zj be the random proportion of customers who are willing to buy product j − 1 who are also willing to buy product j and let γj = E[Zj ] for j = 2, ..., n. The demand for product j is denoted DjN R , where N R stands for Nested preferences with Random partitioning. D1N R = D £ NR ¤+ DjN R = Dj−1 − qj−1 Zj " #+ j j−1 X Y qi Zi = D− Qi k=2 Zk i=1 i=2

j = 2, ..., n

Lemma 8. DjN F ≤icx DjN R for j = 1, ..., n Proof. (by induction on j) For j = 1, we have D1N F = D1N R = D. NF ≤ NR Now suppose that Dj−1 icx Dj−1 . By definition of the increasing convex order, this is equivalent to: NR NF E[Dj−1 − qj−1 ]+ ≥ E[Dj−1 − qj−1 ]+ for all qj−1

So we have:

h ¡ ¢+ N R i NR E[DjN R − qj ]+ = E E[ Dj−1 − qj−1 )+ Zj − qj |Dj−1 ] ¤+ £ NR ≥ E (Dj−1 − qj−1 )+ γj − qj ¤+ £ NF ≥ E (Dj−1 − qj−1 )+ γj − qj = E[DjN F − qj ]+

The first inequality is an application of Jensen’s inequality while the second inequality uses Theorem + 3.4.9 of Shaked and Shanthikumar (1994) and the fact that [(x − qj−1 )+ αj − qj ] is an increasing convex function of x. This result implies that E[DjN F ] ≤ E[DjN R ] for j = 1, ..., n. However for j = 2, we have the following stronger result. Lemma 9. If n = 2, E[ΠN F (q)] ≥ E[ΠN R (q)] for every vector q Proof. We have E[D2N R ] = E[(D − q1 )+ Z2 ] = E[D − q1 ]+ α2 = E[D2N F ]

(22)

so that D2N F ≤cx D2N R . This in turn implies that E[q2 − D2N F ]+ ≤ E[q2 − D2N R ]+ and therefore E[ΠN F (q)] ≥ E[ΠN R (q)] for every vector q. For n > 2, we obtain the same result given a certain condition on the price and cost parameters of the products that are stocked in a positive quantity in q. 22

Proposition 4. For a given vector q, let J = {j1 , ..., jm } ⊆ {1, ..., n} be the set such that qj > 0 if j ∈ J and qj = 0 otherwise. Assume that j1 < j2 < ... < jm . If for l = 1, ..., m − 1, ujl + ojl

>

βjl+1 (ujl+1 + ojl+1 ) βjl

(23)

then EΠN F (q) ≥ EΠN R (q) Proof. We can write: EΠN R (q) = =

n X £ j=1 m X

£

uj qj − (uj + oj )E[qj − DjN R ]+

¤

ujl qjl − (ujl + ojl )E[qjl − DjNl R ]+

¤

l=1

=

m X £

¡ ¢ ¤ (ujl + ojl ) E[DjNl R ] − E[DjNl R − qjl ]+ − ojl qjl

l=1

¡ ¢ = (uj1 + oj1 ) E[D] − E[D − qj1 ]+ µ ¶¸ X m · m X NR + βjl NR + (ujl + ojl ) E[Djl−1 − qjl−1 ] + ojl qjl − E[Djl − qjl ] − βjl−1 l=2 l=1 ¸ n · X βjl R = (uj1 + oj1 )E[D] − (ujl−1 + ojl−1 ) − (uj + ojl ) E[DjNl−1 − qjl−1 ]+ βjl−1 l j=2

−(ujm +

ojm )E[DjNmR

+

− qjm ] −

m X

ojl qjl

l=1

Since we have E[DjN R − qj ]+ ≥ E[DjN F − qj ]+ for all j and q, we get the desired result if (ujl−1 + ojl−1 ) −

βjl βjl−1 (ujl

+ ojl ) ≥ 0 for l = 1, ..., m − 1.

Corollary 1. The optimal inventory vector q ∗N F satisfies condition (23) ˜, o ˜ ). Proof. By Lemma 7, EΠN F (q, u, o) = EΠH (˜ q, u By Lemma 3, the optimal inventory vector for the second expression satisfies the condition u ˜jl + o˜jl > (˜ ujl+1 + o˜jl+1 ) for jl ∈ J and J defined as in Proposition 5. Given the transformation given by 21, this is equivalent to condition (23). Therefore condition (23) is naturally satisfied for the optimal inventory vector under fixed proportion. In all the numerical examples we considered, we found that the condition is also satisfied by q∗N R , as estimated by the Sample Path Gradient Algorithm2 . This allows us to obtain an upper bound on the optimal expected profit under random proportion, as shown in the following proposition. 2

Note that in §, we can guarantee that the condition is satisfied for every numerical example we consider.

