On Recoverable and Two-Stage Robust Selection Problems with Budgeted Uncertainty

TL;DR

This paper studies robust two-stage and recoverable selection problems under budgeted uncertainty, providing polynomial algorithms and mixed integer formulations for different uncertainty models.

Contribution

It introduces polynomial algorithms and mixed integer formulations for robust selection problems with budgeted uncertainty, addressing both continuous and discrete cases.

Findings

Polynomial algorithms for continuous uncertainty models.

Mixed integer formulations for discrete uncertainty models.

Effective handling of budgeted uncertainty in selection problems.

Abstract

In this paper the problem of selecting $p$ out of $n$ available items is discussed, such that their total cost is minimized. We assume that costs are not known exactly, but stem from a set of possible outcomes. Robust recoverable and two-stage models of this selection problem are analyzed. In the two-stage problem, up to $p$ items is chosen in the first stage, and the solution is completed once the scenario becomes revealed in the second stage. In the recoverable problem, a set of $p$ items is selected in the first stage, and can be modified by exchanging up to $k$ items in the second stage, after a scenario reveals. We assume that uncertain costs are modeled through bounded uncertainty sets, i.e., the interval uncertainty sets with an additional linear (budget) constraint, in their discrete and continuous variants. Polynomial algorithms for recoverable and two-stage selection problems with continuous bounded uncertainty, and compact mixed integer formulations in the case of discrete bounded uncertainty are constructed.

Tables

Table 1. Table 1: The known results for 𝒰 D superscript 𝒰 𝐷 \mathcal{U}^{D} and 𝒰 I superscript 𝒰 𝐼 \mathcal{U}^{I} obtained in  [ 20 ] and new results for 𝒰 c superscript 𝒰 𝑐 \mathcal{U}^{c} and 𝒰 d superscript 𝒰 𝑑 \mathcal{U}^{d} shown in this paper.

$𝒰$	IREC	AREC	RREC	I2ST	A2ST	R2ST
	$O\left(n\right)$	$O\left(Kn\right)$	NP-hard for const. $K$;	$O\left(n\right)$	$O\left(Kn\right)$	NP-hard for const. $K$;
			str. NP-hard			str. NP-hard
${𝒰}^D$			not at all appr.			appr. $O\left(\log K+\log n\right)$
			for unbounded $K$			not appr. $\left(1-ϵ\right)\log n$
						for unbounded $K$
${𝒰}^I$	$O\left(n\right)$	$O\left(n\right)$	$O\left(\left(p-k\right)n^2\right)$	$O\left(n\right)$	$O\left(n\right)$	$O\left(n\right)$
${𝒰}^c$	$O\left(n\right)$	$O\left(n^2\right)$	poly. sol.	$O\left(n\right)$	$O\left(n^2\right)$	poly. sol.
		$O\left(n\log n\right)$	compact MIP		$O\left(n\log n\right)$	compact MIP
${𝒰}^d$	$O\left(n\right)$	$O\left(n^3\right)$	compact MIP	$O\left(n\right)$	$O\left(n^2\right)$	compact MIP

Table 2. Table 2: The coefficient matrix of ( 16 )

	$x_1$	$x_2$	$\cdots$	$x_n$	$z_1$	$z_2$	$\cdots$	$z_n$	${\overline{z}}_1$	${\overline{z}}_2$	$\cdots$	${\overline{z}}_n$	$z_1^{\prime }$	$z_2^{\prime }$	$\cdots$	$z_n^{\prime }$	${\overline{z}}_1^{\prime }$	${\overline{z}}_2^{\prime }$	$\cdots$	${\overline{z}}_n^{\prime }$
$a_1$	1	1	$\cdots$	1	0	0	$\cdots$	0	0	0	$\cdots$	0	0	0	$\cdots$	0	0	0	$\cdots$	0
$a_2$	0	0	$\cdots$	0	1	1	$\cdots$	1	0	0	$\cdots$	0	0	0	$\cdots$	0	0	0	$\cdots$	0
$a_3$	0	0	$\cdots$	0	0	0	$\cdots$	0	1	1	$\cdots$	1	0	0	$\cdots$	0	0	0	$\cdots$	0
$a_4$	0	0	$\cdots$	0	0	0	$\cdots$	0	0	0	$\cdots$	0	1	1	$\cdots$	1	0	0	$\cdots$	0
$a_5$	0	0	$\cdots$	0	0	0	$\cdots$	0	0	0	$\cdots$	0	0	0	$\cdots$	0	1	1	$\cdots$	1
$b_1$	1	0	$\cdots$	0	0	0	$\cdots$	0	0	0	$\cdots$	0	-1	0	$\cdots$	0	0	0	$\cdots$	0
$b_2$	0	1	$\cdots$	0	0	0	$\cdots$	0	0	0	$\cdots$	0	0	-1	$\cdots$	0	0	0	$\cdots$	0
$⋮$	$⋮$	$⋮$		$⋮$	$⋮$	$⋮$		$⋮$	$⋮$	$⋮$		$⋮$	$⋮$	$⋮$		$⋮$	$⋮$	$⋮$		$⋮$
$b_n$	0	0	$\cdots$	1	0	0	$\cdots$	0	0	0	$\cdots$	0	0	0	$\cdots$	-1	0	0	$\cdots$	0
$c_1$	0	0	$\cdots$	0	1	0	$\cdots$	0	0	0	$\cdots$	0	-1	0	$\cdots$	0	0	0	$\cdots$	0
$c_2$	0	0	$\cdots$	0	0	1	$\cdots$	0	0	0	$\cdots$	0	0	-1	$\cdots$	0	0	0	$\cdots$	0
$⋮$	$⋮$	$⋮$		$⋮$	$⋮$	$⋮$		$⋮$	$⋮$	$⋮$		$⋮$	$⋮$	$⋮$		$⋮$	$⋮$	$⋮$		$⋮$
$c_n$	0	0	$\cdots$	0	0	0	$\cdots$	1	0	0	$\cdots$	0	0	0	$\cdots$	-1	0	0	$\cdots$	0
$d_1$	1	0	$\cdots$	0	0	0	$\cdots$	0	0	0	$\cdots$	0	0	0	$\cdots$	0	-1	0	$\cdots$	0
$d_2$	0	1	$\cdots$	0	0	0	$\cdots$	0	0	0	$\cdots$	0	0	0	$\cdots$	0	0	-1	$\cdots$	0
$⋮$	$⋮$	$⋮$		$⋮$	$⋮$	$⋮$		$⋮$	$⋮$	$⋮$		$⋮$	$⋮$	$⋮$		$⋮$	$⋮$	$⋮$		$⋮$
$d_n$	0	0	$\cdots$	1	0	0	$\cdots$	0	0	0	$\cdots$	0	0	0	$\cdots$	0	0	0	$\cdots$	-1
$e_1$	0	0	$\cdots$	0	0	0	$\cdots$	0	1	0	$\cdots$	0	0	0	$\cdots$	0	-1	0	$\cdots$	0
$e_2$	0	0	$\cdots$	0	0	0	$\cdots$	0	0	1	$\cdots$	0	0	0	$\cdots$	0	0	-1	$\cdots$	0
$⋮$	$⋮$	$⋮$		$⋮$	$⋮$	$⋮$		$⋮$	$⋮$	$⋮$		$⋮$	$⋮$	$⋮$		$⋮$	$⋮$	$⋮$		$⋮$
$e_n$	0	0	$\cdots$	0	0	0	$\cdots$	0	0	0	$\cdots$	1	0	0	$\cdots$	0	0	0	$\cdots$	-1

Table 3. Table 3: coefficient matrix of problem ( 23 ).

	$x_1$	$x_2$	$x_3$	$\cdots$	$x_n$	$z_1$	$z_2$	$z_3$	$\cdots$	$z_n$	${\overline{z}}_1$	${\overline{z}}_2$	${\overline{z}}_3$	$\cdots$	${\overline{z}}_n$
$a_1$	1	1	1	$\cdots$	1	0	0	0	$\cdots$	0	0	0	0	$\cdots$	0
$a_2$	0	0	0	$\cdots$	0	1	1	1	$\cdots$	1	0	0	0	$\cdots$	0
$a_3$	0	0	0	$\cdots$	0	0	0	0	$\cdots$	0	1	1	1	$\cdots$	1
$b_1$	1	0	0	$\cdots$	0	1	0	0	$\cdots$	0	0	0	0	$\cdots$	0
$b_2$	0	1	0	$\cdots$	0	0	1	0	$\cdots$	0	0	0	0	$\cdots$	0
$b_3$	0	0	1	$\cdots$	0	0	0	1	$\cdots$	0	0	0	0	$\cdots$	0
$⋮$	$⋮$	$⋮$	$⋮$		$⋮$	$⋮$	$⋮$	$⋮$		$⋮$	$⋮$	$⋮$	$⋮$		$⋮$
$b_n$	0	0	0	$\cdots$	1	0	0	0	$\cdots$	1	0	0	0	$\cdots$	0
$c_1$	1	0	0	$\cdots$	0	0	0	0	$\cdots$	0	1	0	0	$\cdots$	0
$c_2$	0	1	0	$\cdots$	0	0	0	0	$\cdots$	0	0	1	0	$\cdots$	0
$c_3$	0	0	1	$\cdots$	0	0	0	0	$\cdots$	0	0	0	1	$\cdots$	0
$⋮$	$⋮$	$⋮$	$⋮$		$⋮$	$⋮$	$⋮$	$⋮$		$⋮$	$⋮$	$⋮$	$⋮$		$⋮$
$c_n$	0	0	0	$\cdots$	1	0	0	0	$\cdots$	0	0	0	0	$\cdots$	1

Equations

$$\begin{array}[]{lll}\min&\sum_{i\in[n]}c_{i}x_{i}\\ \text{s.t. }&\sum_{i\in[n]}x_{i}=p\\ &x_{i}\in\{0,1\}&\forall i\in[n].\end{array}$$

$${\mathcal{U}}^c=\{({\underline{c}}_i+{\delta }_i{)}_{i\in [n]}:{\delta }_i\in [0,d_i],\sum _{i\in [n]}{\delta }_i\le \Gamma \}\subseteq {\mathbb{R}}_{+}^n$$

$${\mathcal{U}}^d=\{({\underline{c}}_i+{\delta }_i{)}_{i\in [n]}:{\delta }_i\in \{0,d_i\},|\{i\in [n]:{\delta }_i=d_i\}|\le \Gamma \}\subseteq {\mathbb{R}}_{+}^n$$

$$\text{textsc}MinMax:\underset{\mathbf{x}\in \Phi }{\min }\underset{\mathbf{c}\in \mathcal{U}}{\max }\mathbf{c}\mathbf{x},$$

$$\text{textsc}MinMax-Regret:\underset{\mathbf{x}\in \Phi }{\min }\underset{\mathbf{c}\in \mathcal{U}}{\max }\underset{\mathbf{y}\in \Phi }{\max }(\mathbf{c}\mathbf{x}-\mathbf{c}\mathbf{y}).$$

$$\text{textsc}RREC:\underset{\mathbf{x}\in \Phi }{\min }\left(\mathbf{C}\mathbf{x}+\underset{\mathbf{c}\in \mathcal{U}}{\max }\underset{\mathbf{y}\in {\Phi }_{\mathbf{x}}^k}{\min }\mathbf{c}\mathbf{y}\right).$$

$$\text{textsc}R2ST:\underset{\mathbf{x}\in {\Phi }_1}{\min }\left(\mathbf{C}\mathbf{x}+\underset{\mathbf{c}\in \mathcal{U}}{\max }\underset{\mathbf{y}\in {\Phi }_{\mathbf{x}}}{\min }\mathbf{c}\mathbf{y}\right),$$

$$\text{textsc}AREC(A2ST):\underset{\mathbf{c}\in \mathcal{U}}{\max }\underset{\mathbf{y}\in {\Phi }_{\mathbf{x}}^k({\Phi }_{\mathbf{x}})}{\min }\mathbf{c}\mathbf{y}.$$

$$\text{textsc}IREC(I2ST):\underset{\mathbf{y}\in {\Phi }_{\mathbf{x}}^k({\Phi }_{\mathbf{x}})}{\min }\mathbf{c}\mathbf{y}.$$

$$\begin{array}[]{lllll}opt_{1}=\min&\displaystyle\sum_{i\in[n]}c_{i}y_{i}\\ \text{s.t. }&\displaystyle\sum_{i\in[n]}y_{i}=p\\ &\displaystyle\sum_{i\in[n]}x_{i}y_{i}\geq p-k\\ &y_{i}\in[0,1]&i\in[n]\end{array}$$

$$\begin{array}[]{lllll}\max&\displaystyle p\alpha+(p-k)\beta-\sum_{i\in[n]}\gamma_{i}\\ \text{s.t. }&\displaystyle\alpha+x_{i}\beta\leq\gamma_{i}+c_{i}&i\in[n]\\ &\beta\geq 0\\ &\gamma_{i}\geq 0&i\in[n]\end{array}$$

$$\underset{\alpha ,\beta \ge 0}{\max }f(\alpha ,\beta )=\underset{\alpha ,\beta \ge 0}{\max }\left\{p\alpha +(p-k)\beta -\sum _{i\in [n]}[\alpha +\beta x_i-c_i{]}_{+}\right\},$$

$$f(\alpha ,\beta )$$

$$=p{c}_{\varsigma (k)}+(p-k)(c_{\nu (p-k)}-c_{\varsigma (k)})-\sum _{i\in X_{\mathbf{x}}}[c_{\nu (p-k)}-c_i{]}_{+}-\sum _{i\in {\overline{X}}_{\mathbf{x}}}[c_{\varsigma (k)}-c_i{]}_{+}$$

$$=p{c}_{\varsigma (k)}+(p-k)(c_{\nu (p-k)}-c_{\varsigma (k)})-(p-k)c_{\nu (p-k)}+\sum _{i\in [p-k]}{c}_{\nu (i)}-k{c}_{\varsigma (k)}+\sum _{i\in [k]}{c}_{\varsigma (i)}$$

$$=\sum _{i\in [p-k]}{c}_{\nu (i)}+\sum _{i\in [k]}{c}_{\varsigma (i)}=op{t}_1$$

$${\mathbf{c}}_1{\mathbf{y}}_1=\mathbf{c}\mathbf{y}+\Delta -(p-k-k_1)\frac{\Delta }{|A|}.$$

$${\mathbf{c}}_1{\mathbf{y}}_1=\mathbf{c}\mathbf{y}+(p-k-k_1)\frac{\Delta }{|A|}-(k-k_2)\frac{\Delta }{|B|},$$

$${\mathbf{c}}_2{\mathbf{y}}_2=\mathbf{c}\mathbf{y}-(p-k-k_1)\frac{\Delta }{|A|}+(k-k_2)\frac{\Delta }{|B|}.$$

$$\begin{array}[]{llll}\max&p\alpha+(p-k)\beta-\sum_{i\in[n]}\gamma_{i}\\ \text{s.t. }&\alpha+x_{i}\beta\leq\gamma_{i}+\underline{c}_{i}+\delta_{i}&\forall i\in[n]\\ &\sum_{i\in[n]}\delta_{i}\leq\Gamma\\ &\delta_{i}\leq d_{i}&\forall i\in[n]\\ &\beta\geq 0\\ &\gamma_{i},\delta_{i}\geq 0&i\in[n]\end{array}$$

