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A Nonconvex Min-Max Framework And ItsApplications to Robust Assortment Optimization

Antoine DesirTechnology and Operations Management Area, INSEAD, [email protected]

Vineet GoyalDepartment of Industrial Engineering and Operations Research, Columbia University, [email protected]

Bo JiangResearch Institute for Interdisciplinary Sciences, School of Information Management and Engineering, Shanghai University of

Finance and Economics, [email protected]

Tian XieResearch Institute for Interdisciplinary Sciences, School of Information Management and Engineering, Shanghai University of

Finance and Economics, [email protected]

Jiawei ZhangStern School of Business, New York University, [email protected]

Assortment optimization problems arise widely in many practical applications such as retailing and online

advertising. In this problem, the goal is to select a subset from a universe of substitutable items to offer to

customers in order to maximize the expected revenue. The demand for each item is captured by a choice

model. Like in any estimation problem, statistical errors are unavoidable when going from data to model.

Motivated by this fundamental issue, we take a natural description of the estimates as an uncertainty set

and aim to make decisions that are robust for any parameters in this set. In other words, we solve a robust

assortment optimization problem.

We present a framework that captures the robust assortment problem under a general class of choice

models (both the Multinomial logit (MNL) and Markov chain choice models fall in this framework). Our main

result is to show that, under certain reasonable assumptions, a min-max duality for the robust assortment

optimization for this class of choice models holds. This is surprising as our objective function does not satisfy

the properties usually needed for known saddle point results. Furthermore, we give efficient algorithms to

compute the optimal robust assortment under both the Markov chain and MNL models. Inspired by the

min-max relation, these algorithms are iterative in nature. Moreover, they yield operational insights towards

the effect of changing the uncertainty set on the optimal robust assortment. In particular, consistent with

previous literature, we find that bigger uncertainty set always lead to bigger assortment, and a firm should

offer larger assortments to hedge against uncertainty.

1

: A Nonconvex Min-Max Framework And Its Applications to Robust Assortment Optimization2 Article submitted to Operations Research; manuscript no. (Please, provide the manuscript number!)

1. Introduction

Assortment optimization problems arise widely in many practical applications such as retailing

and online advertising. In this problem, the goal is to select a subset from a universe of substi-

tutable items to offer to customers in order to maximize the expected revenue. The demand of

any item depends on the substitution behavior of the customers that is captured mathematically

by a choice model that specifies the probability a random consumer selects a particular item from

any given offer set. The objective of the decision maker is to identify an offer set that maximizes

expected revenue. Many parametric choice models have extensively been studied in the literature

in diverse areas including marketing, transportation, economics, and operations management. The

Multinomial logit (MNL) model is by far the most popular model in practice due to its tractability

(McFadden (1978), Talluri and Van Ryzin (2004)). However, some of the simplifying assumptions

behind this model, such as the Independence of Irrelevant Alternatives property, make it inade-

quate for many applications. To alleviate this limitation, a recent paper (Blanchet et al. (2016))

considers a Markov chain based choice model. In this model, customer substitution is captured by

a Markov chain, where each product (including the no-purchase option) corresponds to a state, and

substitutions are modeled using transitions in the Markov chain. The authors show that this model

provides a good approximation in choice probabilities to a large class of existing choice models.

We would like to note that a similar idea was already considered by Zhang and Cooper (2005)

in the context of airline revenue management. This alternative way to model customer behavior

has recently received a lot of attention. Feldman and Topaloglu (2017) study the network revenue

management problem under the Markov chain model. Simsek and Topaloglu (2017) propose a

method to estimate the parameters of the model from data. Desir et al. (2015) show that the con-

strained assortment problem is APX-hard and design efficient approximation algorithms to tackle

the constrained variant of the assortment problem. This paper adds to this stream of work and is

motivated by a robust variant of the assortment problem under the Markov chain model.

This paper is motivated by a fundamental issue that exists in using parameters estimated from

data for any underlying model. Statistical errors in such estimated parameters are unavoidable.

Therefore, “optimal” decisions based on the point estimators could potentially be highly sub-

optimal for the true parameters even when the true parameters are a small perturbation of our

estimate. This motivates a natural description of the estimates as a “confidence set” or an “uncer-

tainty set” corresponding high confidence region for the true parameters based on the statistical

estimation procedure. Given such an uncertainty set, the goal is to select decisions that are robust

to the setting of the true parameters in this set. Consequently, we consider the robust assortment

problem where the goal is to choose an assortment that maximizes the worst case expected revenue,

: A Nonconvex Min-Max Framework And Its Applications to Robust Assortment OptimizationArticle submitted to Operations Research; manuscript no. (Please, provide the manuscript number!) 3

where the worst case is taken over all likely parameter values. The next section gives the precise

mathematical framework that we use to study this problem. Note that our work is closely related to

Rusmevichientong and Topaloglu (2012) who consider a similar variant of the assortment problem

for the MNL model.

1.1. Model framework

Although our motivation is to study the robust assortment optimization problem under the Markov

chain model, we present a more general framework that captures the robust assortment optimiza-

tion problem under a broader class of choice models. Denote N = 1,2, . . . , n to be the set of n

products and product 0 as the no-purchase alternative with the convention that N+ =N ∪ 0.

Each of the n products is associated with a revenue (or price) rj ≥ 0. Let U be the uncertainty set

(possibly nonconvex) that the model parameter u is adversarially selected from. Given an assort-

ment S ⊆N and the model parameter u∈ U , the expected revenue generated by this assortment is

given by

R(S,u) =∑j∈N

λ(u)jvj

s.t. vj = rj,∀j ∈ S

A(u)Tj v= b(u)j,∀j ∈ SC ,

(1)

where A(u)Tj ∈ Rn is the jth row of A(u), b(u)j ∈ R, λ(u)j ∈ R for all j ∈ N and SC denotes the

complement of S, i.e. SC = j : j /∈ S.

We would like to contrast (1) with the traditional expression for the revenue function R(S) =∑i∈S riπ(i,S) where π(i,S) is the choice probability, i.e. the probability that product i is chosen

when the assortment S is offered. In light of this comparison, our setting can be thought of as

one where the choice probabilities are given as solutions to a system of linear equations. To be

more concrete and build some intuition, we briefly explain how this framework can capture the

Markov chain choice model. The exact mappings to the Markov chain and MNL choice models are

given in Section 3 and 4 respectively. For a fixed parameter u, it is useful to think of λ(u)j ∈ R

as the fraction of customers whose most preferred product is j, i.e., who would purchase product

j if present in the assortment. Using this clustering of the population, vj represents the revenue

from the customers who want to buy product j. If present in the assortment, vj is equal to rj,

the revenue of product j. Otherwise, the customers substitute to other products (possibly to the

outside option). The revenue generated by these substitutions is captured by the system of linear

equations A(u)Tj v = b(u)j for all j ∈ SC . Note that in the Markov chain model, the substitution

behavior is captured by the transitions in an underlying Markov chain. This is what the matrix

A(u) encodes in this case.


The explicit dependence on u is specified for each of the models (Markov chain and MNL) in

subsequent sections. We are interested in the robust assortment optimization problem where the

decision maker wants to maximize worst-case revenue. In other words, we are interested in the

following optimization problem:

maxS⊆N

minu∈U

R(S,u). (2)

As mentioned, our main result is to show that under suitable assumptions, a max-min relation

holds, i.e. that the order of the operators can be switched in (2). We next detail our contributions.

1.2. Our contributions

Our main contributions are as follows.

A general framework. As explicited in Section 1.1, we present a more general framework that

captures the robust assortment optimization problem under a broader class of choice models. In

particular, we consider a setting where the choice probabilities. and therefore the revenue function,

are given as solutions to a system of linear equations. A special case is the Markov chain choice

model, where the substitution behavior is captured by the transitions in an underlying Markov

chain and where the system of linear equations naturally arises as the balance equations in the

Markov chain. The matrix defining the system of linear equations is parametrised by an uncertainty

parameter belonging to an uncertainty set. The robust assortment optimization problem is to select

the assortment that achieves the best worst-case expected revenue over all possible parameters in

the uncertainty set.

Min-max duality. Our main result is to show a min-max duality for the robust assortment

optimization for this class of choice models. In particular, we show that under certain reasonable

assumptions, the order of the max and min operators can be interchanged in (2). The optimal

expected revenue where the decision maker first chooses an assortment to maximize and then an

adversary picks the model parameters to minimize the expected revenue of the selected assortment

is equal to the expected revenue where the adversary makes the model parameter selection first to

minimize and then the decision maker chooses the assortment to maximize (Theorem 1). We would

like to mention that the objective function defined in (1) does not satisfy the properties (such as

the objective function is convex and concave with respect to the minimization and maximization

part respectively) for any known saddle point results to hold.

