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 in-ventory 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 defi-nition 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 optimal-ity. 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.
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 heteroge-neous 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 pro-viding 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 differ-ent individuals prefer differdiffer-ent 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.
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
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. Sec-ond, 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 t16= t2 6= ... 6= tm.
In theory, the number of possible consumer types in a product category with n variants can be as large asPnj=0(n−j)!n! . 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.
• 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 mod-els:
• 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
propor-tions.
• 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
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
Dj for product j is defined as the number of customers who attempt to buy product j. Note that Pn
j=1Dj ≥ 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, EΠ(q) =
n
X
j=1
©
rjE[min(qj, Dj)] − cjqj+ sjE[qj− Dj]+− pjE[Dj − qj]+
ª
Let uj = rj+
Pj
k=1pk− 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 EΠ(q) =
n
X
j=1
©
rjE[min(qj, Dj)] − cjqj+ sjE[qj− Dj]+− pjE[Dj− qj]+
ª = n X j=1 " (rj+ j X k=1 pk− cj)qj − (rj+ j X k=1 pk− sj)E[qj− Dj]+ # − µ n X j=1 pj = n X j=1 £ ujqj− (uj+ oj)E[qj − Dj]+ ¤ − µ n X j=1 pj
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
q≥0 EΠ(q)
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, DH
j where H stands for Homogeneous preferences, depends on the
inventory levels of product 1 to j − 1, and is given by:
D1H = D DHj (q1, ..., qj−1) = " D − j−1 X i=1 qi #+ for j = 2, ..., n
The one-period (newsvendor) expected profit for the product category given by EΠ(q) = n X j=1 £ ujqj− (uj+ oj)E[qj− DjH]+ ¤ = n X j=1 " ujqj − (uj+ oj) à qjF Ãj−1 X i=1 qi ! + Z Pj i=1qi Pj−1 i=1qi à j X i=1 qi− x ! f (x)dx !# (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: EΠH(q1, q2) = u1q1+ (u1+ o1) Z q1 0 (q1− x) f (x)dx +u2q2− (u2+ o2) µ q2F (q1) + Z q1+q2 q1 (q1+ q2− x) f (x)dx ¶ (2)
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 ¤+ (3) where θ2 ≡ u2u+o2 2.
Plugging this value back into (2), we obtain:
EΠH(q1) = (u1− u2)q1− [(u1+ o1) − (u2+ o2)] Rq1 0 (q1− x) f (x)dx + u2F−1(θ2) − (u2+ o2) RF−1(θ 2) 0 ¡ F−1(q 2) − x ¢ f (x)dx if q1< F−1(θ2) u1q1− (u1+ o1)Rq1 0 (q1− x) f (x)dx if q1≥ F−1(θ2) (4)
Note that EΠH(q
1) is continuously differentiable in q1. Taking derivative with respect to q1, we
get: ∂EΠH(q 1) ∂q1 = ½ (u1− u2) − [(u1+ o1) − (u2+ o2)]F (q1) if q1 < F−1(θ2) u1− (u1+ o1)F (q1) if q1 ≥ F−1(θ 2) (5)
Figure 1: Solution to the 2-product problem
θ1 - θ2
u1 – u2
Stock only product 2
Stock only product 1 Stock both
Stock only product 1 or only product 2
) ( ), ( 2 1 2 12 1 1 θ θ − − = =F q F q (1), 2 0 1 1= = − q F q θ ) ( , 0 2 1 2 1 θ − = = q F q 0 ), ( ) ( , 0 2 1 1 1 2 1 2 1 = = = = − − q F q or F q q θ θ Case 3 Case 1 Case 2 Case 4
First, note that EΠH is not concave in q1if 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(q
1) 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 equal, respectively, to F−1(θ
12) and F−1(θ1), where θ12 = (u1+ou11)−(u−u22+o2) and θ1 = u1u+o1 1. It
follows that, because EΠH(q
1) is differentiable everywhere, it can have at most one interior local
maximum, either at F−1(θ
12) 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(q
1) 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(θ
12). Since this value is less than
F−1(θ
2), (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(θ