$$\underset{\alpha ,\beta \ge 0}{\max }\left\{\alpha p+\beta (p-k)-\max \left\{\sum _{i\in [n]}[\alpha +\beta x_i-{\underline{c}}_i{]}_{+}-\Gamma ,\sum _{i\in [n]}[\alpha +\beta x_i-{\overline{c}}_i{]}_{+}\right\}\right\}$$

$$\begin{array}[]{lllll}\max&\displaystyle p\alpha+(p-k)\beta-\sum_{i\in[n]}[\alpha+\beta x_{i}-\underline{c}_{i}-\delta_{i}]_{+}\\ \text{s.t. }&\displaystyle\sum_{i\in[n]}\delta_{i}\leq\Gamma&\\ &0\leq\delta_{i}\leq d_{i}&i\in[n]\\ &\beta\geq 0\\ \end{array}$$

$$\begin{array}[]{lllll}z=\min&\displaystyle\sum_{i\in[n]}[\alpha+\beta x_{i}-\underline{c}_{i}-\delta_{i}]_{+}\\ \text{s.t. }&\displaystyle\sum_{i\in[n]}\delta_{i}\leq\Gamma\\ &0\leq\delta_{i}\leq d_{i}&i\in[n]\\ \end{array}$$

$$z$$

$$=\max \left\{\sum _{i\in [n]}[\alpha +\beta x_i-{\underline{c}}_i{]}_{+}-\Gamma ,\sum _{i\in [n]}[\alpha +\beta x_i-{\overline{c}}_i{]}_{+}\right\},$$

$$\underset{\alpha }{\max }f(\alpha )=\underset{\alpha }{\max }\left\{\alpha p-\max \left\{\sum _{i\in [n]}[\alpha -{\underline{c}}_i{]}_{+}-\Gamma ,\sum _{i\in [n]}[\alpha -{\underline{c}}_i-d_i{]}_{+}\right\}\right\}$$

$$\underset{\alpha \in [4,5]}{\max }\min \{-5\alpha +44,4\alpha +4\}.$$

$$\begin{array}[]{lll}\displaystyle\max_{\alpha,\gamma\geq\alpha}g(\alpha,\gamma)=\max_{\alpha,\gamma\geq\alpha}\{\alpha k+\gamma(p-k)-\\ \displaystyle\left.\max\left\{\sum_{i\in X_{\boldsymbol{x}}}[\gamma-\underline{c}_{i}]_{+}+\sum_{i\in\overline{X}_{\boldsymbol{x}}}[\alpha-\underline{c}_{i}]_{+}-\Gamma,\sum_{i\in X_{\boldsymbol{x}}}[\gamma-\underline{c}_{i}-d_{i}]_{+}+\sum_{i\in\overline{X}_{\boldsymbol{x}}}[\alpha-\underline{c}_{i}-d_{i}]_{+}\right\}\right\}.\end{array}$$

$$\underset{\alpha \in [2,4]}{\max }g(\alpha ,6)=\underset{\alpha \in [2,4]}{\max }\min \{28-2\alpha ,2\alpha +17\}.$$

$$U_1=[\gamma -{\underline{c}}_k{]}_{+}-[\gamma -{\underline{c}}_l{]}_{+}+[\alpha -{\underline{c}}_l{]}_{+}-[\alpha -{\underline{c}}_k{]}_{+}\ge 0$$

$$U_2=[\gamma -{\overline{c}}_k{]}_{+}-[\gamma -{\overline{c}}_l{]}_{+}+[\alpha -{\overline{c}}_l{]}_{+}-[\alpha -{\overline{c}}_k{]}_{+}\ge 0$$

$$\mathbf{C}{\mathbf{x}}^{\prime }+\underset{\mathbf{c}\in {\mathcal{U}}^c}{\max }\underset{\mathbf{y}\in {\Phi }_{{\mathbf{x}}^{\prime }}^k}{\min }\mathbf{c}\mathbf{y}\le \mathbf{C}\mathbf{x}+\underset{\mathbf{c}\in {\mathcal{U}}^c}{\max }\underset{\mathbf{y}\in {\Phi }_{\mathbf{x}}^k}{\min }\mathbf{c}\mathbf{y},$$

Full text

Robust recoverable and two-stage models of this selection problem are analyzed. In the two-stage problem, up to $p$ items is chosen in the first stage, and the solution is completed once the scenario becomes revealed in the second stage. In the recoverable problem, a set of $p$ items is selected in the first stage, and can be modified by exchanging up to $k$ items in the second stage, after a scenario reveals.

We assume that uncertain costs are modeled through bounded uncertainty sets, i.e., the interval uncertainty sets with an additional linear (budget) constraint, in their discrete and continuous variants. Polynomial algorithms for recoverable and two-stage selection problems with continuous bounded uncertainty, and compact mixed integer formulations in the case of discrete bounded uncertainty are constructed.

**Keywords: ** combinatorial optimization; robust optimization; selection problem; budgeted uncertainty; two-stage robustness; recoverable robustness

1 Introduction

In this paper we consider the following Selection problem. We are given a set of $n$ items with cost $c_{i}$ for each $i\in[n]:=\{1,\dots,n\}$ and an integer $p\in[n]$. We seek a subset $X\subseteq[n]$ of $p$ items, $|X|=p$, whose total cost $\sum_{i\in X}c_{i}$ is minimum. It is easy to see that an optimal solution is composed of $p$ items of the smallest cost. It can be found in $O(n)$ time by using the well-known fact, that the $p$th smallest item can be found in $O(n)$ time (see, e.g., [10]). Selection is a basic resource allocation problem [17]. It is also a special case of 0-1 knapsack, 0-1 assignment, single machine scheduling, and minimum matroid base problems (see [19] for an overview). It can be formulated as the following integer linear program:

[TABLE]

We will use $\Phi\subseteq\{0,1\}^{n}$ to denote the set of all feasible solutions to (1). Given $\boldsymbol{x}\in\{0,1\}^{n}$, we also define $X_{\boldsymbol{x}}=\{i\in[n]:x_{i}=1\}$, and $\overline{X}_{\boldsymbol{x}}=[n]\setminus X_{\boldsymbol{x}}$, i.e. $X_{\boldsymbol{x}}$ is the item set induced by vector $\boldsymbol{x}$ and $\overline{X}_{\boldsymbol{x}}$ denotes its complement.

Consider the case when the item costs are uncertain. As part of the input, we are given a scenario set $\mathcal{U}$, containing all possible vectors of the item costs, called scenarios. Several methods of defining $\mathcal{U}$ have been proposed in the existing literature (see, e.g., [3, 4, 18, 23, 26]). Under discrete uncertainty (see, e.g., [23]), the scenario set contains $K$ distinct scenarios i.e. $\mathcal{U}^{D}=\{\boldsymbol{c}_{1},\dots,\boldsymbol{c}_{K}\}$, $\boldsymbol{c}_{i}\in\mathbb{R}^{n}_{+}$. Under interval uncertainty, the cost of each item $i\in[n]$ belongs to a closed interval $[\underline{c}_{i},\overline{c}_{i}]$, where $d_{i}:=\overline{c}_{i}-\underline{c}_{i}\geq 0$ is the maximal deviation of the cost of $i$ from its nominal value $\underline{c}_{i}$. In the traditional interval uncertainty representation, $\mathcal{U}^{I}$ is the Cartesian product of all the intervals (see, e.g., [23]). In this paper we will focus on the following two generalizations of scenario set $\mathcal{U}^{I}$, which have been examined in [3, 4, 26]:

• •

Continuous budgeted uncertainty:

[TABLE]

• •

Discrete budgeted uncertainty:

[TABLE]

The fixed parameter $\Gamma\geq 0$ is called a budget and it controls the amount of uncertainty which an adversary can allocate to the item costs. For a sufficiently large $\Gamma$, $\mathcal{U}^{c}$ reduces to $\mathcal{U}^{I}$, and $\mathcal{U}^{d}$ reduces to the extreme points of $\mathcal{U}^{I}$.

In order to compute a solution, under a specified scenario set $\mathcal{U}$, we can follow a robust optimization approach. For general overviews on robust optimization, see, e.g., [1, 14, 23, 21, 27]. In a typical, single-stage robust model we seek a solution minimizing the total cost in a worst case. This leads to the following minmax and minmax regret problems:

[TABLE]

[TABLE]

The minmax (regret) versions of the Selection problem have been discussed in the existing literature. For scenario set $\mathcal{U}^{D}$ both problems are NP-hard even for $K=2$ (the number of scenarios equals 2) [2]. If $K$ is part of the input, then MinMax and MinMax-Regret are strongly NP-hard and not approximable within any constant factor [19]. On the other hand, MinMax is approximable within $O(\log K/\log\log K)$ [12] but MinMax-Regret is only known to be approximable within $K$, which is due to the results given in [1]. The MinMax problem under scenario sets $\mathcal{U}^{c}$ and $\mathcal{U}^{d}$ is polynomially solvable, according to the results obtained in [3]. Also, MinMax-Regret, under scenario set $\mathcal{U}^{I}$, is polynomially solvable by the algorithms designed in [2, 9].

The problems which arises in practice often have a two-stage nature. Namely, a partial solution is computed in the first stage and completed in the second stage, or a complete solution is formed in the first stage and modified to some extent in the second stage. Typically, the costs in the first stage are precisely known, while the costs in the second stage are uncertain. Before we formally define the two-stage models, let us introduce some additional notation:

• •

$\Phi_{1}=\{\boldsymbol{x}\in\{0,1\}^{n}:\sum_{i\in[n]}x_{i}\leq p\}$,

• •

$\Phi_{\boldsymbol{x}}=\{\boldsymbol{y}\in\{0,1\}^{n}:\sum_{i\in[n]}(x_{i}+y_{i})=p,x_{i}+y_{i}\leq 1,i\in[n]\},\;\;\boldsymbol{x}\in\Phi_{1}$,

• •

$\Phi^{k}_{\boldsymbol{x}}=\{\boldsymbol{y}\in\{0,1\}^{n}:\sum_{i\in[n]}y_{i}=p,\ \sum_{i\in[n]}x_{i}y_{i}\geq p-k\},\;\;\boldsymbol{x}\in\Phi,k\in[p]\cup\{0\}.$

If $\boldsymbol{y}\in\Phi_{\boldsymbol{x}}$, then $X_{\boldsymbol{x}}\cap X_{\boldsymbol{y}}=\emptyset$ and $|X_{\boldsymbol{x}}\cup X_{\boldsymbol{y}}|=p$. Hence $\Phi_{\boldsymbol{x}}$ encodes all subsets of the item set $[n]$, which added to $X_{\boldsymbol{x}}$ form a complete solution of cardinality $p$. Set $\Phi^{k}_{\boldsymbol{x}}$ is called a recovery set, $k$ is a given recovery parameter. If $\boldsymbol{y}\in\Phi^{k}_{\boldsymbol{x}}$, then $|X_{\boldsymbol{x}}\setminus X_{\boldsymbol{y}}|=|X_{\boldsymbol{y}}\setminus X_{\boldsymbol{x}}|\leq k$, so $\Phi^{k}_{\boldsymbol{x}}$ encodes all solutions which can be obtained from $X_{\boldsymbol{x}}$ by exchanging up to $k$ items. Let $\boldsymbol{C}=(C_{1},\dots,C_{n})$ be a vector of the first stage item costs, which are assumed to be precisely known. Let scenario set $\mathcal{U}$ contain all possible vectors of the uncertain second stage costs. Given $k\in[p]\cup\{0\}$, we study the following recoverable selection problem:

[TABLE]

In RREC a complete solution (exactly $p$ items) is chosen in the first stage. Then, after a scenario from $\mathcal{U}$ reveals, one can exchange optimally up to $k$ items in the second stage. Notice that if $k=0$ and $C_{i}=0$ for each $i\in[n]$, then RREC becomes the MinMax problem. The robust recoverable model for linear programming, together with some applications, was discussed in [24]. It has been also recently applied to the shortest path [5], spanning tree [15, 16], knapsack [6] and traveling salesman problems [8]. The RREC problem under scenario sets $\mathcal{U}^{D}$ and $\mathcal{U}^{I}$ has been recently discussed in [20]. Under $\mathcal{U}^{D}$ it turned out to be NP-hard for constant $K$, strongly NP-hard and not at all approximable when $K$ is part of the input (this is true even if $k=1$). On the other hand, under scenario set $\mathcal{U}^{I}$, a polynomial $O((p-k)n^{2})$ time algorithm for RREC has been proposed in [20]. No results for scenario rests $\mathcal{U}^{c}$ and $\mathcal{U}^{d}$ have been known to date.

We also analyze the following robust two-stage selection problem:

[TABLE]

In R2ST we seek a first stage solution, which may contain less than $p$ items. Then, after a scenario from $\mathcal{U}$ reveals, this solution is completed optimally to $p$ items. The robust two-stage model was introduced in [22] for the bipartite matching problem. The R2ST problem has been recently discussed in [20]. It is polynomially solvable under scenario set $\mathcal{U}^{I}$. For scenario set $\mathcal{U}^{D}$, the problem is strongly NP-hard and hard to approximate within $(1-\epsilon)\log n$ for any $\epsilon>0$, but it has an $O(\log K+\log n)$ randomized approximation algorithm. No results for scenario sets $\mathcal{U}^{c}$ and $\mathcal{U}^{d}$ have been known to date.

Given a first stage solution $\boldsymbol{x}\in\Phi$ (resp. $\boldsymbol{x}\in\Phi_{1}$), we will also study the following adversarial problem:

[TABLE]

If, additionally, scenario $\boldsymbol{c}\in\mathcal{U}$ is fixed, then we get the following incremental problem:

[TABLE]

The adversarial and incremental versions of some network problems were discussed in [26, 11]. The incremental versions of the shortest path and the spanning tree problems are polynomially solvable [11], whereas the adversarial versions of these problems under scenario set $\mathcal{U}^{d}$ are strongly NP-hard [26, 13, 25].

New results.

All new results for scenario sets $\mathcal{U}^{c}$ and $\mathcal{U}^{d}$, obtained in this paper, are summarized in Table 1. In particular, we show that all the considered problems are polynomially solvable under scenario set $\mathcal{U}^{c}$. The polynomial algorithms for RREC and R2ST under $\mathcal{U}^{c}$ are based on solving a polynomial number of linear programming subproblems. We also provide polynomial time combinatorial algorithms for AREC and A2ST under both $\mathcal{U}^{c}$ and $\mathcal{U}^{d}$. The complexity of RREC and R2ST under $\mathcal{U}^{d}$ remains open. For these problems we construct compact MIP formulations and propose approximation algorithms.

2 Continuous Budgeted Uncertainty

In this section we address the RREC and R2ST problems under scenario set $\mathcal{U}^{c}$. We will show that both problems can be solved in polynomial time.