We show that our framework encompasses two interesting applications. In particular, we show

that our result implies a min-max relation for the robust assortment optimization problem under

the Markov chain model when the uncertainty set is row-wise but possibly non-convex. We also

obtain a min-max relation for the MNL model under general uncertainty set. Note that because


the MNL model is a special case of the Markov chain model (Blanchet et al. (2016)), we are able

to obtain a stronger result (i.e. for general uncertainty set) for the MNL model.

Optimal algorithms for Robust MC and MNL and Insights. In addition, we also give

efficient algorithms to compute the optimal robust assortment under both the Markov chain and

MNL model (Propositions 5 and 9). These algorithms are both iterative procedure solving a fixed

point equation inspired by the min-max relation. We also present operational insights towards the

effect of changing the uncertainty set on the optimal robust assortment under the Markov chain

model. In particular, we find that bigger uncertainty set always lead to bigger assortment, and a

firm should offer larger assortments to hedge against uncertainty. These are in the same spirit as

Rusmevichientong and Topaloglu (2012) who study the robust assortment optimization problem

for the MNL model.

Finally, we conduct some numerical experiments that help quantify the tradeoffs between the

uncertainty set and the conservativeness of the associated optimal robust assortment.

1.3. Related literature

Our work is closely to related to the growing literature on choice models and assortment optimiza-

tion. In order to overcome the MNL model limitation and capture a richer class of substitution,

many choice models have been proposed. All of them increase the model complexity and therefore

makes the estimation and assortment optimization problems significantly more difficult. One of

the key challenges in assortment planing is choosing a model that strikes a good balance between

its predictability and tractability, as there is a fundamental tradeoff between these desirable prop-

erties. The interested readers are referred to Kok et al. (2015) for a comprehensive back ground

reading for the assortment optimization.

In the class of random utility models, generalizations of the MNL model include the nested

logit model (Williams (1977)) and the mixture of MNL model (McFadden and Train (2000)).

The assortment optimization problem has been studied under both of these extensions. Under the

mixture of MNL model, the assortment optimization problem becomes NP-hard, even when the

number of mixtures is 2 (Rusmevichientong et al. (2014a)). Under the nested logit model, several

variant of the assortment problem are tractable (Davis et al. (2014), Gallego and Topaloglu (2014),

Li et al. (2015)). One main limitation of this model is that it depends on a predefined nest structure

that the decision maker must know in advance.

Another approach to model customer preferences is through the lens of distribution over pref-

erence lists. This approach leads to very expressive model (Farias et al. (2013)). However, the

assortment optimization problem becomes untractable under such a general model (Aouad et al.


(2015b)). Several special cases of this model lead to more tractable optimization problems (Honhon

et al. (2010), Aouad et al. (2015a), Jagabathula and Rusmevichientong (2016)).

Finally, our paper relates to the robust optimization literature (Ben-Tal et al. (2009)) which

incorporate the uncertainty in the model parameters into the decision making process. The tech-

niques that we use is robust optimization, which is a handle tailored for dealing with uncertainty

(see Ben-Tal and Nemirovski (2000), Bertsimas and Sim (2003), Xu and Burer (2016)). Recently

this paradigm has found some application in the operations management and revenue management

literature such as airline revenue management (Birbil et al. (2009), Perakis and Roels (2010)),

pricing (Thiele (2009)), portfolio selection (Chen et al. (2011), Zhu and Fukushima (2009)), process

flexibility (Wang and Zhang (2015)), appointment scheduling (Mak et al. (2014)) and our paper

adds to that stream of work.

1.4. Outline

The remainder of this paper is organized as follows. In Section 2, we prove the main min-max

result. We then show how to apply the framework to Markov chain choice model, give efficient

algorithm to compute the corresponding optimal robust assortment and some operational insights

in Section 3. Section 4 is devoted to the consequnces of applying the framework to MNL choice

model. Finally, we conduct some numerical experiments in Section 5.

2. Min-Max Duality

In this section, we state and prove our main min-max duality result. Before doing so, we describe

the set of assumptions that are needed for our result to hold.

2.1. Assumptions

We present a set of four assumptions that are needed for our main result. The first two assumptions

are conditions that the underlying choice model should satisfy under any uncertainty parameters.

The last two assumptions concern the uncertainty set.

Assumption 1. For any u∈ U , λ(u)> 0.

Assumption 2. For any u ∈ U , A(u) is a strictly row diagonally dominant matrix with positive

diagonal elements and non-positive off-diagonal elements, i.e.,

• A(u)ii > 0,∀i∈N ,

• A(u)ij ≤ 0,∀i 6= j,

•∑

jA(u)ij > 0,∀i∈N .

Moreover, A(U) is irreducible for any U = [u1, . . . , un] ∈ Un, where A(U) is the matrix whose ith

row is the ith row of A(ui) for i= 1, . . . , n.


Note that we do not assume any structure on U . Rather, Assumption 2 imposes some structure

on the constraint matrix A(u). In particular, we require not only that A(u) is irreducible for any

parameter u ∈ U but also that A(U), which is constructed by picking one row from each of n

matrices A(ui), is irreducible for any family u1, . . . , un ∈ U .

For the Markov chain model, as we discuss in Section 3, these assumptions are not related to the

structure of the uncertainty set but rather are assumptions on the underlying model parameters

(for each choice of uncertainty parameters) and are standard for this model (Blanchet et al. 2016).

Note that, under the above assumptions, the linear system in (1) has a unique solution, and thus,

R(S,u) is well defined for all S ⊆N and u∈ U .

To proceed with our discussion, we consider two related problems:

minu∈U,v

∑j∈N

λ(u)jvj

s.t. vj ≥ rj,∀j ∈N

A(u)Tj v≥ b(u)j,∀j ∈N

(3)

and,

minu∈U,v

∑j∈N

λ(u)jvj


A(u)Tj v= b(u)j,∀j ∈ SC .

(4)

Note that (3) and (4) are optimization problems over both v and u. As we will see in Section 2.2,

these minimization problems are closely related to (2). In particular, we show in Proposition 1 that

when U is a singleton, (2) and (3) are equivalent. Before stating the next assumptions, we need an

additional piece of notation. For all v, let

w(v) = minu∈U

∑j∈N

λ(u)jvj.

We can now state the final assumptions which assume that the problems can be reformulated as

optimization problems over v alone and that the uncertainty in the parameter u can be captured

through the objective function and the constraints for the two related problems. In particular,

the assumptions assumes that the minimization operator over the uncertainty parameters can be

pushed around. At this point, this assumption may be a bit hard to parse. However, we show in sub-

sequent sections that under reasonable assumptions on the MNL and Markov chain models, these

assumptions are naturally satisfied. In particular, for the Markov chain model, these assumptions

are closely related to a row-wise assumption on the uncertainty set, i.e. one where the uncertainty

across different rows is unrelated.


Assumption 3. Problem (3) is equivalent to the following optimization problem:

minv

w(v)

s.t. vj ≥ rj,∀j ∈N

vj ≥minu∈U

[∑i 6=j

−A(u)jiA(u)jj

vi +1

A(u)jjb(u)j

],∀j ∈N .

Assumption 4. Given S ⊆N , (4) is equivalent to the following optimization problem.

minv

w(v)


vj = minu∈U

[∑i 6=j

−A(u)jiA(u)jj

vi +1

A(u)jjb(u)j

],∀j ∈ SC

2.2. Min-max duality result

Under the specific nonconvex framework introduced in Section 1.1, we can now state our main

result which establishes a min-max relation.

Theorem 1. Let R(S,U) be defined in (1). Under Assumptions 1, 2, 3 and 4,

maxS⊆N

minu∈U

R(S,u) = minu∈U

maxS⊆N

R(S,u).

Furthermore, the optimal robust assortment can be characterized as follows,

S∗ = j ∈N |v∗j = rj,

where v∗ is the unique fixed point of the mapping f , which is defined as

fj(v) = max

rj,min

u∈U

[∑i 6=j

−A(u)jiA(u)jj

vi +1

A(u)jjb(u)j

], ∀ j ∈N . (5)

The rest of this section is devoted to proving Theorem 1 . We discuss its applications to robust

assortment optimization under Markov chain choice model and MNL choice model in the subse-

quent sections respectively.

We start by relating (2), our maximization problem of interest, to (3), a minimization problem,

when there is no uncertainty.

