When Advertising Meets Assortment Planning: Joint Advertising and Assortment Optimization Under Multinomial Logit Model

Abstract

While the topic of assortment optimization has received a significant amount of attention, the relationship between advertising and its impact on this issue has not been well-explored. This paper aims to fill the gap in research by addressing the joint advertising and assortment optimization problem. We propose that advertising can influence product selection by increasing preference for certain products, and the extent of this effect is determined by the product-specific effectiveness of advertising and the resources allocated to advertising for that product. Our goal is to find an optimal solution, which comprises of a combination of advertising strategy and product assortment, that maximizes revenue, taking into account budget constraints on advertising. In this paper, we examine the characteristics of this problem and present efficient methods to solve it under various scenarios. Both the unconstraint and cardinality constraint settings are studied and the joint assortment, pricing, and advertising problem is also examined. We further extend our findings to account for consumer decision-making patterns.

Appendices

Appendix

A Joint Assortment, Pricing, and Advertising Optimization

In this section, we study the case when the price of each product is also a decision variable. Formally, we assume that the preference weight of each product $i\in \mathcal {N}$ can be represented as $e^{q_i + f(c_i x_i) - p_i}$, whose value is jointly decided by i’s initial utility $q_i$, i’s price $p_i$, and the advertising efforts $x_i$ received from the platform. Hence, the revenue $r_i$ of each product $i\in \mathcal {N}$ is $r_i = p_i - d_i$, where $d_i$ is the production cost of i. Based on the above notations, we can represent the expected revenue $R(S, \textbf{p}, \textbf{x})$ of an assortment S as

$$\begin{aligned} R(S, \textbf{p}, \textbf{x}) = \frac{\sum _{i \in S} (p_i - d_i) e^{q_i + f(c_i x_i) - p_i}}{1 + \sum _{i \in S} e^{ q_i +f(c_i x_i) -p_i}} = \frac{\sum _{i \in S} (p_i - d_i) e^{q_i - p_i}g(c_i x_i) }{1 + \sum _{i \in S} e^{ q_i -p_i}g(c_i x_i)}. \end{aligned}$$

(A.1)

1.1 A.1 Unconstrained Case

If there is no cardinality constraint, our goal is to solve the following joint advertising, pricing, and assortment optimization problem:

$$\begin{aligned} \begin{array}{cl} &{}\mathop {\textrm{max}}_{\textbf{p, x}, S} \quad R(S, \textbf{p}, \textbf{x})\\ &{}\mathop {\mathrm {s.t.}}\quad \sum _{j \in S} x_j \le B. \end{array}\end{aligned}$$

(A.2)

Before describing our solution, we first present a useful lemma from [16].

Lemma A.1

[16]. Given any assortment S, the optimal price for each product $i \in S$ is $p_i = W(\sum _{i \in S} e^{q_i - d_i -1})g(c_i x_i) + d_i + 1$, where $W(\cdot )$ is the Lambert W function; that is, $W(\cdot )$ is the value of x that satisfies $x e^x = z$. Moreover, the revenue of the optimal solution is $W(\sum _{i \in S} e^{q_i - d_i - 1}g(c_i x_i))$.

For any given advertising strategy $\textbf{x}$ and assortment S, the optimal price and corresponding expected revenue are explicitly given by Lemma A.1. Because $W(\cdot )$ is an increasing function, Lemma A.1 implies that the optimal assortment must include all products. Hence, we can transform (A.2) into

$$\begin{aligned} \begin{array}{cl} \mathop {\textrm{max}}_{\textbf{x}} &{} \qquad \sum _{i=1}^n e^{q_i - d_i -1 }g(c_i x_i) \\ \mathop {\mathrm {s.t.}}&{} \qquad \sum _{i=1}^n x_i \le B, \\ &{} \qquad x_i \ge 0 \quad \forall i \in \mathcal {N}. \end{array}\end{aligned}$$

(A.3)

Denote $\alpha _i = e^{q_i - d_i - 1}$, and rewrite the above problem as

$$\begin{aligned} \begin{array}{cl} \mathop {\textrm{max}}_{\textbf{x}} &{} \qquad \sum _{i=1}^n \alpha _i g(c_i x_i)\\ \mathop {\mathrm {s.t.}}&{} \qquad \sum _{i=1}^n x_i \le B, \\ &{} \qquad x_i \ge 0 \quad \forall i \in \mathcal {N}. \end{array} \end{aligned}$$

(A.4)

Because $g(\cdot )$ is a concave function (Assumption 1) and $\sum _{i = 1}^n x_i \le B$ is a linear constraint, (A.4) is a concave maximization problem with convex constraints. Hence, (A.4) is a convex minimization problem over a convex set, and the problem has efficient solutions [5].

A special case where $g(\cdot )$ is a linear function: If $g(\cdot )$ is a linear function, that is, $g(x) = ax + 1$ for some $a\ge 0$, then the optimal advertising strategy is to allocate the entire advertising budget to a single product.

Lemma A.2

When $g(\cdot )$ is a linear function, the optimal solution to (A.4) is to allocate the entire advertising budget to the product with the largest $\alpha _i c_i$.

1.2 A.2 Cardinality-Constrained Case

We next consider a case where the size of the assortment is at most $K\ge 0$. Lemma A.1 indicates that the optimal assortment contains the top K products that have the largest $e^{q_i +f(c_i x_i) - d_i-1}$. We next show that if $e^{f(\cdot )}$ is a linear function, then we only need to consider two possible advertising strategies. Hence, this problem can be solved efficiently.