2.1 Recoverable Robust Selection

2.1.1 The incremental problem

Given $\boldsymbol{x}\in\Phi$ and $\boldsymbol{c}\in\mathcal{U}$, the incremental problem, IREC, can be formulated as the following linear program (notice that the constraints $y_{i}\in\{0,1\}$ can be relaxed):

[TABLE]

It is easy to see that the IREC problem can be solved in $O(n)$ time. Indeed, we first choose $p-k$ items of the smallest cost from $X_{\boldsymbol{x}}$ and then $k$ items of the smallest cost from the remaining items. We will now show some additional properties of (2), which will be used extensively later. The dual to (2) is

[TABLE]

From now on, we will assume that $k>0$ (the case $k=0$ is trivial, since $\boldsymbol{y}=\boldsymbol{x}$ holds). Let $b(\boldsymbol{c})$ be the $p$th smallest item cost for the items in $[n]$ under $\boldsymbol{c}$ (i.e. if $c_{\sigma(1)}\leq\dots\leq c_{\sigma(n)}$ is the ordered sequence of the item costs under $\boldsymbol{c}$, then $b(\boldsymbol{c})=c_{\sigma(p)}$). Similarly, let $b_{1}(\boldsymbol{c})$ be the $(p-k)$th smallest item cost for the items in $X_{\boldsymbol{x}}$ and $b_{2}(\boldsymbol{c})$ be the $k$th smallest item cost for the items in $\overline{X}_{\boldsymbol{x}}$ under $\boldsymbol{c}$. The following proposition characterizes the optimal values of $\alpha$ and $\beta$ in (3), and is fundamental in the following analysis:

Proposition 1**.**

Given scenario $\boldsymbol{c}\in\mathcal{U}$, the following conditions hold:

if $b_{1}(\boldsymbol{c})\leq b(\boldsymbol{c})$, then $\alpha=b(\boldsymbol{c})$ and $\beta=0$ are optimal in (3), 2. 2.

if $b_{1}(\boldsymbol{c})>b(\boldsymbol{c})$, then $\alpha=b_{2}(\boldsymbol{c})$ and $\beta=b_{1}(\boldsymbol{c})-b_{2}(\boldsymbol{c})$ are optimal in (3).

Proof.

By replacing $\gamma_{i}$ by $[\alpha+\beta x_{i}-c_{i}]_{+}$, the dual problem (3) can be represented as follows:

[TABLE]

where $[a]_{+}=\max\{0,a\}$. Let us sort the items in $[n]$ so that that $c_{\sigma(1)}\leq\dots\leq c_{\sigma(n)}$. Let us sort the items in $X_{\boldsymbol{x}}$ so that $c_{\nu(1)}\leq\dots\leq c_{\nu(p)}$ and the items in $\overline{X}_{\boldsymbol{x}}$ so that $c_{\varsigma(1)}\leq\dots\leq c_{\varsigma(n-p)}$. We distinguish two cases. The first one: $c_{\nu(p-k)}\leq c_{\sigma(p)}$ ($b_{1}(\boldsymbol{c})\leq b(\boldsymbol{c})$). Then it is possible to construct an optimal solution to (2) with the cost equal to $\sum_{i\in[p]}c_{\sigma(i)}$. Namely, we choose $p-k$ items of the smallest costs from $X_{\boldsymbol{x}}$ and $k$ items of the smallest cost from the remaining items. Fix $\alpha=c_{\sigma(p)}$ and $\beta=0$, which gives the case 1. By using (4), we obtain $f(\alpha,\beta)=\sum_{i\in[p]}c_{\sigma(i)}=opt_{1}$ and the proposition follows from the weak duality theorem. The second case: $c_{\nu(p-k)}>c_{\sigma(p)}$ ($b_{1}(\boldsymbol{c})>b(\boldsymbol{c})$). The optimal solution to (2) is then formed by the items $\nu(1),\dots,\nu(p-k)$ and $\varsigma(1),\dots,\varsigma(k)$. Fix $\alpha=c_{\varsigma(k)}$ and $\beta=c_{\nu(p-k)}-c_{\varsigma(k)}$, which gives the case 2. By (4), we have

[TABLE]

and the proposition follows from the weak duality theorem. ∎

2.1.2 The adversarial problem

Consider the adversarial problem AREC for a given solution $\boldsymbol{x}\in\Phi$. We will again assume that $k>0$. If $k=0$, then all the budget $\Gamma$ is allocated to the items in $X_{\boldsymbol{x}}$. Scenario $\boldsymbol{c}\in\mathcal{U}^{c}$ which maximizes the objective value in this problem is called a worst scenario for $\boldsymbol{x}$ (worst scenario for short). We now give a characterization of a worst scenario.

Proposition 2**.**

There is a worst scenario $\boldsymbol{c}=(\underline{c}_{i}+\delta_{i})_{i\in[n]}\in\mathcal{U}^{c}$ such that

$b_{1}(\boldsymbol{c})\leq b(\boldsymbol{c})$* or* 2. 2.

$b_{1}(\boldsymbol{c})$* or $b_{2}(\boldsymbol{c})$ belongs to $\mathcal{D}=\{\underline{c}_{1},\dots,\underline{c}_{n},\overline{c}_{1},\dots,\overline{c}_{n}\}$.*

Proof.

Assume that $b_{1}(\boldsymbol{c})>b(\boldsymbol{c})$ and both $b_{1}(\boldsymbol{c})$ and $b_{2}(\boldsymbol{c})$ do not belong to $\mathcal{D}$. The main idea of the proof is to show that there is a worst scenario satisfying condition 1 or 2. Note that $b_{1}(\boldsymbol{c})>b(\boldsymbol{c})$ implies $b_{1}(\boldsymbol{c})>b_{2}(\boldsymbol{c})$. Let $A=\{i\in X_{\boldsymbol{x}}:\underline{c}_{i}+\delta_{i}=b_{1}(\boldsymbol{c})\}$ and $B=\{i\in\overline{X}_{\boldsymbol{x}}:\underline{c}_{i}+\delta_{i}=b_{2}(\boldsymbol{c})\}$. Observe that $A,B\neq\emptyset$ by the definition of $b_{1}(\boldsymbol{c})$ and $b_{2}(\boldsymbol{c})$. Also, a positive budget must be allocated to each item in $A$ and $B$. In Figures 1a and 1b we have $A=\{2,4,5\}$ and $B=\{7,9\}$. Let $k_{1}$ be the number of items in $X_{\boldsymbol{x}}$ such that $\underline{c}_{i}+\delta_{i}<b_{1}(\boldsymbol{c})$ and $k_{2}$ be the number of items in $\overline{X}_{\boldsymbol{x}}$ such that $\underline{c}_{i}+\delta_{i}<b_{2}(\boldsymbol{c})$. In the sample problem (see Figures 1a and 1b) we have $k_{1}=2$ and $k_{2}=1$. Suppose that there is an item $j$ such that $\underline{c}_{j}+d_{j}>b_{1}(\boldsymbol{c})$ and $\underline{c}_{j}+\delta_{j}<b_{1}(\boldsymbol{c})$ (see the item 3 in Figure 1a). Let us transform scenario $\boldsymbol{c}\in\mathcal{U}^{c}$ into scenario $\boldsymbol{c}_{1}\in\mathcal{U}^{c}$ as follows: $\delta_{j}:=\delta_{j}+\Delta$ and $\delta_{i}:=\delta_{i}-\Delta/|A|$ for each $i\in A$, where $\Delta>0$ is a sufficiently small number (see Figure 1a). Let $\boldsymbol{y}$ be an optimal solution under $\boldsymbol{c}$ and let $\boldsymbol{y}_{1}$ be an optimal solution under $\boldsymbol{c}_{1}$. The following equality holds

[TABLE]

Since $|A|+k_{1}\geq p-k$, $\boldsymbol{c}_{1}\boldsymbol{y}_{1}\geq\boldsymbol{c}\boldsymbol{y}$ and $\boldsymbol{c}_{1}$ is also a worst scenario. We can increase $\Delta$ until $b_{1}(\boldsymbol{c}_{1})\in\mathcal{D}$, or $b_{1}(\boldsymbol{c}_{1})=b_{2}(\boldsymbol{c}_{1})$ (which implies $b_{1}(\boldsymbol{c}_{1})=b(\boldsymbol{c}_{1})$), or $b_{1}(\boldsymbol{c}_{1})=\underline{c}_{j}+\delta_{j}$. In the first two cases the proposition follows and the third case will be analyzed later. The same reasoning can be applied to every item $j\in\overline{X}_{\boldsymbol{x}}$ such that $\underline{c}_{j}+d_{j}>b_{2}(\boldsymbol{c})$ and $\underline{c}_{j}+\delta_{j}<b_{2}(\boldsymbol{c})$ (see the item 8 in Figure 1a). So, it remains to analyze the case shown in Figure 1b. Let us again choose some sufficiently small $\Delta>0$. Define scenario $\boldsymbol{c}_{1}$ by modifying $\boldsymbol{c}$ in the following way $\delta_{i}:=\delta_{i}+\Delta/|A|$ for each $i\in A$ and $\delta_{i}:=\delta_{i}-\Delta/|B|$ for each $i\in B$. Similarly, define scenario $\boldsymbol{c}_{2}$ by modifying $\boldsymbol{c}$ as follows $\delta_{i}:=\delta_{i}-\Delta/|A|$ for each $i\in A$ and $\delta_{i}:=\delta_{i}+\Delta/|B|$ for each $i\in B$. Let $\boldsymbol{y}_{1}$ be an optimal solution under $\boldsymbol{c}_{1}$ and $\boldsymbol{y}_{2}$ be an optimal solution under $\boldsymbol{c}_{2}$. The following equalities hold

[TABLE]

[TABLE]

Hence, either $\boldsymbol{c}_{1}\boldsymbol{y}_{1}\geq\boldsymbol{c}\boldsymbol{y}$ or $\boldsymbol{c}_{2}\boldsymbol{y}_{2}\geq\boldsymbol{c}\boldsymbol{y}$, so $\boldsymbol{c}_{1}$ or $\boldsymbol{c}_{2}$ is also a worst scenario. We can now increase $\Delta$ until $\boldsymbol{c}_{1}$ ($\boldsymbol{c}_{2}$) satisfies condition 1 or 2. ∎

Using (3) and the definition of scenario set $\mathcal{U}^{c}$, we can represent AREC as the following linear programming problem:

[TABLE]

Thus AREC can be solved in polynomial time. In the following we will construct a strongly polynomial combinatorial algorithm for solving AREC. The following corollary is a consequence of Proposition 1 and Proposition 2:

Corollary 3**.**

There is an optimal solution to (5) in which

$\beta=0$* or* 2. 2.

$\alpha$* or $\alpha+\beta$ belongs to $\mathcal{D}=\{\underline{c}_{1},\dots,\underline{c}_{n},\overline{c}_{1},\dots,\overline{c}_{n}\}$.*

Proof.

According to Proposition 1, there is an optimal solution to (5) which induces a worst scenario $\boldsymbol{c}=(\underline{c}_{i}+\delta_{i})_{i\in[n]}\in\mathcal{U}^{c}$, which satisfies conditions 1 and 2 of Proposition 1. If the condition 1 is fulfilled, i.e. $b_{1}(\boldsymbol{c})\leq b(\boldsymbol{c})$, then according to Proposition 2 we get $\beta=0$. If $b_{1}(\boldsymbol{c})>b(\boldsymbol{c})$, then condition 2 from Proposition 1 and condition 2 from Proposition 2 hold. Both these conditions imply the condition 2 of the corollary. ∎

Proposition 4**.**

The optimal values of $\alpha$ and $\beta$ in (5) can be found by solving the following problem:

[TABLE]

Proof.

Let us first rewrite (5) in the following way:

[TABLE]

Let us fix $\alpha$ and $\beta\geq 0$ in (7). Then the optimal values of $\delta_{i}$ can be then found by solving the following subproblem:

[TABLE]

Let $U=\sum_{i\in[n]}[\alpha+\beta x_{i}-\underline{c}_{i}]_{+}$. Observe that $[U-\Gamma]_{+}$ is a lower bound on $z$ as $z\geq 0$ and it is not possible to decrease $U$ by more than $\Gamma$. The subproblem (8) can be solved by applying the following greedy method. For $i:=1,\dots,n$, if $\alpha+\beta x_{i}-\underline{c}_{i}>0$, we fix $\delta_{i}=\min\{\alpha+\beta x_{i}-\underline{c}_{i},d_{i},\Gamma\}$ and modify $\Gamma:=\Gamma-\delta_{i}$. If, at some step, $\Gamma=0$ we have reached the lower bound. Hence $z=[U-\Gamma]_{+}$. On the other hand if, after the algorithm terminates, we still have $\Gamma>0$, then $z=\sum_{i\in[n]}[\alpha+\beta x_{i}-\underline{c}_{i}-d_{i}]_{+}$. In consequence

[TABLE]

which together with (7) completes the proof. ∎

Having the optimal values of $\alpha$ and $\beta$, the worst scenario $\boldsymbol{c}=(\underline{c}_{i}+\delta_{i})_{i\in[n]}$, can be found in $O(n)$ time by applying the greedy method to (8), described in the proof of Proposition 4. We now construct an efficient algorithm for solving (6), which will give us the optimal values of $\alpha$ and $\beta$. We will illustrate this algorithm by using the sample problem shown in Figure 2.

Let $h_{(1)}\leq h_{(2)}\leq\dots\leq h_{(l)}$ be the ordered sequence of the distinct values from $\mathcal{D}$. This sequence defines a family of closed intervals $I_{j}=[h_{(j)},h_{(j+1)}]$, $j\in[l-1]$, which partitions the interval $[\min_{i\in[n]}\underline{c}_{i},\max_{i\in[n]}\overline{c}_{i}]$. Notice that $l\leq 2n$. In the example shown in Figure 2 we have six intervals $I_{1},\dots I_{6}$ which split the interval $[1,8]$.

By Corollary 3, we need to investigate two cases. In the first case, we have $\beta=0$. Then (6) reduces to the following problem:

[TABLE]

Consider the problem of maximizing $f(\alpha)$ over a fixed interval $I_{j}$. It is easy to verify that (9) reduces then to finding the maximum of a minimum of two linear functions of $\alpha$ over $I_{j}$. For example, when $\alpha\in I_{3}=[4,5]$, then after an easy computation, the problem (9) becomes

[TABLE]

It is well known that the maximum value of $\alpha$ is attained at one of the bounds of the interval $I_{j}$ or at the intersection point of the two linear functions of $\alpha$. In this case we compute $\alpha$ by solving $-5\alpha+44=\alpha+4$ which yields $\alpha=4.44$. We can now solve (9) by solving at most $2n$ subproblems consisting in maximizing $f(\alpha)$ over $I_{1},\dots I_{l}$. Notice, however, that in some cases we do not have to examine all the intervals $I_{1},\dots,I_{l}$. We can use the fact that $\alpha$ is the $p$th smallest item cost in the computed scenario. In the example, the optimal value of $\alpha$ belongs to $I_{1}\cup I_{2}\cup I_{3}$. The function $f(\alpha)$ for the sample problem is shown in Figure 2. The optimal value of $\alpha$ is 4.44. The scenario corresponding to $\alpha=4.44$ can be obtained by applying a greedy method and it is also shown in Figure 3.

