# Assortment Optimization with Repeated Exposures and Product-dependent Patience Cost

In this paper, we study the assortment optimization problem faced by many online retailers such as Amazon. We develop a cascade multinomial logit model, based on the classic multinomial logit model, to capture the consumers' purchasing behavior across multiple stages. Different from existing studies, our model allows for repeated exposures of a product, i.e., the same product can be displayed multiple times across different stages. In addition, each consumer has a patience budget that is sampled from a known distribution and each product is associated with a patience cost, which captures the cognitive efforts spent on browsing that product. Given an assortment of products, a consumer sequentially browses them stage by stage. After browsing all products in one stage, if the utility of a product exceeds the utility of the outside option, the consumer proceeds to purchase the product and leave the platform. Otherwise, if the patience cost of all products browsed up to that point is no larger than her patience budget, she continues to view the next stage. We propose an approximation solution to this problem.

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## 1 Introduction

In this paper, we consider the sequential assortment optimization problem with repeated exposures and product-dependent patience cost. The input of our problem is a set of products and a limited number of stages, each stage has a limited capacity, our goal is to find the best assignment of products to stages that maximizes the expected revenue. We develop a variant of multinomial logit model, called cascade multinomial logit model (C-MNL), to capture the consumer’s purchasing behavior across multiple stages. In our model, each consumer has a patience budget which is drawn from a known distribution, and each product is associated with a patience cost that quantifies the cognitive efforts spent on browsing a product. In each stage, the consumer browses all products displayed in that stage, if the utility of some product is larger than the no-purchase option, then she purchases the one with the largest utility and leaves the system. Otherwise, the consumer continues to enter the next stage if and only if her current patience budget is non-negative. Our model generalizes the previous studies on sequential assortment optimization in two ways:

1. Our model allows for repeated exposures, i.e., the same product can be displayed multiple times across different stages. In the filed of marketing, it has been well recognized that a consumer typically must be exposed to an advertisement or a message more than once in order to get familiar with it and take actions. From a consumer cognition perspective, we believe that assortment planning is similar to online advertising in that they both push a set of products’ information to the consumer. We develop a rigorous mathematical model to capture the effect of repeated exposures.

2. We assign a product-dependent patience cost to each product. The patience cost of a product quantifies the amount of efforts needed to read and digest the information about that product. Most of existing studies Ma et al. (2019) assign a fixed and identical patience cost to each stage, e.g., they assume that the patience cost of browsing all products in one stage does not depend on the offered products in that stage. In contrast, our model allows different products to have different patience cost, and the patience cost of viewing one stage is characterized by the summation of the individual patience costs of all products allocated to that stage. Our model is motivated by the observation that browsing different products may require different amount of cognitive efforts.

3. Our problem formulation incorporates a set of practical constraints. For example, there is a constraint on the number of products displayed in each stage, and there is also a limit on the maximum number of exposures of a product. We develop an approximation algorithm with polynomial time complexity when the number of stages is a constant. An interesting research direction is to design an efficient algorithm whose running time is polynomial in the number of stages.

#### Related Works

Our work is closely related to the assortment optimization problems Li et al. (2015); Davis et al. (2014); Blanchet et al. (2016); Farias et al. (2013); Aouad et al. (2015). Among existing studies on assortment optimization, Davis et al. (2013) and Abeliuk et al. (2016) were the first to study this problem under MNL model with position bias. Since then, there is considerable number of studies Aouad and Segev (2015); Ferreira et al. (2019); Aouad et al. (2019) on assortment optimization problem with position bias. However, most of them adopt the consider-then-choose model where the consumer first browses a random number of products and then makes her purchase decision within these products. Our model differs from theirs in that we do not separate “consider” from “choose”, e.g., the list of products browsed by a consumer is jointly decided by her patience budget and the choice model. We build our study on the recent advances of sequential assortment optimization Ma et al. (2019). As mentioned earlier in this section, our model generalizes the previous studies by allowing for repeated exposures and product-dependent patience cost. In addition, our problem formulation incorporates a set of practical constraints.

## 2 Cascade Multinomial Logit Model and Problem Formulation

In the rest of this paper, we use to denote the set for any positive integer .