1) and since this value is greater than F−1(θ2), one should stock only
product 1. In Case 3, EΠH is decreasing in q
1 so that it is optimal to stock only product 2. Finally
in Case 4, EΠH(q
1) has two local maxima at q1 = 0 and q1 = F−1(θ1). In that case the optimal
solution is not completely defined by the underage and overage costs, as it depends on the relative values of EΠH(0) and EΠH¡F−1(θ
1)
¢
, 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 q
2 = 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
Table 1: Example 2 (q1, q2) µ = 40 µ = 50 ¡ F−1(θ 1), 0 ¢ EΠ =69.09 EΠ =89.09 ¡ 0, F−1(θ 2) ¢ 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:
ρ(x) =
½
j such that Pj−1i=1qi < x ≤
Pj
i=1qj if x ≤
Pn
i=1qi
0 otw
Now let ξ(x) be the marginal expected profit associated with the xth potential unit of demand:
ξ(x) =
½
uρ(x)−¡uρ(x)+ oρ(x)¢P [D ≤ x] if x ≤Pni=1qi
0 otw
In other words, the marginal profit associated with the xth unit of demand for x ≤Pnj=1qj 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−2are fixed, then the problem reduces to a
two-product 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
total inventory of products 1, ..., j − 1 is Q ≡Pj−1k=1qk. We have hj(qj, Q) = ujqj− (uj + oj)qjF (Q) − (uj+ oj) Z 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
hj(q) = ujq − (uj+ oj) Z q 0 (q − x) f (x)dx and h0j(q) ≡ ∂hj(q) ∂q = uj− (uj+ oj)F (q) (7) h00j(q) ≡ ∂ 2h j(q) ∂q2 = −(uj+ oj)f (q) < 0 (8)
hj(q) is a strictly concave function of q with a unique maximum at F−1(θj) where θj = ujuj+oj, for
j = 1, ..., n.
We are also interested in the following quantity: for j, k ∈ {1, ..., n} and j 6= k,
hj(q) − hk(q) = (uj− uk)q − [(uj+ oj) − (uk+ ok)] Z q 0 (q − x) f (x)dx and h0j(q) − h0k(q) ≡ ∂[hj(q) − hk(q)] ∂q = (uj− uk) − [(uj+ oj) − (uk+ ok)]F (q) (9) h00j(q) − h00k(q) ≡ ∂2[hj(q) − hk(q)] ∂q2 = −[(uj+ oj) − (uk+ ok)]f (q) (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 h0
k 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 j=2 hj à qj, j−1 X k=1 qk ! = h1(q1) + n X j=2 " hj à j X k=1 qk ! − hj Ãj−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 qj≥0{hj(qj, Q) + Vj+1(Q + qj)} = max qj≥0{hj(qj+ Q) − hj(Q) + Vj+1(Q + qj)} = max q≥Q{hj(q) − hj(Q) + Vj+1(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 q≥QGj(q), so that Vj(Q) = max q≥QGj(q) − hj(Q) (13) = Gj(Q) − hj(Q). Lemma 1. EΠ(q∗) = V 1(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:
V2(q) = h2
³£
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 the following piecewise structure for each
j = 1, ..., n: Vj(Q) = Aj1− hij 1(Q) 0 ≤ Q ≤ a j 1 Aj2− hij 2(Q) a j 1 < Q ≤ aj2 .. AjKj− hiKj(Q) a j Kj−1 < Q ≤ a j Kj 0 Q > ajKj (14)
where Kj is the number of breakpoints in the function, the constants 0 < aj
1 < ... < ajKj are the
values of the breakpoints, the constants Aj1, . . . , AjKj determine the height of each piece function at
To simplify the notation we drop the superscript j for the variables K, i, a and A, where it is clear that they refer to the parameters of Vj.
Proposition 1. For each j = 1, ..., n,
(1) Vj(Q) has the piecewise structure given by (14),
(2) Vj(Q) is continuous in Q,
Proof. (induction) For j = n, we have Vn(Q) = max q≥Qhn(q) − hn(Q) = ½ hn ¡ F−1(θn) ¢ − hn(Q) if Q ≤ F−1(θn), 0 otherwise. (15) Vn is continuous in £ 0, F−1(θ n) ¢
because hn is continuous. Also, Vn
¡
F−1(θ
n)
¢
equals zero, and thus, Vn is continuous at F−1(θn). This satisfies the first part of the proposition. (For j = n,
Vn is also differentiable everywhere. However, we will show that Vj, j < n, may have points of
non-differentiability.) The second part of the proposition is obtained by setting Kn = 1, in
1 = n, An1 = hn ¡ F−1(θn) ¢ and an1 = F−1(θn).
Now, we prove Proposition 1 for Vj, j = 1, . . . , n − 1, assuming that it is true for Vj+1.
Note that Gj(q) = Vj+1(q) + hj(q) is continuous in q. However, it may have multiple local optima because Vj+1 is not concave in q.