We now discuss the second case in Corollary 3. Let us fix $\gamma=\alpha+\beta$ and rewrite (6) as follows:

[TABLE]

According to Corollary 2, the optimal value of $\alpha$ or $\gamma$ belongs to $\mathcal{D}$. So, let us first fix $\gamma\in\mathcal{D}$ and consider the problem $\max_{\alpha}g(\alpha,\gamma)$. The optimal value of $\alpha$ can be found by optimizing $\alpha$ over each interval $I_{j}$, whose upper bound is not greater then $\gamma$ (it follows from the constraint $\alpha\leq\gamma$). Again, the problem $\max_{\alpha\in I_{j}}g(\alpha,\gamma)$ can be reduced to maximizing a minimum of two linear functions of $\alpha$ over a closed interval. To see this consider the sample problem shown in Figure 2. Fix $\gamma=6$ and assume that $\alpha\in I_{2}$. Then, a trivial verification shows that

[TABLE]

The maximum is attained when $28-2\alpha=2\alpha+17$, so for $\alpha=2.75$. The function $g(\alpha,6)$ is shown in Figure 4. It attains the maximum in the interval $I_{2}$ at $\alpha=2.75$. The scenario which corresponds to $\alpha=2.75$ and $\gamma=6$ is also shown in Figure 4. In the same way we can find the optimal value of $\alpha$ for each fixed $\gamma\in\mathcal{D}$. Since $\gamma$ is the $(p-k)$th smallest item cost in $X_{\boldsymbol{x}}$ under the computed scenario, not all values of $\gamma$ in $\mathcal{D}$ need to be examined. In the example we have to only try $\gamma\in\{2,4,5,6\}$.

We can then repeat the reasoning for every fixed $\alpha\in\mathcal{D}$. Namely, we solve the problem $\max_{\gamma\geq\alpha}g(\alpha,\gamma)$ by solving the problem for each interval whose lower bound is not less than $\alpha$. Again, not all values of $\alpha\in\mathcal{D}$ need to be examined. Since $\alpha$ is the $k$th smallest item cost in $\overline{X}_{\boldsymbol{x}}$, we should check only the values of $\alpha\in\{1,2,4,5\}$.

Theorem 5**.**

The problem AREC under scenario set $\mathcal{U}^{c}$ can be solved in $O(n^{2})$ time.

Proof.

We will present a sketch of the $O(n^{2})$ algorithm. We first determine the family of intervals $I_{1},\dots,I_{l}$, which requires $O(n\log n)$ time. Now, the key observation is that we can evaluate first all the sums that appear in (9) and (10) for each interval $I_{j}$. We can compute and store these sums for every $I_{j}$ in $O(n^{2})$ time. Now each problem $\max_{\alpha\in I_{j}}g(\alpha,\gamma)$ for $\gamma\in\mathcal{D}$, $\max_{\gamma\in I_{j}}g(\alpha,\gamma)$ for $\alpha\in\mathcal{D}$, and $\max_{\alpha\in I_{j}}f(\alpha)$ can be solved in constant time by inserting the computed earlier sums into (10) and (9). The number of problems that must be solved is $O(n^{2})$, so the overall running time of the algorithm is $O(n^{2})$. ∎

Using a more refined analysis and data structures such as min-heaps (see, e.g., [10]), this $O(n^{2})$ result can be further improved to $O(n\log n)$. We present the proof in Appendix A.

We now show the following proposition, which will be used later:

Proposition 6** (Dominance rule).**

Let $k,l$ be two items such that $\underline{c}_{k}\leq\underline{c}_{l}$ and $\overline{c}_{k}\leq\overline{c}_{l}$. Let $x_{l}=1$ and $x_{k}=0$ in (5). Then the maximum objective value in (5) will not increase when we change $x_{l}=0$ and $x_{k}=1$.

Proof.

Let $X_{\boldsymbol{x}^{\prime}}=X_{\boldsymbol{x}}\cup\{k\}\setminus\{l\}$. Notice that (9) does not depend on the first stage solution $\boldsymbol{x}$, so it remains to investigate the effect of replacing $X_{\boldsymbol{x}}$ with $X_{\boldsymbol{x}^{\prime}}$ in (10). It is enough to show that for each $\alpha$ and $\gamma\geq\alpha$ the following inequalities hold:

[TABLE]

and

[TABLE]

Inequality (11) can be proven by distinguishing the following cases: if $\alpha\leq\gamma\leq\underline{c}_{k}\leq\underline{c}_{l}$, then $U_{1}=0$; if $\alpha\leq\underline{c}_{k}\leq\gamma\leq\underline{c}_{l}$, then $U_{1}=\gamma-\underline{c}_{k}\geq 0$; if $\alpha\leq\underline{c}_{k}\leq\underline{c}_{l}\leq\gamma$, then $U_{1}=\underline{c}_{l}-\underline{c}_{k}\geq 0$; if $\underline{c}_{k}\leq\alpha\leq\gamma\leq\underline{c}_{l}$, then $U_{1}=\gamma-\alpha\geq 0$; if $\underline{c}_{k}\leq\alpha\leq\underline{c}_{l}\leq\gamma$, then $U_{1}=\gamma-\underline{c}_{k}-\gamma+\underline{c}_{l}-\alpha+\underline{c}_{k}=\underline{c}_{l}-\alpha\geq 0$; if $\underline{c}_{k}\leq\underline{c}_{l}\leq\alpha\leq\gamma$, then $U_{1}=0$. The proof of the fact that $U_{2}\geq 0$ is just the same. ∎

2.1.3 The recoverable robust problem

In this section we study RREC under scenario set $\mathcal{U}^{c}$. We first identify some special cases of this problem, which are known to be polynomially solvable. If $\Gamma$ is sufficiently large, say $\Gamma\geq\sum_{i\in[n]}d_{i}$, then scenario set $\mathcal{U}^{c}$ reduces to $\mathcal{U}^{I}$ and the problem can be solved in $O((p-k)n^{2})$ time [20]. Also the boundary cases $k=0$ and $k=p$ are polynomially solvable. When $k=p$, then we choose in the first stage $p$ items of the smallest costs under $\boldsymbol{C}$. The total cost of this solution can be then computed in $O(n^{2})$ time by solving the corresponding adversarial problem. If $k=0$, then RREC is equivalent to the MinMax problem with cost intervals $[C_{i}+\underline{c}_{i},C_{i}+\overline{c}_{i}]$, $i\in[n]$. Hence it is polynomially solvable due to the results obtained in [3].

Consider now the general case with any $k\in[p]$. We first show a method of preprocessing a given instance of the problem. Given two items $i,j\in[n]$, we write $i\preceq j$ if $C_{i}\leq C_{j}$, $\underline{c}_{i}\leq\underline{c}_{j}$ and $\overline{c}_{i}\leq\overline{c}_{j}$. For any fixed item $l\in[n]$, suppose that $|\{k:k\preceq l\}|\geq p$. Let an optimal solution $\boldsymbol{x}\in\Phi$ to RREC be given, in which $x_{l}=1$. There is an item $x_{k}$ such that $k\preceq l$ and $x_{k}=0$ in $\boldsymbol{x}$. We form solution $\boldsymbol{x}^{\prime}$ by setting $x_{k}=1$ and $x_{l}=0$. From Proposition 6 and inequality $C_{k}\leq C_{l}$, we get

[TABLE]

and $\boldsymbol{x}$ is also an optimal solution to RREC. In what follows, we can remove $l$ from $[n]$ without violating the optimum obtaining (after renumbering the items) a smaller item set $[n-1]$. We can now repeat iteratively this reasoning, which allows us to reduce the size of the input instance. Also, for each $l\in[n]$, if $|\{k:l\preceq k\}|\geq n-p$, then we do not violate the optimum after setting $x_{l}=1$.

We now reconsider the adversarial problem (5). Its dual is the following:

[TABLE]

Using this formulation, we can represent the RREC problem under scenario set $\mathcal{U}^{c}$ as the following compact mixed-integer program:

[TABLE]

The products $x_{i}y_{i}$, $i\in[n]$, can be linearized by using standard methods, which leads to a linear MIP formulation for RREC. Before solving this model the preprocessing described earlier can be applied. We now show that (13) can be solved in polynomial time. Notice that we can assume $\pi\in[0,1]$ and $\rho_{i}=[y_{i}-\pi]_{+}$. Let us split each variable $y_{i}=\mathscr{y}_{i}+\overline{\mathscr{y}}_{i}$, where $\mathscr{y}_{i}\in[0,\pi]$ is the cheaper, and $\overline{\mathscr{y}}_{i}\in[0,1-\pi]$ is the more expensive part of $y_{i}$ (through the additional costs of $\rho_{i}$). The resulting formulation is then

[TABLE]

Observe that if $\overline{\mathscr{y}}_{i}>0$, then $\mathscr{y}_{i}=\pi$ for each $i\in[n]$ in some optimal solution, as the whole cheaper part of each item is taken first. Substituting $z_{i}\pi$ into $\mathscr{y}_{i}$ and $\overline{z}_{i}(1-\pi)$ into $\overline{\mathscr{y}}_{i}$, we can write equivalently

[TABLE]

Again, in some optimal solution, if $\overline{z}_{i}>0$, then $z_{i}=1$ (and if $z_{i}<1$, then $\overline{z}_{i}=0$) for each $i\in[n]$. The following lemma characterizes the optimal solution to (15):

Lemma 7**.**

There is an optimal solution to (15) which satisfies the following properties:

at most one variable in $\{z_{1},\dots,z_{n},\overline{z}_{1},\dots,\overline{z}_{n}\}$ is fractional, 2. 2.

$\pi=\frac{q}{r}$* for $r\in[n+1]$, $q\in[n]\cup\{0\}$.*

Proof.

Let $\boldsymbol{x}^{*}\in\Phi$ be optimal in (15). Fix $\boldsymbol{x}=\boldsymbol{x}^{*}$ in (14) and consider the problem with additional slack variables:

[TABLE]

This problem contains $2n+3$ constraints and $4n+3$ variables. Thus, there is an optimal solution with $2n+3$ basis variables and $2n$ non-basis variables. The following cases are possible.

• •

First let us assume $\pi=0$. Then, $\mathscr{y}_{i}=0$ for all $i\in[n]$ and for the resulting problem

[TABLE]

there exists an optimal integer solution $\boldsymbol{\overline{\mathscr{y}}}$ (by first taking the $p-k$ cheapest items from $X_{\boldsymbol{x}^{*}}$, and completing the solution with the $k$ cheapest items from $[n]$). Since $\pi=0$, we get $\boldsymbol{z}=\boldsymbol{1}$ and $\boldsymbol{\overline{z}}=\boldsymbol{\overline{\mathscr{y}}}$ and the claim is shown. The proof for $\pi=1$ is analogous.

• •

Assume that $0<\pi<1$, so both $\pi$ and $\gamma$ are basis variables. If $\mathscr{y}_{i}$ is a non-basis variable, then $\alpha_{i}$ is a basis-variable (and vice versa). The same holds for $\overline{\mathscr{y}}_{i}$ and $\beta_{i}$. The following cases result:

If $\delta$ is a basis variable, then it follows that the $2n$ non-basis variables are found in $\boldsymbol{\mathscr{y}}$, $\boldsymbol{\overline{\mathscr{y}}}$, $\boldsymbol{\alpha}$ and $\boldsymbol{\beta}$. Hence, $\mathscr{y}_{i}\in\{0,\pi\}$ and $\overline{\mathscr{y}}_{i}\in\{0,1-\pi\}$, so $z_{i},\overline{z}_{i}\in\{0,1\}$ for each $i\in[n]$. Let $\mathscr{Z}=\sum_{i\in[n]}z_{i}$ and $\overline{\mathscr{Z}}=\sum_{i\in[n]}\overline{z}_{i}$. We then get $\pi=(p-\overline{\mathscr{Z}})/(\mathscr{Z}-\overline{\mathscr{Z}})$ (see model (15)) and point 2 of the lemma is proven (note that if $\mathscr{Z}=\overline{\mathscr{Z}}$, we can assume $\pi=1$). 2. 2.

Let us assume that $\delta$ is a non-basis variable. Then one of two cases must hold:

1. (a)

There is $j$ such that both $\mathscr{y}_{j}$ and $\alpha_{j}$ are basis variables. Then, all other $\mathscr{y}_{i}$ are either [math] or $\pi$ ($z_{i}\in\{0,1\}$), and all other $\overline{\mathscr{y}}_{i}$ are either [math] or $1-\pi$ ($\overline{z}_{i}\in\{0,1\}$). In terms of formulation (15), we have $z_{j}\in[0,1]$ and only $z_{j}$ can be fractional (i.e. other than 0 or 1). In order to show the second point of the lemma, let us define $\mathscr{Z}=\sum_{i\in[n]\setminus\{j\}}z_{i}$, $\overline{\mathscr{Z}}=\sum_{i\in[n]}\overline{z}_{i}$, $\mathscr{Z}^{\prime}=\sum_{i\in X_{\boldsymbol{x}^{*}}\setminus\{j\}}z_{i}$, and $\overline{\mathscr{Z}}^{\prime}=\sum_{i\in X_{\boldsymbol{x}^{*}}}\overline{z}_{i}$. Since $z_{i}\geq\overline{z}_{i}$ for each $i\in[n]$ and $\overline{z}_{j}=0$, the inequalities $\mathscr{Z}\geq\overline{\mathscr{Z}}$ and $\mathscr{Z}^{\prime}\geq\overline{\mathscr{Z}}^{\prime}$ hold. We consider now the subproblem of (15) that reoptimizes the solution only in $\pi$ and $z_{j}$:

[TABLE]

where $t=\left(\Gamma+\sum_{i\in[n]\setminus\{j\}}\underline{c}_{i}z_{i}-\sum_{i\in[n]}\overline{c}_{i}\overline{z}_{i}\right)$ is a constant. We remove variable $z_{j}$ using the equality $z_{j}=(p-\overline{\mathscr{Z}}-(\mathscr{Z}-\overline{\mathscr{Z}})\pi)/\pi$ from the problem. The constraint $z_{j}\geq 0$ becomes $\pi\leq(p-\overline{\mathscr{Z}})/(\mathscr{Z}-\overline{\mathscr{Z}})$, while the constraint $z_{j}\leq 1$ is $\pi\geq(p-\overline{\mathscr{Z}})/(\mathscr{Z}-\overline{\mathscr{Z}}+1)$. Hence, the reoptimization problem becomes

[TABLE]

We can conclude that the optimal value of $\pi$ is one of $\frac{p-\overline{\mathscr{Z}}}{\mathscr{Z}-\overline{\mathscr{Z}}}$, $\frac{p-\overline{\mathscr{Z}}}{\mathscr{Z}-\overline{\mathscr{Z}}+1}$, or $\frac{\overline{\mathscr{Z}}-\overline{\mathscr{Z}}^{\prime}-k}{(\mathscr{Z}^{\prime}-\overline{\mathscr{Z}}^{\prime})-(\mathscr{Z}-\overline{\mathscr{Z}})}$. Since $\overline{\mathscr{Z}},\overline{\mathscr{Z}}^{\prime},\mathscr{Z},\mathscr{Z}^{\prime}$ are all integers from [math] to $n$, the second point of the lemma is true. 2. (b)

There is $j$ such that both $\overline{\mathscr{y}}_{j}$ and $\beta_{j}$ are basis variables. This case is analogue to the previous case.