23

Figure 4: Example of Outtree α4 α2

4

2 α5

1

5 α3 3

α6

6

Proposition 5. If q∗N R satisfies equation (23), then EΠN F (q∗N F ) ≥ EΠN R (q∗N R ) Proof. (omitted) We propose using q∗N F as a heuristic for the NR setting. We can estimate the performance of that solution, that is EΠN R (q∗N F ) by comparing it to the upper bound EΠN F (q∗N F ). Numerical results are presented in §6.

5

Outtree-shaped preferences

In this case, the preference structure can be represented by an outtree where the nodes correspond to the products. By definition of an outtree, there is a single initial note representing the first choice product for all consumer types and there is a unique directed path from the initial node to any other node. Each such path corresponds to a consumer type. The following figure shows an example of such a tree: Let the initial node of the tree (the one with no predecessor) correspond to product 1, without loss of generality. For node j, let Sj be the sets of direct successors of node j. By definition of an outtree, nodes have only one immediate predecessor, let p(j) be the predecessor of node j. Finally let Pj be the set of (non-immediate) predecessors of node j.

5.1

Fixed partitioning

For j > 1, let γj be defined as the fixed proportion of customers wanting to buy product p(j) who are also willing to buy product j. We have: X γk ≤ 1, j = 2, ..., n k∈Sj

and without loss of generality we set γ1Q = 1. In the above example we have γ2 + γ3 ≤ 1, γ4 + γ5 ≤ 1 and γ6 ≤ 1. Also let us define βj = i∈Pj ∪{j} γi as the total proportion of customers that are willing to buy product j.

24

The demand for product j is denoted DjOF , where OF stands for Outtree-shaped preferences with Fixed partitioning, and is given by: D1OF DjOF

= D h i+ OF = Dp(j) − qp(j) γj  + X qi  βj = D − βi

j = 2, ..., n

i∈Pj

We carry out the following transformation, similar to that of §4.2,

u ˜j

qj βj = uj βj

o˜j

= oj βj

q˜j

=

Let ΠOF denote the profit in this setting. Expected profit is given by        Z P n q ˜ +˜ q i j X X X i∈Pj u  EΠOF (q) = ˜j q˜j − (˜ uj + o˜j ) q˜j F  q˜i + q˜j − x f (x)dx q˜i  + P j=1

i∈Pj

i∈Pj

q˜i

i∈Pj

To solve this problem, we use the following dynamic programming formulation. Let Vj (Q) be the maximum expected profit obtained from products in Sj ∪ {j} given that total inventory for products in Pj is Q. We have:     X Vk (Q + qj ) Vj (Q) = max hj (qj , Q) +  qj ≥0  k∈Sj

= max Gj (q) − hj (Q) q≥Q

where Gj (q) = hj (q) +

X

Vk (q)

k∈Sj

and we have EΠOF (q) = V1 (0). In this case too, we show that Vj has a piecewise structure given by equation (14). The only difference is that the indices ik do not necessarily belong to the set {j, ..., n} but rather to a larger set that we denote Cj and define as the set of composite products. And we define “j1 + ... + jm ” as composite product with overage cost uj1 + ... + ujm and overage cost oj1 + ... + ojm where j1 6= j2 ... 6= jm and ji ∈ {1, ..., n} for i = 1, ..., m. We have: Z q h“j1 +...+jm ” (q) = (uj1 + ... + ujm )q − (uj1 + ... + ujm + oj1 + ... + ujm ) (q − x) f (x) 0

= hj1 (q) + ... + hjm (q) By (7) and (8), this is a concave function with a maximum at θ“j1 +...+jm ” = 25

uj1 +...+ujm uj1 +...+ujm +oj1 +...+ujm .

And we have: : Cj

= {”j1 + ... + jm ” : ji ∈ {j, ..., n} for i = 1, ..., m, there is not path connecting any two ji } The equivalent to Proposition 1 of §3 is the following:

Proposition 6. For each j = 1, ..., n, (1) Vj (Q) has the piecewise structure given by (14), (2) Vj (Q) is continuous in Q, Proof. The proof of continuity is similar to that of Proposition 1. Following the same arguments as in Proposition 1, we obtain (in Case 1):  P V (Q) if 0 ≤ Q ≤ b1   Pl∈Sj l   V (x ) + h (x ) − h (Q) if b1 < Q ≤ x1  j 1 j l∈Sj l 1    P if x1 < Q ≤ b2 l∈Sj Vl (Q) Vj (Q) =  ...   P   if bm < Q ≤ xm  l∈S Vl (xm ) + hj (xm ) − hj (Q)   P j if xm < Q l∈Sj Vl (Q) P For r = 1, .., m, Cr = k∈Sj Vk (xr ) + hj (xr ) does not depend on Q therefore we can write Vj (Q) = Cr − hj (Q) for Q ∈ (br , xr ]. By induction hypothesis, we have Vl (Q) = Alk − hil (Q), l ∈ Sj k

where ilk ∈ Ck . Therefore: X

Vl (Q) =

l∈Sj

X l∈Sj

Alk −

X l∈Sj

hil (Q) k

= Ak − hik (Q) where ik ∈ Cj and Ak =

P l∈Sj

Alk . Therefore the result holds for Vj (Q).

Algorithm 1 still holds, with the following extra step. Algorithm 3 Given (il1 , ..., ilK l ), (al1 , ..., alK l ) and Al1 corresponding to Vl for every l ∈ Sj . • Step 0: Sort the breakpoints al1 , ..., alK l for l ∈ Sj in increasing order and rename them P (a1 , ..., aK ) where K = l∈Sj K l . Construct the corresponding vector (i1 , ..., iK ) such that ik is a composite product involving product ilr if and only if alr−1 < ak ≤ alr for r = 1, ..., K l and l ∈ Sj . • Steps 1 to 4: See Algorithm 1. It is easy to see that the value function is also convex in this case and that the complexity of Algorithm 3 is also O(n2 ). 26

6

Numerical Results

In this section, we propose two heuristics to solve the assortment planning problem under nested preferences with random proportion, i.e., the problem formulated in §4.3. We numerically evaluate the performance of these heuristics with respect to the upper bound established in Proposition 5. We also benchmark their performance against two previously known heuristics, one based on static substitution, and the other being the Sample Path Gradient Algorithm of Mahajan and van Ryzin (2001). We call our first heuristic as the fixed proportion heuristic (NF). In this heuristic, we solve the assortment planning problem under nested preferences using the same parameters but with fixed proportion, and use the corresponding optimal inventory levels in the assortment planning problem with random proportion. Let q∗N F denote the vector of inventory levels in this solution. We call our second heuristic as the modified fixed proportion heuristic (MNF). Again consider the optimal solution to the assortment planning problem under fixed proportion. Let J = {j1 , ..., jm } ⊆ {1, ..., n} be the set of all products such that qj∗N F > 0, and let (θj1 ,j2 , ..., θjm−1 ,jm , θjm ) be the set of optimal critical fractiles as defined after Lemma 3. In the MNF heuristic, we use the same critical fractiles as in the NF heuristic, but we obtain inventory levels, q∗M N F , from the true distribution of demand based on random proportions. For ease of computation, we estimate the true distribution of demand by simulation. This estimation is done sequentially for all products in the order of the nested preferences. Let Fejl denote the estimated cdf of demand for product jl given inventory levels N F , . . . , q ∗M N F for products j , . . . , j qj∗M 1 l−1 . Then, we have jl−1 1 ½ NF qj∗M l

=

¡ ¢ θjl ,jl+1 l = 1, ..., m − 1, Fj−1 l Fj−1 (θjm ) l = m. m

Under static substitution, the optimal solution is to carry only one product in the assortment, i.e. the one that yields the highest expected profit when stocked alone. This provides the solution under the static substitution heuristic (SS). This solution is denoted as q∗SS . Finally, the solution under the Sample Path Gradient Heuristic (SPGA) of Mahajan and van Ryzin (2001) is denoted as q∗SP . We vary the cost and price parameters of the products, the mean demand, and the proportions of customers who are willing to buy each product in order to evaluate the performance of the heuristics in different cases. The parameters in all problem instances are such that the following conditions are satisfied: uj + oj

≥ γj+1 (uj+1 + oj+1 ) for j = 1, ..., n − 1

This implies that condition (23) is satisfied for every inventory vector, and therefore, by Proposition 5, EΠN F (q∗N F ) constitutes an upper bound on the performance of the four heuristics. In addition, we choose all parameters in such a way that all n products are stocked in the optimal solution in the NF heuristic. Our numerical study is based on a potential assortment of five products. We generate demand D for the product category using a Poisson random variable with mean λ. For each customer, we generate the customer type using a multinomial distribution with parameters α1 , . . . , αn , where αj is the probability that the customer is willing to buy products 1 to j. 27