Proposition 1. For any given u ∈ U , let λ := λ(u),A := A(u), b := b(u). Under Assumptions 1

and 2, the following maximization problem and minimization problem share the same optimum

value.

θ1 :=

maxS⊆N

∑j∈N

λjvj

s.t. vj = rj, j ∈ SATj v= bj, j ∈ SC

= θ2 :=

minv

∑j∈N

λjvj

s.t. vj ≥ rj,∀j ∈NATj v≥ bj,∀j ∈N

. (6)


Proof. We first show that θ1 ≥ θ2. By Assumption 2, for any given j ∈ N , Ajj > 0. We can

therefore write

ATj v= bj ⇔ vj =∑i 6=j

−AjiAjj

vi +1

Ajjbj.

Denote aji =

−AjiAjj

i 6= j

0 i= jand bj = 1

Ajjbj, and let A := [aij]. Under Assumption 2, A has a spectral

radius which is strictly less than 1, and 0<∑i∈N

aji < 1 for any j ∈N . Define the following mapping:

gj(v) = max

rj,∑i∈N

ajivi + bj

,∀j ∈N

and g(v) = [g1(v), . . . , gn(v)]>. Note that with this notation, the minimization problem can be

equivalently rewritten asminv

∑j∈N

λjvj

s.t. v≥ g(v)(7)

We next show that the mapping g has a unique fixed point. To do so, let

d(v1, v2) =∥∥v1− v2

∥∥2,

be the `2 distance between v1 and v2. We have

d(g(v1), g(v2)) =∥∥g(v1)− g(v2)

∥∥2

≤∥∥∥A(v1− v2)

∥∥∥2

<∥∥v1− v2

∥∥2

= d(v1, v2)

where the first inequality is true because for any a, b, c∈R,

|maxa, b−maxa, c| ≤ |b− c| ⇒ |maxa, b−maxa, c|2 ≤ |b− c|2

and the second inequality follows because for any matrix A with spectral radius strictly less than

1,

supx

‖Ax‖2‖x‖2

< 1⇒‖Ax‖2 < ‖x‖2,∀x.

g is therefore a contraction mapping and has a unique fixed point v∗ that satisfies v = g(v) by

Banach’s fixed point theorem. Moreover, v∗ is feasible to (7) .

By Assumption 2, all aij ≥ 0. Therefore, g is monotone increasing, i.e., for v1 ≥ v2, g(v1)≥ g(v2).

Consequently, v≥ g(v) implies that for all k,

g(v)≥ g(g(v))≥ . . .≥ gk−1(v)≥ gk(v).


In addition, gk(v)≥ r for all k. Therefore, the sequence gk(v)k=1,2,... associated with any feasible

solution v of problem (7) is monotonically decreasing and bounded below. It must converges to the

unique fixed point v∗ and

v≥ g(v)≥ g(g(v))≥ . . .≥ v∗.

Therefore, the above fact together with Assumption 1 implies that any feasible solution v 6= v∗ of

(7) has a larger objective value than that of v∗. Consequently, v∗ is the optimum of the minimization

problem in (6). Moreover, S = i | v∗i = rj is feasible for the maximization problem in (6) with

objective value∑j∈N

λjv∗j = θ2, which implies that to θ1 ≥ θ2.

We now prove the reverse inequality, i.e. θ2 ≥ θ1. For any feasible S in the maximization problem

of (6), let

vj(S) :=

rj, j ∈ S∑i∈N

ajivi(S) + bj, j ∈ SC .

In the following, we want to show that∑j∈N

λjvj ≤∑j∈N

λjv∗j , (8)

which in turn implies θ2 ≥ θ1 and completes the proof. From the definition of g(·) and the con-

struction of v, we have g(v)≥ v. Moreover, since g is monotone increasing, it holds that

g0(v) := v≤ g(v)≤ g(g(v))≤ . . .≤ gk−1(v)≤ gk(v),

i.e., the sequence gk(v)k=0,1,... is monotonically increasing. Next, we show that gk(v)k=0,1,... has

a uniform upper bound. To this end, note that V :=vj(S) : S ∈ 0,1|N |

is a finite set and hence

is bounded. Let vmax = maxv∈V

maxj∈N

vj. Furthermore,

gj(v) = max

rj,∑i∈N

ajivi + bj

= max

rj,∑i∈N

ajivi +

(1−

∑i∈N

aji

)bj

1−∑i∈N

aji

≤max

rj, v1, · · · , vn,bj

1−∑i∈N

aji

≤maxvmax, δ

where

δ= max

r1, · · · , rn,b1

1−∑i∈N

a1i

, · · · , bn1−

∑i∈N

ani

.


Therefore, the sequence gk(v)k=0,1,... is bounded above and converges to the unique fixed point

v∗, which shows that any v defined by the feasible solution S of the maximization problem (6)

satisfies v≤ v∗. By Assumption 1, (8) holds and θ2 ≥ θ1 as desired.

Remark 1. The following example shows that our assumptions are necessary. In particular, when

A is a strictly diagonal dominant matrix but with positive off-diagonal elements, Proposition 1

fails. Consider

A=

[1 −0.8

0.8 1

], b=

[0.70.2

], λ=

[0.90.1

].

In this case, the solution of the following three problems are all different.

• The maximizer of maxS⊆N

∑j∈N

λjvj

s.t. vj = rj, j ∈ SAjv= bj, j ∈ SC

,is S∗ = 2 with v∗1 = [0.7,0]>. The objective is 0.63.

• The minimizer of minv

∑j∈N

λjvj

s.t. v≥ rAv≥ b

is v∗2 = [0,0.875]>. The objective is 0.0875.

Proposition 1 fails because the mapping g in Proposition 1 is no longer monotone increasing. In

particular, starting from v∗1 , limk→∞

gk(v∗1) = [0.3293,0.4634]> for which the objective is 0.3427.

We are now ready to prove Theorem 1.

Proof of Theorem 1. The following inequality is easily verified.

maxS⊆N

minu∈U

R(S,u)≤minu∈U

maxS⊆N

R(S,u),

We now prove that the reverse inequality also holds. We start by reformulating the min-max

problem. Denote w(v) = minu∈U

∑j∈N

λ(u)jvj. Using Proposition 1 and Assumption 3, we can rewrite

the min-max problem as

minu∈U

maxS⊆N

R(S,u) =

minv

w(v)

s.t. v≥ r

vj ≥minu∈U

[∑i 6=j

−A(u)jiA(u)jj

vi + 1A(u)jj

b(u)j

],∀j ∈N

. (9)

Recall the mapping f(·) such that for all v,

fj(v) = max

rj,min

u∈U

[∑i 6=j

−A(u)jiA(u)jj

vi +1

A(u)jjb(u)j

].


We can then rewrite (9) as

minv

w(v)

s.t. v≥ f(v).(10)

In fact, we can show that the optimal solution v∗ of the above problem satisfies v∗ = f(v∗). To

see this, we first show that f is monotonically increasing, i.e., f(v′)≥ f(v′′) for any v′ ≥ v′′ (the

inequality is true componentwise). For any u ∈ U , Assumption 2 ensures that−A(u)jiA(u)jj

≥ 0 for all

i, j ∈N . Therefore for any v′ ≥ v′′, denoting u′ = arg minu∈U

[∑i6=j

−A(u)jiA(u)jj

v′i + 1A(u)jj

b(u)j

], we have

minu∈U

[∑i 6=j

−A(u)jiA(u)jj

v′i +1

A(u)jjb(u)j

]=∑i6=j

−A(u′)jiA(u′)jj

v′i +1

A(u′)jjb(u′)j

≥∑i 6=j

−A(u′)jiA(u′)jj

v′′i +1

A(u′)jjb(u′)j

≥minu∈U

[∑i6=j

−A(u)jiA(u)jj

v′′i +1

A(u)jjb(u)j

]

for all j ∈N . As a result, for all j ∈N , we have

fj(v′) = max

rj,min

u∈U

[∑i6=j

−A(u)jiA(u)jj

v′i +1

A(u)jjb(u)j

]

≥max

rj,min

u∈U

[∑i 6=j

−A(u)jiA(u)jj

v′′i +1

A(u)jjb(u)j

]= fj(v

′′)

which means that f is monotonically increasing as desired. We are now in position to show that

(10) is equivalent to

minv

w(v)

s.t. v= f(v)(11)

By contradiction, assume there exists an optimal solution v∗ of (10) such that v∗j > fj(v∗) for some

j. Since f is monotonically increasing and λ(u)j > 0 for any j ∈N and u ∈ U due to Assumption

1, we can decrease v∗j by a small amount while strictly decreasing the objective value and not

violating other constraints. This contradicts the optimality of v∗. In summary, we have shown that

the min-max problem (9) is equivalent to (11).