∎

Lemma 8**.**

Problem RREC under $\mathcal{U}^{c}$ can be solved by solving the problem

[TABLE]

for polynomially many sets $\mathcal{J}$ and values of $\mathscr{Z}$, $\overline{\mathscr{Z}}$, $\mathscr{Z}^{\prime}$, $\overline{\mathscr{Z}}^{\prime}$ and $\pi$.

Proof.

Using Lemma 7, we first consider the case when $z_{i},\overline{z}_{i}\in\{0,1\}$ for all $i\in[n]$. Then we set $\mathcal{J}=[n]$ and guess the values $\mathscr{Z}=\sum_{i\in\mathcal{J}}z_{i}$ and $\overline{\mathscr{Z}}=\sum_{i\in\mathcal{J}}\overline{z}_{i}$. We set $\pi=(p-\overline{\mathscr{Z}})/(\mathscr{Z}-\overline{\mathscr{Z}})$, and further guess all possible values of $\mathscr{X}=\sum_{i\in\mathcal{J}}x_{i}z_{i}$ and $\overline{\mathscr{X}}=\sum_{i\in\mathcal{J}}x_{i}\overline{z}_{i}$ for which the constraint $\pi\mathscr{X}+(1-\pi)\overline{\mathscr{X}}\geq p-k$ is fulfilled. In total, we have to try polynomially many values. For each resulting problem we linearize $z_{i}x_{i}$ and $\overline{z}_{i}x_{i}$ and we get (16).

Assume now that some $z_{j}\in[0,1]$ can be fractional (notice that in this case we can fix $\overline{z}_{j}=0$). We guess the index $j$, the value of $\pi$, and the value of $x_{j}$. We fix then $\mathcal{J}=[n]\setminus\{j\}$ and continue as in the first part of the proof. Namely we guess $\mathscr{Z}$, $\overline{\mathscr{Z}}$, and $\mathscr{X}$, $\overline{\mathscr{X}}$ for the fixed $\pi$, and construct the problem (16). Notice that the value of $z_{j}$ can be retrieved from $\pi z_{j}=(p-\overline{\mathscr{Z}}-(\mathscr{Z}-\overline{\mathscr{Z}})\pi)$. The case where $\overline{z}_{j}\in[0,1]$ can be fractional is analogue. Again, we have to try polynomially many values. To solve problem (15), we then take the best of all solutions. ∎

Lemma 9**.**

For fixed $\mathcal{J}$, $\mathscr{Z}$, $\overline{\mathscr{Z}}$, $\mathscr{Z}^{\prime}$, $\overline{\mathscr{Z}}^{\prime}$ and $\pi$, the problem (16) can be solved in polynomial time.

Proof.

We will show that the coefficient matrix of the relaxation of (16) is totally unimodular. We will use the following Ghouila-Houri criterion [7]: An $m\times n$ integral matrix is totally unimodular, if and only if for each set of rows $R=\{r_{1},\ldots,r_{K}\}\subseteq[m]$ there exists a coloring (called a valid coloring) $l(r_{i})\in\{-1,1\}$ such that the weighted sum of every column restricted to $R$ is $-1$, [math], or $1$. For simplicity, we assume w.l.o.g. that $\mathcal{J}=[n]$. Note that it is enough to show that the coefficient matrix of (16) without the relaxed constraints $x_{i},z_{i},\overline{z}_{i},z^{\prime}_{i},\overline{z}^{\prime}_{i}\leq 1$ is totally unimodular. The matrix, together with a labeling of its rows, is shown in Table 2.

Given a set of rows $R$ we use the following algorithm to color the rows in $R$:

$l(d_{i})=1$ for each $d_{i}\in R$ 2. 2.

If $a_{5}\in R$, set $l(a_{5})=1$, $l(e_{i})=1$ for each $e_{i}\in R$ and $l(a_{3})=-1$. 3. 3.

If $a_{5}\notin R$, set $l(e_{i})=-1$ for each $e_{i}\in R$ and $l(a_{3})=1$. 4. 4.

If $a_{1}\in R$, set $l(a_{1})=-1$ and $l(b_{i})=1$ for each $b_{i}\in R$.

1. (a)

If $a_{4}\in R$, set $l(a_{4})=1$, $l(c_{i})=1$ for each $c_{i}\in R$ and $l(a_{2})=-1$. 2. (b)

If $a_{4}\notin R$, set $l(c_{i})=-1$ for each $c_{i}\in R$ and $l(a_{2})=1$. 5. 5.

If $a_{1}\notin R$, set $l(b_{i})=-1$ for each $b_{i}\in R$.

1. (a)

If $a_{4}\in R$, set $l(a_{4})=-1$, $l(c_{i})=-1$ for each $c_{i}\in R$ and $l(a_{2})=1$. 2. (b)

If $a_{4}\notin R$, set $l(c_{i})=1$ for each $c_{i}\in R$ and $l(a_{2})=-1$.

If $a_{1}\in R$, then $l(a_{1})=-1$ and the rows $b_{i},d_{i}$, $\in R$ have always color 1; if $a_{1}\notin R$, then the rows $b_{i}\in R$ have color -1 and the rows $d_{i}\in R$ have color 1. So the coloring is valid for all columns corresponding to $x_{i}$. In order to prove that the coloring is valid for the columns corresponding to $z_{1},\dots,z_{n}$, $\overline{z}_{1},\dots,\overline{z}_{n}$ it is enough to observe that the algorithm always assigns different colors to $a_{2}$ and the rows $c_{i}\in R$, and $a_{3}$ and the rows in $e_{i}\in R$. If $a_{4}\in R$, then $a_{4}$ has the same color as all $b_{i}\in R$ or all $c_{i}\in R$; if $a_{4}\notin R$, then $b_{i}\in R$ and $c_{i}\in R$ have different color. In consequence, the coloring is valid for the columns corresponding to $z^{\prime}_{1},\dots,z^{\prime}_{n}$. It is also easy to see that the coloring is valid for the columns corresponding to variables $\overline{z}^{\prime}_{1},\dots,\overline{z}^{\prime}_{n}$ (see steps 1-3).

∎

Theorem 10**.**

The RREC problem under scenario set $\mathcal{U}^{c}$ is solvable in polynomial time.

Proof.

This result is a direct consequence of Lemma 8 and Lemma 9. ∎

2.2 Two-Stage Robust Selection

In this section we investigate the two-stage model, namely the I2ST, A2ST and R2ST problems under scenario set $\mathcal{U}^{c}$. In order to solve A2ST we will use the results obtained for AREC. We will also show that R2ST is polynomially solvable as it can be reduced to solving a polynomial number of linear programming problems.

2.2.1 The incremental and adversarial problems

We are given a first stage solution $\boldsymbol{x}\in\Phi_{1}$ with $|X_{\boldsymbol{x}}|=p_{1}$, where $p_{1}\in[p]\cup\{0\}$. Define $\tilde{p}=p-p_{1}$. The incremental problem, I2ST, can be solved in $O(n)$ time. It is enough to choose $\tilde{p}$ items of the smallest costs out of $\overline{X}_{\boldsymbol{x}}$ under the given scenario $\boldsymbol{c}$. On the other hand, the adversarial problem, A2ST, can be reduced to AREC. We first remove from $[n]$ all the items belonging $X_{\boldsymbol{x}}$, obtaining (after an appropriate renumbering) the item set $[n-p_{1}]$. We then fix $k=\tilde{p}$. As we can exchange all the items, the choice of the first stage solution in AREC is irrelevant. Consider the formula (10) for the constructed instance of AREC. The optimal value of $\gamma$ satisfies $\gamma=\alpha$ and (10) becomes:

[TABLE]

which is in turn the same as (9). Problem (17) can be solved in $O(n^{2})$ time, by the method described in Section 2.1.2. This means that A2ST is solvable in $O(n^{2})$ time.

2.2.2 The two-stage robust problem

Given $\boldsymbol{x}\in\Phi_{1}$ and $\boldsymbol{c}\in\mathcal{U}$, the incremental problem, I2ST, can be formulated as the following linear program:

[TABLE]

We can now relax the integrality constraints in (18) and dualize this problem, to find a compact formulation for the adversarial problem, A2ST, under scenario set $\mathcal{U}^{c}$:

[TABLE]

Dualizing (19), we get the following problem:

[TABLE]

which can be used to construct the following mixed-integer program for R2ST:

[TABLE]

We now show that (20) can be solved in polynomial time. We first apply to (20) similar transformation as for the RREC problem (see Section 2.1.3), which results in the following equivalent formulation:

[TABLE]

Again, by setting $z_{i}\pi=\mathscr{y}_{i}$ and $\overline{z}_{i}(1-\pi)=\overline{\mathscr{y}}_{i}$, we rescale the variables and find the following equivalent, nonlinear problem:

[TABLE]

Note that we can write $x_{i}+z_{i}\leq 1$ instead of $x_{i}+\pi z_{i}\leq 1$ and $x_{i}+\overline{z}_{i}\leq 1$ instead of $x_{i}+(1-\pi)\overline{z}_{i}\leq 1$ for each $i\in[n]$.

Lemma 11**.**

There exists an optimal solution to (22) in which

$z_{i},\overline{z}_{i}\in\{0,1\}$* for all $i\in[n]$,* 2. 2.

$\pi=\frac{q}{r}$, where $q\in[p]\cup\{0\}$, $r\in[n]$.

Proof.

We first prove point 1. Let $\boldsymbol{x}^{*}\in\Phi_{1}$ be optimal in (22). Using formulation (21), we consider the following linear program for fixed $\boldsymbol{x}=\boldsymbol{x}^{*}$ and additional slack variables:

[TABLE]

This problem has $4|\overline{X}_{\boldsymbol{x}*}|+2$ variables and $2|\overline{X}_{\boldsymbol{x}^{*}}|+2$ constraints. Thus, there is an optimal solution with $2|\overline{X}_{\boldsymbol{x}^{*}}|+2$ basis variables and $2|\overline{X}_{\boldsymbol{x}^{*}}|$ non-basis variables. If $\pi=0$, then $\mathscr{y}_{i}=0$ for all $i\in\overline{X}_{\boldsymbol{x}^{*}}$ and the problem becomes a selection problem in $\boldsymbol{\overline{\mathscr{y}}}$, for which there is an optimal integer solution. If $\pi=1$, then $\overline{\mathscr{y}}_{i}=0$ for each $i\in[n]$ and the problem becomes a selection problem in $\boldsymbol{\mathscr{y}}$. Hence, in both these cases, there exists an optimal solution to (22) that is integer in $\boldsymbol{z}$ and $\boldsymbol{\overline{z}}$. So let us assume that $\pi>0$ and $\pi<1$, i.e., both $\pi$ and $\gamma$ are basis variables. Note that whenever $\mathscr{y}_{i}$ (resp. $\overline{\mathscr{y}}_{i}$) is a non-basis variable, then $\alpha_{i}$ (resp. $\beta_{i}$) is a basis variable, and vice versa. Hence, all variables $\mathscr{y}_{i}$ are either [math] or $\pi$, and all variables $\overline{\mathscr{y}}_{i}$ are either [math] or $1-\pi$. This corresponds to a solution where all $z_{i}$ and $\overline{z}_{i}$ are either [math] or $1$ in formulation (22).

We now prove point 2. Let $x^{*}_{i},z^{*}_{i},\overline{z}^{*}_{i}\in\{0,1\}$, $i\in[n]$, be optimal in (22). If $\sum_{i\in[n]}z^{*}_{i}=\sum_{i\in[n]}\overline{z}^{*}_{i}$, then there exists an optimal solution with $\pi^{*}\in\{0,1\}$. So let us assume $\sum_{i\in[n]}z^{*}_{i}>\sum_{i\in[n]}\overline{z}^{*}_{i}$ (recall that $z^{*}_{i}\geq\overline{z}^{*}_{i}$ for each $i\in[n]$). By rearranging terms, we obtain

[TABLE]

Write $\mathscr{X}=\sum_{i\in[n]}x^{*}_{i}$, $\mathscr{Z}=\sum_{i\in[n]}z^{*}_{i}$ and $\overline{\mathscr{Z}}=\sum_{i\in[n]}\overline{z}^{*}_{i}$. We have $\mathscr{X}\in\{0,\ldots,p\}$, $\mathscr{Z},\overline{\mathscr{Z}}\in\{0,\ldots,n\}$. Note that if for an item $i$ we have $\overline{z}^{*}_{i}=1$, then also $z^{*}_{i}=1$. Consequently, $\pi$ is of the form described in point 2 of the lemma. ∎

Lemma 12**.**

The R2ST problem under $\mathcal{U}^{c}$ can be solved by solving problem

[TABLE]

for polynomially many values of $\mathscr{X},\mathscr{Z},\overline{\mathscr{Z}}$ and $\pi$.

Proof.

Using Lemma 11, we will try all possible values of $\pi$ and for each fixed $\pi$ we will find an optimal solution to (22) where all $\boldsymbol{x}$, $\boldsymbol{z}$ and $\boldsymbol{\overline{z}}$ are integer. Let the value $\pi=\pi^{*}$ be fixed. The resulting problem is then

[TABLE]

As $\pi^{*}$ is fixed, we can enumerate all possible values of $\mathscr{X}=\sum_{i\in[n]}x_{i}$, $\mathscr{Z}=\sum_{i\in[n]}z_{i}$ and $\overline{\mathscr{Z}}=\sum_{i\in[n]}\overline{z}_{i}$ that generate this value $\pi^{*}$, i.e., we enumerate all possible solutions to $\mathscr{X}+\pi^{*}\mathscr{Z}+(1-\pi^{*})\overline{\mathscr{Z}}=p$. There can be at most $p$ choices of $\mathscr{X}$ and at most $n$ choices of $\mathscr{Z}$ and $\overline{\mathscr{Z}}$. This leads to the problem (23). By choosing the best of the computed solutions, we then find an optimal solution to R2ST. ∎

Lemma 13**.**

For fixed $\mathscr{X},\mathscr{Z},\overline{\mathscr{Z}}$ and $\pi$, the problem (23) can be solved in polynomial time.

Proof.

We prove that the coefficient matrix of the relaxation of (23) is totally unimodular. We will use the Ghouila-Houri criterion [7] (see the proof of Lemma 9). The coefficient matrix of the constraints of (23) is shown in Table 3 (we can skip the relaxed constraints $x_{i},z_{i},\overline{z}_{i}\leq 1$).

Let us choose a subset of the rows $R=A\cup B\cup C$ with $A\subseteq\{a_{1},a_{2},a_{3}\}$, $B\subseteq\{b_{1},\ldots,b_{n}\}$ and $C\subseteq\{c_{1},\ldots,c_{n}\}$. We now determine the coloring for $R$ in the following way:

• •

If $A=\emptyset$, then $l(b_{i})=-1$, $l(c_{i})=1$.

