The Refined Assortment Optimization Problem
aa r X i v : . [ ec on . T H ] F e b The Refined Assortment Optimization Problem
Gerardo Berbeglia ∗ Alvaro Flores † Guillermo Gallego ‡ December 23, 2020
Abstract
We introduce the refined assortment optimization problem where a firm may de-cide to make some of its products harder to get instead of making them unavailableas in the traditional assortment optimization problem. Airlines, for example, offerfares with severe restrictions rather than making them unavailable. This is a moresubtle way of handling the trade-off between demand induction and demand cannibal-ization. For the latent class MNL model, a firm that engages in refined assortmentoptimization can make up to min( n, m ) times more than one that insists on traditionalassortment optimization, where n is the number of products and m the number ofcustomer types. Surprisingly, the revenue-ordered assortment heuristic has the sameperformance guarantees relative to personalized refined assortment optimization as itdoes to traditional assortment optimization. Based on this finding, we construct refine-ments of the revenue-order heuristic and measure their improved performance relativeto the revenue-ordered assortment and the optimal traditional assortment optimizationproblem. We also provide tight bounds on the ratio of the expected revenues for therefined versus the traditional assortment optimization for some well known discretechoice models. ∗ Melbourne Business School, The University of Melbourne, Australia, [email protected]. † Department of Innovation, HIVERY, Sydney, Australia, Australia, [email protected]. ‡ Industrial Engineering and Decision Analytics, Hong Kong University of Science and Technology,Kowloon, Hong Kong, [email protected]. Supported by RGC project 16211619. Introduction
Discrete choice models are of interest to both industry and academia as they can accuratelycapture demand substitution patterns that enable firms to offer a better mix of productsto consumers, see Thurstone [1927], Anderson et al. [1992], Luce [1959], Plackett [1975],McFadden [1978], Guadagni and Little [1983], McFadden and Train [2000], Ben-Akiva and Lerman[1985]. There is a growing body of literature supporting the idea that using discrete choicemodels leads to better sales outcomes [Talluri and Van Ryzin, 2004, Vulcano et al., 2010,Farias et al., 2013]. Parametric discrete choice models yield more accurate estimates thanmachine learning techniques Feldman et al. [2018] unless data is abundant and the groundtruth model cannot be easily captured by a parametric model [Chen et al., 2019b]. Para-metric models have the additional advantage that they are amenable to optimization ofthe firm’s objective such as expected sales or expected revenues.A key problem at the center of e-commerce and revenue management is the traditionalassortment optimization problem (TAOP), which requires finding a subset of products thatthe firm should offer to maximize expected revenues (or profits), see Talluri and Van Ryzin[2004], K¨ok et al. [2005], M´endez-D´ıaz et al. [2014], Vulcano et al. [2010]. The tradeoff isbetween marginal revenues and demand cannibalization. Indeed, it is optimal to excludeproducts when their revenue contribution is lower than the revenue losses due to demandcannibalization so the firm can improve profits by redirecting part of the demand of theexcluded products to other more profitable products in the assortment.In practice, firms may prefer a more subtle approach that make some products harderto get rather than unavailable. This avoids some of the feared cannibalization withoutcompletely losing their revenue potential. As an example, a human resource firm in HongKong that specializes in placing domestic helpers does not disclose its entire list to cus-tomers. Instead, they first offer a list of helpers that have been in the system for a while.Customers that reject the first batch typically get a second list of better qualified helpers.Restricted fares in revenue management are a compromise between offering unrestrictedlow-fares and not offering them at all. By imposing time-of-purchase and travel restrictionsthey make these fares less desirable for people who travel for business allowing the airlineto obtain revenue from seats that otherwise would fly empty without severely cannibalizingdemand for higher-fares. Rolex, makes its steel sport watches extremely difficult to find,steering impatient customers to buy gold watches instead.The examples above are instances of refined assortment optimization , which we willmodel and study in detail in this paper. While our stylized model is general in nature, theexact product modification is context dependent and it can range from temporal delays ondelivery or handover of the product, to obscuring or removing some attributes to make someproducts less desirable. Firms are sometimes forced to display products in a way that makesthem less desirable due to a limit on the number of prime display locations. This is knownas the product framing problem, see Tversky and Kahneman [1989], Feng et al. [2007],Craswell et al. [2008], Kempe and Mahdian [2008], Aggarwal et al. [2008] and Gallego et al.22020]. Although similar to the refined assortment optimization problem, the crucial differ-ence is that in the product framing problem the refinements are a consequence of exogenousconstraints, whereas these are endogenous decisions in the refined assortment optimization.
