On the adaptivity gap in two-stage robust linear optimization under uncertain packing constraints

Abstract

In this paper, we study the performance of static solutions in two-stage adjustable robust packing linear optimization problem with uncertain constraint coefficients. Such problems arise in many important applications such as revenue management and resource allocation problems where demand requests have uncertain resource requirements. The goal is to find a two-stage solution that maximizes the worst case objective value over all possible realizations of the second-stage constraints from a given uncertainty set. We consider the case where the uncertainty set is column-wise and constraint-wise (any constraint describing the set involve entries of only a single column or a single row). This is a fairly general class of uncertainty sets to model constraint coefficient uncertainty. We show that the two-stage adjustable robust problem is $\varOmega (\log n)$-hard to approximate. On the positive side, we show that a static solution is an $O\big (\log n \cdot \min (\log \varGamma , \log (m+n))\big )$-approximation for the two-stage adjustable robust problem where m and n denote the numbers of rows and columns of the constraint matrix and $\varGamma$ is the maximum possible ratio of upper bounds of the uncertain constraint coefficients. Therefore, for constant $\varGamma$, surprisingly the performance bound for static solutions and therefore, the adaptivity gap matches the hardness of approximation for the adjustable problems. Furthermore, in general the static solution provides nearly the best efficient approximation for the two-stage adjustable robust problem.

Appendices

Appendix A: Proof of Theorem 2

In this section, we show that the general two-stage adjustable robust problem $\varPi _\mathsf{AR}^\mathsf{Gen}$ (2.1) is $\varOmega (2^{\log ^{1-\epsilon }m})$-hard to approximate for any constant $0<\epsilon <1$. We prove this by an approximation preserving reduction from the Label-Cover-Problem. The reduction is similar in spirit to the reduction from the Set-Cover-Problem to the two-stage adjustable robust problem.

Label-cover-problem We are given a finite set V ($|V|=m$), a family of subset $\{{\mathcal V}_1,\ldots ,{\mathcal V}_K\}$ of V and graph $G=(V, E)$. Let H be a supergraph with vertices $\{{\mathcal V}_1,\ldots ,{\mathcal V}_K\}$ and edges F where $({\mathcal V}_i, {\mathcal V}_j)\in F$ if there exists $(k,l)\in E$ such that $k\in {\mathcal V}_i, l\in {\mathcal V}_j$. The goal is to find the smallest cardinality set $C\subseteq V$ such that F is covered, i.e., for each $({\mathcal V}_i, {\mathcal V}_j)\in F$, there exists $k\in {\mathcal V}_i\cap C, l\in {\mathcal V}_j\cap C$ such that $(k,l)\in E$.

The label cover problem is $\varOmega (2^{\log ^{1-\epsilon }m})$-hard to approximate for any constant $0<\epsilon <1$, i.e., there is no polynomial time approximation algorithm that give an $O(2^{\log ^{1-\epsilon }m})$-approximation for any constant $0<\epsilon <1$ unless $\mathbf {NP}\subseteq \mathbf {DTIME}(m^{\text {polylog}(m)})$ [1].

Proof of Theorem 2

Consider an instance $\mathcal I$ of Label-Cover-Problem with ground elements V ($|V|=m$), graph $G=(V, E)$, a family of subset of V: $({\mathcal V}_1,\ldots ,{\mathcal V}_K)$ and a supergraph $H=(\{{\mathcal V}_1,\ldots ,{\mathcal V}_K\}, F)$ where $|F|=n$. We construct the following instance $\mathcal{I}'$ of the general adjustable robust problem $\varPi _\mathsf{AR}^\mathsf{Gen}$ (2.1):

$$\begin{aligned} \varvec{A}=\varvec{0},\;\varvec{c}=\varvec{0},\;\varvec{d}=\left( \begin{array}{c}\varvec{e}\\ mb{e}\end{array}\right) \in {\mathbb R}^{n+m},\; \varvec{h}=\varvec{e}\in {\mathbb R}^m,\; \mathcal{U}=\left\{ [\varvec{B}\;-\varvec{I}_m]\;|\;\varvec{B}\in \mathcal{U}_F\right\} \end{aligned}$$