• •

If $A=\{a_{1}\}$, then $l(a_{1})=-1$, $l(b_{i})=l(c_{i})=1$.

• •

If $A=\{a_{2}\}$, then $l(a_{2})=-1$, $l(b_{i})=1$, $l(c_{i})=-1$.

• •

If $A=\{a_{1},a_{2}\}$, then $l(a_{1})=l(a_{2})=-1$, $l(b_{i})=l(c_{i})=1$.

• •

If $A=\{a_{3}\}$, then $l(a_{3})=-1$, $l(b_{i})=-1$, $l(c_{i})=1$.

• •

If $A=\{a_{1},a_{3}\}$, then $l(a_{1})=l(a_{3})=-1$, $l(b_{i})=l(c_{i})=1$.

• •

If $A=\{a_{2},a_{3}\}$, then $l(a_{2})=-1$, $l(a_{3})=1$, $l(b_{i})=1$, $l(c_{i})=-1$.

• •

If $A=\{a_{1},a_{2},a_{3}\}$, then $l(a_{i})=-1$, $l(b_{i})=l(c_{i})=1$.

It is easy to verify that the coloring is valid for each of these cases. ∎

Theorem 14**.**

The R2ST problem under scenario set $\mathcal{U}^{c}$ is solvable in polynomial time.

Proof.

A direct consequence of Lemma 12 and Lemma 13. ∎

3 Discrete Budgeted Uncertainty

In this section we consider the RREC and R2ST problems under scenario set $\mathcal{U}^{d}$. We will use some results obtained for the continuous budgeted uncertainty (in particular Proposition 1). Notice also that the incremental problems IREC and I2ST are the same as for the continuous case.

3.1 Recoverable Robust Selection

3.1.1 The adversarial problem

Let us fix solution $\boldsymbol{x}\in\Phi$. The adversarial problem, AREC, under scenario set $\mathcal{U}^{d}$, can be represented as the following mathematical programming problem:

[TABLE]

Relaxing the integrality constraints $y_{i}\in\{0,1\}$ in (24) for the inner incremental problem, and taking the dual of it, we obtain the following integer linear program for AREC:

[TABLE]

Let $\boldsymbol{\delta}^{*}\in\{0,1\}^{n}$ be optimal in (25). The vector $\boldsymbol{\delta}^{*}$ describes the worst scenario $\hat{\boldsymbol{c}}=(\hat{c}_{i})_{i\in[n]}=(\underline{c}_{i}+d_{i}\delta^{*}_{i})_{i\in[n]}\in\mathcal{U}^{d}$. When we fix this scenario in (25), then we get the problem (3), discussed in Section 2.1, to which Proposition 1 can be applied. Since $\hat{c}_{i}$ is either $\underline{c}_{i}$ or $\overline{c}_{i}$ for each $i\in[n]$, only a finite number of values of $\alpha$ and $\beta$ need to be considered as optimal to (25) (see Proposition 1). In the following, we will show how to find these values efficiently. We can fix $\gamma_{i}=[\alpha+x_{i}\beta-d_{i}\delta_{i}-\underline{c}_{i}]_{+}$ for each $i\in[n]$ in (25). Hence, $\gamma_{i}=[\alpha+x_{i}\beta-d_{i}-\underline{c}_{i}]_{+}$ if $\delta_{i}=1$, and $\gamma_{i}=[\alpha+x_{i}\beta-\underline{c}_{i}]_{+}$ if $\delta_{i}=0$. In consequence, (25) can be rewritten as follows:

[TABLE]

It is easily seen that for fixed $\alpha$, $\beta$, and $\boldsymbol{x}$, (26) is the Selection problem, which can be solved in $O(n)$ time. We now find the sets of relevant values of $\alpha$ and $\beta$. Let us order the elements in $[n]$ according to their costs $\hat{c}_{i}$ and the cost bounds $\underline{c}_{i}$ and $\overline{c}_{i}$ for $i\in[n]$ in the following way:

[TABLE]

Similarly, let us order the elements in $X_{\boldsymbol{x}}$ so that

[TABLE]

and in $\overline{X}_{\boldsymbol{x}}$, namely

[TABLE]

According to Proposition 1, $\alpha=\hat{c}_{\sigma(p)}$ and $\beta=0$, or $\alpha=\hat{c}_{\varsigma(k)}$, and $\beta=\hat{c}_{\nu(p-k)}-\hat{c}_{\varsigma(k)}$ are optimal in (26). Thus we have

[TABLE]

For simplicity of notation, we write $\underline{c}_{\underline{\sigma}(n)}$ instead of $\underline{c}_{\underline{\sigma}(p+\Gamma)}$, when $p+\Gamma>n$. The same holds for $\underline{c}_{\underline{\nu}(p-k+\Gamma)}$ and $\underline{c}_{\underline{\varsigma}(k+\Gamma)}$. Observe that we can assume that $\Gamma\leq n/2$. Indeed, if $\Gamma>n/2$, then it suffices to substitute variables $z_{i}$ by $1-w_{i}$, $w_{i}\in\{0,1\}$, $i\in[n]$. Now the constraint $\sum_{i\in[n]}\delta_{i}\leq\Gamma$ and the costs $\hat{c}_{i}$ become $\sum_{i\in[n]}w_{i}\geq n-\Gamma$ and $\hat{c}_{i}=\underline{c}_{i}+d_{i}(1-w_{i})$, respectively. From Proposition 1 and the above, it follows that $(\alpha,\beta)\in\mathcal{S}_{\boldsymbol{x}}$, where $\mathcal{S}_{\boldsymbol{x}}$ is defined as follows:

[TABLE]

Finally, (26) becomes

[TABLE]

Accordingly, it now suffices to solve (27) for each $(\alpha,\beta)\in\mathcal{S}_{\boldsymbol{x}}$ and choose the best of the computed solutions which encodes a worst scenario. Solving (27) for fixed $(\alpha,\beta)$ can be done in $O(n)$. Since the cardinality of the sets $\mathcal{C}_{\sigma(p)}$, $\mathcal{C}_{\nu(p-k)}$ and $\mathcal{C}_{\varsigma(k)}$ is $O(n)$, the cardinality of the set $\mathcal{S}_{\boldsymbol{x}}$ is $O(n^{2})$. This leads to the following theorem:

Theorem 15**.**

The problem AREC under scenario set $\mathcal{U}^{d}$ can be solved in $O(n^{3})$ time.

3.1.2 The recoverable robust problem

We first identify some special cases of RREC which are polynomially solvable.

Observation 16**.**

The following special cases of RREC under $\mathcal{U}^{d}$ are polynomially solvable:

1. (i)

if $k=0$, then RREC is solvable in $O(n^{2})$ time, 2. (ii)

if $\Gamma=n$, then RREC is solvable in $O((p-k+1)n^{2})$ time, 3. (iii)

if $k\geq\Gamma$ and $C_{i}=0$, $i\in[n]$, then RREC is solvable in $O(n)$ time.

Proof.

1. (i)

If $k=0$, then RREC is equivalent to the MinMax problem under scenario set $\mathcal{U}^{d}$ with the cost intervals $[C_{i}+\underline{c}_{i},C_{i}+\overline{c}_{i}]$, $i\in[n]$. This problem is solvable in $O(n^{2})$ according to the results obtained in [3]. 2. (ii)

If $\Gamma=n$, then RREC can be reduced to the recoverable robust problem under scenario set $\mathcal{U}^{I}$, which can be solved in $\mathcal{O}((p-k+1)n^{2})$ time [20]. 3. (iii)

Consider the case $k\geq\Gamma$. Let $\boldsymbol{x}^{*}\in\Phi$ be an optimal solution to the Selection problem for the costs $\underline{c}_{i}$, $i\in[n]$, and $\boldsymbol{\delta}\in\{0,1\}^{n}$ stands for any vector that encodes scenario $\underline{c}_{i}+\delta_{i}d_{i}$, $i\in[n]$, $\sum_{i\in[n]}\delta_{i}\leq\Gamma$. Let $\boldsymbol{y}$ be an optimal solution to the Selection problem with respect to $\underline{c}_{i}+\delta_{i}d_{i}$, $i\in[n]$. Now $\boldsymbol{x}^{*}$ and $\boldsymbol{y}$ have at least $p-\Gamma$ elements in common, which is due to the fact that $\sum_{i\in[n]}\delta_{i}\leq\Gamma$. Since $k\geq\Gamma$, $\boldsymbol{x}^{*}$ can be recovered to $\boldsymbol{y}$. Hence, no better solution can exist.

∎

We will now construct a compact MIP formulation for the general RREC problem under scenario set $\mathcal{U}^{d}$. In order to do this we will use the formulation (27). Observe that the sets $\mathcal{C}_{\nu(p-k)}$ and $\mathcal{C}_{\varsigma(k)}$, defined in Section 3.1.1, depend on a fixed solution $\boldsymbol{x}$ (the set $\mathcal{C}_{\sigma(p)}$ does not depend on $\boldsymbol{x}$). We now define $\boldsymbol{x}$-independent sets of possible values of the $(p-k)$th smallest element in $X_{\boldsymbol{x}}$ and the $k$th smallest element in $\overline{X}_{\boldsymbol{x}}$ under any scenario in $\mathcal{U}^{d}$ by

[TABLE]

Again, from Proposition 1 and the above, we get a new set of relevant values of $\alpha$ and $\beta$

[TABLE]

Obviously $\mathcal{C}_{\nu(p-k)}\subseteq\mathcal{C}_{\sigma(p-k)}$, $\mathcal{C}_{\varsigma(k)}\subseteq\mathcal{C}_{\sigma(k)}$, $\mathcal{S}_{\boldsymbol{x}}\subseteq\mathcal{S}$ for any $\boldsymbol{x}\in\Phi$ and the cardinality of $\mathcal{S}$ remains $\mathcal{O}(n^{2})$. Let us represent the adversarial problem (27) as follows:

[TABLE]

Dualizing the inner Selection problem in (28), we get:

[TABLE]

For every pair $(\alpha^{\ell},\beta^{\ell})\in\mathcal{S}$, we introduce a set of variables $\pi^{\ell}$, $\boldsymbol{\rho}^{\ell}$, $\ell\in[S]$ with $S=|\mathcal{S}|$. The RREC problem under scenario set $\mathcal{U}^{d}$ is then equivalent to the following mathematical programming problem:

[TABLE]

Finally, we can linearize all the nonlinear terms $[a+bx_{i}]_{+}=[a]_{+}+([a+b]_{+}-[a]_{+})x_{i}$, where $a,b$ are constant. In consequence, we obtain a compact MIP formulation for RREC, with $O(n^{3})$ variables and $O(n^{3})$ constraints.

We now present an approximation algorithm, which can be applied for larger problem instances. Suppose that $\underline{c}_{i}\geq\alpha\overline{c}_{i}$ for each item $i\in[n]$, where $\alpha\in(0,1]$ is a constant. This inequality means that for each item $i$ the nominal cost $\underline{c}_{i}$ is positive and $\overline{c}_{i}$ is at most $1/\alpha$ greater than $\underline{c}_{i}$. It is reasonable to assume that this condition will be true in many practical applications for not very large value of $1/\alpha$. Consider scenario set $\mathcal{U}^{\prime}=\{(\underline{c}_{i})_{i\in[n]}\}$, so $\mathcal{U}^{\prime}$ contains only one scenario composed of the nominal item costs. Let $\hat{\boldsymbol{x}}$ be an optimal solution to the RREC problem under $\mathcal{U}^{\prime}$. This solution can be computed in polynomial time [20]. Using the same reasoning as in [16], one can show that the cost of $\hat{\boldsymbol{x}}$ is at most $1/\alpha$ greater than the optimum. Hence there is an $1/\alpha$ approximation algorithm for RREC under scenario set $\mathcal{U}^{d}$.

3.2 Two-Stage Robust Selection

3.2.1 The adversarial problem

Let $\boldsymbol{x}\in\Phi_{1}$ be a fixed first stage solution, with $|X_{\boldsymbol{x}}|=p_{1}$. Using the same reasoning as in Section 2.2.1, we can represent the A2ST problem as a special case of AREC with $\tilde{p}=p-p_{1}$ and $k=\tilde{p}$. In this case $\beta=0$ in the formulation (25) and there are only $O(n)$ candidate values for $\alpha$. Hence the problem can be solved in $O(n^{2})$ time under scenario set $\mathcal{U}^{d}$.

We now show some additional properties of the adversarial problem, which will then be used to solve the more general R2ST problem. By dualizing the linear programming relaxation of the incremental problem (18), we find the following MIP formulation for the adversarial problem A2ST:

[TABLE]

The following lemma characterizes an optimal solution to (29):

Lemma 17**.**

There is an optimal solution to (29) in which $\alpha=0$, $\alpha=\underline{c}_{j}$ or $\alpha=\overline{c}_{j}$ for some $j\in[n]$.

Proof.

Let us fix $\boldsymbol{\delta}$ in (29). Then the resulting linear program with additional slack variables is

[TABLE]

Note that we only consider nonnegative values of dual variable $\alpha$ associated with the cardinality constraint in (18), since replacing this constraint in (18) by $\sum_{i\in[n]}(y_{i}+x_{i})\geq p$ does not change the set of optimal solutions. The problem has $2n+1$ variables and $n$ constraints. If $\alpha$ is a non-basis variable in an optimal solution, then $\alpha=0$. So, let us assume that $\alpha$ is a basis variable. As there are $n-1$ remaining basis variables, there is at least one $j\in[n]$ where both $\epsilon_{j}$ and $\gamma_{j}$ are non-basis variables. Hence, $\alpha=\underline{c}_{j}+d_{j}\delta_{j}$ and the lemma follows since $\delta_{j}\in\{0,1\}$. ∎

Define $\mathcal{S}=\{0\}\cup\{\underline{c}_{i}:i\in[n]\}\cup\{\overline{c}_{i}:i\in[n]\}$ and write $\mathcal{S}=\{\alpha^{1},\ldots,\alpha^{S}\}$ with $S=|\mathcal{S}|\in O(n)$. Using Lemma 17, problem (29) is then equivalent to

[TABLE]

Let $(\boldsymbol{\delta}^{*},\boldsymbol{\gamma}^{*})$ be an optimal solution to the inner maximization problem. Note that we can assume that if $\delta^{*}_{i}=0$, then $\gamma^{*}_{i}=[\alpha-\underline{c}_{i}]^{+}$ and if $\delta^{*}_{i}=1$, then $\gamma^{*}_{i}=[\alpha-\underline{c}_{i}-d_{i}]^{+}$. Hence, the inner problem is equivalent to

[TABLE]

As this is the Selection problem, we state the following result:

Theorem 18**.**

The problem A2ST under scenario set $\mathcal{U}^{d}$ can be solved in $O(n^{2})$ time.

3.2.2 The two-stage robust problem

The R2ST problem is polynomially solvable when $\Gamma\geq n$. Scenario set $\mathcal{U}^{d}$ can be then replaced by $\mathcal{U}^{I}$ and the problem is solvable in $O(n)$ time [20].