This paper is, to our knowledge, the first to study refined assortment optimization. Ourcontributions are as follows. • We introduce the
Refined Assortment Optimization Problem (RAOP) and the
Per-sonalized Assortment Optimization Problem (p-RAOP) as two continuous optimiza-tion problems (Section 2). • We show that when consumers follow the latent class multinomial logit (LC-MNL)with m segments and n products, the seller can make up to a factor min { m, n } morewith the RAOP than with the TAOP. As a consequence, when consumers follow anyrandom utility model, the seller could make up to n times more by using RAOPinstead of TAOP. For the Random Consideration Set (RCS) model, the firm canmake up to a factor of 2 more with the RAOP relative to the TAOP. We provideexamples to show that these bounds are tight (Section 3). • Perhaps surprisingly, we show that the tight revenue guarantees for revenue-orderedassortments against the TAOP obtained in Berbeglia and Joret [2020] hold verbatimfor the RAOP and even for the personalized RAOP where the firm can use a separaterefine assortment for each consumer segment (Section 3). • We introduce three heuristics for the RAOP which are refinements of the traditionalrevenue-ordered assortments (Section 4). We performed a series of computationalexperiments using the latent-class MNL (LC-MNL) instances available from the lit-erature. Our numerical results show that these polynomial heuristic often yield higherexpected revenues than the optimal solution for the TAOP (which is NP-hard) indi-cating that there are many situations where firms may benefit from a strategic use ofthe RAOP where the RAOP identifies the products that need to be refined, and oncethey are refined the problem can be solved as a TAOP until the prevailing conditionschange and new choice models are fitted at which point the RAOP may need to becalled again.Besides the main contributions stated above, we introduce the sequential assortmentcommitment problem (SACP) in which a seller wishes to sell items to consumers over afinite time horizon by committing to an assortment schedule. We show that there is astrong connection between this problem and the RAOP.3
Model
A discrete choice model is a function that maps a mean utility vector u = ( u , u , . . . , u n )to non-negative numbers ( q ( u ) , . . . , q n ( u )) such that P ni =0 q i ( u ) = 1. We interpret q i ( u )as the probability that the consumer selects product i ∈ N := { , . . . , n } , and q ( u ) asthe probability that the consumer selects the outside alternative. We will assume withoutloss of generality that u = 0, so from now on when we refer to the vector u we willdrop the component corresponding to the outside alternative. A discrete choice model issaid to be regular if q i ( u ) is decreasing in u k , k = i . The class of regular discrete choicemodels includes the class of random utility models (RUM). The representative agent model(RAM) assumes that q ( u ) is as a solution to S ( u ) := max q ∈Q [ u ′ q − C ( q )], where Q is thesimplex, and C is a function that penalizes the concentration of probabilities [Spence, 1976,Hofbauer and Sandholm, 2002]. Feng et al. [2017] shows that the RAM contains the classof all RUMs and that q ( u ) is regular when S is sub-modular.Let { , } ⊆ Θ i ⊆ [0 ,
1] for all i ∈ N , and let Θ := Θ × . . . × Θ n . For fixed u let˜ u i := u i + ln( x i ). For each x ∈ Θ, let p ( x ) := q (˜ u ) = q ( u + ln( x )). Notice that if q isregular then p i ( x ) is decreasing in x k , k = i . Let r be the unit profit contribution vector.We will refer to r as the vector of revenues, keeping in mind the broader interpretation asthe profit contribution vector. Let R ( x ) := r ′ p ( x ). The refined assortment optimizationproblem (RAOP) is given by: R ( u | Θ) := max x ∈ Θ R ( x ) . (1)The case Θ i = { , } for all i ∈ N reduces to the traditional assortment optimizationproblem (TAOP), while Θ i = [0 ,
1] for all i ∈ N is the fully flexible RAOP. For economy ofnotation, we write R ∗ ( u ) as the optimal expected revenue for TAOP and ¯ R ( u ) for the fullyflexible RAOP. We will write R ∗ and ¯ R except when we need to make the dependence on u explicit.When there are multiple customer types, the firm’s objective is to maximize R ( x ) = P j ∈ M θ j R j ( x ) where M := { , . . . , m } is the set of consumer types, and θ j is the proportionof type j customers. If the firm can engage in personalized RAOP (p-RAOP) the firm canearn ¯ R p = P j ∈ M θ j ¯ R j where ¯ R j is the optimal expected revenue for type j customers.Assume without loss of generality that the products are sorted in decreasing order of the r i s, so r ≥ r ≥ · · · r n ≥ r n +1 := 0. Let e k be the k th unit vector and for each i ∈ N ,define e i := P k ≤ i e k . Let R o := max i ∈ N R ( e i ) be the maximum expected revenue amongthe class of revenue-ordered assortments. Clearly R o ≤ R ∗ ≤ R ( u | Θ) ≤ ¯ R ≤ ¯ R p . As we will see, RAOP can significantly improve revenues for the firm, and in some casesit can also increase consumer surplus.
Example 1.
Consider a firm that has two products to offer with profit contributions r =3 . , r = 1 . There are two customer types, each following a maximum utility model. Type customers have utility vector u = (1 , , . , while Type 2 customers have utility vector u = ( − , . . Let θ = . , θ = . be the market shares of the two customer types. Theoptimal solution for the TAOP is x = (1 , , resulting in p ( x ) = ( θ , , R ∗ = 3 . × . . , and expected consumer surplus . , with type 2 customers left out. The RAOPwith Θ = { , } and Θ = { , . , } results in x = (1 , . , p ( x ) = ( θ , θ ) , ¯ R = 1 . ,and expected consumer surplus . . This represents a 66.6% improvement in profits forthe firm and a 214.7% increase in consumer surplus. All of the gains come from type 2customers that buy product 2 at x = 0 . , a level that does not cannibalize the demand forproduct 1 from type 1 customers. Example 2.