Lemma A.3

Let $\alpha _i = e^{q_i - d_i - 1}$ and $g(x) = ax + 1$ for some $a\ge 0$. Assume all products are indexed in non-increasing order of $\alpha _i$. Let $t_1 = \mathop {\textrm{argmax}}_{i\in \{1,\cdots , K\}} \{\alpha _i c_i \}$ and $t_2 = \mathop {\textrm{argmax}}_{j \in \{ K+1, ..., n\}} \{\alpha _j (a c_j B + 1)\}$, then the optimal advertising strategy is $\textbf{x}_{t_1}$ or $\textbf{x}_{t_2}$.

We next discuss a case with a general response function. For each product $i\in \mathcal {N}$, let $t_j=1$ if a product j is offered in the assortment and let $t_j=0$ otherwise. Our problem can be formulated as the following mixed-integer programming problem:

$$\begin{aligned} \mathop {\textrm{max}}_{\textbf{x}, \textbf{t}} & \qquad \sum _{i=1}^n \alpha _i t_i g(c_i x_i)\\ \mathop {\mathrm {s.t.}}& \qquad \sum _{i=1}^n x_i \le B, \nonumber \\ & \qquad \sum _{i=1}^n t_i \le K, \nonumber \\ & \qquad x_i \ge 0 \quad \forall i \in \mathcal {N}, \nonumber \\ & \qquad t_i \in \{0,1\} \quad \forall i \in \mathcal {N}. \nonumber \end{aligned}$$

(A.5)

If all products have the same advertising effectiveness, that is, $c_i = c,$ for all $i\in \mathcal {N}$, the optimal assortment is to select the top K products that have the largest $\alpha _i$.

Lemma A.4

Assume all products are indexed in non-increasing order of $\alpha _i$ and $c_i = c$ for all $i\in \mathcal {N}$. The optimal assortment is $S^* = \{1, \ldots , K\}$, and the optimal advertising strategy $\textbf{x}^*$ satisfies $x^*_i \ge x^*_j \forall i \le j$.

To find the optimal advertising strategy under $S^* = \{1, \ldots , K\}$, we need to solve an optimization problem that is similar to (A.4). Because this problem is a concave maximization problem with convex constraints, it can be solved efficiently.

For a general case, where advertising effectiveness is heterogeneous, the objective function of (A.5) contains the bilinear terms $t_i g(c_i x_i)$. We linearize each of these terms by relaxation. Specifically, for each $t_i e^{f(c_i x_i)}$, we introduce a new continuous variable $w_i = t_i g(c_i x_i)$ and add the inequalities: $g(c_i x_i) - w_i \le g(c_i B) (1- t_i)$, $0 \le w_i \le g(c_i x_i)$, and $w_i \le g(c_i B) t_i$. This leads to the following mixed-integer nonlinear programming problem:

$$\begin{aligned} \begin{array}{cl} \mathop {\textrm{max}}_{\textbf{x}, \textbf{t}, \textbf{w}} &{} \qquad \sum _{i=1}^n \alpha _i w_i\\ \mathop {\mathrm {s.t.}}&{} \qquad \sum _{i=1}^n x_i \le B, \\ &{} \qquad \sum _{i=1}^n t_i \le K, \\ &{} \qquad g(c_i x_i) - w_i \le g(c_i B) (1- t_i) \quad \forall i \in \mathcal {N}, \\ &{} \qquad 0 \le w_i \le g(c_i x_i) \quad \forall i \in \mathcal {N}, \\ &{} \qquad w_i \le g(c_i B) t_i \quad \forall i \in \mathcal {N}, \\ &{} \qquad x_i \ge 0 \quad \forall i \in \mathcal {N}, \\ &{} \qquad t_i \in \{0,1\} \quad \forall i \in \mathcal {N}. \end{array}\end{aligned}$$

(A.6)

B Sequential Joint Advertising and Assortment Optimization

In this section, we extend our study to consider a sequential joint advertising and assortment problem. The model put forward by [15] examines the behavior of consumers who may visit multiple product assortments before making a purchase or leaving the store. The consumer is assumed to progress through a sequence of m stages, each featuring a different assortment ($\mathcal {S} = (S_1, \ldots , S_m)$). If the consumer chooses to buy a product in stage i, they will leave the store, but if they do not make a purchase, they will proceed to the next stage. If no product is selected after visiting all m assortments, the consumer exits the store without making a purchase. This choice model is referred to as the sequential multinomial logit (SMNL) choice model. For the purpose of simplicity, we assume that the consumer will continue visiting subsequent assortments if they do not make a purchase in the current stage. However, it should be noted that this assumption can be relaxed to include the factor of consumer patience.

We will now provide a detailed explanation of the SMNL model. Given a sequence of assortments $\mathcal {S}$, the consumer will purchase product i in stage k with a probability of

$$\begin{aligned} \phi _i^k(\mathcal {S}) = \frac{v_i}{(1+\sum _{\ell =1}^{k-1} V(S_\ell ))(1+ \sum _{\ell =1}^k V(S_{\ell }))}. \end{aligned}$$

Let $V(S) = \sum _{i \in S} v_i$ and $W(S) = \sum _{i\in S} r_i v_i$. The expected revenue is represented as

$$\begin{aligned} R(\mathcal {S}) = \sum _{k=1}^m \frac{W(S_k)}{(1+\sum _{\ell =1}^{k-1} V(S_\ell ))(1+ \sum _{\ell =1}^k V(S_{\ell }))}. \end{aligned}$$

Under the advertising strategy $\textbf{x}$, the expected revenue increases to

$$\begin{aligned} R(\mathcal {S}, \textbf{x}) = \sum _{k=1}^m \frac{ W(S_k, \textbf{x})}{(1+\sum _{\ell =1}^{k-1} V(S_\ell , \textbf{x}))(1+ \sum _{\ell =1}^k V(S_{\ell }, \textbf{x}))}, \end{aligned}$$

(B.1)

where $V(S, \textbf{x}) = \sum _{i \in S} v_i g(c_i x_i)$ and $W(S, \textbf{x})= \sum _{i \in S} r_i v_i g(c_i x_i)$.