where $d_1=d_2=\cdots =d_n=1$, $\varvec{I}_m$ is the m-dimensional identity matrix and each column set of $\mathcal{U}_F\subseteq {\mathbb R}^{m\times n}_+$ corresponds to an edge $({\mathcal V}_i, {\mathcal V}_j)\in F$ with

$$\begin{aligned} \mathcal{U}_{({\mathcal V}_i, {\mathcal V}_j)}=\mathsf{conv}\left( \{\varvec{0}\}\bigcup \left\{ \left. \frac{1}{2}(\varvec{e}_k+\varvec{e}_l)\;\right| \;(k,l)\in E, k\in {\mathcal V}_i, l\in {\mathcal V}_j\right\} \right) \subseteq {\mathbb R}^m_+. \end{aligned}$$

Therefore, $\mathcal U$ is column-wise with column sets $\mathcal{U}_{({\mathcal V}_i, {\mathcal V}_j)}, \forall ({\mathcal V}_i, {\mathcal V}_j)\in F$ and $\mathcal{U}_j,j\in [m]$ where $\mathcal{U}_j=\{-\varvec{e}_j\}$, i.e., there is no uncertainty in $\mathcal{U}_j$. The instance $\mathcal{I}'$ of $\varPi _\mathsf{AR}^\mathsf{Gen}$ can be formulated as

$$\begin{aligned} \begin{aligned} z_\mathsf{AR}^\mathsf{Gen}&=\min _{\varvec{B}\in \mathcal{U}_F} \max _{\varvec{y}\ge \varvec{0},\varvec{z}\ge \varvec{0}} \{\varvec{e}^T \varvec{y}-\varvec{e}^T\varvec{z}\;|\;\varvec{B}\varvec{y}-\varvec{z} \le \varvec{e}, \varvec{y}\ge \varvec{0}, \varvec{z}\ge \varvec{0}\}\\&=\min _{\varvec{b}_{({\mathcal V}_i, {\mathcal V}_j)}\in \mathcal{U}_{({\mathcal V}_i, {\mathcal V}_j)}} \max _{\varvec{y}\ge \varvec{0},\varvec{z}\ge \varvec{0}} \left\{ \varvec{e}^T \varvec{y}-\varvec{e}^T\varvec{z}\;\left| \;\sum _{({\mathcal V}_i, {\mathcal V}_j)\in F} y_{({\mathcal V}_i, {\mathcal V}_j)}\varvec{b}_{({\mathcal V}_i, {\mathcal V}_j)}-\varvec{z} \le \varvec{e}, \varvec{y}\ge \varvec{0}, \varvec{z}\ge \varvec{0}\right. \right\} . \end{aligned} \end{aligned}$$

Suppose $(\hat{\varvec{y}},\hat{\varvec{z}}, {\hat{\varvec{b}}}_{({\mathcal V}_i, {\mathcal V}_j)}, ({\mathcal V}_i, {\mathcal V}_j)\in F)$ is a feasible solution for instance $\mathcal{I}'$. Then, we can compute a label cover of instance $\mathcal I$ with cardinality at most $\varvec{e}^T\hat{\varvec{y}}-\varvec{e}^T\hat{\varvec{z}}$. From strong duality, there exists an optimal solution ${\hat{\varvec{\mu }}}$ for

$$\begin{aligned} \min \left\{ \varvec{e}^T\varvec{\mu }\;|\;{\hat{\varvec{b}}}_{({\mathcal V}_i, {\mathcal V}_j)}^T\varvec{\mu }\ge 1, \quad \forall ({\mathcal V}_i, {\mathcal V}_j)\in F, \mu \in [0,1]^{m}\right\} \end{aligned}$$

and $\varvec{e}^T{\hat{\varvec{\mu }}}=\varvec{e}^T\hat{\varvec{y}}-\varvec{e}^T\hat{\varvec{z}}$. For each $({\mathcal V}_i, {\mathcal V}_j)\in F$, consider a basic optimal solution $({\tilde{\varvec{b}}}_{({\mathcal V}_i, {\mathcal V}_j)},({\mathcal V}_i, {\mathcal V}_j)\in F)$ where

$$\begin{aligned} {\tilde{\varvec{b}}}_{({\mathcal V}_i, {\mathcal V}_j)}\in \arg \max \left\{ \varvec{b}^T{\hat{\varvec{\mu }}}\;|\;\varvec{b}\in \mathcal{U}_{({\mathcal V}_i, {\mathcal V}_j)}\right\} . \end{aligned}$$