We construct Gj ≡ maxq≥QGj(q) as follows. Let x1 denote the highest local maximum of Gj,
x2 denote the next highest local maximum of Gj that is located to the right of x1, and so on until
xm, where m is the number of local maxima such that there is no higher local maximum on the
right of xm. Assume that x0= 0.
For each local maximum, xr, let br be such that xr−1 < br < xr and Gj(br) = Gj(xr). By the
continuity of Gj, there might not exist a point b1, however, b2, ..., bm always exist and are uniquely
defined. Let Case 1 be the case when b1 exists, and Case 2 when it does not exist. Figure 3.2
illustrates the construction of the values of xr and br in both cases.
Gj(Q) is now defined as follows. In Case 1,
Gj(Q) = Vj+1(Q) + hj(Q) if x0≤ Q ≤ b1, Vj+1(Q) + hj(Q) if xr−1< Q ≤ br for r = 2, . . . , m, Vj+1(xr) + hj(xr) if br < Q ≤ xr for r = 1, . . . , m, Vj+1(Q) + hj(Q) if xm< Q. (16) In Case 2, Gj(Q) = Vj+1(x1) + hj(x1) if x0≤ Q ≤ x1 Vj+1(Q) + hj(Q) if xr−1< Q ≤ br for r = 2, . . . , m, Vj+1(xr) + hj(xr) if br< Q ≤ xr for r = 2, . . . , m, Vj+1(Q) + hj(Q) if xm< Q.
In words, Gj(Q) is decreasing and equal to Gj(Q) is segments (xr−1, br), r = 1, . . . , m as well as in
Figure 2: Case 1 and Case 2 0 10 20 30 40 50 510 520 530 540 550 560 570 580 590 q Case 1 Gj max Gj b1 x1 b2 x2 0 5 10 15 20 25 30 12.5 13 13.5 14 14.5 15 15.5 16 16.5 q Case 2 Gj max Gj x1 x2 b2
Thus, the value of Vj is obtained in Case 1 as:
Vj(Q) = Vj+1(Q) if 0 ≤ Q ≤ b1 or xr−1< Q ≤ br for r = 2, . . . , m, Vj+1(x1) + hj(x1) − hj(Q) if br< Q ≤ xr for r = 2, . . . , m, Vj+1(Q) if xm < Q. (17) Case 2 can be treated similarly.
Let Cr = Vj+1(xr) + hj(xr), r = 1, ..., m, and note that Crdoes not depend on Q. Therefore, we can write Vj(Q) = Cr− hj(Q) for Q ∈ (br, xr], r = 1, . . . , m. By the induction hypothesis, Vj+1(Q)
is also decomposed into pieces of the type Ak− hik(Q). Combining these together, we obtain a
decomposition of Vj into pieces of the type Ak− hik(Q), where the breakpoints are br and xr for
r = 1, . . . , m, and the breakpoints of Vj+1 located in (xr−1, br), r = 1, . . . , m and in (xm, ∞).
It remains to be shown that Vj is continuous. We prove this result by contradiction. Assume
that Gj(Q) = maxq≥QGj(q) ≡ maxq≥Q{Vj+1(q) + hj(q)} is not continuous at some Q1. Gj is non-increasing in Q because the feasibility set {q : q ≥ Q} gets smaller as Q increases. Therefore, at
Q1, we must have1
Gj(Q−1) > Gj(Q+1)
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(Q+1) 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.
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 akotherwise 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+1cannot 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),
Vj0(a−k) ≤ Vj0(a+k). (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 hjis 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).
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(θ
j,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 maximum of Gj then G0
j(x) = 0. By Proposition 1, we have Gj(q) = A1− hi1(q) + hj(q) 0 ≤ q ≤ a1 A2− hi2(q) + hj(q) a1 < q ≤ a2 ... AK− hiK(q) + hj(q) aK−1 < q ≤ aK hj(q) q > aK
It follows that at any q where Gj is differentiable, G0j(q) = h0j(q) − h0ik(q) for some k. And h0j(q) −
h0
ik(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 q∗
j for j = 1, ..., n can be recovered from (ij1, ..., ijKj), (aj1, ..., ajKj) for j = 1, ..., n, in the
following way:
Algorithm 2
• Set Q = 0, j = 1 • while Q ≤ ajKj,
– if ijk= j then q∗ j = ajk− Q and Q → Q + qj∗. Otherwise q∗ j = 0. – j → j + 1 • if j ≤ n, q∗ j = ... = q∗n= 0.