Based on the transformation in (5), the optimization problem can be written as

$$\begin{aligned} \mathop {\textrm{max}}_{\textbf{u}, \mathcal {S}} & \qquad \sum _{k=1}^m \frac{\sum _{i \in S_k} r_i u_i}{(1 + \sum _{l=1}^{k-1} \sum _{i \in S_l} u_i)(1 + \sum _{l=1}^k \sum _{i \in S_l} u_i)} \\ \mathop {\mathrm {s.t.}}& \qquad \textbf{u} \in \mathcal {U}. \nonumber \end{aligned}$$

(B.2)

We focus on the unconstrained setting. Given an arbitrary advertising strategy, [15] demonstrated that the optimal assortments are sequential revenue-ordered assortments. Specifically, there exists a set of decreasing thresholds $\{t_1^*, t_2^*, \ldots , t_{m+1}^*\}$, such that $S_{k}^{*}=\left\{ i \in \mathcal {N}: t_{k+1}^{*} \le r_{i}<t_{k}^{*}\right\}$ for $k \in \mathcal {M} = [1, 2, \ldots , m]$. The values of $\{t_i^*\}_{i=1}^{m+1}$ are given in the following lemma.

Lemma B.1

[15, Theorem 3.1]. There exists an optimal solution $\left( S_{1}^{*}, \ldots , S_{m}^{*}\right)$ such that for $i \in S_k^*$, we have $t_{k+1}^{*} \le r_{i}<t_{k}^{*}$. Let $R_k(S^{*}_1, \ldots , S^{*}_m) = \frac{ W(S^{*}_k)}{(1+\sum _{\ell =1}^{k-1} V(S^{*}_\ell ))(1+ \sum _{\ell =1}^k V(S^{*}_{\ell }))}$. The value of $t^*_k$ can be chosen as follows:

$$\begin{aligned} t^{*}_1 = + \infty , \quad t^{*}_k = \frac{R_{k-1}(S^{*}_1, \ldots , S^{*}_m) + R_k(S^{*}_1, \ldots , S^{*}_m) }{\frac{1}{1+\sum _{\ell =1}^{k-2} V(S^{*}_\ell )} - \frac{1}{1+\sum _{\ell =1}^k V(S^{*}_\ell )}} \qquad \forall k \in \mathcal {M} \backslash \{1\}, \quad t^{*}_{m+1} = \frac{R_m(S^{*}_1, \ldots , S^{*}_m)}{\frac{1}{1+\sum _{\ell =1}^{m-1} V(S^{*}_{\ell })}}. \end{aligned}$$

Based on this lemma, we analyze the structure of the optimal assortments and the advertising strategy. We denote the optimal solution of B.2 as $\textbf{u}^*$ and $\mathcal {S}^*$.

Lemma B.2

For the optimization problem (B.2), we have $\frac{\partial R(\mathcal {S}^*, \textbf{u}^*)}{\partial u^*_i} \ge \frac{\partial R(\mathcal {S}^*, \textbf{u}^*)}{\partial u^*_j} \ge 0$ for all products $i,j \in \mathcal {N}$ and $i < j$.

[4] showed that in the MNL choice model, the partial derivative $h_i^1 \ge 0$, indicating that the seller has no incentive to reduce the utilities of products in order to maximize their expected revenue. In Lemma B.2, we extend this result to the SMNL choice model. Moreover, due to the sequential revenue-ordered property stated in Lemma B.1 being maintained for any feasible set of products, this result remains valid even under capacity constraints, meaning that the seller has no incentive to decrease product utilities in the capacity-constrained scenario either. If the seller has the ability to enhance product utilities, the optimal advertising strategy would be to allocate the entire budget to the product that generates the highest revenue.

Lemma B.3

Denote the optimal solution of the following optimization problem as $(\textbf{x}^*, \mathcal {S}^*)$. $x_1^* = B$ and $x_i^* = 0$ for all $i \in \mathcal {N}\setminus \{1\}$.

$$\begin{aligned} \mathop {\textrm{max}}_{\textbf{x}, \mathcal {S}} & \qquad \sum _{k=1}^m \frac{\sum _{i \in S_k} r_i (v_i+x_i)}{(1 + \sum _{l=1}^{k-1} \sum _{i \in S_l} (v_i+x_i))(1 + \sum _{l=1}^k \sum _{i \in S_l} (v_i+x_i))} \\ \mathop {\mathrm {s.t.}}& \qquad \sum _{i=1}^n x_i \le B \nonumber \end{aligned}$$

(B.3)

In our setting, the allocation of budget $x_i$ to product i increases its utility to $v_i e^{f(c_i x_i)}$, where f is the nonlinear response function. Due to the heterogeneous advertising effectiveness, utility and nonlinear response function, the optimal advertising strategy may be more complex than a single-product advertising strategy. Given a specific sequence of assortments, finding the optimal advertising strategy is equivalent to solving the following optimization problem.

$$\begin{aligned} \mathop {\textrm{max}}_{\textbf{u}} & \qquad \sum _{k=1}^m \frac{\sum _{i \in S_k} r_i u_i}{(1 + \sum _{l=1}^{k-1} \sum _{i \in S_l} u_i)(1 + \sum _{l=1}^k \sum _{i \in S_l} u_i)} \\ \mathop {\mathrm {s.t.}}& \qquad \textbf{u} \in \mathcal {U} \nonumber \end{aligned}$$

(B.4)

where $\mathcal {U} = \{\textbf{u} | \sum _{i=1}^n m_i(u_i) \le {B}, u_i \ge v_i, i = 1, \ldots , n\}$ and $m_i (\cdot ) = g^{-1}(\frac{\cdot }{v_i})/c_i$. When $m=1$, this problem is a single-ratio FP problem, which can be solved efficiently. However, the sum-of-ratio problem is generally NP-complete [14]. Hence, even though the optimal assortments may be sequential revenue-ordered assortments, finding the optimal advertising strategy may not be straightforward. As a result, we propose a heuristic method as an alternative approach.