We first explain the idea of C-MNL. Assume there is a set of products and a set of stages . The capacity of each stage is , e.g., we can assign at most products to each stage. Each product can be displayed in at most stages and the same product can be displayed at most once in each stage. The utility of the -th exposure of product is a random value drawn from the Gumbel distribution with location-scale parameters . The utility of the no-purchase option, denoted by , is a random value drawn from the Gumbel distribution with location-scale parameters

. The patience budget of a consumer is given by the random variable

. Let

denote the probability that

. Each product is associated with a patience cost : Browsing a product consumes the consumer amount of patience budget. In addition, let denote the revenue of product : The platform earns revenue if the consumer purchases . We made two innocuous assumptions in this paper.

###### Assumption 1

For any two non-negative numbers and , where denotes the probability that conditioned on that .

This assumption states that the patience budget of a customer declines rapidly as she browses more stages. We believe that as more stages browsed without a purchase, it is more likely that the customer will run out of her patience budget sooner.

###### Assumption 2

.

This assumption states that the expected utility of a product reaches its maximum point at the first exposure and then declines with each additional exposure. This is called burnout effect in the field of online advertising Naik et al. (1998). In the context of assortment optimization, because the platform still pushes the product to a customer, we expect a similar repetition effect: the probability of purchasing a product declines with each additional exposure of that product.

In our choice model, an arriving customer sequentially browses the assortments in each stage. In each stage, if the largest utility for a product is larger than the no-purchase option, she purchases it and leaves the systems. Otherwise, if her remained patience budget is non-negative, she enters the next stage, otherwise, she leaves the system.

#### Feasible Assortment

We use to denote one assortment, where indicates whether the -th exposure of product is displayed in stage , e.g., if the -th exposure of product is displayed in stage , and otherwise, for all . We say an assortment is feasible if and only if it satisfies the following two conditions: (1) , and (2) . The first condition ensures that each product is displayed at most once in each stage, and the second condition ensures that the resulting assortment is implementable, e.g., the -th exposure of a product can only be displayed after the first exposures of that product have been displayed.

#### Problem Formulation

Before introducing our problem, we first present a closed form expression of choice probabilities.

###### Lemma 1

Let denote the set of all feasible assortments, if we offer , then the customer purchases product in stage with probability

 F(t−1∑z=1w∑k=1n∑i=1xi,k,zci)w∑k=1xi,k,zeμi,k(1+t−1∑z=1w∑k=1n∑i=1xi,k,zeμi,k)(1+t∑z=1w∑k=1n∑i=1xi,k,zeμi,k)

The proof of the above lemma is similar to the proof of Theorem 2.1 in Ma et al. (2019), thus omitted here to save space.

We next derive the expected revenue of any feasible assortment .

###### Lemma 2

The expected revenue of a feasible assortment is

 m∑t=1F(t−1∑z=1w∑k=1n∑i=1xi,k,zci)w∑k=1n∑i=1xi,k,zrieμi,k(1+t−1∑z=1w∑k=1n∑i=1xi,k,zeμi,k)(1+t∑z=1w∑k=1n∑i=1xi,k,zeμi,k)

This lemma follows immediately from Lemma 1 and the fact that the revenue of product is . For simplicity of notation, we use to denote in the rest of this paper.

Now we are ready to introduce the assortment optimization problem with repeated exposures and product-dependent patience cost. The objective of our problem P.0 is to find the best feasible assortment that maximizes the expected revenue. A formal definition of our problem is listed as follows.

P.0 Maximize

## 3 Technical Lemma

In this section, we will present one technical lemma that will be used in our latter algorithm design and analysis. For ease of presentation, we first introduce the concept of reachability. Given a solution, we define the reachability of a stage or a product as the probability that the customer has enough patience to browse that stage or that product. A formal definition of reachability is provided in Definition 1.

###### Definition 1

Given a solution , we define the reachability of any stage or any product that is displayed in stage as .

We next provide an important observation about the optimal solution , which we will use to design our solution.

###### Lemma 3

For any , there is a solution of expected revenue at least

 f(y)≥(1−ρ)f(xopt)

such that the reachability of all products under is at least .

Proof: Assume is the last stage in whose reachability is no smaller than , e.g., . We next construct two assortments based on : The first assortment, denoted by , is constructed by removing all products displayed after from the optimal solution. The second assortment, denoted by , is constructed by removing all products scheduled earlier than from the optimal solution and moving the rest of products stages ahead. It is easy to verify that the reachability of every product in the first assortment is no smaller than .