Therefore, ${\tilde{\varvec{b}}}_{({\mathcal V}_i, {\mathcal V}_j)}$ is a vertex of $\mathcal{U}_{({\mathcal V}_i, {\mathcal V}_j)}$ for each $({\mathcal V}_i, {\mathcal V}_j)\in F$, which implies that ${\tilde{\varvec{b}}}_{({\mathcal V}_i, {\mathcal V}_j)}=\frac{1}{2}(\varvec{e}_{k_i}+\varvec{e}_{l_j})$ for some $(k_i,l_j) \in E$ and $k_i\in {\mathcal V}_i, l_j\in {\mathcal V}_j$. Also, ${\tilde{\varvec{b}}}_{({\mathcal V}_i, {\mathcal V}_j)}^T{\hat{\varvec{\mu }}}\ge 1, \forall ({\mathcal V}_i, {\mathcal V}_j)\in F$. Now, let $\tilde{\varvec{\mu }}$ the optimal solution of the following LP:

$$\begin{aligned} \min \left\{ \varvec{e}^T\varvec{\mu }\;|\;{\tilde{\varvec{b}}}_{({\mathcal V}_i, {\mathcal V}_j)}^T\varvec{\mu }\ge 1, \quad \forall ({\mathcal V}_i, {\mathcal V}_j)\in F,\varvec{0}\le \varvec{\mu }\le \varvec{e}\right\} . \end{aligned}$$

Clearly, $\varvec{e}^T\tilde{\varvec{\mu }}\le \varvec{e}^T{\hat{\varvec{\mu }}}$. Also, since ${\tilde{\varvec{b}}}_{({\mathcal V}_i, {\mathcal V}_j)}=\frac{1}{2}(\varvec{e}_{k_i}+\varvec{e}_{l_j})$ and ${\tilde{\varvec{b}}}_{({\mathcal V}_i, {\mathcal V}_j)}^T\tilde{\varvec{\mu }}\ge 1$, $\tilde{\mu }_{k_i}=\tilde{\mu }_{l_j}=1$. Therefore, $\tilde{\varvec{\mu }}\in \{0,1\}^{m}$. Let

$$\begin{aligned} C=\{j\;|\;\tilde{\mu }_j=1\}. \end{aligned}$$

Clearly, C is a valid label cover for F and $|C|=\varvec{e}^T\tilde{\varvec{\mu }}\le \varvec{e}^T{\hat{\varvec{\mu }}}=\varvec{e}^T\hat{\varvec{y}}-\varvec{e}^T\hat{\varvec{z}}$.

Conversely, given a label cover C of instance $\mathcal I$, for any $j\in [m]$, let $\bar{\mu }_j=1$ if $j\in C$ and zero otherwise. This implies that $\varvec{e}^T\bar{\varvec{\mu }}=|C|$. For any ${({\mathcal V}_i, {\mathcal V}_j)}\in F$, let $\bar{\varvec{b}}_{({\mathcal V}_i, {\mathcal V}_j)}=\frac{1}{2}(\varvec{e}_{k_i}+\varvec{e}_{l_j})$ where $k_i\in {\mathcal V}_i\cap C, l_j\in {\mathcal V}_j\cap C$ such that $(k_i,l_j) \in E$. Then, let $\varvec{\mu }'$ be an optimal solution for the following LP

$$\begin{aligned} \min \left\{ \varvec{e}^T\varvec{\mu }\;|\;\bar{\varvec{b}}_{({\mathcal V}_i, {\mathcal V}_j)}^T\varvec{\mu }\ge 1, \quad \forall ({\mathcal V}_i, {\mathcal V}_j)\in F,\varvec{0}\le \varvec{\mu }\le \varvec{e}\right\} . \end{aligned}$$