1.1 B.1 Heuristic Method

The design of our heuristic method (listed in Algorithm 4) is based on two key observations. Firstly, given an advertising strategy, the optimal sequence of assortments can be found efficiently in polynomial time. Secondly, given the set of products to be displayed, the single-stage optimal advertising strategy is computationally tractable. Specifically, Algorithm 4 iteratively updates the assortments and advertising strategy until the expected revenue cannot be improved any further.

Algorithm 4

Heuristic for Unconstrained Multi-stage JAAOP

By exploring the structure of the objective function in (B.2), we next show that our heuristic method achieves an approximation ratio of 50%.

Lemma B.4

Let $(\mathcal {S}^*, \textbf{x}^*), (\mathcal {S}^h, \textbf{x}^h)$ be the optimal values of (B.2) and our heuristic method. We have $R(\mathcal {S}^h, \textbf{x}^h) \ge \frac{1}{2} R(\mathcal {S}^*, \textbf{x}^*)$.

C Numerical Study

In this section, we explore the effect of advertising on assortment optimization and validate the superiority of our algorithms compared with several heuristic methods on randomly generated instances and different response functions. The revenue of each product is drawn uniformly from the interval [1, 10]. For the preference weight $v_i$ of product i, we first sample $\gamma _i$ uniformly from the interval [1, 10] and then assign $v_i = \gamma _i/\varDelta$, where $\varDelta = P_0 \sum _{i \in \mathcal {N}} \gamma _i /(1-P_0)$. In this case, we guarantee the no-purchase probability when providing all products is exactly $P_0$. We consider three types of response functions: $g_1(x) = \sqrt{x} + 1$, $g_2(x) = \log (x+1) + 1$, $g_3(x) = 2 - e^{-x}$. For advertising effectiveness, we consider the following settings.

• Setting A: The advertising effectiveness $c_i$ of each product $i\in \mathcal {N}$ is drawn uniformly from the interval [0, 1].

• Setting B: The advertising effectiveness $c_i$ of each product $i\in \mathcal {N}$ is drawn independently from a standard log-normal distribution and rescaled by a factor of $\frac{1}{2\sqrt{e}}$ to make sure the same mean as setting A. In this case, there is more dispersion in advertising effectiveness.

We choose the number of products from $\{50, 100, 200\}$, the cardinality constraint K from $\{5,10,20\}$, and the value of $P_0$ from $\{0.1,0.3\}$. For the multi-stage problem, the stage m is chosen from $\{3,5,8\}$. For each setting, we randomly generate 10 instances and calculate the average percentage of improvement over the non-advertising strategy. Finally, we denote our heuristic algorithm as HA.

1.1 C.1 Compared Heuristics

For the cardinality-constrained single-stage problem, the main challenge lies in finding the optimal advertising strategy as the optimal assortment for a given advertising strategy can be found efficiently in polynomial time. In order to tackle this difficulty, we propose two practical advertising strategies.

• Uniform advertising (UA) strategy: for any assortment S, we have $x_i = B/|S|$ if $i \in S$.

• Revenue advertising (RA) strategy: for any assortment S, we have $x_i = B\cdot \frac{r_i}{\sum _{i\in S}r_i}$ if $i \in S$.

We start with the optimal assortment with no advertising strategy $S^1$. After allocating the budget according to the heuristic method, we recompute the optimal assortment $S^2$; if $S^1 \ne S^2$, then we reallocate the budget and compute the new assortment. This process continues until the assortment is unchanged with advertising (Table 2).

Table 1. Average Performance of Tested Heuristic Algorithm on Single Stage Problem

Table 2. Average Performance of Tested Heuristic Algorithm on Multiple Stage Problem

1.2 C.2 Performance Evaluation

Table 1 presents the average performance of three heuristic algorithms for the single-stage joint advertising and assortment problem, evaluated over 36 different parameter settings. Algorithm 3 demonstrates superior performance compared to the other heuristic algorithms, particularly when $P_0$ is large and the cardinality constraint is small. In most cases, the RA strategy performs slightly better than the UA strategy. The performance of each heuristic algorithm does not vary significantly with an increase in the dispersion of advertising effectiveness. When the set of products is less attractive and the cardinality constraint is small, advertising has a more significant impact, and the gap between our algorithm and the compared heuristic algorithms is even larger.

The multi-stage setting has a decreasing impact on advertising as the seller is given more stages. Even without a cardinality constraint, the revenue improvement can still be significant when the utility of the no-purchase option is relatively high. The improvement under the UA strategy can be less than 0.5%, while the improvement using the heuristic method is at least 2%. This shows the importance of advertising strategy on expected revenue (Fig. 2).

Finally, we evaluate the computational efficiency of our Algorithm 3. Table 3 shows its average running time for different parameters. Our algorithm has a low computational complexity as it only requires solving two linear programming problems to find the optimal assortment and a few convex optimization problems to find the corresponding advertising strategy. The results in Table 3 demonstrate that our algorithm has a running time of less than 2 s for all cases, making it highly efficient.

Table 3. Average Running Time of Algorithm 3

1.3 C.3 Effect of Budget on Expected Revenue

In practicality, the seller must also decide on the advertising budget. Since the return per budget investment can reduce with increasing budget, this subsection examines the relationship between expected revenue and invested budget. The experiment has 100 products and the budget is varied from 0 to 50 while the revenue and advertising effectiveness are kept constant. 100 preference weights are sampled for each budget and response function. The results of the expected revenue for each of these settings are displayed in Fig. 1.