Table 3: Underage and P1 P2 u 15 14 o 10.15 6.15

overage cost parameters P3 P4 P5 13 12 11 4.29 3.29 2.75

Given the inventory levels from the four heuristics, we use the distribution of demand to estimate the expected profits under all four heuristics as well as under the upper bound. This estimation is done by simulation using a common set of 10,000 sample paths of random numbers in all cases. Let EΠN R (q) denote the expected profit for the assortment planning problem under random proportion for an inventory vector q. The optimality gaps (OG) of the four heuristics with respect to the upper bound are computed as EΠN F (q∗N F ) − EΠN R (q) , EΠN F (q∗N F ) where q is set equal to q∗N F , q∗M N F , q∗SS , and q∗SP . The inventory levels in the four heuristics are computed as follows. Since the NF heuristic requires a continuous distribution, we use the normal approximation of the Poisson distribution to compute q∗N F . We also use this same approximation in the SS heuristic to compute q∗SS . For the MNF heuristic, we use the Poisson and multinomial distributions to compute q∗M N F given (θj1 ,j2 , ..., θjm−1 ,jm , θjm ) obtained from the NF heuristic. Finally, for SPGA, we use the Poisson and multinomial distributions to compute q∗SP . Since SPGA is a simulation-based algorithm, we use the following parameters for its implementation: number of iterations=10,000, starting inventory vector, qj0 = λβj /n, and step size = 1/ iteration index. The step size values chosen by us are the same as used by Mahajan and van Ryzin (2001). We note that the inventory levels are discrete for the MNF heuristic, but are real numbers for all other heuristics. For our purposes, we do not round the inventory levels to the nearest integers. We present the results of the numerical study by grouping the problem instances into three scenarios, based on which variable is varied while everything else is kept constant. In Scenario 1, we study the impact of the βj parameters on the performance of the heuristics. In Scenario 2, we study the impact of the amount of safety stock. Finally, in Scenario 3, we study the impact of mean demand. In Scenario 1, we fix the underage and overage parameters as shown in Table 3 and set mean demand λ to 30. The corresponding optimal critical fractiles are: (θ1,2 , ..., θ4,5 , θ5 ) = (0.2, 0.35, Pn 0.5, 0.65, 0.8). We generate 11 problem instances by varyingPthe proportions βj in such a way that j=1 βj den creases from 5 to 3 in steps of 0.2. When j=1 βj = 5 all customers are willing to buy all 5 products, i.e. this is the homogeneous population case. As the value of this sum decreases, the proportion of customers willing to buy the least preferred products decreases and eventually when the value reaches 3, all customers are equally likely of being of type (1), (1,2), (1,2,3), (1,2,3,4) or (1,2,3,4,5). Table 4 reports the value of the optimal quantity under Pn all 4 heuristics as well as their optimality gaps with respect to the upper bound. We see that as j=1 βj increases, the optimality gap of the NF heuristic decreases from 0.14% to 0%, and that of the MNF heuristic decreases from 0.12% to 0%. The MNF heuristic performs better than the NF heuristic in all cases. The optimality gaps 28

of the SS and SPGA heuristics are both much larger than those of the NF and MNF heuristics. Further, the optimality P gap of the SS heuristic behaves differently from the NF and MNF heuristics because it decreases as nj=1 βj decreases; however, it remains larger in all cases. In Scenario 2, we keep the same values for the underage costs of each product (see Table 3), but vary the overage cost vector so as to have the optimal critical fractiles be equidistant with θ12 = 0.1 and θ5 which varies between 0.2 and 0.9. This has the effect of varying the total amount of safety stock. We also set mean demand λ to 50 and αj = 0.7 for j = 2, ..., 5. In Table 5, we see that the performance of the SS heuristic deteriorates as the amount of safety stock increases while the optimality gap of the other heuristics are fairly constant. We also see that the MNF heuristic significantly improves on the NF heuristics and that both always do better than the SPGA. Note that as the amount of safety stock increases, the SS heuristic increases the inventory level of the first product only, while the remaining three heuristics increase inventory levels of all products, with total inventories of products 2-5 increasing more than the inventory of product 1. Thus, the SS heuristic is unable to exploit the differences between critical fractiles of products, which the other heuristics are able to. In Scenario 3, we use the same cost parameters as in Table 3 and set αj = 1/5 for j = 1, .., n so that the customers are equally likely to be of each possible type. We vary mean demand λ between 30 and 110 by steps of 20. In Table 6, we see that as mean demand increases, the optimality gaps of the NF, MNF and SS heuristics decrease while that of the SPGA increases. We also see that the NF and MNF heuristics perform very well, with an average optimality gap of 0.18% and 0.22% respectively. The poor performance of the SPGA indicates that the algorithm may need more iterations to converge for a larger mean demand than for a smaller mean demand. In total (including the 3 scenarios) we generated 230 problem instances. The average optimality gaps of the heuristics were equal to 0.18% and 0.14% respectively for the NF and MNF heurtics, compared to 0.54% and 1.57% for the SPGA and SS heuristics. In conclusion, we see that assuming static substitution can lead to a substantial loss in expected profit when customers actually dynamically substitute, in particular, when a large amount of safety stock is required, and when the total proportion of customers buying each product is high. We also see that the NF heuristic performs well, especially when mean demand and the proportion of customers willing to substitute are high. The heuristic often does better than the SPGA, especially when mean demand is large. Moreover, it significantly reduces the computing time because it is not simulation-based but instead uses an efficient DP algorithm. Finally, in our results, the MNF heuristic performs better than the NF heuristic in all cases. This may not be true in general, and a decision-maker may consider computing both heuristics and taking the higher value between them. Note that the MNF heuristic is slightly more computationally intensive because one needs to compute the true distribution of demand under random proportion. It remains, however, significantly faster than the SPGA, given the parameters that we chose. While our heuristics perform better than the SPGA for the chosen preference structure, it should be noted that SPGA is a very general algorithm which can handle any type of preference structure whereas the NF and MPF heuristics can only deal with nested (and outtree-like) preferences. Moreover it offers the guarantee of converging to a stationary point of the expected profit function, while our heuristics do not. In contrast, our NF heuristic presents the advantage of being optimal 29