Then, $\varvec{e}^T\varvec{\mu }'\le \varvec{e}^T\bar{\varvec{\mu }}$ as $\bar{\varvec{\mu }}$ is feasible for the above LP. From strong duality, there exists $\bar{\varvec{y}}\in {\mathbb R}^n_+$ and $\bar{\varvec{z}}\in {\mathbb R}^m_+$ such that $(\bar{\varvec{y}},\bar{\varvec{z}}, \bar{\varvec{b}}_{({\mathcal V}_i, {\mathcal V}_j)},{({\mathcal V}_i, {\mathcal V}_j)}\in F)$ is a feasible solution for instance $\mathcal{I}'$ of $\varPi _\mathsf{AR}^\mathsf{Gen}$ with cost $\varvec{e}^T\bar{\varvec{y}}-\varvec{e}^T\bar{\varvec{z}}=\varvec{e}^T\varvec{\mu }'\le \varvec{e}^T\bar{\varvec{\mu }}=|C|$. $\square$

Appendix B: Approximate separation to optimization

For any $\varvec{x} \in {\mathbb R}^n_+$, let

$$\begin{aligned} Q^*(\varvec{x})=\min _{\varvec{B}\in \mathcal{U}}\max _{\varvec{y}\ge \varvec{0}}\left\{ \varvec{d}^T\varvec{y}\;|\;\varvec{B}\varvec{y}\le \varvec{h}-\varvec{A}\varvec{x}\right\} . \end{aligned}$$

We show that if we can approximate the separation problem, we can also approximate $\varPi _\mathsf{AR}$. Let $\mathcal{A}$ be a $\gamma$-approximate algorithm for the separation problem (3.1), i.e., $\mathcal{A}$ computes a $\gamma$-approximation algorithm for the min-max problem in (3.1). For any $\varvec{x} \in {\mathbb R}^n_+$, let $\varvec{B}^\mathcal{A}(\varvec{x})$ denote the matrix returned by $\mathcal{A}$ and let

$$\begin{aligned} Q^\mathcal{A}(\varvec{x})=\max _{\varvec{y}\ge \varvec{0}}\left\{ \varvec{d}^T\varvec{y}\;|\;\varvec{B}^\mathcal{A}(\varvec{x})\varvec{y}\le \varvec{h}-\varvec{A}\varvec{x}\right\} . \end{aligned}$$

Therefore, the approximate separation based on Algorithm $\mathcal{A}$ is as follows: for any $(\varvec{x}, z)$, return feasible if $Q^\mathcal{A}(\varvec{x}) \ge z$. Otherwise give a violating hyperplane corresponding to $\varvec{B}^\mathcal{A}(\varvec{x})$. Now, we prove the following theorem.

Theorem 12

Suppose we have an Algorithm $\mathcal{A}$ that is a $\gamma$-approximation for the separation problem (3.1). Then we can compute a $\gamma$-approximation for the two-stage adjustable robust problem $\varPi _\mathsf{AR}$ (1.1).

Proof

Since $\mathcal{A}$ is a $\gamma$-approximation to the min-max problem in (3.1), for any $\varvec{x} \in {\mathbb R}^n_+$,

$$\begin{aligned} Q^*(\varvec{x})\le Q^\mathcal{A}(\varvec{x})\le \gamma \cdot Q^*(\varvec{x}). \end{aligned}$$

Let $(\varvec{x}^*, z^*)$ be an optimal solution for $\varPi _\mathsf{AR}$ and let

$$\begin{aligned} \mathsf{OPT} = \varvec{c}^T \varvec{x}^* + z^*. \end{aligned}$$

Consider the optimization algorithm based on the approximate separation algorithm $\mathcal{A}$ and suppose it returns the solution $(\hat{\varvec{x}}, \hat{z})$. Note that $(\varvec{x}^*, z^*)$ is feasible according to the approximate separation algorithm $\mathcal{A}$ as $Q^\mathcal{A}(\varvec{x}^*)\ge Q^*(\varvec{x}^*)=z^*$. Therefore,

$$\begin{aligned} \varvec{c}^T\hat{\varvec{x}}+\hat{z}\ge \varvec{c}^T\varvec{x}^*+z^*. \end{aligned}$$

(B.1)

Note that $\hat{z}$ is an approximation for the worst case second-stage objective value when the first stage solution is $\hat{\varvec{x}}$. The true objective value for the first stage solution $\hat{\varvec{x}}$ is given by

$$\begin{aligned} \varvec{c}^T\hat{\varvec{x}}+ Q^*(\hat{\varvec{x}})\ge & {} \varvec{c}^T\hat{\varvec{x}}+ \frac{1}{\gamma } Q^\mathcal{A}(\hat{\varvec{x}})\nonumber \\\ge & {} \varvec{c}^T\hat{\varvec{x}}+ \frac{1}{\gamma } \hat{z} \nonumber \\\ge & {} \frac{1}{\gamma }(\varvec{c}^T\hat{\varvec{x}}+\hat{z}) \nonumber \\\ge & {} \frac{1}{\gamma }\mathsf{OPT}, \end{aligned}$$