Proof. From (13) we get that the optimal quantity of product j, as function of Q = q∗
1+ ... + qj−1∗
is given by:
q∗j = arg max
q≥QGj(q) − Q (19)
Let ijk be such that Vj(Q) = Aij k− hi
j
k(Q) for k ∈ {1, ..., K j}.
From (16), we know that maxq≥QGj(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 ijk6= 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 q∗j > 0 if j ∈ J and qj∗ = 0
otherwise. Assume that j1< j2 < ... < jm. Then
q∗jl = ( F−1¡θ jl,jl+1 ¢ −Pl−1k=1q∗ jk l = 1, ..., m − 1 F−1(θ jm) − Pm−1 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 opti-mal 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: ξ(x) = uj1 − (uj1+ oj1)F (x) x ≤ qj1 uj2 − (uj2+ oj2)F (x) qj1 < x ≤ qj1 + qj2 ... ujm− (ujm+ ojm)F (x) Pm−1 j=1 qj < x ≤ Pm j=1qj 0 x >Pmj=1qj
Table 2: Example 3 P1 P2 P3 P4 P5 u 6.67 0.46 16.06 16.65 14.00 o 15.16 0.47 22.68 30.19 15.08 Figure 3: Example 3: V2(q1) 0 5 10 15 20 25 30 35 40 45 0 50 100 150 200 250 300 350 400 450 500 q1 V2 stock 4&5 stock 3&5 stock 5 stock 2 NDBP NDBP a1 a2 a3 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
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 Kj, the
number of breakpoints in the value function.
Lemma 4. Kj ≤ 2(n − j) + 1 where Kj is defined in (14).
The intuition behind this result is the following. The indices ij1, ..., ijKj 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 ≤ Kj. In order to obtain a bound
on Kj, 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 Kj 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 consid-ering dynamic substitution:
g = EΠH(q∗H) − EΠSS(q∗SS)
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 max
jhj ¡ F−1(θ j) ¢ , q∗SSj = ½ F−1(θj) if j = j∗ 0 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
g = h1 ¡ F−1(θ12) ¢ − h2 ¡ F−1(θ12) ¢ h1(F−1(θ1)) ≥ 0
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) ¢ = (u1− u2)F−1(θ12) − [(u1+ o1) − (u2+ o2)] Z F−1(θ 12) 0 (q − x) f (x)dx → u1F−1(θ1) − (u1+ o1) Z F−1(θ 1) 0 (q − x) f (x) = h1 ¡ F−1(θ1) ¢ 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 α2 they are of
type (1, 2),... and with probability αn they are of type (1, 2, ..., n). Assume that
Pn
βj =
Pn
k=jαk 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 DN Tj , where N T stands for Nested preferences with Trend-following partitioning, and is given by:
D1N T = D DjN T = ( h D −Pj−1i=1qi i+ w.p. βj 0 w.p. (1 − βj) j = 2, ..., n
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: Lemma 6. EΠN T(q, u, o) = EΠH(q, ˜u, o) where ˜u
j = βjuj− (1 − βj)oj for j = 1, .., n Proof. EΠN T(q, u, o) = n X j=1 £ ujqj− (uj + oj)E[qj− Dj]+ ¤ = n X j=1 " ujqj− (uj+ oj)qj " (1 − βj) + βjF Ãj−1 X i=1 qi !# −(uj+ oj)βj Z Pj i=1qi Pj−1 i=1qi à j X i=1 qi− x ! f (x)dx # = n X j=1 [(βjuj− (1 − βj)oj)qj −βj(uj+ oj) à qjF Ãj−1 X i=1 qi ! + Z Pj i=1qi Pj−1 i=1qi à j X i=1 qi− x ! f (x)dx !# If we set ˜uj = βjuj − (1 − βj)oj, we get EΠN T(q, u, o) = n X j=1 " ˜ ujqj− (˜uj+ oj) à qjF Ãj−1 X i=1 qi ! + Z Pj i=1qi Pj−1 i=1qi à j X i=1 qi− x ! f (x)dx !# = EΠH(q, ˜u, o) 4.2 Fixed partitioning
In this case we assume that a fixed proportion α1 of customers are of type (1), α2are of type (1, 2),... and αn are of type (1, 2, ..., n). Again we assume that that