in one setting, namely, the homogeneous population case.

7

Conclusion

We have obtained the optimal assortment and inventory levels under dynamic substitution when customers have homogeneous, nested and outtree-shaped preferences with trend-following or fixed partitioning of demand. The dynamic program algorithm used to compute the solution involves the maximization of the sum of a concave and a convex function which may lead to points of non-differentiability in the value function. Despite this fact, we were able to bound the number of breakpoints in the value function and prove that the complexity of the algorithm is O(n2 ). Under nested preferences with random partitioning of demand, we show that the algorithm provides a good heuristic which gives results similar or better than those obtained with the SPGA, and significantly reduces computation time. The key managerial insights of our paper are as follows. First, we showed that ignoring dynamic substitution can lead to a substantial loss in expected profit because the retailer does not take advantage of the differences in return and risk among products. Second, we showed that inventory cost economics should be considered, along with customer heterogeneity and competition, as a driver of product variety. This insight is relevant for applications in product design and pricing problems under competition. Finally, we showed that contrary to previous research under static substitution, it is not necessarily optimal to stock the most preferred or the most profitable product in the assortment. The next steps of this research consist in studying the case of acyclic and general preferences using an approach similar to the one we used for nested and outtree-shaped preferences. Our results are derived under the assumption that the prices of the products are exogenous. An interesting extension of our work would be to consider prices are decision variables which would influence not only profit but also the proportions of customers willing to buy each product.

References Anupindi, R., M. Dada, S. Gupta. 1998. Estimation of Consumer Demand with Stockout based Substitution: An Application to Vending Machine Products. Marketing Sci. 17 406-423. Alptekinoglu, A. 2004. Mass Customization vs. Mass Production: Variety and Price Competition. M&SOM 6(1) 98-103. Alptekinoglu, A., C.J. Corbett 2005. Mass Customization versus Mass Production: Variety and Price Competition. Working paper. Anderson School of Management. Aydin, G., J.K. Ryan. 2000. Product Line Selection and Pricing under the Multinomial Logit Choice model. Working paper. Stanford University. Baumol, W. J., E. A. Ide. 1956. Variety in Retailing. Management Sci. 3(1) 93-101. Bassok, Y., R. Anupindi, R. Akella. 1999. Single-Period Multiproduct Inventory Models with Substitution. Oper. Res. 47 632-642. 30