We now investigate the max-min problem. Its inner minimization problem minu∈U R(S,u) is

minu∈U,v

∑j∈N

λ(u)jvj

s.t. vj = rj, j ∈ S

A(u)Tj v= b(u)j, j ∈ SC .


Therefore, according to Assumption 4, it can further be written as

minu∈U

R(S,u) =

minv

w(v)


vj = minu∈U

[∑i 6=j

−A(u)jiA(u)jj

vi + 1A(u)jj

b(u)j

],∀j ∈ SC

. (12)

Before proceeding, we state two claims which are needed to complete the proof. Their proofs is

presented after the main proof.

Claim 1. For any given S ⊆N , the problem (12) has a unique feasible solution.

Claim 2. The mapping f defined in (5) has a unique fixed point.

Let S∗ ⊆N be the optimal assortment to maxS∈N minu∈U R(S,u) with v∗ being the corresponding

optimum of the inner problem (12). By feasibility of v∗, we have

v∗j = minu∈U

[∑i 6=j

−A(u)jiA(u)jj

v∗i +1

A(u)jjb(u)j

],∀j ∈ (S∗)C and v∗j = rj, ∀j ∈ S∗.

It then follows that for any j ∈N ,

fj(v∗) = max

rj,min

u∈U

[∑i6=j

−A(u)jiA(u)jj

v∗i +1

A(u)jjb(u)j

]≥ v∗j .

On the other hand, take any u0 ∈ U , then

fj(v∗)≤max

rj,∑i6=j

−A(u0)jiA(u0)jj

v∗i +

∑iA(u0)jiA(u0)jj

b(u0)j∑iA(u0)ji

≤B,

where B := maxr1, · · · , rn, v∗1 , · · · , v∗n,

∑iA(u0)1iA(u0)11

b(u0)1∑iA(u0)1i

, · · · ,∑iA(u0)niA(u0)nn

b(u0)n∑iA(u0)ni

. f being mono-

tonically increasing, we can recursively prove that fk−1(v∗)≤ fk(v∗)≤B for any k. It follows that

the sequence fk(v∗)k=1,2,... is monotonically increasing, bounded above, and converges to a fixed

point v of f(·), i.e f(v) = v ≥ v∗. Consider S = j ∈N | vj = rj. Note that v is a feasible solution

of (12) associated with S. Combining this fact with Claim 1 implies that v is also optimal to (12)

with respect to S. Consequently, we have w(v)≤w(v∗). On the other hand, v≥ v∗ implies that

w(v) = minu∈U

∑j∈N

λ(u)j vj ≥minu∈U

∑j∈N

λ(u)jv∗j =w(v∗)

Furthermore, recall that Assumption 1 guarantees that λ(u)j > 0 for any j ∈N and u ∈ U , which

in turn implies v∗ = v. Hence, v∗ = f(v∗) and v∗ is a feasible solution to (11). Consequently, the

optimal value of the max-min problem is no less than that of (11), which yields

maxS⊆N

minu∈U

R(S,u) =w(v∗)≥minu∈U

maxS⊆N

R(S,u).

Since v∗ = v = f(v∗) is the unique fixed point of f by Claim 2, the associated optimal assortment

is S∗ = S = j ∈N | (v∗)j = rj. This completes the proof.


Remark 2. According to the proof of Theorem 1, the optimal (S∗, v∗) is only dependent on Uρ

and not Uλ. The latter affects the value of the objective function but not the optimal assortment.

We now prove the two claims used in the previous proof.

Proof of Claim 1.

We fix some S ⊆N and construct a new mapping h(·) defined for all v by,

hj(v) = rj, j ∈ S;hj(v) = minu∈U

[∑i 6=j

−A(u)jiA(u)jj

vi +1

A(u)jjb(u)j

], j ∈ SC .

To prove the claim, it suffices to show this mapping has a unique fixed point. Suppose on the

contrary that h has two different fixed points v1 and v2. For j ∈ SC , hj(v) is the minimum of affine

function and is therefore concave. Consequently, by letting a(u)ji =

−A(u)jiA(u)jj

i 6= j

0 i= j, we can find

y1 ∈ ∂hj(v1)⊆Conva(u)j: | u∈ U and y2 ∈ ∂hj(v2)⊆Conva(u)j: | u∈ U such that

(y1)>(v1− v2)≤ hj(v1)−hj(v2)≤ (y2)>(v1− v2).

For j ∈N , let

uj0 := arg maxu∈U

∑i6=j

−A(u)jiA(u)jj

|v1i − v2

i |.

For all j ∈ SC ,

|hj(v1)−hj(v2)| ≤max

∑i∈N

y1i |v1

i − v2i |,∑i∈N

y2i |v1

i − v2i |

≤∑i∈N

a(uj0)ji|v1i − v2

i |.

When j ∈ S, |hj(v1)−hj(v2)|= 0≤∑i∈N

a(uj0)ji|v1i − v2

i |.

Consider the matrix A(U0) with U0 = [u10, · · · , un0 ] and its row j equals to a(uj0)j:, and let z be

the left eigenvector associated with the largest absolute eigenvalue τ . Assumption 2 implies that

all components of A(U0) are nonnegative, that A(U0) is irreducible as the off-diagonal elements

of A(U0) and A(U0) considered in Assumption 2 are simultaneously to be zeros or nonzeros, and

thus its spectral radius is less than 1. It follows from Perron-Frobenius Theorem that z > 0 and

0< τ < 1. As a result, ∑j∈N

zj|v1j − v2

j | =∑j∈N

zj|hj(v1)−hj(v2)|

≤∑j∈N

zj∑i∈N

a(uj0)ji|v1i − v2

i |

=∑i∈N

|v1i − v2

i |

(∑j∈N

zj a(uj0)ji

)=∑i∈N

|v1i − v2

i |τzi

= τ ·∑j∈N

zj|v1j − v2

j |,


which contradicts the fact that v1 6= v2 and completes the proof.

Proof of Claim 2.

Define the mapping h such that for all v,

hj(v) = minu∈U

[∑i6=j

−A(u)jiA(u)jj

vi +1

A(u)jjb(u)j

],∀ j ∈N .

With this notation, we have fj(v) = maxrj, hj(v) for all v and j ∈N . Suppose by contradiction

that f has two different fixed points v1 and v2. In the proof of Claim 1, we have shown that

|hj(v1)− hj(v2)| ≤ τ |v1j − v2

j | with 0< τ < 1. Using this together with the following inequality,

|maxa, b−maxa, c| ≤ |b− c|,

we have that

|fj(v1)− fj(v2)| ≤ |hj(v1)− hj(v2)| ≤ τ |v1j − v2

j |.

This is a contradiction and concludes the proof.

3. Application to Markov Chain Choice Model: Row-Wise Uncertainty Set

3.1. Recovering the Markov chain choice model

In the Markov chain context, every product is treated as a state of some underlying Markov chain.

To compute the choice probabilities under this model, we assume that the customers arrive to each

of the Markov chain state j with some initial arrival probabilities λj. Upon arrival, the customer

either buys the product if is offered (i.e. j ∈ S) or substitutes to another product i according to the

underlying transition probabilities ρj,i. The customer continues this random walk until she either

land on a product in the offer set or in the no-purchase state, at which point she leaves the system

without purchasing anything. The model parameters are:

• The initial arrival probabilities λj for all states j ∈N , which are the probability that a customer

wants to purchase product j when arriving into the system;

• The transition probabilities ρji for all i∈N , j ∈N+, which can be thought of as the probability

that the customer transits into considering purchasing product i when her favorite product j is

not available.

Let λ= [λ1, · · · , λn]>, ρj = [ρj1, · · · , , ρjn]>, and ρ+j = [ρj0, ρ

>j ]>. Note that ρ+

j is a probability vector,

i.e.,n∑i=0

ρji = 1. Therefore, once ρj is fixed, ρ+j is determined as well.

In practice, the parameters are estimated from data and estimation is prone to error. To account

for this uncertainty in the parameters, we assume that the parameters belong to some uncertainty

set rather than being fixed. In particular, we make the following assumptions on the structure of

the uncertainty set. To simplify notations, we consider uncertainty on ρj rather than ρ+j and let

Uρ (resp. Uλ) be the uncertainty set ρ (resp. λ) belongs to.


Assumption 5. (Row-wise uncertainty set) In the Markov chain based choice model, we assume

that the transition probabilities ρ belong to a row-wise uncertainty set Uρ, i.e. Uρ can be written as

a cartesian product of uncertainty sets, Uρ = Uρ1 × · · ·×Uρn.