We now present a compact MIP formulation for R2ST under scenario set $\mathcal{U}^{d}$. We can use the dual of the linear relaxation of $\eqref{advdisc2}$ and the set $\mathcal{S}$ of candidate values for $\alpha$ to arrive at the following formulation:

[TABLE]

This formulation has $O(n^{2})$ variables and $O(n^{2})$ constraints.

We now propose a fast approximation algorithm for the problem. The idea is the same as for the robust recoverable problem (see Section 3.1.2 and also [16]). Let us fix scenario $\underline{\boldsymbol{c}}=(\underline{c}_{i})_{i\in[n]}\in\mathcal{U}^{d}$, which is composed of the nominal item costs. Consider the following problem:

[TABLE]

Problem (31) can be solved in $O(n)$ time in the following way. Let $c_{i}^{\prime}=\min\{C_{i},\underline{c}_{i}\}$ for each $i\in[n]$ and let $\boldsymbol{x}\in\Phi$ be an optimal solution to the Selection problem for the costs $c_{i}^{\prime}$. The optimal solution $\hat{\boldsymbol{x}}\in\Phi_{1}$ to (31) is then formed by fixing $\hat{x}_{i}=1$ if $x_{i}=1$ and $c_{i}^{\prime}=C_{i}$, and $\hat{x}_{i}=0$ otherwise. We now prove the following result:

Proposition 19**.**

Let $\underline{c}_{i}\geq\alpha\overline{c}_{i}$ for each $i\in[n]$ and $\alpha\in(0,1]$. Let $\hat{\boldsymbol{x}}\in\Phi_{1}$ be an optimal solution to (31). Then $\hat{\boldsymbol{x}}$ is an $\frac{1}{\alpha}$- approximate solution to R2ST.

Proof.

Let $\boldsymbol{x}^{*}$ be an optimal solution to R2ST with the objective value $OPT$. It holds

[TABLE]

because $\underline{c}_{i}\leq c^{*}_{i}$ for each $i\in[n]$. Since $\hat{\boldsymbol{x}}$ is an optimal solution to (31) we get

[TABLE]

where $\hat{\boldsymbol{y}}=\min_{\boldsymbol{y}\in\Phi_{\hat{\boldsymbol{x}}}}\underline{\boldsymbol{c}}\boldsymbol{y}$. By the assumption that $\underline{c}_{i}\geq\alpha\overline{c}_{i}$ for each $i\in[n]$, we obtain

[TABLE]

Finally, as $\alpha\in(0,1]$

[TABLE]

and the proposition follows. ∎

4 Conclusion

The Selection problem is one of the main objects of study for the complexity of robust optimization problems. While the robust counterpart of most combinatorial optimization problems is NP-hard, its simple structure allows in many cases to construct efficient polynomial algorithms. As an example, the MinMax-Regret Selection problem was the first MinMax-Regret problem for which polynomial time solvability could be proved [2].

In this paper we continue this line of research by considering recoverable and two-stage robust problems combined with discrete and budgeted uncertainty sets. All four problem combinations have not been analyzed before, and little is known about other problems of this kind.

We showed that the continuous uncertainty problem variants allow polynomial-time solution algorithms, based on solving a set of linear programs. Additionally, we derived strongly polynomial combinatorial algorithms for the adversarial subproblems and discussed ways to preprocess instances.

For the discrete uncertainty case, we also presented strongly polynomial combinatorial algorithms for the adversarial problems, and constructed mixed-integer programming formulations of polynomial size. It remains an open problem to analyze in future research if the problems with discrete uncertainty are NP-hard or also allow for polynomial-time solution algorithms.

Further research includes the application of our setting to other combinatorial optimization problems, such as Spanning Tree or Shortest Path.

Appendix A An Improved Algorithm for AREC

We now discuss how the $O(n^{2})$ result from Section 2.1.2 can be further improved to $O(n\log n)$. Using the same arguments as before, this result then also applies to A2ST.

Let a solution $\boldsymbol{x}\in\Phi$ and two numbers $L_{\boldsymbol{x}}$,$L_{\overline{\boldsymbol{x}}}$ be given. We call a solution $\boldsymbol{c}$ of problem AREC $(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})-$compatible, if

[TABLE]

[TABLE]

Note that for each tuple $(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})$ there is a unique solution (scenario in $\mathcal{U}^{c}$) which is $(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})-$compatible. We denote this solution by $\boldsymbol{c}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})$.

Lemma 20**.**

Each solution $\boldsymbol{c}$ of AREC can be transformed to a solution $\boldsymbol{c}^{\prime}$ which is $(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})-$compatible for some $L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}}$ and has no lower objective value.

Proof.

The proof contains similar arguments as the proof of Proposition 2. Shifting the budget from items which are above or at the corresponding levels to items which are not yet at its cost upper bound but below the corresponding levels does not decrease the objective function. ∎

Denote by $\boldsymbol{c}_{\boldsymbol{x}}$ ($\boldsymbol{c}_{\overline{\boldsymbol{x}}}$) the projection of $\boldsymbol{c}$ onto $X_{\boldsymbol{x}}$ ($\overline{X}_{\boldsymbol{x}}$). The objective value, $F(\boldsymbol{c})$, of solution $\boldsymbol{c}$ for problem AREC can be expressed as follows:

[TABLE]

where $\langle\boldsymbol{v}\rangle_{(l)}$ denotes the sum of the $l$ smallest entries of vector $\boldsymbol{v}$ ($\langle\boldsymbol{v}\rangle_{0}:=0$).

Lemma 21**.**

$F$* is a concave function on $\mathbb{R}^{n}_{+}$.*

Proof.

Note that $h(\boldsymbol{v})=\langle\boldsymbol{v}\rangle_{(l)}$ is a concave function on $\mathbb{R}^{n}_{+}$. Hence, since the sum of concave functions is concave, $f_{j}(\boldsymbol{c}):=\langle\boldsymbol{c}_{\boldsymbol{x}}\rangle_{(j)}+\langle\boldsymbol{c}_{\overline{\boldsymbol{x}}}\rangle_{(p-j)}$ is a concave function. Since the minimum of concave functions is again a concave function, it follows that $F:=\min_{j\in\{p-k,p-k+1,\dots,p\}}f_{j}$ is a concave function. ∎

For two levels ($L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}}$) we define the set of active items:

[TABLE]

[TABLE]

Recall that in the incremental problem, exactly $p$ items need to be selected. After the $p-k$ cheapest items are chosen from $X_{\boldsymbol{x}}$, the $k$ cheapest items are chosen from the remaining items. If two items have the same cost, we differentiate between two tie-breaking rules: Prefer items from $X_{\boldsymbol{x}}$ (rule 1) and prefer items from $\overline{X}_{\boldsymbol{x}}$ (rule 2). We solve the incremental problem with cost $\boldsymbol{c}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})$ with tie-breaking rule 2. The set of selected items $I_{\boldsymbol{x}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})$ is defined as the set of all items from $X_{\boldsymbol{x}}$ which are part of the incremental solution. Analogously, we define the set $I_{\overline{\boldsymbol{x}}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})$ as the set of all items from $\overline{X}_{\boldsymbol{x}}$ which are part of the incremental solution if tie-breaking rule 1 is used. Further, we define the set $I(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})\subseteq[n]$ as the complete solution of the incremental problem. For the definition of this set, the choice of the tie-breaking rule is not important (without loss of generality we may assume that rule 1 is used).

Assume we are given a small budget $\epsilon$ which we can invest to transform $\boldsymbol{c}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})$ to a new solution $\boldsymbol{c}(L_{\boldsymbol{x}}^{\prime},L_{\overline{\boldsymbol{x}}}^{\prime})$ with $L_{\boldsymbol{x}}\leq L_{\boldsymbol{x}}^{\prime}$ and $L_{\overline{\boldsymbol{x}}}\leq L_{\overline{\boldsymbol{x}}}^{\prime}$. Further assume that the sets of active items and the sets of selected items are the same for $(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})$ and $(L_{\boldsymbol{x}}^{\prime},L_{\overline{\boldsymbol{x}}}^{\prime})$, and that we decide to keep $L_{\overline{\boldsymbol{x}}}^{\prime}=L_{\overline{\boldsymbol{x}}}$ (the case that $L_{\boldsymbol{x}}=L_{\boldsymbol{x}}^{\prime}$ is analogous). To increase $L_{\boldsymbol{x}}$ we have to equally distribute the budget on all items from $A_{\boldsymbol{x}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})$. Hence, we get that

[TABLE]

The objective value of the solution increases only for the selected items which were increased. Hence, we have that

[TABLE]

We see that the gain in the objective function is determined by the ratio of selected and active items. We denote these ratios in the following as

[TABLE]

Ratio $R_{\boldsymbol{x}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})$ ($R_{\overline{\boldsymbol{x}}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})$) defines how efficient it is to increase $L_{\boldsymbol{x}}$ ($L_{\overline{\boldsymbol{x}}}$) and the ratio $R(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})$ computes the efficiency of increasing $L_{\boldsymbol{x}}$ and $L_{\overline{\boldsymbol{x}}}$ simultaneously.

The algorithm performs a sequence of greedy update steps which are locally optimal. We prove later that the so found solution is indeed optimal. The algorithm starts with solution $\boldsymbol{c}^{0}:=\underline{\boldsymbol{c}}$. The initial levels $L_{\boldsymbol{x}}^{0}$ and $L_{\overline{\boldsymbol{x}}}^{0}$ are equal to the smallest value of $\boldsymbol{c}_{\boldsymbol{x}}^{0}$ and $\boldsymbol{c}_{\overline{\boldsymbol{x}}}^{0}$. The algorithm computes $R_{\boldsymbol{x}}:=R_{\boldsymbol{x}}(L_{\boldsymbol{x}}^{0},L_{\overline{\boldsymbol{x}}}^{0})$, $R_{\overline{\boldsymbol{x}}}:=R_{\overline{\boldsymbol{x}}}(L_{\boldsymbol{x}}^{0},L_{\overline{\boldsymbol{x}}}^{0})$, and $R:=R(L_{\boldsymbol{x}}^{0},L_{\overline{\boldsymbol{x}}}^{0})$.

The idea is to change the values of $\boldsymbol{c}^{0}$ to $\boldsymbol{c}^{1}$ such that either $L_{\boldsymbol{x}}^{0}$ or $L_{\overline{\boldsymbol{x}}}^{0}$ (or both) are increased to $L_{\boldsymbol{x}}^{1}$ or $L_{\overline{\boldsymbol{x}}}^{1}$, depending on which update has the highest efficiency.

First, assume that $L_{\boldsymbol{x}}^{0}\neq L_{\overline{\boldsymbol{x}}}^{0}$. Depending on whether $R_{\boldsymbol{x}}$ or $R_{\overline{\boldsymbol{x}}}$ is higher, we increase either $L_{\boldsymbol{x}}^{0}$ or $L_{\overline{\boldsymbol{x}}}^{0}$, until one of the efficiency ratio changes or if the complete budget $\Gamma$ is spent. A change of the efficiency ratio means that either the set of active or the set of selected items changes. We define the set of interesting levels as $\mathcal{L}:=\{\underline{c}_{1},\dots,\underline{c}_{n},\}\cup\{\overline{c}_{1},\dots,\overline{c}_{n},\}$. Note that the set of active items can only change at levels in $\mathcal{L}$. The set of selected items can only change at interesting levels and if both levels $L_{\boldsymbol{x}}$ and $L_{\overline{\boldsymbol{x}}}$ become equal. Therefore, the amount for which we can increase either $L_{\boldsymbol{x}}^{0}$ or $L_{\overline{\boldsymbol{x}}}^{0}$ before one of the efficiency ratio changes is strictly greater zero.

Next, consider the case that $L_{\boldsymbol{x}}^{0}=L_{\overline{\boldsymbol{x}}}^{0}$. In this case, it can happen that increasing $L_{\boldsymbol{x}}^{0}$ and $L_{\overline{\boldsymbol{x}}}^{0}$ simultaneously is more efficient than increasing only one of these two levels. Hence, if $R\geq\max\{R_{\boldsymbol{x}},R_{\overline{\boldsymbol{x}}}\}$, we change $\boldsymbol{c}^{0}$ to $\boldsymbol{c}^{1}$ such that both $L_{\boldsymbol{x}}^{0}$ and $L_{\overline{\boldsymbol{x}}}^{0}$ are increased. Otherwise we increase only $L_{\boldsymbol{x}}^{0}$ or $L_{\overline{\boldsymbol{x}}}$, as described before. Investing a budget of $\epsilon$ in a simultaneous update of $L_{\boldsymbol{x}}^{0}$ and $L_{\overline{\boldsymbol{x}}}$ increases the objective function by $\epsilon R$.

We present in Figures 5, 6, and 7 all three possible situations. The sample instance consists of $10$ items, the left items are $X_{\boldsymbol{x}}$ and the right items are $\overline{X}_{\boldsymbol{x}}$. Further, $p=5$ and $k=2$. The bars represent the actual costs of each item and the boxes visualize the upper bound $\overline{c}_{i}$ for the cost of each item. The items which are active have a grid pattern, the other items diagonal lines. In Figure 5, the best improvement is done by increasing $L_{\boldsymbol{x}}$. In Figure 6, the best improvement is done by increasing $L_{\overline{\boldsymbol{x}}}$. In Figure 7, the best improvement is done by increasing both $L_{\boldsymbol{x}}$ and $L_{\overline{\boldsymbol{x}}}$.

After the costs and levels are updated to $\boldsymbol{c}^{1}$ and $L_{\boldsymbol{x}}^{1},L_{\overline{\boldsymbol{x}}}^{1}$, the algorithm updates $R_{\boldsymbol{x}}:=R_{\boldsymbol{x}}(L_{\boldsymbol{x}}^{1},L_{\overline{\boldsymbol{x}}}^{1})$, $R_{\overline{\boldsymbol{x}}}:=R_{\overline{\boldsymbol{x}}}(L_{\boldsymbol{x}}^{1},L_{\overline{\boldsymbol{x}}}^{1})$, and $R:=R(L_{\boldsymbol{x}}^{1},L_{\overline{\boldsymbol{x}}}^{1})$ an repeats the process until the complete budget $\Gamma$ is spent or each item is at its maximum cost level.

Since the efficiency ratios can only change at interesting levels or if $L_{\boldsymbol{x}}$ and $L_{\overline{\boldsymbol{x}}}$ become equal, we conclude that the algorithm can perform at most $O(n)$ many steps before either the complete budget $\Gamma$ is spent or the cost of each item $c_{i}$ is raised to its maximum cost $\overline{c}_{i}$. We discuss at the end of this section how to efficiently implement the algorithm to have an amortized complexity of $O(\log(n))$ in each step. This gives the claimed $O(n\log(n))$ time complexity for the complete algorithm.

The following lemma is important for the analysis of the algorithm.