The expected revenue is shown to increase with an increase in advertising budget as illustrated by Fig. 1. For the first response function $g_1$, when the budget is adequate and $P_0 = 0.1$, the difference in revenue between the different cardinality constraints becomes small, as indicated by Fig. 1(a). Hence, when the seller has an adequate budget, limiting their focus to a small group of products does not result in a significant reduction in revenue. For the same response function, the trend of increasing expected revenue remains consistent across different values of $P_0$, with lower values leading to higher expected revenue. For the third response function, $g_3(x) = 2 - e^{-x}$, the increase in expected revenue becomes insignificant when more than 20 units of the budget are allocated to advertising.

Fig. 1.

The relationship between budget and expected revenue for 100 products in different settings

D Omitted Proofs

Proof of Lemma 1: Consider an arbitrary advertising strategy $\textbf{x}$ that satisfies $\sum _{i \in S^*} x_i = B_1 < B$, define $\varDelta = B - B_1$ and $\tau =\arg \max _{i\in S^*} r_i$. To prove this lemma, we can increase the expected revenue by allocating the remaining budget $\varDelta$ to the product $\tau$. Let $S^*$ denote the optimal assortment, and $\textbf{x}'$ denote this new strategy. Because $\textbf{x}'$ and $\textbf{x}$ only differs in entry $\tau$, we rewrite $R(S^*, \textbf{v}, \textbf{x}')$ as $\frac{\beta + r_{\tau } v_{\tau } (g(c_{\tau } (x_{\tau } + \varDelta )) - g(c_{\tau } x_{\tau }))}{\alpha + v_{\tau }(g(c_{\tau } (x_{\tau } + \varDelta )) - g(c_{\tau } x_{\tau })) }$ and rewrite $R(S^*, \textbf{v} , \textbf{x})$ as $\frac{\beta }{\alpha }$, where $\beta = \sum _{i \in S^* }r_i v_i g(c_i x_i)$ and $\alpha = 1 + \sum _{i \in S^* } v_i g(c_i x_i)$. Notice that

$$\begin{aligned} R(S^*, \textbf{v}, \textbf{x}') - R(S^*, \textbf{v} , \textbf{x}) = \frac{(r_{\tau } \alpha - \beta ) \cdot v_{\tau } (g(c_{\tau } (x_{\tau } + \varDelta )) - g(c_{\tau } x_{\tau })) }{\alpha \cdot (\alpha + v_{\tau }(g(c_{\tau } (x_{\tau } + \varDelta )) - g(c_{\tau } x_{\tau })) )}. \end{aligned}$$

Because $\tau =\arg \max _{i\in S^*} r_i$, we have $r_{\tau } - r_i \ge 0, \forall i \in S^*$. Thus, $r_{\tau } + \sum _{i \in S^* } (r_{\tau } - r_i) (v_i g(c_i x_i)) > 0$, which is equivalent to $r_{\tau } (1 + \sum _{i \in S^*} v_i g(c_i x_i)) > \sum _{i \in S^* }r_i v_i g(c_i x_i)$. Hence, $r_{\tau } \alpha > \beta$. Because $g(\cdot )$ is an increasing function, we have $g(c_{\tau }(\varDelta + x_{\tau }))- g(c_{\tau } x_{\tau }) \ge 0$. Moreover, because both $r_{\tau } \alpha - \beta$ and $g(c_{\tau }(\varDelta + x_{\tau }))- g(c_{\tau } x_{\tau })$ are non-negative, we obtain that $R(S^*, \textbf{v}, \textbf{x}') \ge R(S^*, \textbf{v} , \textbf{x})$. $\Box$

Proof of Lemma 3:

Proof: Because $(S^*,\textbf{x}^*)$ is optimal solution, we have $R(S^*, \textbf{v} , \textbf{x}^*) \ge R(S_{\textbf{v}}, \textbf{v})$. According to Lemma 2, there exists an $S^*$ such that $S^* = \{i \in \mathcal {N} | r_i > R(S^*, \textbf{v} , \textbf{x}^*)\}$. The following chain proves this lemma: $S^* = \{i \in \mathcal {N} | r_i > R(S^*, \textbf{v} , \textbf{x}^*)\} \subseteq \{i \in \mathcal {N} | r_i > R(S_{\textbf{v}}, \textbf{v} )\}= S_{\textbf{v}}$. $\Box$

Proof of Lemma 4:

Proof: Because $g(\cdot )$ is an increasing concave function, its inverse function $g^{-1}(x)$ is a convex function. Moreover, because $\frac{u}{v_i}$ is a linear function, its composition with $g^{-1}(\cdot )$ is also a convex function. Finally, because $\mathcal {U}_1 = \{\textbf{u} | \sum _{i=1}^n m_i(u_i) \le {B}\}$, which is the level set of $\sum _{i=1}^n m_i(u_i)$, is a convex set, its intersection with the convex set $\mathcal {U}_2 = \{\textbf{u} | u_i \ge v_i, i = 1, \ldots , n \}$ is also a convex set. $\Box$

Proof of Lemma 5:

Proof: For a given assortment S, we can represent any feasible advertising strategy $\textbf{y}$ that satisfies $\sum _{i \in S} y_i = B$ as a convex combination of $\textbf{x}_i$; that is, $\textbf{y} = \sum _{i \in S} \lambda _i \textbf{x}_i$, where $\lambda _i = y_i/B$ and $\sum _{i \in S} \lambda _i = 1$. Assume $k = \mathop {\textrm{argmax}}_{j \in S} L(S, \textbf{x}_j)$. We have $L(S, \textbf{y}) \le L(S, \textbf{x}_k)$ based on the following observation:

$$\begin{aligned} L(S, \textbf{y}) = L(S, \sum _{i \in S} \lambda _i \textbf{x}_i) = \frac{\beta + a B\sum _{i \in S} \lambda _i r_i v_i c_i }{\alpha + a B\sum _{i \in S} \lambda _i v_i c_i }, \end{aligned}$$

where $\alpha =1 + \sum _{i \in S} v_i$ and $\beta = \sum _{i \in S} r_i v_i$. By the definition of k, we have $L(S, \textbf{x}_k) \ge L(S, \textbf{x}_j), \forall j \in S$. Moreover, $L(S, \textbf{x}_k) \ge L(S, \textbf{x}_j)$ is equivalent to

$$\begin{aligned} \alpha c_k v_k r_k - \beta c_k v_k \ge \alpha c_j v_jr_j - \beta c_j v_j + a B c_k v_k (r_j - r_k) c_j v_j, \end{aligned}$$