Cachon, G., C. Terwiesch, Y. Xu. 2004. Retail Assortment Planning in the Presence of Consumer Search, Working paper. The Wharton School, University of Pennsylvania. Chen, F., J. Eliashberg, P. Zipkin. 1998. Customer Preferences, Supply-Chain Costs, and ProductLine Design. In Product Variety Management: Research Advances, T.-H. Ho and C. S. Tang (Eds.), Kluwer Academic Publishers. de Groote, X. 1994. Flexibility and Marketing/Manufacturing Coordination. International Journal of Production Economics 36 153-167. Gaur, V., D. Honhon 2005. Product Variety and Inventory Decisions Under a Locational Consumer Choice Model, Working paper. Stern School of Business, New York University. Ho, T-H., C. Tang. 1998. Research Advances in Product Variety Management, Kluwer Academic Publishers. Hopp, W.J., X. Xu. 2005. Product Line Selection and Pricing with Modularity in Design, Manufacturing and Service Operations Management 7(3) 172-187. Kok, A.G., M.L. Fisher. 2004. Demand Estimation and Assortment Optimization under Substitution: Methodology and Application. Working Paper. Fuqua School of Business, Duke University. Lancaster, K. 1998. Markets and Product Variety Management, Research Advances in Product Variety Management, Kluwer Academic Publishers. Lippman, S. A., K. F. McCardle. 1997. The Competitive Newsboy. Oper. Res. 45 54-65. Mahajan, S., G. van Ryzin. 2001. Stock Retail Assortment under Dynamic Consumer Substitution. Oper. Res., 49 334-351. Netessine, S., N. Rudi. 2003. Centralized and competitive inventory models with demand substitution. Oper. Res. 51(2) 329-335. Rajaram, K. 2001. Assortment Planning in Fashion Retailing: Methodology, Application and Analysis. European Journal of Operations Research 129 186-208. Rajaram, K., C. S. Tang (2001) The Impact of Product Substitution on Retail Merchandising, European Journal of Operations Research, Vol. 135, pp 582-620. Shaked, M., J. G. Shanthikumar. 1994. Stochastic Orders and Their Applications. Academic Press, San Diego, CA. Smith, S.A, N. Agrawal. 2000. Management of Multi-Item Retail Inventory Systems with Demand Substitution. Oper. Res., 48 50-64. van Ryzin, G., S. Mahajan. 1999. On the Relationship between Inventory Costs and Variety Benefits in Retail Assortments. Management Sci. 45 1496-1509.

31

8

Appendix

Proof of Lemma 3.4 Proof. (by induction) For Vn , we see in (15) that: ¡ ¢ ¡ ¢ ¡ ¢ Vn0 F −1 (θn )− = h0n F −1 (θn ) = 0 = Vn0 F −1 (θn )+ Assume that the result is true for Vj+1 . Since hj (q) is differentiable everywhere, the points where Gj (q) = Vj+1 (q) + hj (q) is not differentiable are breakpoints as of Vj+1 such that 0 0 Vj+1 (as −) < Vj+1 (a+ s)

Equivalently, using (12), 0 + G0j (a− s ) < Gj (as ).

These points as cannot be local maxima of Gj because the right derivative is greater than the left derivative. It follows that if x is a local maximum of Gj then G0j (x) = 0. Therefore in (16), G0j (xr ) = 0 for r = 1, ..., m, and 0 + Gj0 (x− r ) = Gj (xr ) = 0.

which implies that 0 + Vj0 (x− r ) = Vj (xr )

As shown in Proposition 1, Gj is decreasing at br , r = 1, ..., m and Gj is constant in the segments (br , xr ), r = 1, ..., m, therefore we have 0 + Gj0 (b− r ) < Gj (br ) = 0

which implies that 0 + Vj0 (b− r ) < Vj (br )

As discussed in Proposition 1, the points xk and bk for k = 1, ..., m are the breakpoints of Vj that are not breakpoints of Vj+1 therefore the result is true for Vj . Proof of Proposition 2 Proof. From (14) we see that Vj is convex between breakpoints as hj is concave for every j. Let ak1 and ak2 be the first two NDBP of Vj (if there are less than two than the proof is simpler). By convexity of Vj and continuity of its derivative in [0, ak1 ), Vj (q) ≥ Vj (ak1 ) + (q − ak1 )Vj0 (a− k1 ) 32

for q ∈ [0, ak1 )

(24)

Similarly we have, Vj (q) ≥ Vj (ak1 ) + (q − ak1 )Vj0 (a+ k1 )

for q ∈ (ak1 , ak2 )

(25)

By Lemma 10 along with (24) and (25) implies that: Vj (q) ≥ Vj (ak1 ) + (q − ak1 )Vj0 (a− k1 )

for q ∈ [0, ak2 )

Thus by induction, all subgradients of Vj lie below Vj and therefore the curve is strictly convex in [0, aK ]. By Lemma (3.4), at the last breakpoint aK we have: 0 + Vj0 (a− K ) ≤ Vj (aK ) = 0

Since Vj is strictly convex in [0, aK ], Vj0 is negative for Q ≤ aK so that Vj is decreasing. Proof of Lemma 4 We first need to establish the following Lemma: Lemma 10. For Vj given by (14), h0ik (q) ≤ −Vj0 (q − ) for q ≤ ak , k = 1, ..., K Proof. By Lemma 3.4, we have that 0 − h0ik (ak−1 ) = −Vj0 (a− k−1 ) ≤ −Vj (ak−1 )