Lemma 22**.**

For $L_{\boldsymbol{x}}\leq L_{\boldsymbol{x}}^{\prime}$ and $L_{\overline{\boldsymbol{x}}}\leq L_{\overline{\boldsymbol{x}}}^{\prime}$, we have that

• (i)

$R_{\boldsymbol{x}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})\geq R_{\boldsymbol{x}}(L_{\boldsymbol{x}}^{\prime},L_{\overline{\boldsymbol{x}}})$**

• (ii)

$R_{\overline{\boldsymbol{x}}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})\leq R_{\overline{\boldsymbol{x}}}(L_{\boldsymbol{x}}^{\prime},L_{\overline{\boldsymbol{x}}})$**

• (iii)

$R_{\overline{\boldsymbol{x}}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})\geq R_{\overline{\boldsymbol{x}}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}}^{\prime})$**

• (iv)

$R_{\boldsymbol{x}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})\leq R_{\boldsymbol{x}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}}^{\prime})$**

Proof.

Point $(ii)$ follows from the facts that $A_{\overline{\boldsymbol{x}}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})=A_{\overline{\boldsymbol{x}}}(L_{\boldsymbol{x}}^{\prime},L_{\overline{\boldsymbol{x}}})$ and $I_{\overline{\boldsymbol{x}}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})\subseteq I_{\overline{\boldsymbol{x}}}(L_{\boldsymbol{x}}^{\prime},L_{\overline{\boldsymbol{x}}})$ (point $(iv)$ is analogous). Next consider point $(i)$ (point $(iii)$ is analogous). We assume that $R_{\boldsymbol{x}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})<1$, since otherwise the inequality is trivial. First, observe that $I_{\boldsymbol{x}}(L_{\boldsymbol{x}}^{\prime},L_{\overline{\boldsymbol{x}}})\subseteq I_{\boldsymbol{x}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})$. Assume we start at $L_{\boldsymbol{x}}$ and increase it stepwise until it reaches $L_{\boldsymbol{x}}^{\prime}$. During each step $R_{\boldsymbol{x}}$ should not change. Three different causes may lead to a change of $R_{\boldsymbol{x}}$. First, one item from $X_{\boldsymbol{x}}$ might become an active item, which was before not an active item. This increases $|A_{\boldsymbol{x}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})|$ by one, whereas $|I_{\boldsymbol{x}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})\cap A_{\boldsymbol{x}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})|$ stays constant (since $R_{\boldsymbol{x}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})<1$). Second, an item $i$ which was active may reach its upper bound $\overline{c}_{i}$. If item $i$ is one of the selected items, we get that both $|A_{\boldsymbol{x}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})|$ and $|I_{\boldsymbol{x}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})\cap A_{\boldsymbol{x}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})|$ are reduced by one. Note that if $i$ is not among the selected items, we must have that $|I_{\boldsymbol{x}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})\cap A_{\boldsymbol{x}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})|=0$, and hence $R_{\boldsymbol{x}}=0$ and can not decrease any further. Third, an item $i$ leaves the set of selected items, in this case $|I_{\boldsymbol{x}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})\cap A_{\boldsymbol{x}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})|$ is reduced by one. Note that all three cases, lead to a reduction of $R_{\boldsymbol{x}}$ (except for the case where $R_{\boldsymbol{x}}=0$). ∎

Lemma 23**.**

Solution $\tilde{\boldsymbol{c}}$ computed by the algorithm is optimal.

Proof.

Denote by $\tilde{L}_{\boldsymbol{x}}$ and $\tilde{L}_{\overline{\boldsymbol{x}}}$ the corresponding levels produced by the algorithm. Let $\boldsymbol{c}^{*}$ be the optimal solution of problem AREC. Assume that $F(\tilde{\boldsymbol{c}})<F(\boldsymbol{c}^{*})$, i.e. $\tilde{\boldsymbol{c}}$ is not an optimal solution. For a small $\epsilon>0$, we consider solution $\hat{\boldsymbol{c}}:=(1-\epsilon)\tilde{\boldsymbol{c}}+\epsilon\boldsymbol{c}^{*}$. Since $F$ is concave (Lemma 21), it must hold that $F(\tilde{\boldsymbol{c}})<F(\hat{\boldsymbol{c}})$. Due to Lemma 20, we can transform solution $\hat{\boldsymbol{c}}$ to a solution $\boldsymbol{c}^{\prime}$ which is $(L_{\boldsymbol{x}}^{\prime},L_{\overline{\boldsymbol{x}}}^{\prime})-$compatible and has no smaller objective value, i.e. $F(\hat{\boldsymbol{c}})\leq F(\boldsymbol{c}^{\prime})$. We consider three different cases:

Case 1: $L_{\boldsymbol{x}}^{\prime}=\tilde{L}_{\boldsymbol{x}}$

We can assume without loss of generality that the complete budget $\Gamma$ is spent to increase $\underline{\boldsymbol{c}}$ to $\tilde{\boldsymbol{c}}$ respectively $\boldsymbol{c}^{\prime}$. Hence, we are able to conclude that $L_{\overline{\boldsymbol{x}}}^{\prime}=\tilde{L}_{\overline{\boldsymbol{x}}}$. In consequence, solution $\boldsymbol{c}^{\prime}$ and $\tilde{\boldsymbol{c}}$ are identical, contradicting that $F(\tilde{\boldsymbol{c}})<F(\boldsymbol{c}^{\prime})$.

Case 2: $L_{\boldsymbol{x}}^{\prime}<\tilde{L}_{\boldsymbol{x}}$

With the same reasoning as in Case 1, we conclude that $L_{\overline{\boldsymbol{x}}}^{\prime}>\tilde{L}_{\overline{\boldsymbol{x}}}$. To arrive at a contradiction we have to take into account the last step of the algorithm, where $L_{\boldsymbol{x}}$ was increased to $\tilde{L}_{\boldsymbol{x}}$ (note that if $L_{\boldsymbol{x}}$ was never increased during the algorithm, it must be at its initial level and, hence, $L_{\boldsymbol{x}}^{\prime}<\tilde{L}_{\boldsymbol{x}}$ cannot be true). We denote by $\boldsymbol{c}^{-}$ the solution of the algorithm before this step. The corresponding levels of $\boldsymbol{c}^{-}$ are denoted by $L_{\boldsymbol{x}}^{-}$ and $L_{\overline{\boldsymbol{x}}}^{-}$. We can assume that $L_{\boldsymbol{x}}^{\prime}$ is arbitrary close to $\tilde{L}_{\boldsymbol{x}}$, since the $\epsilon$ that was used in the definition of $\hat{\boldsymbol{c}}$ can be arbitrary small. So we get that $L_{\boldsymbol{x}}^{-}<L_{\boldsymbol{x}}^{\prime}<\tilde{L}_{\boldsymbol{x}}$ and $L_{\overline{\boldsymbol{x}}}^{-}\leq\tilde{L}_{\overline{\boldsymbol{x}}}<L_{\overline{\boldsymbol{x}}}^{\prime}$. Two different cases were possible for this step of the algorithm.

Case 2a: $L_{\overline{\boldsymbol{x}}}$ stayed constant during this step.

Note that solution $\boldsymbol{c}^{\prime}$ and solution $\tilde{\boldsymbol{c}}$ can both be created from $\boldsymbol{c}(L_{\boldsymbol{x}}^{\prime},\tilde{L}_{\overline{\boldsymbol{x}}})$ by increasing $L_{\boldsymbol{x}}$ or $L_{\overline{\boldsymbol{x}}}$. Denote by $\Gamma_{\boldsymbol{x}}$ the budget which is necessary to increase $L_{\boldsymbol{x}}^{\prime}$ to $\tilde{L}_{\boldsymbol{x}}$ and by $\Gamma_{\overline{\boldsymbol{x}}}$ the budget which is necessary to increase $\tilde{L}_{\boldsymbol{y}}$ to $L_{\boldsymbol{y}}^{\prime}$. The objective values of both solutions are given by

[TABLE]

[TABLE]

Observe that we must have $\Gamma_{\boldsymbol{x}}=\Gamma_{\overline{\boldsymbol{x}}}$ since we can assume without loss of generality that the complete budget is spent to create solution $\tilde{\boldsymbol{c}}$ and solution $\boldsymbol{c}^{\prime}$. The following estimations show that $R_{\boldsymbol{x}}(L_{\boldsymbol{x}}^{\prime},\tilde{L}_{\overline{\boldsymbol{x}}})\geq R_{\overline{\boldsymbol{x}}}(L_{\boldsymbol{x}}^{\prime},\tilde{L}_{\overline{\boldsymbol{x}}})$, which leads to the desired contradiction.

[TABLE]

Case 2b: $L_{\overline{\boldsymbol{x}}}$ was increased during this step.

In this case, we must have that $L_{\boldsymbol{x}}^{-}=L_{\overline{\boldsymbol{x}}}^{-}$. Instead of performing a full update step from $L_{\boldsymbol{x}}^{-}$ to $\tilde{L}_{\boldsymbol{x}}$, we perform only a partial step and increase $L_{\boldsymbol{x}}^{-}$ and $L_{\overline{\boldsymbol{x}}}^{-}$ to $L_{\boldsymbol{x}}^{\prime}$. After this partial step we arrive at solution $\boldsymbol{c}(L_{\boldsymbol{x}}^{\prime},L_{\boldsymbol{x}}^{\prime})$. Note that $\tilde{\boldsymbol{c}}$ and $\boldsymbol{c}^{\prime}$ can both be created from $\boldsymbol{c}(L_{\boldsymbol{x}}^{\prime},L_{\boldsymbol{x}}^{\prime})$ by increasing $L_{\boldsymbol{x}}$ or $L_{\overline{\boldsymbol{x}}}$. Denote by $\Gamma_{\boldsymbol{x}}$ the budget, which is necessary to increase $L_{\boldsymbol{x}}$ from $L_{\boldsymbol{x}}^{\prime}$ to $\tilde{L}_{\boldsymbol{x}}$ and by $\Gamma_{\overline{\boldsymbol{x}}}$ the budget which is necessary to increase $L_{\overline{\boldsymbol{x}}}$ from $L_{\boldsymbol{x}}^{\prime}$ to $\tilde{L}_{\boldsymbol{x}}$. After the last step of the algorithm where $L_{\boldsymbol{x}}$ was increased to $L_{\overline{\boldsymbol{x}}}$, the algorithm performs a sequence of $k$ updating moves increasing $L_{\overline{\boldsymbol{x}}}$, denote by $\Gamma_{\overline{\boldsymbol{x}}}^{j}$ the budget which is spent in the $j$th updating move and by $L_{\overline{\boldsymbol{x}}}^{j}$ the level $L_{\overline{\boldsymbol{x}}}$ before the $j$th updating move. Finally, denote by $\Gamma_{\overline{\boldsymbol{x}}}^{\prime}$ the budget which is necessary to update $L_{\overline{\boldsymbol{x}}}$ from $\tilde{L}_{\overline{\boldsymbol{x}}}$ to $L_{\boldsymbol{y}}^{\prime}$. This allows us to relate the objective values of $\tilde{\boldsymbol{c}}$ and $\boldsymbol{c}^{\prime}$ to the objective value of $\boldsymbol{c}(L_{\boldsymbol{x}}^{\prime},L_{\boldsymbol{x}}^{\prime})$.

[TABLE]

Consider the following estimations for the different efficiency ratios.

[TABLE]

Using these estimations it is straightforward to show that $F(\tilde{L}_{\boldsymbol{x}},\tilde{L}_{\overline{\boldsymbol{x}}})\geq F(L_{\boldsymbol{x}}^{\prime},L_{\overline{\boldsymbol{x}}}^{\prime})$, which leads to the desired contradiction.

Case 3: $L_{\boldsymbol{x}}^{\prime}>\tilde{L}_{\boldsymbol{x}}$

This case is completely analogous to Case 2, just exchange $\boldsymbol{x}$ and $\overline{\boldsymbol{x}}$.

All cases lead to contradictions, which proves that the assumption $F(\tilde{\boldsymbol{c}})<F(\boldsymbol{c}^{*})$ is wrong. Hence, the solution provided by the algorithm is indeed optimal. ∎

Lemma 23 completes the analysis of the proposed $O(n\log(n))$ algorithm. We summarize the findings of this section in the following theorem.

Theorem 24**.**

Problem AREC can be solved in $O(n\log(n))$.

Problem A2ST can be seen as a simpler version of problem AREC. All items which were already selected in the first stage are removed from the instance. The remaining problem is a special instance of problem AREC (where $\overline{X}_{\boldsymbol{x}}$ is empty).

Corollary 25**.**

Problem A2ST can be solved in $O(n\log(n))$.

We only sketch the idea how to efficiently implement the algorithm, since the exact details are cumbersome. For an efficient implementation of the algorithm we renounce an explicit representation of $A_{\boldsymbol{x}},A_{\overline{\boldsymbol{x}}},I_{\boldsymbol{x}},$ and $I_{\overline{\boldsymbol{x}}}$. Instead we store these sets only implicitly and compute the values which are important to compute the different efficiency ratios. To represent the set $A_{\boldsymbol{x}}$ ($A_{\overline{\boldsymbol{x}}}$ is analogous) we store two pointers representing the leftmost and rightmost element of the set (sorted with respect to the actual cost). If the level $L_{\boldsymbol{x}}$ is raised such that one other item becomes active we increase the right pointer by one. If the cost of an active item is raised to its upper bound the item becomes inactive and we increase the left pointer by one. Since we have only an implicit representation of $A_{\boldsymbol{x}}$, it is not trivial to efficiently check whether the cost of an item reaches its upper bound. To handle this problem we create a min-heap which contains the upper bound cost of all active items. Note that each item becomes at most once active and at most once inactive. Inserting and removing an item into the heap requires $O(\log(n))$ (see, e.g., [10]). Hence, building and maintaining the heap requires overall $O(n\log(n))$ time.

The set of selected items $I_{\boldsymbol{x}}$ and $I_{\overline{\boldsymbol{x}}}$ is represented by two pointers, which indicate at which positions the levels $L_{\boldsymbol{x}}$ and $L_{\overline{\boldsymbol{x}}}$ lie with respect to the actual cost of the items in $X_{\overline{\boldsymbol{x}}}$ and $X_{\boldsymbol{x}}$. Further we keep track of all items which became inactive since they have reached its cost upper bound. Note that these items are certainly part of the selected items. This information allows us to compute $|I_{\boldsymbol{x}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})\cap A_{\boldsymbol{x}}|$ and $|I_{\overline{\boldsymbol{x}}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})\cap A_{\overline{\boldsymbol{x}}}(L_{\boldsymbol{x}},L_{\overline{\boldsymbol{x}}})|$ in constant time in each step of the algorithm. Updating the pointers requires overall $O(n)$ time.

For choosing the correct update step length for $L_{\boldsymbol{x}}$ ($L_{\overline{\boldsymbol{x}}}$ is analogous), we take the minimum of four values. First, the cost of the item of $X_{\boldsymbol{x}}$, which is the next item that becomes active, if $L_{\boldsymbol{x}}$ is increased. Second, the minimum of the upper bound cost of all active items of $A_{\boldsymbol{x}}$ (provided by the min-heap). Third, the level $L_{\overline{\boldsymbol{x}}}$. Fourth, the cost of the cheapest item of $X_{\overline{\boldsymbol{x}}}$ which has cost strictly higher than $L_{\boldsymbol{x}}$. Note that we can obtain each value in constant time. Using these update step sizes, it is ensured that after each step all values required to compute the efficiency ratios can be updated efficiently.