Suppose a firm has three products to offer with profit contributions r =100 , r = 65 , r = 58 . Consumers follow a latent class MNL model with 2 segments ofequal weight (i.e. θ = θ = 0 . ). Let u = (0 . , , . and u = (100 , , . denote the mean utility vector for segments 1 and 2 respectively. The optimal solution forthe TAOP is x = (1 , , , p ( x ) = (0 . , . , , and R ∗ = 66 . . The RAOP with Θ = Θ = { , } and Θ = [0 , . results in x = (1 , . , , p ( x ) = (0 . , . , . ,and ¯ R = 71 . . This represents an improvement of over 7% in profits for the firm. Notethat product 3 was added in full under the RAOP. The RAOP may refine products takenin full by the TAOP and fully add products rejected by the TAOP. Notice that the flexibility afforded by the RAOP may have an impact in unit costs asa result in the change in utility. In many applications we expect the change in unit coststo be negligible. This is often true when modifying services, where imposing purchase ortravel restrictions have zero or close to zero marginal costs. For applications where thechange in unit costs are not negligible, we assume that any changes in costs are passedto the consumer so that the unit revenues (profit contributions) are unchanged. Moreprecisely, if the firm modifies the utility of product i by δu i at a cost of δc i , we assume thatthe firm adjusts prices by δc i , resulting in δr i = 0. Then ˜ u i = u i + δu i − β i δc i , where β i isthe price-sensitivity of product i . The price adjustment δc i may be suboptimal, but it isa reasonable approximation if the passthrough rate is close to one and/or δc i /c i is small.As TAOP, the RAOP is a lower bound on what the firm can make if it had more pricingfreedom.We end this section by introducing the Sequential Assortment Commitment Problem (SACP) as a special case of the RAOP. In the SACP a firm commits to sequentially of-fer, possibly empty, assortments A s ⊂ N , s ∈ T := { , . . . , t } . The objective of the firmis to maximize expected profits from consumers who follow a discrete choice model andhave mean valuations u is , i ∈ N, s ∈ T with u is decreasing in s . Unlike recent sequentialassortment problems proposed where consumers are impatient and buy the first satisfy-ing product they find [Chen et al., 2019a, Fata et al., 2019, Flores et al., 2019, Liu et al.,2020], the SACP considers forward-looking consumers who optimally time their purchaseto maximize their utility. The SACP is a special case of the RAOP as the firm can solvea RAOP with Θ i = { u is , s ∈ T } ∀ i ∈ N to obtain a solution for the SACP.5 Tight bounds
The purpose of this section is two fold. First, we will show that the well-known revenue-ordered assortments heuristic possesses the same revenue guarantees for the RAOP asfor the TAOP. Surprisingly, these revenue guarantees (which are already tight under theTAOP) also apply to the personalized version of the RAOP (p-RAOP) in which the firmcan customize the refine assortments to each of the consumer segments. Second, we presenttight bounds on how much more revenue the firm can earn under the RAOP with respectto the TAOP for regular choice models, the MNL, the LC-MNL, and the Random Consid-eration Set model.
Before showing a tight bound on the performance guarantees of the revenue-ordered as-sortment heuristic, we need the following technical lemma.
Lemma 1.
Let x ∈ [0 , n and y ∈ { , } n . If p is regular, then y ′ p ( x ) ≤ y ′ p ( y ) .Proof. Let P ( x ) := P i ∈ N p i ( x ) = 1 − p ( x ). By regularity, p ( x ) is decreasing in x ,implying that P ( x ) is increasing in x . Then y ′ p (min( x, y )) = P (min( x, y )) ≤ P ( y ) = y ′ p ( y ) , so it is enough to show that y ′ p ( x ) ≤ y ′ p (min( x, y )) but this follows from regularity as p i ( x ) ≤ p i (min( x, y )) for all i such that y i = 1. Theorem 1. If p j is a regular discrete choice model for all j ∈ M , then R ∗ ≤ ¯ R ≤ ¯ R p ≤ ω n R o ≤ ω n R ∗ , where ω n := n − ( n − α / ( n − and α = r n /r . Moreover, ω n is increasingin n , with ω n → − ln( α ) as n → ∞ .Proof. The only non-trivial inequality is ¯ R p ≤ ω n R o . Let x j ∈ [0 , n be an optimalsolution of RAOP for segment j , and let p jk ( x ) denote the probability that customer j selects product k under the refined assortment x j . Then6 R p = X j ∈ M θ j ¯ R j = X i ∈ N r i X j ∈ M θ j p ji ( x j ) = X i ∈ N ( r i − r i +1 ) X j ∈ M θ j X k ≤ i p jk ( x j ) ≤ X i ∈ N ( r i − r i +1 ) X j ∈ M θ j X k ≤ i p jk ( e i ) ≤ X i ∈ N r i − r i +1 r i X j ∈ M θ j X k ≤ i r k p jk ( e i )= X i ∈ N r i − r i +1 r i R ( e i ) ≤ X i ∈ N r i − r i +1 r i R o ≤ ω n R o The first inequality follows from Lemma 1. The second from the ordering of the r i s andthe third one from R ( e i ) ≤ R o . The last inequality follows from maximizing P i ∈ N ( r i − r i +1 ) /r i subject to r n +1 = 0 < r n ≤ . . . ≤ r subject to r n /r = α . This yields r k = r α ( k − / ( n − resulting in ω n = P i ∈ N ( r i − r i +1 ) /r i = n − ( n − α / ( n − = 1 + ( n − − α / ( n − ] as claimed. We leave it to the reader to verify that the term ω n − n − − α / ( n − ] increases with n and converges to − ln( α ).Theorem 1 extends results by Berbeglia and Joret [2020] from the TAOP to the p-RAOP. In the proof of Theorem 1 we allowed the r i s to be all different. If the vector r has only k < n different values, the bound can be improved to ω k = k − ( k − α / ( k − with α unchanged resulting in a tighter bound. Similar arguments can be used to showthat ¯ R p ≤ η R o where η = (1 + ln( Q n /Q )), Q is the probability that product 1 sellsunder the optimal p-RAOP, and Q n is the probability that something sells under the samepolicy. All of these bounds were shown to be exactly tight relative to the TAOP so theyare automatically tight for the RAOP and for the p-RAOP. We say that R satisfies the monotone-utility property if R ∗ ( u ) = R ( u |{ , } n ) is increasingin u . This implies that ¯ R = R ∗ , so refined assortment optimization is not useful when themonotone-utility property holds. Lemma 2.