(B.2)

where the first inequality follows as $\mathcal{A}$ is a $\gamma$-approximation and $Q^\mathcal{A}(\hat{\varvec{x}}) \le \gamma \cdot Q^*(\hat{\varvec{x}})$. Inequality (B.2) follows as $(\hat{\varvec{x}}, \hat{z})$ is feasible according to $\mathcal{A}$ and therefore, $\hat{z} \le Q^\mathcal{A}(\hat{\varvec{x}})$ and the last inequality follows from (B.1). Therefore, the optimization problem based on algorithm $\mathcal{A}$ computes a $\gamma$-approximation for $\varPi _\mathsf{AR}$. $\square$

Appendix C: Transformation of the adjustable robust problem

Let $\varvec{x}^*$ be the optimal first-stage solution for $\varPi _\mathsf{AR}$, i.e.,

$$\begin{aligned} z_\mathsf{AR}=\varvec{c}^T\varvec{x}^*+\min _{\varvec{B}}\max _{\varvec{y}}\left\{ \varvec{d}^T\varvec{y}\;|\;\varvec{B}\varvec{y}\le \varvec{h}-\varvec{A}\varvec{x}^*, \varvec{B}\in \mathcal{U}, \varvec{y}\ge \varvec{0}\right\} . \end{aligned}$$

Note that $(\varvec{x}^*, \varvec{0})$ is a feasible solution for $\varPi _\mathsf{Rob}$. We have

$$\begin{aligned} z_\mathsf{Rob}\ge \varvec{c}^T\varvec{x}^*+\max _{\varvec{y}\ge \varvec{0}}\left\{ \varvec{d}^T\varvec{y}\;|\;\varvec{B}\varvec{y}\le \varvec{h}-\varvec{A}\varvec{x}^*, \quad \forall \varvec{B}\in \mathcal{U}\right\} . \end{aligned}$$

Since $\varvec{c}$ and $\varvec{x}^*$ are both non-negative, to prove Theorem 7, it suffice to show

$$\begin{aligned}&\min _{\varvec{B}\in \mathcal{U}}\max _{\varvec{y}\ge \varvec{0}}\left\{ \varvec{d}^T\varvec{y}\;|\;\varvec{B}\varvec{y}\le \varvec{h}-\varvec{A}\varvec{x}^*\right\} \\&\quad \le O(\log (m+n)\log n)\cdot \max _{\varvec{y}\ge \varvec{0}}\left\{ \varvec{d}^T\varvec{y}\;|\;\varvec{B}\varvec{y}\le \varvec{h}-\varvec{A}\varvec{x}^*, \quad \forall \varvec{B}\in \mathcal{U}\right\} . \end{aligned}$$

In this section, we show that we can assume without loss of generality that $(\varvec{h}-\varvec{A}\varvec{x}^*)>\varvec{0}$, as otherwise the static solution is optimal for the two-stage adjustable robust problem $\varPi _\mathsf{AR}$ (1.1), i.e., $z_\mathsf{AR}=z_\mathsf{Rob}$: Note that $(\varvec{h}-\varvec{A}\varvec{x}^*)\ge \varvec{0}$, since otherwise the inner problem becomes infeasible. Now, suppose that $(\varvec{h}-\varvec{A}\varvec{x})_i=0$ for some $i\in [m]$. Since $\mathcal U$ is a full-dimensional convex set, there exist $\varvec{B}^*\in \mathcal{U}$ such that $B_{ij}^*>0$ for all $j\in [n]$. Therefore,

$$\begin{aligned} \min _{\varvec{B}\in \mathcal{U}}\max _{\varvec{y}\ge \varvec{0}}\left\{ \varvec{d}^T\varvec{y}\;|\;\varvec{B}\varvec{y}\le \varvec{h}-\varvec{A}\varvec{x}\}\le \max _{\varvec{y}\ge \varvec{0}}\{\varvec{d}^T\varvec{y}\;|\;\varvec{B}^*\varvec{y}\le \varvec{h}-\varvec{A}\varvec{x}\right\} =0, \end{aligned}$$

which implies that $z_\mathsf{AR}=\varvec{c}^T\varvec{x}^*$ since $\varvec{d},\varvec{y}$ are non-negative. On the other hand, $(\varvec{x}^*, \varvec{0})$ is a feasible solution for $\varPi _\mathsf{Rob}$. Therefore,

$$\begin{aligned} z_\mathsf{Rob}\ge \varvec{c}^T\varvec{x}^*=z_\mathsf{AR}. \end{aligned}$$