(D.1)

based on the following observation:

$$\begin{aligned} L(S, \textbf{x}_k) - L(S, \textbf{x}_j) &= \frac{\beta + aB r_k c_k v_k}{\alpha + aB c_k v_k} - \frac{\beta + aB r_j c_j v_j}{\alpha + aB c_j v_j} \\ &= \frac{(\beta + aB r_k c_k v_k)(\alpha + aB c_j v_j) - (\beta + aB r_j c_j v_j)(\alpha + aB c_k v_k)}{(\alpha + aB c_k v_k)(\alpha + aB c_j v_j)} \\ &= aB \cdot \frac{\alpha (c_k v_k r_k - c_j v_j r_j) + \beta (c_j v_j-c_k v_k)+ a B c_k v_k (r_k -r_j) c_j v_j }{(\alpha + aB c_k v_k)(\alpha + aB c_j v_j)}. \end{aligned}$$

By multiplying $\lambda _j$ by both sides of (D.1) for all $j \in S$ and summing up all inequalities, we have

$$\begin{aligned} \alpha c_k v_k r_k - \beta c_k v_k \ge \alpha \sum _{j \in S} \lambda _j c_j v_j r_j - \beta \sum _{j \in S} \lambda _j c_j v_j +a B c_k v_k \sum _{j \in S} \lambda _j (r_j - r_k) c_j v_j. \end{aligned}$$

(D.2)

Using a similar argument as the one used to prove the equivalence of $L(S, \textbf{x}_k) \ge L(S, \textbf{x}_j)$ and (D.1), we show that (D.2) is equivalent to $L(S, \textbf{x}_k) \ge L(S, \sum _{i \in S} \lambda _i \textbf{x}_i) = L(S, \textbf{y})$. $\Box$

Proof of Lemma 6:

Proof: Let $Z = R(S_{\textbf{v}}, \textbf{v})$ denote the expected revenue of $S_{\textbf{v}}$ when $B=0$, we have $\sum _{i \in S_{\textbf{v}} } (r_i - Z) v_i = Z$. If there exists a product $i\in S_{\textbf{v}}$ such that $r_i < Z$, then removing this product from $S_{\textbf{v}}$ would increase the expected revenue, which contradicts the assumption that $S_{\textbf{v}}$ is the optimal assortment when $B=0$. Thus, we have $S_{\textbf{v}} \subseteq \{1, \ldots , T\}$, where $T =\max \{i | i \in S_{\textbf{v}}\}$. Similarly, let $Z^*= R(S^*, \textbf{v}, \textbf{x}^*)$, we have $S^* \subseteq \{1, \ldots T^*\}$ where $T^* = \max _i \{i | r_i \ge Z^* \}$. Since $Z^* \ge Z$, we conclude that $r_{T^*} \ge r_T$, otherwise we have $Z^* \le r_{T^*} < Z$ or $Z \le r_{T^*} < r_T$. Thus we have $S^* \subseteq \{1, \ldots T^*\} \subseteq \{1, \ldots , T\}$. $\Box$

Proof of Lemma A.2:

Proof: When $g(x) = ax + 1$ for some $a\ge 0$, the objective function of (A.4) can be written as $\sum _{i=1}^n \alpha _i + a \sum _{i=1}^n \alpha _i c_i x_i$. Allocating the entire advertising budget to the product that has the largest $\alpha _i c_i$ maximizes $\sum _{i=1}^n \alpha _i + a \sum _{i=1}^n \alpha _i c_i x_i$. $\Box$

Proof of Lemma A.3:

Proof: Consider a fixed feasible assortment S. If $f(x) = \log (ax + 1)$ for some $a\ge 0$, then the objective function $R(S, \textbf{p}, \textbf{x})$ can be written as $\sum _{i\in S} \alpha _i + a \sum _{i\in S} \alpha _i c_i x_i$. It is easy to verify that the optimal advertising strategy for S must come from $\{\textbf{x}_0, \textbf{x}_1, \ldots , \textbf{x}_n\}$, where $\textbf{x}_0$ is an all-zero vector. Let $S(\textbf{x})$ be the optimal assortment under the advertising strategy $\textbf{x}$. Thus $S(\textbf{x})$ contains the top K products that have the largest $\alpha _i (a c_i x_i +1)$. Because $a c_i x_i\ge 0$ for all $i\in \{1,\ldots , n\}$, we have $S(\textbf{x}_i)=S(\textbf{x}_0) = \{1, \ldots , K\}$, and the expected revenue for $S(\textbf{x}_i)$ is $W(\sum _{j = 1}^K \alpha _j + a \alpha _i c_i B)$ for all $i \in \{1, ..., K\}$. When $j \in \{ K+1, ..., n\}$, there are two possible cases: $S(\textbf{x}_j) \setminus S(\textbf{x}_0) = \{\emptyset \}$ or $S(\textbf{x}_j) \setminus S(\textbf{x}_0) = \{ j \}$.