The MNL satisfies the monotone-utility property. R ( x ) = P j ∈ M θ j R j ( x ) does not inherit the monotone-utility property from the R j , j ∈ M .The next theorem gives an upper bound on ¯ R , and the Proposition below shows that it istight. Theorem 2. If R j satisfies the monotone-utility property for all j ∈ M , then ¯ R ≤ m R ∗ .Proof. Let x ∗ be an optimal solution for the RAOP and let ˜ u j = u j + ln( x ∗ ) for all j ∈ M .Then ¯ R ( u ) = R ( x ∗ )= X j ∈ M θ j R j ( x ∗ ) ≤ X j ∈ M θ j R ∗ j (˜ u j ) ≤ X j ∈ M θ j R ∗ j ( u j ) ≤ m max j ∈ M θ j R ∗ j ( u j ) ≤ m R ∗ . The first inequality follows by solving a RAOP for each segment starting from ˜ u j = u j +ln( x ). The second from the assumption that R ∗ j ( · ) is an increasing function. The third isby picking the best segment and the fourth from max j ∈ M θ j R ∗ j ( u ) ≤ R ∗ ( u ).By Lemma 2, Theorem 2 applies to the LC-MNL. The next result shows that the boundis arbitrarily tight for the LC-MNL. Since the LC-MNL is regular, we see that the boundfrom Theorem 1 is arbitrarily close by setting m = n . Proposition 1.
For every n and every m , there exists a LC-MNL instance with n productsand m segments such that ¯ R is arbitrarily close to min { m, n }R ∗ . Since all RUM can be approximated by the LC-MNL our results imply that over thatclass of discrete choice models, the bound ˆ
R ≤ m R ∗ is arbitrarily close for all m ≤ n . Sincethe MNL is a regular model, the bound for the previous section applies, so we concludethat for all RUMs, the bound ˆ R ≤ min( m, n ) R ∗ is arbitrarily close. It is worth to mentionthat similarly to the TAOP, the RAOP under the LC-MNL is NP-hard [D´esir et al., 2020] . We now provide tight bounds on the benefits of using RAOP with respect to TAOP whenconsumers follow the random consideration set (RCS) model introduced by Manzini and Mariotti In their study of the TAOP under the LC-MNL, D´esir et al. [2020] considered its continuous relaxationand showed that there is no efficient algorithm that can achieve an approximation factor guarantee of atleast O ( m − δ ) for any constant δ > NP ⊂ BP P . ≺ ≺ . . . ≺ n ,and product i has attention probability λ i ∈ [0 , S is offered, then product i is selected with probability q i ( S ) = λ i Π j>i (1 − λ j δ ( j ∈ S )) , where δ ( j ∈ S ) = 1 if j ∈ S and 0 otherwise, and products over empty sets are defined as 1.Gallego and Li [2017] showed that the RCS can be approximated exactly by a the MarkovChain model of Blanchet et al. [2016] and developed an algorithm to find an optimal as-sortment that runs in O ( n ) time. Here we consider the refined assortment optimizationversion of the RCS model. A reduction in the utility of product i can result both in areduction of the attention probability λ i and may bring product i lower in the preferenceordering. Those two issues need to be considered carefully in solving the RAOP for theRCS model. We show that the RAOP can at most double expected revenues relative tothe TAOP under the mild assumption that attention probabilities are non-decreasing inthe products utilities. This means that the RAOP under the RCS model has a 2-factorapproximation algorithm. Theorem 3.
For every RCS, ¯ R ≤ R ∗ . Here we briefly summarize the key steps of the proof, and provide some managerialinsights. First, we show that if the preference orders are unchanged, then the optimalexpected revenue of the RCS model is increasing in the attention probabilities. Second,we show that if the products are sorted in increasing order of the r i s, we can identify thebest and the worst preference orderings in terms of the optimal expected revenue undertraditional assortment optimization. The best preference order is 1 ≺ ≺ . . . ≺ n , and theworst is n ≺ n − ≺ . . . ≺
1. This result is intuitive as consumers have preferences for theproducts with the highest revenues in the former preference order and for the lowest in thelatter. This suggest that refined assortment optimization is about reducing the attentionprobability of low revenue products to make them less attractive in the preference ordering.We end this section by showing that the factor 2 bound is tight fort the RCS model.
Proposition 2.
Theorem 3 is tight.Proof.
Consider a RCS model with n = 2, λ = ǫ , λ = 1, r = ǫ − and r = 1 forsome small ǫ >
0. Suppose that, if products are shown in full, consumers prefer product2 over product 1: 1 ≺
2. Thus, under the TAOP, the firm expected revenue is then R ∗ = max { λ r + (1 − λ ) λ r , λ r } = max { , } = 1. Now suppose that if the firmslightly decreases the utility of product 1, and as a consequence the consumer preferenceorder is reversed to 2 ≺ . Then, we havethat ˆ R ∗ = λ r + (1 − λ ) λ r = 1 + (1 − ǫ )1 = 2 − ǫ . The tightness follows by taking thelimit ǫ → Alternatively one can imagine that product 1 attention probability is reduced by some very smallpositive number and take the limit to zero. Refined Revenue-Ordered Heuristics
In this section we propose three heuristics for the RAOP that are refined versions of therevenue-ordered heuristic. They all enjoy the guarantees from Theorem 1 as they areat least as good as the best revenue-ordered assortment. The first heuristic consists offinding the best revenue-ordered assortment in which the utility of the lowest revenueproduct offered is optimized. We called this, the
Revenue Ordered Heuristic with onepartial product . RO1: Revenue-Ordered Heuristic with one partial product
1. For each i ∈ N let x ∗ i := arg max R ( e i − + x i e i ) and compute R ( e i − + x ∗ i e i ).2. Let k = arg max i ∈ N R ( e i − + x ∗ i ).3. Return e k − + x ∗ k e k and R ( e k − + x ∗ k e k ).This heuristic involves solving n optimization problems over a single real variable over[0 , R ( e k − + x ∗ k e k ) ≥ R ( e k ) for all k ∈ N so its performance is at least asgood as the RO heuristic. The second heuristic builds upon the first by potentially addingmore partial products. RO2: Revenue Ordered Heuristic with several partial products
1. For each i ∈ N , compute x ∗ ii = arg max R ( e i − + x i e i )2. For k = i + 1 , . . . , n compute x ∗ ik = arg max R ( e i − + x ∗ ii e i + . . . + x ∗ i,l − e l − + x ik e k ) .