(26)

Assume that we have h0ik (q) ≤ −Vj0 (q − ) for q ∈ (as , ak ] but (contradiction) that h0ik (q) > −Vj0 (q − ) for some q ∈ (as−1 , as ]. This, along with (26), imply that there exists as−1 < q1 ≤ as , such that h0ik (q1 ) = −Vj0 (q1 ).3 Since −Vj0 (q1 ) = h0is (q1 ) and h0ik and h0is can only cross once by the at-most-one-time-crossing property, we have: h0ik (q) > h0is (q)

for q < q1

h0ik (q)

for q > q1


−Vj (as ). So that − ≤ −Vj (as ) and therefore there exists q1 defined as above.

h0ik (as )

33

-V’j h’ik

F(q1)

h’is

F(as-1)

F(as)

F(ak-1)

F(ak)

F(q)

Proof. We say that the index k is a “repetition” if in (14) there exists is = ik and s < k. Otherwise we say that k is an “original”. Let k be a repetition and let z be the unique index such that iz = ik , z < k and il 6= ik for l = z + 1, ..., k − 1. Claim: z + 1 is an original. We must have z < k − 1 as there should be at least one other product between two occurrences of the same product. By Lemma 10, we have that h0iz+1 (q) < h0iz (q) = −Vj0 (q) for q ∈ (az−1 , az ) h0ik (q) < h0iz+1 (q) = −Vj0 (q) for q ∈ (az , az+1 ) Since iz = ik , this implies that h0iz (az ) = h0ik (az ) = h0iz +1 (az ). The at-most-one-time-crossing property implies that: h0iz+a (q) < h0iz (q) for q < az We cannot have ir = iz+1 with r < z because this would imply that h0ir (q) = h0iz+1 = −Vj0 (q) < h0iz (q) for q ∈ (ar−1 , ar ) and this contradicts Lemma 10. This proves that z + 1 is an original. We say that original z + 1 is associated to repetition k. Let x be the number of originals. Let y be the number of repetitions. The number of breakpoints in Vj is given by x + y. To find the maximum number of breakpoints we solve the following optimization problem: max

x+y

s.t.

x≤n−j+1 y ≤x−1

The first constraint comes from the fact that only products {j, ..., n} can appear in Vj . The second constraint comes from the above claim as for each repetition, there must exist an original associated with it and this original is located on the right of the previous occurrence of that product.This implies that each original can be associated to at most one repetition, except for the first index which cannot be associated with any repetition. 34

The number of breakpoints of Vj is maximized when x = n − j + 1 and y + 1 = x such that x + y = 2(n − j) + 1.

35

β

NF

(1,1,1,1) (1,1,1,1,0.8) (1,1,1,0.8,0.8) (1,1,0.8,0.8,0.8) (1,.8,0.8,0.8,0.8) (1,0.8,0.8,0.8,0.6) (1,0.8,0.8,0.6,0.6) (1,0.8,0.6,0.6,0.6) (1,0.8,0.6,0.6,0.4) (1,0.8,0.6,0.4,0.4) (1,0.8,0.6,0.4,0.2) mean

MNF

q (44.0,3.2,2.7,2.7,3.2) (44.0,3.2,2.7,4.7,1.0) (44.0,3.2,5.8,0.0,2.3) (44.0,6.9,0.0,1.4,2.6) (48.3,0.0,1.3,2.2,2.6) (48.3,0.0,1.3,3.9,0.7) (48.3,0.0,4.0,0.0,1.6) (48.6,2.2,0.0,0.8,1.9) (48.6,2.2,0.0,2.2,0.4) (48.6,2.4,1.2,0.0,0.9) (48.6,2.4,1.3,0.6,0.1)

OG (%) 0.00 0.01 0.03 0.03 0.07 0.08 0.10 0.11 0.13 0.13 0.14 0.08

q (44,3,3,3,3) (44,3,3,5,1) (44,3,6,0,2) (44,7,0,1,3) (48,0,2,2,2) (48,0,2,3,1) (48,0,4,0,2) (49,2,0,1,2) (49,2,0,2,0) (49,0,1,0,1) (49,2,1,1,0)

SS OG (%) 0.00 0.00 0.02 0.03 0.07 0.08 0.10 0.10 0.10 0.12 0.12 0.07

q (51.7,0,0,0,0) (51.7,0,0,0,0) (51.7,0,0,0,0) (51.7,0,0,0,0) (51.7,0,0,0,0) (51.7,0,0,0,0) (51.7,0,0,0,0) (51.7,0,0,0,0) (51.7,0,0,0,0) (51.7,0,0,0,0) (51.7,0,0,0,0)