Case 1 ::

When $S(\textbf{x}_j) \backslash S(\textbf{x}_0) = \{\emptyset \}$ for all $j \in \{ K+1, \ldots , n\}$, the optimal assortment is $\{1, \ldots , K\}$. Because the expected revenue for $S(\textbf{x}_j)$ is $W(\sum _{j = 1}^K \alpha _j)$ for all $j \in \{ K+1, \ldots , n\}$, which is the same as $S({\textbf{x}_0})$, and $t_1 = \mathop {\textrm{argmax}}_{i \in \{1, \ldots , K\}} W(\sum _{j = 1}^K \alpha _j + a \alpha _i c_i B)$, the optimal advertising strategy is $\textbf{x}_{t_1}$.

Case 2 ::

When $S(\textbf{x}_j) \backslash S(\textbf{x}_0) = \{ j \}$ for some $j \in \{K+1, ..., n\}$, the expected revenue for $S(\textbf{x}_j)$ is $W(\sum _{i= 1}^{K-1} \alpha _i + \alpha _j (ac_j B + 1))$. We denote this subset as $S_c$. Because $t_2 = \mathop {\textrm{argmax}}_{j \in S_c} W(\sum _{i= 1}^{K-1} \alpha _i + \alpha _j (ac_j B + 1))$, $\textbf{x}_{t_2}$ is the best advertising strategy in $\{\textbf{x}_{K+1}, \ldots , \textbf{x}_n \}$. Moreover, because $\textbf{x}_{t_1}$ is the best advertising strategy in $\{\textbf{x}_0, \textbf{x}_{1}, \ldots , \textbf{x}_K \}$, the better strategy between $\textbf{x}_{t_1}$ and $\textbf{x}_{t_2}$ must be the optimal advertising strategy. $\Box$

Proof of Lemma A.4: When $c_i = c$ for all $i\in \mathcal {N}$, the objective function of (A.5) can be simplified to $h(\textbf{x}, S) = \sum _{i\in S} \alpha _i g(c x_i)$. To prove the first part of this lemma, we show that for any optimal solution $(S, \textbf{x})$, we can construct a new solution $(S', \textbf{x}')$, where $S' = \{1, \ldots , K\}$, which is no worse than $(S, \textbf{x})$. Due to the monotonicity of the objective function, $|S|=K$ can be assumed. We construct such $\textbf{x}'$ as follows: for each $i \in \{1, \ldots , K\}$, let $x'_i = x_{L(i)}$, where L(i) represents the product that has the i-th largest $\alpha _i$ in S. Therefore $h(\textbf{x}', S') - h(\textbf{x}, S) = \sum _{i \in S'} ( \alpha _i - \alpha _{L(i)} ) g(c x_i) \ge 0$; the inequality exists because $S'$ contains the top K products that have the largest $\alpha _i$. Hence, $(\textbf{x}', S')$ is no worse than $(S, \textbf{x})$.

We next prove that the optimal advertising strategy $\textbf{x}^*$ satisfies $x^*_i \ge x^*_j \forall i \le j$ through contradiction. Assume there exist two products $i, j \in S^*$ such that $x_i^* < x_j^*$ and $i < j$. We can construct a new advertising strategy $\textbf{x}$ such that $x_k = x_k^*$ for $k \notin \{i, j\}$, and $x_i = x_j^*, x_j = x_i^*$. The following chain proves that $h(\textbf{x}, S^*)- h(\textbf{x}^*, S^*) =(\alpha _i - \alpha _j) \cdot (g(c x_j^*) - g(c x^*_i) )$:

$$\begin{aligned} h(\textbf{x}, S^*)- h(\textbf{x}^*, S^*) &= \alpha _i g(c x_i) + \alpha _j g(c x_j) - \alpha _i g(c x^*_i) - \alpha _j g(c x^*_j) \\ &= \alpha _i (g(c x_j^*) - g(c x^*_i) ) + \alpha _j (g(c x_i^*) - g(c x^*_j) ) \\ &= (\alpha _i - \alpha _j) \cdot (g(c x_j^*) - g(c x^*_i) ). \end{aligned}$$

Because $\alpha _i \ge \alpha _j$ and $x_i^* < x_j^*$, $\textbf{x}$ is a better solution than $\textbf{x}^*$ which contradicts to the assumption that $\textbf{x}^*$ is the optimal solution. $\Box$

Proof of Lemma B.2:

Proof: For simplicity, let $h_i^k = \frac{\partial R(\mathcal {S}^*, \textbf{u}^*)}{\partial u^*_i}$ be the partial derivative for product i in assortment $S^*_k$, and denote $A_k = \sum _{l=1}^k V(S_l^*)$ and $B_k = \frac{ W(S_k^*)}{V(S_k^*)}$. We have

$$\begin{aligned} h_i^k &= \frac{(r_i - \frac{W(S_k^*)}{1+A_k})}{(1+A_{k-1})(1+A_k)} - \sum _{j=k+1}^m B_j \cdot \left( \frac{1}{(1+A_{j-1})^2} - \frac{1}{(1+A_{j})^2}\right) . \end{aligned}$$

According to Lemma B.1, $S_k^* = \{i \in \mathcal {N}| t_{k+1}^* \le r_i < t_k^*\}$. We first consider the products in the same stage, that is $i, j \in S_k^*$, and $i < j$. $h_i^k \ge h_j^k$ because $r_i \ge r_j$. Then, we consider the cases i and j in two different stages. The difference between $h_i^k$ and $h_j^{k+1}$ is