3. Let k = arg max i ∈ N R ( e i − + x ∗ ii e i + . . . + x ∗ in e n )4. Return e k − + x ∗ kk e k + . . . + x ∗ kn e n and R ( e k − + x ∗ kk e k + . . . + x ∗ kn e n ).This heuristic requires to solve O ( n ) optimization problems over a single [0 ,
1] variable.Clearly R ( e k − + x ∗ kk e k + . . . + x ∗ kn e n ) ≥ R ( e k − + x ∗ k e k ) so this heuristic is the never worsethan the RO1.The third heuristic is similar to the previous one, but starting from a revenue orderedassortment, instead of evaluating sequentially in a revenue decreasing way among the can-didates for being considered as partial, we greedily select the product that by partiallyadding it to the current solution, increases the revenue the most. We call this exten-sion the Revenue-Ordered Greedy-Heuristic with several partial products) . This heuristicrequires to solve O ( n ) optimization problems over a single [0 ,
1] variable, and for thatreason we refer to it as
RO3 . Observe that while is clear that this heuristic is better thanRO1 it can be worst than RO2 since the greedy nature might make the algorithm to selecta product in the sub-routine that in hindsight was not the best choice. The performanceof these heuristics is reported in Section 5. 10 .1 Upper Bounds for the LC-MNL model
In this section we propose two easy to compute upper bounds for the LC-MNL. The LC-MNL is important because every random utility model can be approximated as accuratelyas desired by a LC-MNL. Since there is no polynomial algorithm to compute R ∗ we presenthere two easy to compute upper-bounds that can help evaluate heuristics. The first oneis the solution to the p-RAOP, ¯ R ≤ ¯ R p = P j ∈ M θ j R j which is easy to compute since aRO assortment is optimal for each market segment. The second upper bound is based onwriting the objective function of R ∗ as a fractional program, which results in a bi-linearprogram that can be linearized to obtain the following linear programming upper bound:¯ R u := max x,y,z X j ∈ M θ j X i ∈ N r i v ij z ij subject to the constraints v j y j + X i ∈ N v ij z ij = 1 ∀ j ∈ Mz ij ≥ x i ∀ i ∈ N, ∀ j ∈ Mz ij ≤ x i /v j ∀ i ∈ N, ∀ j ∈ Mz ij ≤ x i + y j − ∀ i ∈ N, ∀ j ∈ Mz ij ≥ ( x i − /v j + y j ∀ i ∈ N, ∀ j ∈ M ≤ x i ≤ ∀ i ∈ N ≤ y j ∀ j ∈ M ≤ z ij ∀ i ∈ N, ∀ j ∈ M If the solution satisfies z ij = x i y j for all i ∈ N, j ∈ M then the upper bound yieldsan optimal solution. This is similar to the mixed-integer linear formulation studied in[Bront et al., 2009, M´endez-D´ıaz et al., 2014, Feldman and Topaloglu, 2015], but in ourcase x i is also allowed to take continuous values. S¸en et al. [2018] proposed a conic for-mulation for the LC-MNL for the TAOP which relies, as we do above, on McCormickinequalities [McCormick, 1976]. 11 Numerical Results
In this section, we evaluate the performance of the heuristics proposed in Section 4. Sinceany random utility model can be approximate by a LC-MNL to any arbitrary precision[McFadden and Train, 2000, Chierichetti et al., 2018] it is natural to study the performanceof the heuristic on the LC-MNL. We used synthetic instances of the LC-MNL for ourexperiments testing instances with n ∈ { , , , , } and m ∈ { , , , , } . Togenerate the product utilities, we applied a procedure used by Berbeglia et al. [2018] whichdepends on a parameter ǫ > j ∈ M , we generate arandom permutation σ j over the products including the outside option, that we associate toclass j . The exponentiated utility of product i ∈ N ∪ { } for a type j customer is modeledas: v ij = ǫ σ − j ( i ) , where σ − m ( i ) denotes the position of product i in the permutation σ j .The value of ǫ ∈ [0 ,
1] measures how similar or different the utilities are for consumersof a specific class. A value of ǫ close to zero, implies different utilities are very different,whereas a value of ǫ close to 1 makes them very similar.Product prices are sampled from different distributions: uniform distribution U (1 , N (50 , α = r n /r , we scale the resultsobtained from the sampled prices, and set the minimum price to match 1, and then themaximum price to be either 5, 10 or 100 resulting in α ∈ { . , . , . } .For each instance we first assign prices at random to products, without considering theirrelation with product attractiveness across segments. We also consider instances whererandom prices are aligned with the exponentiated utilities and aligned in the oppositeorder, so products with higher prices have lower exponentiated utilities.For each instance, we report the average (and the maximum) expected revenue forthe three RAOP heuristics and the optimal revenue R ∗ for the TAOP obtained throughenumeration . We present the performance of the heuristics relative to the best revenue-ordered assortment. Cells marked with an asterisk indicate that the heuristic outperformedthe optimal solution for the TAOP, suggesting that polynomial heuristics for the RAOP canoften improve over the TAOP without having to solve a difficult combinatorial problem. We report this only for values of n <
50, as the solution by enumeration was too computationallyexpensive to compute. O1 RO2 RO3 R ∗ uniformnormalmulti-modalexpsknorm n=5, m=2 RO1 RO2 RO3 R ∗ n=5, m=10 RO1 RO2 RO3 R ∗ n=5, m=50 RO1 RO2 RO3 R ∗ n=5, m=100 RO1 RO2 RO3 R ∗ uniformnormalmulti-modalexpsknorm n=10, m=2 RO1 RO2 RO3 R ∗ n=10, m=10 RO1 RO2 RO3 R ∗ n=10, m=50 RO1 RO2 RO3 R ∗ n=10, m=100 RO1 RO2 RO3 R ∗ uniformnormalmulti-modalexpsknorm n=15, m=2 RO1 RO2 RO3 R ∗ n=15, m=10 RO1 RO2 RO3 R ∗ n=15, m=50 RO1 RO2 RO3 R ∗ n=15, m=100 RO1 RO2 RO3 R ∗ uniformnormalmulti-modalexpsknorm n=50, m=2 RO1 RO2 RO3 R ∗ n=50, m=10 RO1 RO2 RO3 R ∗ n=50, m=50 RO1 RO2 RO3 R ∗ n=50, m=100 RO1 RO2 RO3 R ∗ uniformnormalmulti-modalexpsknorm n=100, m=2 RO1 RO2 RO3 R ∗ n=100, m=10 RO1 RO2 RO3 R ∗ n=100, m=50 RO1 RO2 RO3 R ∗ n=100, m=100 Average performance (and Maximum) of heuristics against the best revenue ordered assortment.( ǫ = 0 . , ρ = 100 and random price to attractiveness relation). Heuristics D i s t r i bu t i o n s Figure 1: Mean (max) performance of heuristics and the optimal of TAOP relative to RO.For each heuristic, the value is highlighted with an asterisk if it outperformed the optimalof TAOP on average.Figure 1 shows the performance relative to the RO as a baseline for a set-up withhigh variation among products attractiveness ( ǫ = 0 .