However, suppose $(\bar{\varvec{x}}, \bar{\varvec{y}})$ is an optimal solution for $\varPi _\mathsf{Rob}$, then $\varvec{x}=\bar{\varvec{x}}, \varvec{y}(\varvec{B})=\bar{\varvec{y}}$ for all $\varvec{B}\in \mathcal{U}$ is feasible for $\varPi _\mathsf{AR}$. Therefore, $z_\mathsf{AR}\ge z_\mathsf{Rob}$.

Appendix D: Proof of Theorem 3

Let $\varvec{y}^*$ be such that ${\hat{\varvec{B}}}\varvec{y}^*\le \varvec{h}$. For any $\varvec{B}\in \mathcal{U}$, we have $\varvec{B}\le {\hat{\varvec{B}}}$ component-wise by construction. Note that $\varvec{y}^*\ge \varvec{0}$, this implies $\varvec{B}\varvec{y}^*\le {\hat{\varvec{B}}}\varvec{y}^*\le \varvec{h}$ for all $\varvec{B}\in \mathcal{U}$.

Conversely, suppose $\tilde{\varvec{y}}$ satisfies $\varvec{B}\tilde{\varvec{y}}\le \varvec{h}$ for all $\varvec{B}\in \mathcal{U}$. For each $i\in [m]$, note that $\mathsf{diag}(\varvec{e}_i){\hat{\varvec{B}}}\in \mathcal{U}$ by construction. Therefore, $\varvec{e}_i^T{\hat{\varvec{B}}}\tilde{\varvec{y}}\le h_i$ for all $i\in [m]$, which implies that ${\hat{\varvec{B}}}\tilde{\varvec{y}}\le \varvec{h}$.

Appendix E: Proof of Lemma 1.

Let

$$\begin{aligned} {\hat{B}}_{ij}=\frac{1}{(n+i-j+1)\bmod {m}}. \end{aligned}$$

From Theorem 3, $\varPi _\mathsf{Rob}$ is equivalent to

$$\begin{aligned} z_\mathsf{Rob}=\max \left\{ \varvec{e}^T\varvec{y}\;|\;{\hat{\varvec{B}}}\varvec{y}\le \varvec{e},\varvec{y}\ge \varvec{0}\right\} .\end{aligned}$$

The dual problem is

$$\begin{aligned} z_\mathsf{Rob}=\min \left\{ \varvec{e}^T\varvec{z}\;|\;{\hat{\varvec{B}}}^T\varvec{z}\ge \varvec{e},\varvec{z}\ge \varvec{0}\right\} .\end{aligned}$$

Let

$$\begin{aligned} s=\sum _{i=1}^n \frac{1}{i} = \varTheta (\log n). \end{aligned}$$

It is easy to observe that $\frac{1}{s}\varvec{e}$ is a feasible solution for both the primal and the dual formulations of $z_\mathsf{Rob}$. Moreover, they have the same objective value. Therefore,

$$\begin{aligned} z_\mathsf{Rob}=\frac{n}{s}. \end{aligned}$$

On the other hand, for each $j\in [n]$, denote

$$\begin{aligned} \mathcal{U}_j=\left\{ \varvec{b}\in {\mathbb R}^n_+\;\left| \;\sum _{i=1}^{n} [(n+i-j+1)\bmod {n}]\cdot b_i\le 1\right. \right\} . \end{aligned}$$