We first show that

 ρf(xopt>tρ)≥f(xopt)−f(xopt≤tρ) (1)

We first derive the expected revenue of the optimal solution .

 f(xopt) = f(xopt≤tρ) (3) +m∑t=tρ+1F(t−1∑z=1w∑k=1n∑i=1xopti,k,zci)w∑k=1n∑i=1riβi,kxopti,k,t(1+t−1∑z=1w∑k=1n∑i=1βi,kxopti,k,z))(1+t∑z=1w∑k=1n∑i=1βi,kxopti,k,z)

We next analyze the expected revenue of . For ease of presentation, let denote the number of exposures of product in the optimal solution before stage . For ease of presentation, define and .

 f(xopt>tρ) = m∑t=tρ+1F(t−1∑z=tρ+1w∑k=1n∑i=1xopti,k,zci)w∑k=1n∑i=1riβi,k−σixopti,k,t(1+t−1∑z=tρ+1w∑k=1n∑i=1βi,k−σixopti,k,z)(1+t∑z=tρ+1w∑k=1n∑i=1βi,k−σixopti,k,z) (4) = m∑t=tρ+1F(t−1∑z=tρ+1w∑k=1n∑i=1xopti,k,zci)w∑k=1n∑i=1riβi,k−σixopti,k,t(1+t−1∑z=tρ+1w∑k=1n∑i=1βi,k−σixopti,k,z)(1+t∑z=tρ+1w∑k=1n∑i=1βi,k−σixopti,k,z) (5)

To prove inequality 1, it suffice to prove that the value of (3) is upper bounded by times the value of (5). We next prove a stronger result, that is, for every :

 F(t−1∑z=1w∑k=1n∑i=1xopti,k,zci)w∑k=1n∑i=1riβi,kxopti,k,t(1+t−1∑z=1w∑k=1n∑i=1βi,kxopti,k,z))(1+t∑z=1w∑k=1n∑i=1βi,kxopti,k,z)≤ρF(t−1∑z=tρ+1w∑k=1n∑i=1xopti,k,zci)w∑k=1n∑i=1riβi,k−σixopti,k,t(1+t−1∑z=tρ+1w∑k=1n∑i=1βi,k−σixopti,k,z)(1+t∑z=tρ+1w∑k=1n∑i=1βi,k−σixopti,k,z) (7)

We first prove that the denominator of LHS of (7) is no smaller than the denominator of RHS of (7). This is true because and .

We next focus on proving that

 F(t−1∑z=1w∑k=1n∑i=1xopti,k,zci)w∑k=1n∑i=1riβi,kxopti,k,t≤ρF(t−1∑z=tρ+1w∑k=1n∑i=1xopti,k,zci)w∑k=1n∑i=1riβi,k−σixopti,k,t

Due to Assumption 1, we have for every , it follows that . Moreover, due to Assumption 2, we have

 F(t−1∑z=tρ+1w∑k=1n∑i=1xopti,k,zci) ≥ F(t−1∑z=1w∑k=1n∑i=1xopti,k,zci)/F(tρ∑z=1w∑k=1n∑i=1xopti,k,zci) (8) ≥ F(t−1∑z=1w∑k=1n∑i=1xopti,k,zci)/ρ (9)

The second inequality is due to .

This finishes the proof of (7), which implies (1), that is, . Now we are ready to prove this lemma. Because is the optimal solution, we have . Together with (1), we have . According to the definition of , the reachability of all products in is at least . Thus, is such a solution that satisfies .

## 4 Approximate Solution

In this section, we develop an approximate solution to our problem. For ease of presentation, given any , define

 g(x)=m∑t=1w∑k=1n∑i=1riβi,kxi,k,t(1+t−1∑z=1w∑k=1n∑i=1βi,kxi,k,z)(1+t∑z=1w∑k=1n∑i=1βi,kxi,k,z)

Note that is the utility of when the reachability of all stages are 1, e.g., this happens when the patience budget of the consumer is always infinity. Before presenting our algorithm, we first introduce a new problem P.1 whose solution is a key ingredient of algorithm.

P.1 Maximize subject to:

The objective of P.1 is to identify the best feasible assortment that maximizes subject to (C1). The condition (C1) ensures that the reachability of all non-empty stages must be no smaller than . As compared with the original problem P.0, we move the variables of patience cost from the objective function to the constraint (C1), which makes P.1 easier to tackle.