SP OG (%) 2.13 2.10 1.99 1.74 1.23 1.20 1.09 0.87 0.85 0.77 0.76 1.34

q (25.2,14.5,8.4,4.0.2.6) (25.5,16.7,8.7,5.0,0.6) (26.5,17.8,10.7,0.0,0.9) (30.4,21.6,0.0,0.9,2.0) (40.8,7.3,2.8,0.8,1.3) (40.9,7.4,3.1,1.5,0.0) (39.8,8.2,4.9,0.0,0.0) (42.1,10.0,0.0,0.0,0.8) (41.9,10.1,0.0,0.0,0.0) (40.7,11.3,.0.0,0.0,0.0) (40.7,11.3,.0.0,0.0,0.0)

OG (%) 2.46 2.40 2.13 1.47 2.16 2.15 2.68 1.78 1.87 2.34 2.32 2.16

Table 4 – Scenario 1 θn 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 mean

NF q (43,0.3,0.1,0.1,0.0) (44.1,0.6,0.3,0.1,0.1) (44.9,1.0,0.4,0.2,0.1) (45.5,1.5,0.6,0.3,0.1) (46.0,1.9,0.7,0.3,0.2) (46.3,2.4,0.9,0.4,0.2) (46.6,3.0,1.1,0.6,0.3) (46.9,3.5,1.4,0.8,0.5)

OG (%) 0.25 0.27 0.27 0.24 0.24 0.23 0.21 0.18 0.24

MNF OG (%) q (43,0,0,0,0) 0.03 (44,1,0,0,0) 0.10 (45,1,0,0,0) 0.15 (45,2,0,0,0) 0.21 (46,2,0,0,0) 0.30 (46,2,1,0,0) 0.29 (47,3,1,0,0) 0.21 (47,3,2,0,1) 0.18 0.18

SS q (43.4,0,0,0,0) (45.0,0,0,0,0) (46.3,0,0,0,0) (47.3,0,0,0,0) (48.2,0,0,0,0) (48.9,0,0,0,0) (49.7,0,0,0,0) (50.3,0,0,0,0)

Table 5- Scenario 2

OG (%) 0.08 0.14 0.31 0.52 0.81 1.11 1.50 1.95 0.80

SP q (40.0,2.8,0.0,0.0,0.0) (42.1,2.6,0.0,0.0,0.0) (43.6,2.4,0.0,0.0,0.0) (43.3,3.7,0.0,0.0,0.0) (43.0,5.1,0.0,0.0,0.0) (44.4,5.2,0.1,0.1,0.0) (45.3,5.1,0.2,0.4,0.0) (45.2,5.9,1.0,0.0,1.0)

OG (%) 1.07 0.58 0.32 0.55 0.83 0.56 0.38 0.41 0.59

λ 10 30 50 70 90 110 130 mean

NF q (9.4,1.1,0.6,0.3,0.1) (28.9,1.9,1.0,0.5,0.1) (48.6,2.4,103,0.6,0.1) (68.3,2.9,1.5,0.7,0.2) (88.1,3.3,1.7,0.8,0.2) (107.9,3.6,1.9,0.9,0.2) (127.7,3.9,2.1,1.0,0.2)

OG (%) 0.84 0.27 0.15 0.11 0.08 0.07 0.04 0.22

MNF OG (%) q (9,1,1,0,0) 0.67 (29,2,1,0,0) 0.20 (49,2,1,1,0) 0.13 (68,3,1,1,0) 0.10 (88,3,2,1,0) 0.07 (108,3,2,1,0) 0.07 (128,4,2,1,0) 0.03 0.18

SS q (10.8,0,0,0,0) (31.3,0,0,0,0) (51.7,0,0,0,0) (72.0,0,0,0,0) (92.3,0,0,0,0) (112.6,0,0,0,0) (132.8,0,0,0,0)

Table 6- Scenario 3

OG (%) 1.89 1.02 0.74 0.61 0.57 0.49 0.45 0.82

SP q (9.3,1.4,0.5,0.0,0) (27.9,3.8,0.4,0.0,0) (39.8,12.2,0.0,0.0,0) (51.1,19.1,0.8,0,0) (60.4,28.3,1.3,0,0) (36.9,36.0,6.1,0,0) (69.7,42.5,15.5,0,2.0)

OG (%) 0.76 0.39 2.88 4.66 6.56 8.81 11.47 5.08