$$\begin{aligned} h_i^k - h_j^{k+1} &= \frac{(r_i - \frac{W(S_k^*)}{1+A_k})}{(1+A_{k-1})(1+A_k)} - B_{k+1} \cdot \left( \frac{1}{(1+A_{k})^2} - \frac{1}{(1+A_{k+1})^2}\right) - \frac{(r_j - \frac{W(S_{k+1}^*)}{1+A_{k+1}})}{(1+A_{k})(1+A_{k+1})} \\ &\ge \left[ \frac{1}{(1+A_{k-1})(1+A_k)} - \frac{1}{(1+A_{k})(1+A_{k+1})} \right] t_{k+1}^* - \frac{W(S_{k}^*)}{(1+A_{k-1})(1+A_k)^2} \\ \nonumber & \quad - \frac{2(1+A_k)+V(S_{k+1}^*)}{(1+A_{k})^2(1+A_{k+1})^2} \cdot W(S_{k+1}^*) + \frac{W(S_{k+1}^*)}{(1+A_{k})(1+A_{k+1})^2} \\ &= \frac{W(S_k^*)}{(1+A_{k-1})(1+A_k)^2} + \frac{W(S_{k+1}^*)}{(1+A_k)^2(1+A_{k+1})}- \frac{W(S_{k}^*)}{(1+A_{k-1})(1+A_k)^2} \\ & \quad - \frac{2(1+A_k)+V(S_{k+1}^*)}{(1+A_{k})^2(1+A_{k+1})^2} \cdot W(S_{k+1}^*) + \frac{W(S_{k+1}^*)}{(1+A_{k})(1+A_{k+1})^2} \\ & = \frac{2+A_k+A_{k+1} - 2(1+A_k)+V(S_{k+1}^*)}{(1+A_{k})^2(1+A_{k+1})^2}\cdot W(S_{k+1}^*) \\ & = 0. \end{aligned}$$

The first inequality uses the fact that $r_i \ge t_{k+1}^* \ge r_j$. Lastly, for the product in the last assortment $S_m$, we have $h_i^m = \frac{r_i - \frac{W(S_m^*)}{1+A_m}}{(1+A_{m-1})(1+A_m)}$. In this case, $r_i \ge t_{m+1}^* = \frac{W(S_m^*)}{1+A_m}$, which means $h_i^m \ge 0$. $\Box$

Proof of Lemma B.3:

Proof: Let $Q(\textbf{x}) = R(\mathcal {S}^*, \textbf{x})$. Based on the analysis in Lemma B.2, $\frac{\partial Q(\textbf{x})}{\partial x_i} \ge \frac{\partial Q(\textbf{x})}{\partial x_j} \ge 0$ for all $i<j$. For any $\textbf{x}$ satisfying the budget constraint, through the mean value theorem, we have

$$\begin{aligned} Q(\textbf{x}) - Q(\textbf{x}^*) &= \nabla Q(\textbf{x}+(1-c)\textbf{x}^*)^T \cdot (\textbf{x} - \textbf{x}^*) \\ &= Q(\textbf{x}^c)^T \cdot (\textbf{x} - \textbf{x}^*) \\ &= \sum _{i=2}^n \frac{\partial Q(\textbf{x}^c)}{\partial x_i} x_i^c + \frac{\partial Q(\textbf{x}^c)}{\partial x_1} (x_1^c - B) \\ &\le \frac{\partial Q(\textbf{x}^c)}{\partial x_1} \sum _{i=2}^n x_i^c + \frac{\partial Q(\textbf{x}^c)}{\partial x_1} (x_1^c - B) \\ &= \frac{\partial Q(\textbf{x}^c)}{\partial x_1} ( \sum _{i=1}^n x_i^c - B) \\ &\le 0. \end{aligned}$$

Here, $c \in (0,1)$, and we denote $\textbf{x}+(1-c)\textbf{x}^*$ as $\textbf{x}^c$. The first inequality exists because $\frac{\partial Q(\textbf{x}^c)}{\partial x_i} \le \frac{\partial Q(\textbf{x}^c)}{\partial x_1}$, and the last inequality is due to the budget constraint. $\Box$

Proof of Lemma B.4:

Proof: Let $T_k^* = \cup _{i=1}^k S_i^*$. We have

$$\begin{aligned} R(\mathcal {S}^*, \textbf{x}^*) &= \sum _{k=1}^m \cdot \frac{W(T_k^*,\textbf{x}^*) - W(T_{k-1}^*,\textbf{x}^*)}{(1+V(T_{k-1}^*,\textbf{x}^*))(1+V(T_{k}^*,\textbf{x}^*))} \nonumber \\ &= \sum _{k=1}^{m-1} \frac{W\left( T_{k}^{*},\textbf{x}^*\right) }{1+V\left( T_{k}^{*},\textbf{x}^*\right) }\left\{ \frac{1}{1+V\left( T_{k-1}^{*},\textbf{x}^*\right) }-\frac{1}{1+V\left( T_{k+1}^{*},\textbf{x}^*\right) }\right\} +\frac{ W\left( T_{m}^{*},\textbf{x}^*\right) }{\left( 1+V\left( T_{m-1}^{*},\textbf{x}^*\right) \right) \left( 1+V\left( T_{m}^{*},\textbf{x}^*\right) \right) } \nonumber \\ &\le \max _{S,\textbf{x}} \frac{W(S,\textbf{x})}{1+V(S,\textbf{x})} \left[ \sum _{k=1}^{m-1} \left\{ \frac{1}{1+V\left( T_{k-1}^{*},\textbf{x}^*\right) }-\frac{1}{1+V\left( T_{k+1}^{*},\textbf{x}^*\right) }\right\} +\frac{1 }{1+V\left( T_{m-1}^{*},\textbf{x}^*\right) }\right] \nonumber \\ &= \max _{S,\textbf{x}} \frac{W(S,\textbf{x})}{1+V(S,\textbf{x})} \left[ 1 + \frac{1}{1+V\left( T_{1}^{*},\textbf{x}^*\right) } -\frac{1}{1+V\left( T_{m}^{*},\textbf{x}^*\right) }\right] \\ &\le 2 \max _{S,\textbf{x}} \frac{W(S,\textbf{x})}{1+V(S,\textbf{x})} \\ &\le 2 R(\mathcal {S}^h, \textbf{x}^h). \end{aligned}$$

he last inequality holds because our heuristic method starts with the optimal solution of the single-stage problem and iteratively improves upon it. $\Box$