01) and α = 0 .
01. This set up hastremendous variability in the utilities and a large ratio of the largest to the lowest valueof r . The heuristics consistently found refined assortments for the RAOP that outperformoptimal assortments for the TAOP. RO2 and RO3 present a growing gap with respect toRO as the number of products grows, while RO1 stays close to RO as it can only refineone product. When the number of customer types is large, the gap between RO and13 ∗ tightens, while the other heuristics significantly outperform R ∗ on average, with thegap growing in n . The heuristics thrive in settings where there is higher concentration ofproducts with high revenue since there are more opportunities of refining lower revenueproducts that cannibalize products with higher revenues. In addition to the observationsconcerning the average performance of the heuristic, we remark that the maximum overthe instances can significantly outperform the RO heuristic and are often equal or betterthan the maximum for the TAOP. RO1 RO2 RO3 R ∗ uniformnormalmulti-modalexpsknorm n=5, m=2 RO1 RO2 RO3 R ∗ n=5, m=10 RO1 RO2 RO3 R ∗ n=5, m=50 RO1 RO2 RO3 R ∗ n=5, m=100 RO1 RO2 RO3 R ∗ uniformnormalmulti-modalexpsknorm n=10, m=2 RO1 RO2 RO3 R ∗ n=10, m=10 RO1 RO2 RO3 R ∗ n=10, m=50 RO1 RO2 RO3 R ∗ n=10, m=100 RO1 RO2 RO3 R ∗ uniformnormalmulti-modalexpsknorm n=15, m=2 RO1 RO2 RO3 R ∗ n=15, m=10 RO1 RO2 RO3 R ∗ n=15, m=50 RO1 RO2 RO3 R ∗ n=15, m=100 RO1 RO2 RO3 R ∗ uniformnormalmulti-modalexpsknorm n=50, m=2 RO1 RO2 RO3 R ∗ n=50, m=10 RO1 RO2 RO3 R ∗ n=50, m=50 RO1 RO2 RO3 R ∗ n=50, m=100 RO1 RO2 RO3 R ∗ uniformnormalmulti-modalexpsknorm n=100, m=2 RO1 RO2 RO3 R ∗ n=100, m=10 RO1 RO2 RO3 R ∗ n=100, m=50 RO1 RO2 RO3 R ∗ n=100, m=100 Average performance (and Maximum) of heuristics against the best revenue ordered assortment.( ǫ = 0 . , ρ = 100 and random price to attractiveness relation). Heuristics D i s t r i bu t i o n s Figure 2: Mean (max) performance of heuristics and the optimal of TAOP relative to RO.For each heuristic, the value is highlighted with an asterisk if it outperformed the optimalof TAOP on average. 14igure 2 shows results for cases with lower variation of the exponentiated utilities( ǫ = 0 . R ∗ in most cases, but the gap is tighter. Forhigher values of m , the average and maximum uplift of the heuristics is slightly higher thanin Figure 1. For lower values of m , we see an increasing benefit of using the heuristics.At a coarser level, Table 5 shows that RO2 and RO3 perform on average better than R ∗ .RO3 performs better with 5 products, but for higher values of n is on average dominatedby RO2, which dominates R ∗ across the board. We also evaluated the effect on marketmarket share and consumer surplus. RO2 had on average a mild increase of 0 .
38% inmarket relative to R ∗ . The difference in market share decreased with n and grows with m . Note also that RO3, being a greedy heuristic, over-performed RO2 for m = 2 acrossthe board. This can be explained as it is less likely that the greedy heuristic goes wrongwhen m is small. RO2 is more conservative doing modifications in strict order by revenue,which protects the current solution being assessed causing a better performance overall for m >
2. RO2 is also computationally faster and recommended except when m is small.Table 1: Average performance of heuristics and the optimal TAOP against revenue orderedassortment, across all instances, while varying n and m . The maximum on each row ishighlighted with an asterisk. n m RO1 RO2 RO3 R ∗ R ∗ whenever they agree on the set of products that are offered in full.This is because consumers benefit from increased product availability under the RAOP.This agreement happens with relatively high frequency for the RO2 and RO3 (69 .