In other words, for each product j ∈N , the transition probabilities (ρj,i)i∈N belong to an uncer-

tainty set Uρj . This implies that the outgoing probabilities from each state are unrelated from each

other. We further make the following assumption.

Assumption 6. We further assume that the uncertainty set satisfies the following assumptions:

1. Every transition matrix ρ ∈ Uρ is irreducible and has a spectral radius which is stricly less

than 1,

2. ρii = 0 for any i∈N ,

3. Uλ ⊆Rn++.

For any S ⊆ N and ρ ∈ Uρ, let vj be the expected revenue from a customer whose currently

considering purchasing product j. This customer could either be in state j because it is the first

state she visits or because she transitioned to state j after having visited several products which

were not offered. Note that by the Markov assumption, vj does not depend on the history of the

customer. Moreover, v satisfies the following set of balance equations

vj =

rj j ∈ S∑

i∈N ρjivi j ∈ SC

Note that according to Assumption 6, this system has unique solution for any S ⊆N and ρ ∈ Uρ.

Therefore, given transition matrix ρ ∈ Uρ and arrival probabilities λ ∈ Uλ, the expected revenue

achieved by assortment S can be described as:

RMC(S,ρ,λ) =∑j∈N

λj vj

s.t. vj =

rj j ∈ S∑

i∈N ρjivi j ∈ SC.

(13)

Now our robust optimization problem can be expressed as

maxS⊆N

minρ∈Uρ,λ∈Uλ

RMC(S,ρ,λ). (14)

This robust assortment optimization problem (13) is in fact a special case of the framework intro-

duced in (1). Indeed, by letting u= λ× ρ and λ(u) = λ, vj = vj, A(u) = I − ρ and b(u) = 0 in (1),

one recovers (13).

It is immediate to verify that Assumptions 1 and 2 are satisfied under our assumptions.

Proposition 2. Under Assumptions 5 and 6, both Assumptions 1 and 2 hold.


Under Assumption 5, the rows in the uncertainty set Uρ are unrelated and under Assumption 6,

λj > 0 for all j ∈N . We use this facts to show that Assumption 3 holds.

Proposition 3. Under Assumptions 5, the model satisfies Assumption 3, i.e.

minv,λ∈Uλ,ρ∈Uρ

∑j∈N

λjvj

s.t. vj ≥ rj,∀j ∈Nvj ≥

∑i∈N

ρjivi,∀j ∈N⇔

minv,λ∈Uλ

∑j∈N

λjvj

s.t. vj ≥ rj,∀j ∈Nvj ≥ min

ρj∈Uρj

∑i∈N

ρjivi,∀j ∈N.

Proof. Suppose (v∗, ρ∗, λ∗) and (v, λ) are the optimal solutions to left and right problems respec-

tively. We have,

v∗j ≥∑i∈N

ρ∗jiv∗i ≥ min

ρj∈Uρjρjiv

∗i ,∀j ∈N .

This means that (v∗, λ∗) is feasible to the right hand side problem, and thus∑

j∈N λ∗jv∗j ≥∑

j∈N λj vj. On the other hand, letting

ρj = arg minρj∈U

ρj

∑i∈N

ρjivi,∀j ∈N ,

(v, ρ, λ) is feasible for the left hand side problem. Note that we can construct such ρ because of the

row-wise structure of the uncertainty set, i.e. Assumption 5. Consequently,∑

j∈N λ∗jv∗j ≤

∑j∈N λj vj

and the conclusion follows.

We now show that Assumption 4 also holds.

Proposition 4. Under Assumptions 5, for any assortment S ⊆N , the optimization problem

minv,λ∈Uλ,ρ∈Uρ

∑j∈N

λjvj (15)

s.t. vj =

rj j ∈ S∑

i∈N ρjivi j ∈ SC .

is equivalent to

minv, λ∈Uλ

∑j∈N

λjvj (16)

s.t. vj = rj, j ∈ S

vj = minρj∈U

ρj

∑i∈N

ρjivi, j ∈ SC .

Proof. Suppose (v∗, ρ∗, λ∗) and (v, λ) are the optimal solutions to problem (15) and problem (16)

respectively. Let ρ= [ρ>1 ; · · · , ρ>n ] and ρj = arg minρj∈Uρj∑

i∈N ρjivi for all j ∈ SC . Again, note that

we can construct such ρ because of the row-wise structure of the uncertainty set, i.e. Assumption

5. (v, ρ, λ) is feasible for (15) , and thus∑j∈N

λ∗jv∗j ≤

∑j∈N

λj vj.


Next, we shall claim v∗j =∑

i∈N ρ∗jiv∗i = minρj∈Uρj

∑i∈N ρjiv

∗i for all j ∈ SC . Suppose by contradic-

tion that we have v∗j >minρj∈Uρj∑

i∈N ρjiv∗i for some j ∈ SC . Then we can decrease the value of v∗j

by a small amount, while keeping the feasibility of the solution. This leads to a solution providing a

strictly smaller objective value than the optimal solution and therefore a contradiction. Note that

this is true because of Assumption 6, i.e. λj > 0 for all j ∈ N . Therefore, (v∗, λ∗) is also feasible

for (16), and ∑j∈N

λ∗jv∗j ≥

∑j∈N

λj vj,

which completes the proof.

Consequently, using Theorem 1, we have the following result.

Corollary 1. Under Assumptions 5 and 6,

maxS⊆N


RMC(S,ρ,λ) = minρ∈Uρ,λ∈Uλ

maxS⊆N

RMC(S,ρ,λ).

3.2. Algorithm to find the optimal robust assortment

The techniques we use to prove our min-max result reformulate the robust assortment optimization

problem as a fixed point problem. Inspired by Theorem 1, we present an iterative algorithm that

aims to compute the unique fixed point of the following mapping

f(v)j = max

(rj, min

ρj∈Uρj

∑i∈N

ρjivi

), ∀j ∈N .

Note that f(·) is the mapping presented in (5) specialized for the Markov chain model. Therefore, an

optimal solution for robust Markov chain model with row-wise uncertainty set can be constructed

by S∗ = j ∈N | v∗j = rj where v∗ = f(v∗) is the fixed point.

Algorithm 1 Iterative algorithm for computing the optimal robust assortment under the Markovchain choice model

1: In The uncertainty set Uρj for all j ∈N

2: Out The optimal assortment S∗

3: For t= 1,2, · · ·

4: For j = 1,2, · · · , n

5: vtj←max

(rj, min

ρj∈Uρj

∑i∈N

ρjivt−1i

)= f(vt−1)j

6: If vt = vt−1

7: return S∗ = j ∈N | vtj = rj


This iterative procedure converges in polynomial time when the no-purchase probability ρj0 =

1−∑i∈N

ρji is polynomially bounded away from zero for any ρj ∈ Uρj and j ∈N . More formally, we

need the following to hold true,

δ= minj∈N

minρj∈U

ρjρj0 = Ω(1/nα)

for some constant α.

Proposition 5. Suppose δ = Ω(1/nα) for some constant α and the uncertainty set Uρ satisfies

Assumption 6. Then Algorithm 1 converges to the fixed point v∗ of f in polynomially many steps.

Proof. In Theorem 1, we have proved that f is monotonically increasing and bounded above.

Therefore, f t(v0)t=1,2,... converges to v∗ for any starting point v0. We now prove that the algorithm

terminates in polynomially many steps. Observe that in Algorithm 1 since vt is increasing, for

every j ∈N once vtj exceeds rj, it never goes back to rj again. Denote

rmax = maxj∈N

rj, rmin = minj∈N

rj, j0 = arg minj∈N

rj.

Observe that in each iteration, there is a probability of at least δ of being absorbed by state 0.

Therefore, after t steps, the maximum possible expected revenue for j0 is (1−δ)t rmax. Consequently,

the maximum possible iteration number would not be larger than T with (1 − δ)T rmax ≥ rmin.

Therefore, the algorithm terminates in at most log(rmax/rmin)/δ= Ω(nα) log(rmax/rmin) steps.

3.3. Operational insights

In this subsection, we study how the robust optimal assortment changes with respect to the uncer-

tainty set and the revenue of each product under the Markov chain model. Rusmevichientong and

Topaloglu (2012) provide similar results for the MNL model and we are able to extend them for

the Markov chain model here.