Given the formulation of P.1, we are now ready to present our algorithm, called assortment optimization under cascade multinomial logit model (ACME), for finding an approximate solution to P.0.

Description of ACME.

1. Solve P.1 approximately and get a solution .

2. Solve P.0 with optimally and get a solution .

3. Return the better solution between and as the final solution.

We first discuss the second step of ACME. It was worth noting that when there is only one stage, e.g, , P.0 is reduced to the classic assortment optimization problem subject to a cardinality constraint. We can solve it optimally based on Rusmevichientong et al. (2010) and obtain . We next present the main theorem of this paper. It says that if we can find an approximation algorithm for P.1, we can solve P.0 approximately.

###### Theorem 1

If there exists an -approximate solution to P.1, ACME achieves approximation ratio to P.0.

Proof: Recall that is the last stage in whose reachability is no smaller than , e.g., . We first prove that . Let denote a “sub” schedule of , removing all products scheduled after from .

 f(xopt≤tρ) = f(xopt

Inequality 11 is due to , and . Inequality 12 is due to is a feasible solution to P.1 and is the optimal solution to P.1. Inequality 13 is due to for all and , and . Inequality 14 is due to is the optimal solution to the assortment optimization problem with a single stage.

Assume , based on inequality 14, we have . Because the reachability of all stages under is lower bounded by , we have . It follows that . Because , we have . It follows that . Because ACME picks the better one between and as the final solution, this theorem holds.

In the next subsection, we propose solution to P.1 based on dynamic program, and we prove in Lemma 5 that it achieves approximation ratio for any . By setting in Theorem 1, we have the following performance bound for ACME.

###### Corollary 2

Given that we develop a -approximate solution to P.1 for any in subsection 4.1, ACME achieves approximation ratio to P.0.

In the rest of this paper, we introduce a -approximate solution to P.1 based on dynamic program. We build our solution on the recent advances in the assortment optimization problem subject to one capacity constraint (Désir et al., 2014), we generalize their idea and provide an approximate algorithm for the assortment optimization problem subject to a capacity constraint, a cardinality constraint, and a matroid-type feasibility constraint.

### 4.1 A Dynamic Program based Solution to p.1

We first introduce some notations. Let be the minimum revenue of a single product and let be the maximum revenue of a single product. Define and . Let , , and .

For a given , we first construct a geometric grid where and are defined as follows.

 I={γmin(1+ϵ)a∣a∈[⌈lndγmaxϵγmin⌉]},J={βmin(1+ϵ)b∣b∈[⌈lndβmaxϵβmin⌉]}

Then we build a group of guesses and . We go through all guesses and check whether or not there exists a solution such that is approximately equal to and is approximately equal to for all .

For a given guess , we discretize the values of and for all , , and as follows,

 ~γi,k,z=⌈γi,kuzϵ/d⌉,~βi,k,z=⌊βi,kvzϵ/d⌋

Denote by function for any guess and the optimal solution value of the following problem:

 h(j,u,v,l):=minx∈X{m∑z=1w∑k=1j∑i=1cixi,k,z:m∑z=1w∑k=1j∑i=1~γi,k,zxi,k,z=uz,m∑z=1w∑k=1j∑i=1~βi,k,zxi,k,z=vz,m∑z=1w∑k=1j∑i=1xi,k,z=lz}

We set the initial values as follows: we first set when or there exists some such that or or , and then set when the following conditions are satisfied: , and for all , , , and .

For ease of presentation, we next introduce an alternative way to represent the schedule of a product: For every product

, we use a vector

to represent the schedule of such that if is displayed on stage , and otherwise. Given a schedule of , define as a vector that replaces the -th non-zero element of with for all , and define as a vector that replaces the -th non-zero element of with for all . Let denote the norm of , then we fill up the dynamic program table using the following recurrence function.

 h(j,u,v,l)=minyjh(j−1,u−yj~γj,v−yj~βj,l−yj)+|yj|1cj

One way to compute is to enumerate all possible schedules of and find the one that minimizes . Because each product can only be displayed at most times and there are stages, the time complexity of enumerating all is .

###### Lemma 4

The time complexity of the dynamic program is .

Proof: Our proof is based on the following two observations. First, the total number of guesses is bounded by . Second, the size of and is bounded by . Third, the time complexity of computing is is , e.g., this is done by enumerating all possible . It follows that the total time complexity of the dynamic program is