97% and68 .
83% respectively for n ≤
15) where their output offers the R ∗ set in full. We also noticedthat while this trend increases with the number of customer segments, it decreased withthe number of products. If the firm is determined to use the RAOP to improve consumersurplus, it can first solve the R ∗ heuristically or by enumeration and then determine whichproducts left out can be brought in partially. In this paper we proposed the refined assortment optimization problem and demonstratedthat it can substantially improve revenues for the firm. Moreover, if the refinement is usedonly on products excluded by the traditional assortment optimization problem, then theexpected consumer surplus goes up resulting in a win-win policy for the firm and its cus-tomers. We also developed refined revenue-ordered heuristics and showed that their worstcase performance relative to the personalized refined assortment optimization has the sameperformance guarantees that were previously known relative to the traditional assortmentoptimization problem. For special demand classes, such as the multinomial logit, and therandom consideration set model we showed that the benefits from personalized assortmentsrelative to traditional assortment optimization have a factor of 1 and 2, respectively. Someinteresting future research directions are: (1) quantifying the benefits of using RAOP withrespect to TAOP in other important choice models such as the Exponomial and the MarkovChain model; (2) a detailed study of the RAOP in the case where Θ i is a discrete set foreach i ∈ N ; (3) a best response analysis when a firm uses refined assortment optimizationunder competition; and (4) the study of the RAOP with cardinality constraints.16 eferences Gagan Aggarwal, Jon Feldman, Shanmugavelayutham Muthukrishnan, and Martin P´al.Sponsored search auctions with markovian users. In
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Proofs for Section 3 (Tight bounds)
Lemma 2.
The MNL satisfies the monotone-utility property.Proof.
We need to show that if x ∈ { , } n is an optimal assortment, then R ∗ ( u ) = R ∗ ( u, x ) is increasing in u . Set v i = exp( u i ) , i ∈ N . The choice probabilities can bewritten as p i ( v, x ) := v i x i / [ v + v ′ x ] and the optimal expected revenue can be written as R ∗ ( v ) := P i ∈ N r i p i ( v, x ). Elementary calculus yields ∂ R ∗ ( v ) /∂v i = p i ( v, x )[ r i − R ∗ ( v )] forall i such that x i = 1. Now r i ≥ R ∗ ( v ) for all i such that x i = 1 as otherwise it is optimalto eliminate i from the assortment. Thus R ∗ ( v ) and therefore R ∗ ( u ) is increasing in u i for all i such that x i = 1. If x i = 0 then r i < R ∗ ( v ) and ∂ R ∗ ( v ) /∂v i = 0 completing theproof. Proposition 1.
For every n and every m , there exists a LC-MNL instance with n productsand m segments such that ¯ R is arbitrarily close to min { m, n }R ∗ .Proof. We begin with the construction of a latent class MNL instance with k ≤ n consumersegments. Consider γ < ǫ < i for consumer type j to be u ji = i (1 + ǫ ) ln( γ ) + β ij + ε ij where i (1 + ǫ ) ln( γ ) is a segment independent utility component for product i , β ij is a segmentdependent utility component; and ε ij are independent Gumbel random variables. For j ≥ i we set β ij = ( j − i ) · ln( γ ). If j < i , β ij = M ln( γ ) where M = γ − is a large positivenumber. The no-purchase alternative 0 has utility u = 3 n ln( γ ).Observe that when γ is small enough, the product nominal utilities for consumer seg-ment j satisfy: u j > u j > . . . > u jj > u > Θ( M ) ln( γ ) and u jh = Θ( M ) ln( γ ) for all h = j + 1 , . . . , n .The revenue of product i is set to r i = ǫ i − for some 0 < ǫ <
1. Thus, the revenueincrease as a function of the product index with r = 1, r = ǫ − , r = ǫ − , etc.Customer segment j with j < k has a probability mass of λ j = ǫ j − − ǫ j . Observethat P k − j =1 λ j = 1 − ǫ + ǫ − ǫ + ǫ − ǫ + . . . + ǫ k − − ǫ k − = 1 − ǫ k − . Thus, ǫ k − is theprobability mass of customer segment k .Suppose we offer an assortment S without refining the utility of the products. Let ℓ ( S ) = min { i : i ∈ S } denote the product with smallest index in S . Consider now consumersegment j . When ℓ ( S ) ≤ j , we have that the probability that consumer type j buys ℓ ( S )when γ tends to zero is:lim γ → P j ( ℓ ( S ) , S ) ≥ lim γ → γ j + ǫ ℓ ( S ) P ji = ℓ ( S ) γ j + ǫ i + P ni = j +1 γ i (1+ ǫ )+ γ − + γ n = 1 . Above, we use the fact that P j ( i, S ) ≥ P j ( i, { , . . . , n } ) and that to calculate the limitin the right we only need to determine the term that has the smallest exponent. When20 ( S ) > j , the probability that consumer type j buys nothing when γ and ǫ tends to zerois lim γ → ,ǫ → P j (0 , S ) ≥ lim γ → ,ǫ → P j (0 , { j + 1 , . . . , n } ) = lim γ → γ n P ni = j +1 γ γ − + γ n = 1 . Therefore in the limit, for any non-empty assortment S ⊆ N , ℓ ( S ) is the only productthat has a non-zero probability of being purchased. Moreover, the probability of purchasing ℓ := ℓ ( S ) is P kj = ℓ λ j = ǫ ℓ − − ǫ ℓ + ǫ ℓ − ǫ ℓ +1 + . . . + ǫ k − = ǫ ℓ − . Thus, any assortmentachieves the optimal revenue of R ∗ = ( ǫ ℓ ( S ) − ) · ǫ ℓ ( S ) − = 1.Suppose now that the firm refines the segment-independent component of the product’s i utility from i (1 + ǫ ) ln( γ ) to n ln( γ ) for every i = 1 , . . . , n . Now the nominal utility forthe customer segment j becomes ˜ u ji = ( n + j − i ) ln( γ ) if j ≥ i . If j < i , the nominal utilityfor the customer segment j becomes ˜ u ji = ( n + M ) ln( γ ).Thus, the nominal utilities for consumer segment j now satisfy ˜ u jj > ˜ u jj − > . . . > ˜ u j > > ˜ u jj +1 = . . . = ˜ u jn = Θ( M ) ln( γ ). In the limit when γ →
0, by offering an assortment S with j ∈ S the probability that a consumer from segment j buys product j is:lim γ → P j ( j, S ) ≥ lim γ → P j ( j, N ) = γ n P ji =1 γ n + j − i + P ni = j +1 γ n + M + γ n = 1 . As a result, if the firm offers the assortment S = { , . . . , k } it can obtain k X j =1 λ j · r j = k − X j =1 ǫ j − − ǫ j ǫ j − + ǫ k − ǫ k − = ( k − − ǫ ) + 1 ≥ k (1 − ǫ ) R ∗ . When m ≤ n , setting k = m shows that the bound is tight. If m > n , the bound isalso tight since one can construct an LC-MNL instance with m segments where the first k = n segments are those in the above construction and the remaining m − k segments areassigned a zero-weight. Theorem 3.