Recall that we still work under Assumption 5, that is the uncertainty set can be represented as

(Uλ,Uρ) where Uρ = Uρ1 × · · · ×Uρn . Let S∗(Uλ,Uρ) denote the largest robust optimal assortment

and Z∗(Uλ,Uρ) the corresponding objective value. We first present a sensitivity analysis with

respect to the uncertainty set.

Proposition 6. For any Uλ ⊆ Uλ and Uρ ⊆ Uρ,

Z∗(Uλ, Uρ)≤Z∗(Uλ,Uρ) and S∗(Uλ,Uρ)⊆ S∗(Uλ, Uρ).

Proof. Given (Uλ,Uρ) and (Uλ, Uρ), denote v∗ and v the corresponding fixed point defined in

Theorem 1, i.e.

v∗j = max

rj, min

ρj∈Uρj

∑i∈N

ρjiv∗i

and vj = max

rj, min

ρj∈Uρj

∑i∈N

ρjivi

for any j ∈N . (17)


Let f be the mapping associated with Uρ,

f(v)j = max

rj, min

ρj∈Uρj

∑i∈N

ρjivi

, j ∈N .

Since Uρ ⊆ Uρ and using the monotonicity of f , we have

v≤ f(v)≤ f2(v)≤ . . .≤ fd(v)→ v∗.

Therefore, v≤ v∗. Moreover, since Uλ ⊆ Uλ, we obtain

Z∗(Uλ, Uρ) = minλ∈Uλ

∑j∈N

λj vj ≤ minλ∈Uλ

∑j∈N

λjv∗j =Z∗(Uλ,Uρ).

Additionally,

minρj∈U

ρj

∑i∈N

ρjiv∗i ≥ min

ρj∈Uρj

∑i∈N

ρjivi ≥ minρj∈U

ρj

∑i∈N

ρjivi.

Consequently, it follows from Theorem 1 that

S∗(Uλ,Uρ) =

j ∈N : v∗j = rj ≥ min

ρj∈Uρj

∑i∈N

ρjiv∗i

⊆

j ∈N : vj = rj ≥ min

ρj∈Uρj

∑i∈N

ρjivi

= S∗(Uλ, Uρ).

This concludes the proof.

The above proposition states that when the degree of uncertainty increases the optimal worse case

revenue will decrease and the decision maker should offer more products to protect against larger

uncertainty. We next provide an alternative characterization of S∗(Uλ,Uρ) which relate the optimal

robust assortment to the optimal assortments when the parameters are given. In particular, for

a given ρ ∈ Uρ, let S∗ρ = S∗(Uλ,ρ). We show that the robust optimal assortment is the largest

optimal assortment among S∗ρ : ρ∈ Uρ.

Proposition 7. For any (Uλ,Uρ), S∗(Uλ,Uρ) =⋃ρ∈Uρ S

∗ρ .

Proof If follows from Proposition 6 that S∗ρ = S∗(Uλ,ρ)⊆ S∗(Uλ,Uρ) for any ρ ∈ Uρ. There-

fore,⋃ρ∈Uρ S

∗ρ ⊆ S∗(Uλ,Uρ). To prove the converse inclusion, let (ρ∗, λ∗) denote the optimal solution

to minρj∈U

ρj ,λ∈UλRMC(S∗(Uλ,Uρ), ρ,λ) and let S∗ρ∗ = arg max

S⊆NRMC(S,ρ∗, λ∗). By Theorem 1, we have

maxS⊆N


RMC(S,ρ,λ) =RMC(S∗(Uλ,Uρ), ρ∗, λ∗)

= minρ∈Uρ,λ∈Uλ

maxS⊆N

RMC(S,ρ,λ) =RMC(S∗ρ∗ , ρ∗, λ∗)

Due to the uniqueness of the largest optimal assortment, we have S∗ρ∗ = S∗(Uλ,Uρ). This completes

the proof.


As a consequence of Proposition 7, the decision marker should focus on the customer types with

transition probabilities that lead to the largest optimal assortment to protect against worst case

scenario. Finally, we present a result showing that the robust optimal assortment shrinks as we

decrease the product revenues. This type of result is particularly helpful in the context of network

revenue management problems (see Feldman and Topaloglu (2017)).

Proposition 8. For any (Uλ, Uρ), let S∗η(Uλ,Uρ) be the optimal robust assortment for the revenues

rη where rηj = rj + η, for all j ∈N . For any η > 0,

S∗(Uλ,Uρ)⊆ S∗η(Uλ,Uρ).

In words, additive incremental revenues lead to larger robust assortments.

Proof. Let vη be the fixed point associated with S∗η(Uλ,Uρ). We define v by letting vj = vηj − η

for all j ∈N . For all j ∈ S∗η(Uλ,Uρ), we have vj = rj. For j /∈ S∗η(Uλ,Uρ), we have

minρj∈U

ρj

∑i∈N

ρjivi = minρj∈U

ρj

∑i∈N

ρji(vηi − η)> min

ρj∈Uρj

∑i∈N

ρjivηi − η= vηj − η= vj > rj,

where the first inequality is due to 0<∑

i ρji < 1 for all ρj ∈ Uρj . Consider the mapping f(·) such

that for all v,

fj(v) = max

rj, min

ρj∈Uρj

∑i∈N

ρjivi

, j ∈N .

Let v∗ be the unique fixed point of f . Using the monotonicity of f ,

v≤ f(v)≤ f2(v)≤ . . .≤ fd(v)→ v∗.

Therefore, for any j /∈ S∗η(Uλ,Uρ), v∗j ≥ vj > rj. Consequently, j /∈ S∗(Uλ,Uρ) by Theorem 1. Con-

sequently, (S∗η(Uλ,Uρ))C ⊆ (S∗(Uλ,Uρ))C and in turn S∗(Uλ,Uρ)⊆ S∗η(Uλ,Uρ).

4. Application to MNL Choice Model: General Uncertainty Set

4.1. Recovering the MNL choice model

Let p= [p0, p1, . . . , pn] be the MNL parameters such that∑n

i=0 pi = 1. The MNL model is a special

case of the Markov chain model by letting, λ(p)i = pi and ρ(p)ji = pi ∀ i= 1, . . . , n (see Blanchet

et al. (2016)). Thus, for any p= [p1, p2, . . . , pn], we can write

RMNL(S,p) =∑j∈N

pj vj

s.t. vj =

rj j ∈ S∑i6=j

pi1−pj

vi j ∈ SC .

Note that

vj =∑i 6=j

pi1− pj

vi⇔ vj =∑i∈N

pivi,


which yields the following equivalent formulation of RMNL(S,p):

RMNL(S,p) =∑j∈N

pjvj

s.t. vj =

rj j ∈ S∑i∈N

pivi j ∈ SC ,

and thus our robust optimization problem can be expressed as

maxS⊆N

minp∈Up

RMNL(S,p). (18)

It is straightforward to see that this is again a special case of (1) by letting parameters u = p,

λ(u) = p, b(u) = 0 and

A(u)ji =

1− pi j = i−pi j 6= i

.

In addition, we make the following regularization assumptions.

Assumption 7. In the MNL choice model, we assume that for all p ∈ Up, we have 0< pj < 1 for

all j ∈N and∑i∈N

pi < 1.

Assumption 7 directly implies Assumptions 1 and Assumption 2. Moreover,

minp∈Up,v

∑i∈N

pivi

s.t. vj ≥ rj,∀j ∈Nvj ≥

∑i∈N

pivi,∀j ∈N⇔

minv

minp∈Up

∑i∈N

pivi

s.t. vj ≥ rj,∀j ∈Nvj ≥ min

p∈Up

∑i∈N

pivi,∀j ∈N

since the terms minp∈Up

∑i∈N

pivi in the objective and constraint are identical. As a result, Assumption

3 holds and we can similarly show that Assumption 4 also holds. Consequently, using Theorem 1,

we have the following result.

Corollary 2. Under Assumptions 7,

maxS⊆N


RMNL(S,ρ,λ) = minρ∈Uρ,λ∈Uλ

maxS⊆N

RMNL(S,ρ,λ).

Note that because MNL is a special case of Markov chain model, we obtain a stronger result under

the MNL model. In particular, Assumption 7 allows for much more general uncertainty set than

Assumptions 5 and 6.

4.2. Algorithm to find the optimal robust assortment

A similar procedure Algorithm 1 also applies to the MNL model, which is presented in Algorithm

2. The convergence result is similar to that in Algorithm 1, and thus is omitted.

This procedure also helps uncover some interesting structural property of the optimal robust

assortment. In particular, in every iteration t, vtj = max(rj, γt) for all j ∈N . Importantly, γt does

not depend on j. At every iteration t, let

St = j ∈N | vtj = rj= j ∈N | rj ≥ γt.