By writing the dual of the inner maximization problem of $\varPi _\mathsf{AR}$, we have

$$\begin{aligned}\begin{aligned} z_\mathsf{AR}&=\min \left\{ \varvec{e}^T\varvec{\alpha }\;|\;\varvec{B}^T\varvec{\alpha }\ge \varvec{e},\varvec{\alpha }\ge \varvec{0},\varvec{B}\in \mathcal{U}\right\} \\&=\min \left\{ \lambda \;|\;\lambda \varvec{B}^T\varvec{\mu }\ge \varvec{e},\varvec{e}^T\varvec{\mu }=1,\varvec{\mu }\ge \varvec{0},\varvec{B}\in \mathcal{U}\right\} \\&\\&=\min \left\{ \left. \frac{1}{\theta }\; \right| \;\varvec{b}_j^T\varvec{\mu }\ge \theta ,\varvec{b}_j\in \mathcal{U}_j,\varvec{e}^T\varvec{\mu }=1,\varvec{\mu }\ge \varvec{0}\right\} . \end{aligned} \end{aligned}$$

Therefore, we just need to solve

$$\begin{aligned} \frac{1}{z_\mathsf{AR}}=\max \left\{ \theta \;|\;\varvec{b}_j^T\varvec{\mu }\ge \theta ,\varvec{b}_j\in \mathcal{U}_j,\varvec{e}^T\varvec{\mu }=1,\varvec{\mu }\ge \varvec{0}\right\} \end{aligned}$$

(E.1)

Suppose $({\hat{\theta }},{\hat{\varvec{\mu }}}, {\hat{\varvec{b}}}_j, j\in [m])$ is an optimal solution for (E.1). For each $j\in [n]$, consider a basic optimal solution ${\tilde{\varvec{b}}}_j$ of the following LP:

$$\begin{aligned} {\tilde{\varvec{b}}}_j\in \arg \max \left\{ \varvec{b}^T{\hat{\varvec{\mu }}}\;|\;\varvec{b}\in \mathcal{U}_j\right\} . \end{aligned}$$

Therefore, ${\tilde{\varvec{b}}}_j$ is a vertex of $\mathcal{U}_j$, which implies that ${\tilde{\varvec{b}}}_j={\hat{B}}_{i_j j}\varvec{e}_{i_j}$ for some $i_j\in [n]$ and ${\tilde{\varvec{b}}}_j^T{\hat{\varvec{\mu }}}\ge {\hat{\theta }}$. For each $i\in [n]$, let ${\mathcal S}_i=\{j\;|\;i_j=i\}$. We have $\sum _{i=1}^n |{\mathcal S}_i|=n$. For each $i\in [n]$ such that ${\mathcal S}_i\ne \emptyset$, ${\hat{B}}_{ij}$ can only take values in $\{1, 1/2,\ldots , 1/n\}$ for $j\in {\mathcal S}_i$. Moreover, ${\hat{B}}_{ij}\ne {\hat{B}}_{ik}$ for $j\ne k$. Therefore, there exists $l_i\in {\mathcal S}_i$ such that

$$\begin{aligned} {\hat{B}}_{i l_i}\le \frac{1}{|{\mathcal S}_i|}, \text{ and } {\tilde{\varvec{b}}}_{l_i}^T{\hat{\varvec{\mu }}}={\hat{B}}_{i l_i}{\hat{\mu }}_{i}\ge {\hat{\theta }}. \end{aligned}$$

We have

$$\begin{aligned} 1=\sum _{i:{\mathcal S}_i\ne \emptyset } {\hat{\mu }}_i \ge \sum _{i:{\mathcal S}_i\ne \emptyset } \frac{{\hat{\theta }}}{{\hat{B}}_{i l_i}}\ge \sum _{i:{\mathcal S}_i\ne \emptyset }{\hat{\theta }}|{\mathcal S}_i|={\hat{\theta }}n. \end{aligned}$$

Therefore, ${\hat{\theta }}\le \frac{1}{n}$, which implies that $z_\mathsf{AR}\ge n$.

On the other hand, it is easy to observe that $z_\mathsf{AR} \le n$: $\varvec{b}_j=\varvec{e}_j$, $\varvec{\mu } = 1/n \cdot \varvec{e}$ and $\theta = 1/n$ is a feasible solution for (E.1). Therefore,

$$\begin{aligned} z_\mathsf{AR}= n=\sum _{i=1}^n \frac{1}{i}\cdot z_\mathsf{Rob}=\varTheta (\log n)\cdot z_\mathsf{Rob}, \end{aligned}$$

which completes the proof.