For every RCS, ¯ R ≤ R ∗ .Proof. We omit the proof that the best and the worse preference orderings are, respectively,in the order and in the reverse order of the revenues, and a lemma that shows that therevenue of the RCS is monotonic on the attention probabilities. The details are availablefrom the authors upon request. We next proceed to the more difficult part of the resultthat bounds the expected revenue with the best order in terms of the expected revenue ofthe worst order.Assume, that r i , i = 1 , , . . . , n is a non-decreasing sequence of non-negative real num-bers and let λ i , i = 1 , , . . . , n be an arbitrary sequence of attention probabilities in210 , ≺ ≺ . . . , ≺ n . Let H := 0 and for k ≥
1, let H k := H k − + λ k ( r k − H k − ) for k = 1 , . . . , n . Then H n is the optimal expected revenue under this preference ordering, andan upper bound on any other preference ordering.We now compute the optimal expected revenue when the preference order goes in theopposite direction. Set G n = 0 and for k ∈ { , . . . , n } do G nk := G nk − + λ n +1 − k ( r n +1 − k − G nk − ) + and define G n := G nn . Then G n is the optimal expected revenue under this ordering,and a lower bound under any other preference ordering. Clearly G n ≤ H n . We now showthat H n G n ≤ ∀ n ≥ . Let f (1) := 0, f ( k ) := (1 − λ k )( λ k − + f ( k − k >
1. Define also the sequenceˆ f (1) = 0, ˆ f ( k ) := (1 − λ k +1 )( λ k + ˆ f ( k − k >
1. Notice that both f and ˆ f are of thesame form except for a shift in the index. We next show by induction that f ( k ) = ˆ f ( k −
1) + k Y j =2 (1 − λ j ) λ ∀ k ≥ . Moreover, f ( k ) ≤ − λ k , and ˆ f ( k ) ≤ − λ k +1 for all k . For k = 2, the left hand side is f (2) = (1 − λ ) λ while the right hand side is ˆ f (1) + (1 − λ ) λ = (1 − λ ) λ so the resultholds for k = 2. Suppose the result holds for k , so f ( k ) = ˆ f ( k −
1) + Q kj =2 (1 − λ j ) λ . Then f ( k + 1) = (1 − λ k +1 )( λ k + f ( k ))= (1 − λ k +1 )( λ k + ˆ f ( k −
1) + k Y j =2 (1 − λ j ) λ )= (1 − λ k +1 )( λ k + ˆ f ( k − k +1 Y j =2 (1 − λ j ) λ )= ˆ f ( k ) + k +1 Y j =2 (1 − λ j ) λ , where the first equality follows from the definition of f , the second from the inductivehypothesis, the third by the distributive property, and the fourth from the definition ofˆ f ( k ) = (1 − λ k +1 )( λ k + ˆ f ( k − f ( k ) ≤ − λ k . This holds for k = 1 as 0 ≤ − λ follows from λ ≤
1. Suppose it holds for k − f ( k − ≤ − λ k − , then f ( k ) = (1 − λ k )( λ k − + f ( k − ≤ − λ k completing the proof. A similar argumentapplies for ˆ f and is omitted.We will show that by induction that H n /G n ≤ f ( n ) ≤ − λ n . Assume theresult holds for all instances of size n −
1. Then the result holds for the instance of size22 − H n − and ˆ G n − be the optimal expected revenuescorresponding to the best and the worst orderings of { , . . . , n } , we see that the inductivehypothesis correspinds to ˆ H n − / ˆ G n − ≤ f ( n − n that includes product 1 under the assumption that r ≥ G n . Then r ≥ G n ≥ G nn − and G n = (1 − λ ) G nn − + λ r .Since ˆ G n − = G nn − we see that G n = (1 − λ ) ˆ G n − + λ r . We can also write H n in terms of ˆ H n − resulting in H n = ˆ H n − + n Y j =2 (1 − λ j ) λ r . Since the ratio H n G n = ˆ H n − + Q nj =2 (1 − λ j ) λ r (1 − λ ) ˆ G n − + λ r is decreasing in r ≥ ˆ G n − , it follows that it is maximized by setting r = ˆ G n − . For thischoice of r we have H n G n = ˆ H n − + Q nj =2 (1 − λ j ) λ ˆ G n − ˆ G n − . By the inductive hypothesis, ˆ H n − ≤ (1 + ˆ f ( n − G n − . Then, H n ≤ f ( n −