Algorithm 2 Iterative algorithm for computing the optimal assortment in robust MNL choicemodel

1: In The uncertainty set Up

2: Out The optimal assortment S∗

3: For t= 1,2, · · ·

4: γt = minp∈Up

∑i∈N

pivt−1i

5: vtj←max(rj, γt), for all j ∈N

6: If vt = vt−1

7: return S∗ = j ∈N | vtj = rj

Note that for every t, St is therefore a revenue ordered assortment, i.e. consists of the highest k

revenues products for some k. Since γt is an increasing sequence, St is a sequence of revenue ordered

assortment such that St+1 ⊆ St. This implies that S∗ is itself a revenue ordered assortment. This

prove the following structural result.

Proposition 9. Algorithm 2 produces a revenue ordered assortment.

This result is consistent with Rusmevichientong and Topaloglu (2012).

5. Numerical Experiments

In this section, we provide numerical results to show the benefits of using a robust approach to

assortment optimization. Rusmevichientong and Topaloglu (2012) conduct numerical experiments

under the MNL model. We present comparative results for both the Markov chain and MNL

models. We first describe the setting and then the corresponding results.

5.1. Settings

5.1.1. Experimental Setup Following Rusmevichientong and Topaloglu (2012), we use a

mixture of MNL model as our ground truth. This is very general as a mixture of MNL can approx-

imate any random utility based model (McFadden and Train (2000)).

In a mixture of MNL model, the customers are partitioned into G heterogeneous types. For each

type g = 1, . . . ,G, customers’ choice follows a MNL model with parameter vg ∈ Rn+1++ . Moreover,

let Θ = (Θ1, . . . ,ΘG) whereG∑g=1

Θg = 1, where a customer belongs to group g with probability Θg.

We use the expected revenue of the ground truth model to evaluate the assortments. Since the

underlying model is a mixture of MNL model, given Θ and v, the expected revenue for assortment

S can be expressed as

Rev(S,Θ, v) =G∑g=1

Θg

∑i∈S

rivgi

vg0 +∑i∈S

vgi.


We assume that Θ is unknown, θ = (θ1, . . . , θG) is the nominal value of Θ estimated from the

observed data and ε is the error of θ biased from the ground truth Θ. Now suppose Θ and ε is

available (this can be done via some regression approach). we obtian the following polyhedron

uncertainty set associated with θ and tolerance ε:

UΘθ,ε =

Θ :

G∑g=1

Θg = 1,Θg ≥ 0, |Θg − θg| ≤ ε, g= 1, . . . ,G

,

where we want to maximize the worst-case revenue among all possible parameters.

In practice, using a mixture of MNL is challenging because even when the parameters are known

the assortment optimization is intractable even G= 2 (Bront et al. 2009, Rusmevichientong et al.

2014b). Therefore, we focus on the MNL model and the Markov chain model as substitute. In

particular, we:

(I) compare the performance for different tolerance level ε for each model and provide the readers;

(II) compare the performance for the same tolerance level ε of both model and illustrate the

tradeoff in modelling power.

5.1.2. The design of uncertainty sets To make the comparison of different models fair, we

assume that for every scenario (θ, v), the uncertainty set for the two models is derived from UΘθ,ε

and v.

For the MNL model, the only parameter is v. Since the underlying model is a mixture of MNL,

the uncertainty set for v can naturally be constructed from UΘθ,ε as follows:

Uvθ,ε =

G∑g=1

Θgvg : Θ∈ UΘθ,ε

The optimal robust assortment can be obtained by solving n linear programs, see Example 3.3 in

Rusmevichientong and Topaloglu (2012) for more details. Alternatively, Algorithm 2 can also be

used.

Constructing the uncertainty set for the Markov chain model is slightly more complex. (Blanchet

et al. 2016) propose a procedure to estimate a Markov chain model from a mixture of MNL. In

particular, given Θ, the parameters of the Markov chain model can be estimated as follows:

λ(Θ)i =G∑g=1

Θgvgin∑k=0

vgk

=G∑g=1

Θgvgi

ρ(Θ)ij =

G∑g=1

Θgvgjn∑k=0

vgk − vgi

−G∑g=1

Θgvgjn∑k=0

vgk

/

G∑g=1

Θgvgin∑k=0

vgk

=

G∑g=1

Θgvgij

G∑g=1

Θgvgi


where vgi =vgi

n∑k=0

vgk

and vgij = vgj

1n∑k=0

vgk−vgi− 1

n∑k=0

vgk

. Based on these expressions, we construct uncer-

tainty sets Uλθ,ε and Uρjθ,ε, j = 1, . . . , n as follows:

Uλθ,ε =λ(Θ)ini=1 : Θ∈ UΘ

θ,ε

Uρjθ,ε =

ρ(Θ)jini=1

: Θ∈ UΘθ,ε

Note that in order for Assumption 5 to hold, we explicitly construct a row-wise uncertainty set for

ρ. In each step of Algorithm 1, we need to solve the following minimization problem for each j,

minρj∈U

ρjθ,ε

n∑i=1

ρjixt−1i ,

for some xt−1 computed in the previous iteration. Given the structure of our uncertainty set, we

can write this problem as a linear-fractional program

minΘ∈UΘ

θ,ε

n∑i=1

ρ(Θ)jixi = minΘ∈UΘ

θ,ε

G∑g=1

Θgcgj

G∑g=1

Θgvgj

where cgj =n∑i=1

xivgji. Using the Charnes-Cooper transformation (Charnes and Cooper 1962), we

can reformulate this as a linear program. In particular, letting yg = Θg

G∑g=1

Θg vgj

and t= 1G∑g=1

Θg vgj

, the

problem is equivalent to the following linear program:

miny,t

G∑g=1

cgjyg

s.t.G∑g=1

yg = t

G∑g=1

vgj yg = 1

(θg − ε)t≤ yg ≤ (θg + ε)t, ∀g= 1, . . . ,Gy, t≥ 0.

Moreover, the optimal solution to our original problem and the optimal solution to the linear

program are related in the following way: (Θg)∗ = (yg)∗/t∗.

5.1.3. Generating scenarios Our experiment is inspired by Rusmevichientong and

Topaloglu (2012). We assume that there is a total of n = 20 products and that the customers

are clustered into G = 3,6,12 groups. In our experiment, we assume Θ follows a G-dimensional

Dirichlet distribution, and let θg and ρg denote the mean and the CV (Coefficients of Variation)

for Θg. We try different values of ε to see how it affects the performance of the corresponding


robust model. To generate instances of (Θ1, . . . ,ΘG) from G-dimensional Dirichlet distribution,

we can start by generating a series of xg i.i.d. according to a gamma distribution Γ(αg,1) and

then let Θg = xg/

G∑i=1

xi for g = 1, . . . ,G, where αg = θg

S

(S−θgθg(ρg)2

− 1)

and S =G∑g=1

θg. We consider

two different cases for θ, which we differentiate by T = (I) or (II): (I) each group has a similar

mean proportion, θ ∝ (1,1, . . . ,1); (II) the mean proportion of each group has a large variance,

θ∝ (12,22, . . . ,G2).

Let U [a, b] denote a uniform distributed random variable in the interval [a, b] and B(p) a Bernoulli

random variable with success rate p. The revenues rj of each product j are generated iid following

U [0,200j]j=1,...,n. Without loss of generality, we rename the revenues so that r1 ≤ . . .≤ rn.

For each g, we generate vg (the parameter of the gth mixture) independently as follows. We

first set vg0 = 1. Let νg = (νg1 , . . . , νgn) where all νgj are i.i.d. from U [0,1]. Define an auxiliary vector

ηg = (ηg1 , . . . , ηgn). With probability 0.5, we sample ηgi according to U

[0, 1+νg

2

]and otherwise sample

ηgi according to U[0, 1−νg

2

]. Finally, we let (vg1 , . . . , v

gn) be the sorted version of νg where we either

sort by increasing or decreasing value each with probability 0.5. In other words, we assume that in

each of the mixture, the utility for the products is either aligned with the price or completely in

the opposite order.

We note that our setting may have a different performance than Rusmevichientong and Topaloglu

(2012). Because we intentionally sort the utilities in each of the mixture, we make the customer

groups significantly more heterogeneous. This leads to a larger bias of modeling error when the

real proportions are not correctly estimated.

5.2. Numerical results

We parametrize our experiments by the triplet (T,G,CV ) where ρ. We assume uniform CV, i.e.

