In this paper, we study two natural generalizations of ordered $k$-median, named robust ordered $k$-median and fault-tolerant ordered $k$-median. In ordered $k$-median, given a finite metric space $(X,d)$, we seek to open $k$ facilities $S\subseteq X$ which induce a service cost vector $\vec{c}=\{d(j,S):j\in X\}$, and minimize the ordered objective $w^\top\vec{c}^\downarrow$. Here $d(j,S)=\min_{i\in S}d(j,i)$ is the minimum distance between $j$ and facilities in $S$, $w\in\mathbb{R}^{|X|}$ is a given non-increasing non-negative vector, and $\vec{c}^\downarrow$ is the non-increasingly sorted version of $\vec{c}$. The current best result is a $(5+\epsilon)$-approximation [CS19]. We first consider robust ordered $k$-median, a.k.a. ordered $k$-median with outliers, where the input consists of an ordered $k$-median instance and parameter $m\in\mathbb{Z}_+$. The goal is to open $k$ facilities $S$, select $m$ clients $T\subseteq X$ and assign the nearest open facility to each $j\in T$. The service cost vector is $\vec{c}=\{d(j,S):j\in T\}$ and $w$ is in $\mathbb{R}^m$. We introduce a novel yet simple objective function that enables linear analysis of the non-linear ordered objective, apply an iterative rounding framework [KLS18] and obtain a constant-factor approximation. We devise the first constant-approximations for ordered matroid median and ordered knapsack median using the same method. We also consider fault-tolerant ordered $k$-median, where besides the same input as ordered $k$-median, we are also given additional client requirements $\{r_j\in\mathbb{Z}_+:j\in X\}$ and need to assign $r_j$ distinct open facilities to each client $j\in X$. The service cost of $j$ is the sum of distances to its assigned facilities, and the objective is the same. We obtain a constant-factor approximation using a novel LP relaxation with constraints created via a new sparsification technique.
翻译:在本文中, 我们研究两个订购的美元中位元的自然常规值 。 命名为强的订购的美元中位元和不折不扣的美元中位值 。 在订购的美元中位值是美元中位值的最小距离 $( X, d), 我们寻求打开的美元中位值 $S\ subseteq X$, 这会引起一个服务成本矢量$\ vec{c\ d( J, S: j\ nxxxxxxxxxxxxxx 美元, 并尽量减少订购的美元中位值 $ 。 目前的最佳结果是 $( j, S) 美元中位值 美元中位值 美元中位值中位值 美元中位值中位值 。