Pn
j=1αj = 1 and let βj =
Pn
k=jαk be
the probability that a customer is willing to buy product j. Also let γj = βj/βj−1 for j = 2, ...n
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:
DN F1 = D DN Fj = [Dj−1− qj−1]+γj = " D − j−1 X i=1 qi βi #+ βj j = 2, ..., n
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: Lemma 7. EΠN F(q, u, o) = EΠH(˜q, ˜u, ˜o) where, for j = 1, ..., n
˜ qj = βjqj ˜ uj = ujβj ˜ oj = ojβj (21) Proof. EΠNF(q, u, o) = n X j=1 £ ujqj − (uj+ oj)E[qj− Dj]+ ¤ = n X j=1 ujqj− (uj+ oj) Z ∞ 0 qj− Ã x − j−1 X i=1 qi βi !+ βj + f (x)dx = n X j=1 " ujqj− (uj+ oj)qjF Ãj−1 X i=1 qi βi ! −(uj + oj) Z ∞ Pj−1 i=1 qiβi " qj− Ã x − j−1 X i=1 qi βi ! βj #+ f (x)dx = n X j=1 " ujqj− (uj+ oj)qjF Ãj−1 X i=1 qi βi ! −(uj + oj) Z Pj i=1qiβi Pj−1 i=1 qiβi " qj− Ã x − j−1 X i=1 qi βi ! βj # f (x)dx #
We do the following transformation of variables in (21), we get: EΠN F(q, u, o) = n X j=1 " ˜ ujq˜j− (˜uj + ˜oj) Ã ˜ qjF Ãj−1 X i=1 ˜ qi ! + Z Pj i=1qi˜ Pj−1 i=1qi˜ Ã j X i=1 ˜ qi− x ! f (x)dx !# = EΠH(˜q, ˜u, ˜o)
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 DN R
j , where N R stands for Nested preferences with
Random partitioning. DN R1 = D DN Rj = £DN Rj−1− qj−1 ¤+ Zj = " D − j−1 X i=1 qi Qi k=2Zk #+ j Y i=2 Zi j = 2, ..., n Lemma 8. DN F j ≤icxDN Rj for j = 1, ..., n
Proof. (by induction on j) For j = 1, we have DN F
1 = D1N R= D.
Now suppose that DN F
j−1 ≤icx Dj−1N R. By definition of the increasing convex order, this is
equiva-lent to:
E[DN Rj−1− qj−1]+ ≥ E[DN Fj−1− qj−1]+ for all qj−1 So we have: E[DjN R− qj]+ = E h E[¡Dj−1N R− qj−1)+Zj − qj ¢+ |DN Rj−1] i ≥ E£(DN Rj−1− qj−1)+γj− qj ¤+ ≥ E£(DN Fj−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[DN F
j ] ≤ E[DN Rj ] 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[DN R2 ] = E[(D − q1)+Z2] = E[D − q1]+α2 = E[D2N F] (22)
so that DN F
2 ≤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.
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 βjl
(ujl+1 + ojl+1) (23)
then
EΠN F(q) ≥ EΠN R(q)
Proof. We can write: EΠN R(q) = n X j=1 £ ujqj− (uj+ oj)E[qj− DjN R]+ ¤ = m X l=1 £ ujlqjl− (ujl+ ojl)E[qjl− DN Rjl ]+ ¤ = m X l=1 £ (ujl+ ojl) ¡ E[DjlN R] − E[DN Rjl − qjl]+ ¢ − ojlqjl ¤ = (uj1 + oj1) ¡ E[D] − E[D − qj1] +¢ + m X l=2 · (ujl+ ojl) µ E[DN Rjl−1− qjl−1]+ββjl jl−1 − E[DjlN R− qjl]+ ¶¸ − m X l=1 ojlqjl = (uj1 + oj1)E[D] − n X j=2 · (ujl−1+ ojl−1) −ββjl jl−1(ujl+ ojl) ¸ E[Djl−1N R − qjl−1]+ −(ujm+ ojm)E[DN Rjm − qjm]+− m X l=1 ojlqjl
Since we have E[DjN R− qj]+ ≥ E[DN Fj − 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)
Proof. By Lemma 7, EΠN F(q, u, o) = EΠH(˜q, ˜u, ˜o).
By Lemma 3, the optimal inventory vector for the second expression satisfies the condition ˜
ujl+ ˜ojl> (˜ujl+1+ ˜ojl+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.
Figure 4: Example of Outtree 1 2 3 4 5 6 α2 α3 α4 α5 α6
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
k∈Sj
γk ≤ 1, j = 2, ..., n
and without loss of generality we set γ1 = 1. In the above example we have γ2+ γ3 ≤ 1, γ4+ γ5 ≤ 1
and γ6 ≤ 1. Also let us define βj =
Q
i∈Pj∪{j}γi as the total proportion of customers that are