ρ1 = . . .= ρG. For each triplet (T,G,CV ), we generate 1,500 test cases. Each case consists of: (i) a

vector r, the revenue for selling out a product, (ii) vectors v1, · · · , vG, the MNL parameter vector

for each group. Recall that Θ = (Θ1, . . . ,ΘG) is the portion of each group in the fixture model

and is assumed to follow a G-dimensional Dirichlet distribution. For the given value of CV , the

distribution of Θ is determined as well. We then we sample 100,000 points (numerically stable

enough) from this distribution and calculate the corresponding expected revenue.

We also calculate the optimal robust assortments under two choice models for four different

values of tolerance levels ε= 0.50,0.25,0.10,0.01. Note that for each setting (T,G,CV ), we only

generate 100,000 scenarios (i.e. value of Θ) once.

Given some assortment S, the revenue under 100,000 scenarios can be easily computed. We

pick the top 1% percentile revenue (i.e., the 1,000-th lowest revenue), the mean revenue and the

standard deviation of the revenue as indicators, which is shown as “1% Per.”, “Mean” and “St.


0 500 1000 1500

Revenue

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

Fre

qu

en

cy

RMC, eps = 0.50

RMC, eps = 0.25

RMC, eps = 0.10

RMC, eps = 0.01

Figure 1 RobustMC with various ε

0 500 1000 1500

Revenue

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

Fre

qu

en

cy

RMNL, eps = 0.50

RMNL, eps = 0.25

RMNL, eps = 0.10

RMNL, eps = 0.01

Figure 2 RobustMNL with various ε


Table 1 Descriptive statistics for the case in Figure 1 and 2 (T = 1, G = 3 and CV = 1.0)

ε= 0.50 ε= 0.25 ε= 0.10 ε= 0.01

1% Per.RMC 470.42 445.11 377.39 316.55

RMNL 445.14 377.42 245.44 91.84

MeanRMC 963.26 970.00 963.14 972.32

RMNL 969.76 962.87 929.69 902.58

St. Dev.RMC 246.45 262.47 292.92 327.94

RMNL 262.84 293.31 342.82 406.21

Dev.” respectively. For ease of exposition, we choose a typical case from our simulated dataset

T = 1, G= 3 and CV = 1.0. Figure 1 and Figure 2 show the revenue distributions for both models

and some descriptive statistics are listed in Table 1. Comparing the two models, we can see that the

Markov chain model has a better performance in this case. It leads to a higher expected revenue

and lower revenue dispersion (quantified by the sample standard deviation). The results of the “1%

Per.” almost indicate that minΘ∈UΘ

θ,ε

R(S∗RMC ,Θ)≥ minΘ∈UΘ

θ,ε

R(S∗RMNL,Θ) for all ε, meaning that Markov

chain model is more protective under extreme cases.

Looking at the revenue distribution of the assortment decision under both models, we observe

that when ε is small, assortment problem almost correspond to the deterministic problem and the

assortment tends to have less products. This leads to a more volatile solution as illustrated by the

value of the standard deviation. When, ε increases, the assortment is more conservative, with more

products, and leads to less variable revenue. In that case, the revenue more concentrated.

We now compare the performances of the Markov chain model and MNL models in Table 2.

In particular, we compute the ratio of the different metric between the two models. For exam-

ple, the top-left number “1.04” in the first matrix, means for this setting, the averaged ratio

Mean(

1% Per.(ε=0.50,MC)

1% Per.(ε=0.50,MNL)

)= 1.04.

We observe that the Markov chain model strongly dominates the MNL model as it allows a

higher expected revenue and a lower revenue dispersion simultaneously. This illustrates the added

modeling power that the Markov chain model provides over the MNL model (recall that the MNL

model is a special case of the Markov chain model). We also note that when G is large, a large

ε leads both models to provide very conservative assortment since the uncertainty set is actually

very large: their decisions are often identical.

6. Conclusion

In this paper, we study the robust assortment optimization problem under a general class of choice

models. This framework assumes that the choice probabilities, capturing the substitution behavior

of customers, are solutions to a system of linear equations. Under reasonable assumptions, we

show that this problem admits a max-min duality relationship, i.e. the two operators in the robust


Table 2 Estimated ratios under same ε between robust Markov chain choice model and robust MNL model (the

benchmark is robust MNL model)

(T,G,CV ) ε 1% Per. 5% Per. Median Mean St. Dev.

(I, 3, 1.0)

0.50 1.04 1.03 0.99 1.00 0.980.25 1.04 1.04 1.00 1.00 0.980.10 1.07 1.06 1.02 1.01 0.970.01 1.05 1.05 1.02 1.01 0.98

(I, 3, 1.2)

0.50 1.04 1.04 0.99 1.00 0.980.25 1.04 1.04 1.01 1.00 0.980.10 1.07 1.07 1.03 1.01 0.970.01 1.05 1.05 1.02 1.01 0.98

(I, 6, 1.2)

0.50 1.00 1.00 1.00 1.00 0.990.25 1.02 1.01 0.99 1.00 0.980.10 1.07 1.05 1.01 1.01 0.950.01 1.05 1.04 1.01 1.01 0.98

(I, 6, 2.0)

0.50 1.01 1.01 1.00 1.00 0.990.25 1.04 1.04 0.99 1.00 0.980.10 1.11 1.11 1.01 1.01 0.950.01 1.06 1.06 1.02 1.01 0.98

(I, 12, 2.0)

0.50 1.00 1.00 1.00 1.00 1.000.25 1.00 1.00 1.00 1.00 1.000.10 1.03 1.01 0.99 1.00 0.980.01 1.05 1.04 1.01 1.01 0.98

(I, 12, 3.0)

0.50 1.00 1.00 1.00 1.00 1.000.25 1.01 1.01 1.00 1.00 1.000.10 1.06 1.04 0.99 1.00 0.980.01 1.07 1.06 1.02 1.01 0.98

(II, 3, 0.5)

0.50 1.03 1.02 1.00 1.00 0.980.25 1.03 1.02 1.00 1.00 0.980.10 1.03 1.03 1.01 1.01 0.980.01 1.03 1.03 1.01 1.01 0.98

(II, 3, 0.7)

0.50 1.04 1.04 0.99 1.00 0.980.25 1.04 1.03 1.01 1.00 0.980.10 1.03 1.03 1.03 1.01 0.980.01 1.03 1.03 1.02 1.01 0.98

(II, 6, 0.7)

0.50 1.00 1.00 1.00 1.00 0.990.25 1.04 1.02 0.99 1.00 0.970.10 1.06 1.05 1.01 1.01 0.970.01 1.04 1.04 1.01 1.01 0.98

(II, 6, 1.0)

0.50 1.01 1.00 1.00 1.00 0.990.25 1.06 1.05 0.99 1.00 0.970.10 1.08 1.07 1.01 1.01 0.970.01 1.05 1.05 1.02 1.01 0.98

(II, 12, 1.0)

0.50 1.00 1.00 1.00 1.00 1.000.25 1.00 1.00 1.00 1.00 1.000.10 1.07 1.04 1.00 1.00 0.940.01 1.05 1.04 1.01 1.01 0.98

(II, 12, 1.5)

0.50 1.00 1.00 1.00 1.00 1.000.25 1.00 1.00 1.00 1.00 1.000.10 1.13 1.09 0.99 1.00 0.940.01 1.07 1.06 1.02 1.01 0.98

optimization problem can be swapped. This is surprising as none of the properties for known saddle

point results are satisfied.

Equipped with this duality result, we obtain min-max duality results for the Markov chain and

MNL choice models, both of which fall in our framework. For the Markov chain model, the main

assumption needed on the underlying Markov chain is that the uncertainty across the different

rows of the transition matrix are unrelated. For the MNL model, the result holds under milder

assumptions and thus for more general uncertainty sets. This is consistent with the fact that the


MNL model is a special case of the Markov chain model and therefore we are able to obtain stronger

results for the MNL. Inspired by the duality results, we also develop efficient iterative algorithms

to find the optimal robust assortment.

In this work, we presented a general framework for choice models. It would be interesting to see

if that approach can unify an even broader class of choice models. For instance, Desir et al. (2018)

recently showed that under the Mallows model (a choice model based on a particular probability

distribution over preference lists), the choice probabilities can be obtained by solving a system of

linear equations. If we can develop general estimation and/or optimization techniques for this class

of models, this might give a more parsimonious approach to modeling choice.

Finally, another interesting research direction is to push the results for broader type of uncer-

tainty sets for the Markov chain model, in particular allowing the uncertainty across different

rows of the Makov chain model to be related. Having a budget constraint across rows to limit the

adversary power would be one way to correlates the uncertainty sets.

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