Reallocating Multiple Facilities on the Line

TL;DR

This paper introduces a polynomial-time algorithm for the multistage K-facility reallocation problem on the real line, optimizing facility movements and agent connection costs over multiple stages, and presents an online algorithm for the case of two facilities.

Contribution

It provides the first polynomial-time algorithm for optimal multistage K-facility reallocation and a constant-competitive online algorithm for two facilities, connecting to the K-server problem.

Findings

Polynomial-time algorithm for optimal reallocation for any K.

Constant-competitive online algorithm for K=2.

Connection established with the classical K-server problem.

Abstract

We study the multistage $K$-facility reallocation problem on the real line, where we maintain $K$ facility locations over $T$ stages, based on the stage-dependent locations of $n$ agents. Each agent is connected to the nearest facility at each stage, and the facilities may move from one stage to another, to accommodate different agent locations. The objective is to minimize the connection cost of the agents plus the total moving cost of the facilities, over all stages. $K$-facility reallocation was introduced by de Keijzer and Wojtczak, where they mostly focused on the special case of a single facility. Using an LP-based approach, we present a polynomial time algorithm that computes the optimal solution for any number of facilities. We also consider online $K$-facility reallocation, where the algorithm becomes aware of agent locations in a stage-by-stage fashion. By exploiting an interesting connection to the classical $K$-server problem, we present a constant-competitive algorithm for $K = 2$ facilities.

Equations

$$Cost(x)=\sum_{t=1}^{T}\bigg{[}\sum\limits_{k=1}^{K}|x_{k}^{t}-x_{k}^{t-1}|+\sum\limits_{i=1}^{n}\min_{1\leq k\leq K}|\alpha_{i}^{t}-x_{k}^{t}|\bigg{]}$$

$$x_k^{\ast t}\in C_1\cup \dots \cup C_T\cup x^0$$

$$\sum _{\ell =1}^{j_m^t-1}{c}_{\ell }^t\le m-1\le \sum _{\ell =1}^{j_m^t}{c}_{\ell }^t.$$

$$x_m^t\leftarrow d(j,1)+\underset{p\in C_1\cup \dots \cup C_T\cup x^0}{\min }p.$$

$$Cost(x)=\sum_{t=1}^{T}\bigg{[}\sum\limits_{i\in C}\sum_{j\in V}d(\text{Loc}(i,t),j)x_{ij}^{t}+\sum_{k\in F}S_{k}^{t}\bigg{]}$$

$$j\in V_t^{+}\text{ if and only if }{c}_j^t>0$$

$$\sum _{j\in V}d(\text{Loc}(i,t),j)x_{ij}^t=\frac{1}{N}\sum _{h=\ell }^{\ell +N-1}d(\text{Loc}(i,t),Y_h^t)$$

$$\sum _{t=1}^T\sum _{k\in F}{S}_k^t=\frac{1}{N}\sum _{t=1}^T\sum _{j=1}^{K\cdot N}d(Y_j^{t-1},Y_j^t)$$

$$\frac{1}{N}\sum _{p=1}^{N}MovingCost(So{l}_p)=\sum _{t=1}^T\sum _{k\in F}{S}_k^t$$

$$\frac{1}{N}\sum _{p=1}^{N}MovingCost(So{l}_p)$$

$$\frac{1}{N}\sum _{p=1}^{N}ConCos{t}_i^t(So{l}_p)=\sum _{j\in V}d(\text{Loc}(i,t),j)x_{ij}^t$$

$$\frac{1}{N}\sum _{p=1}^{N}Cost(So{l}_p)$$

$$\sum _{\ell =1}^{j_m^t-1}{c}_{\ell }^t\le m-1<\sum _{\ell =1}^{j_m^t}{c}_{\ell }^t$$

$$\sum _{\ell =1}^{j_m^t-1}{c}_{\ell }^t=\sum _{\ell =1}^{j_m^t-1}\sum _{{\ell }^{\prime }\in \text{Copies}(\ell )}{c}_{{\ell }^{\prime }}^{{}^{\prime }t}\le (m-1)N\frac{1}{N}=m-1$$

$$\sum _{\ell =1}^{j_m^t}{c}_{\ell }^t=\sum _{\ell =1}^{j_m^t}\sum _{{\ell }^{\prime }\in \text{Copies}(\ell )}{c}_{{\ell }^{\prime }}^{{}^{\prime }t}=((m-1)N+1)\frac{1}{N}>m-1$$

$$\sum _{j\in V}\sum _{\ell \in V}d(j,\ell )S_{kj\ell }^t$$

$$\sum _{j\in V}d(\text{Loc}(i,t),j)x_{ij}^t$$

$$Cost(x)\le 63\cdot Cost(x^{\ast })+|x_1^0-x_2^0|$$

$$y_{k}^{t}=\left\{\begin{array}[]{ll}M_{C_{kt}^{*}}&\text{ if }C_{kt}^{*}\neq\emptyset\\ x_{k}^{*t}&\text{ if }C_{kt}^{*}=\emptyset\\ \end{array}\right.$$

$$\sum_{t=1}^{T}\bigg{[}\sum_{k=1}^{2}[H(C_{kt}^{*})+|y_{k}^{t}-y_{k}^{t-1}|]\bigg{]}\leq 3\cdot Cost(x^{*})$$

$$\sum _{t=1}^T\sum _{k=1}^2|y_k^t-y_k^{t-1}|\le 2\sum _{t=1}^T\sum _{k=1}^2\left[H(C_{kt}^{\ast })+|x_k^{\ast t}-x_k^{\ast t-1}|\right]$$

$$\sum _{t=1}^T\sum _{k=1}^2|y_k^t-y_k^{t-1}|\le \sum _{t=1}^T\sum _{k=1}^2[|y_k^t-x_k^{\ast t}|+|y_k^{t-1}-x_k^{\ast t-1}|+|x_k^{\ast t}-x_k^{\ast t-1}|]$$

$$\sum _{t=1}^T\sum _{k=1}^2\sum _{a\in C_{kt}^{\ast }}|x_k^{\ast t}-a|+\sum _{t=1}^T\sum _{k=1}^2\sum _{a\in C_{k(t-1)}^{\ast }}|x_k^{\ast t-1}-a|\le 2\sum _{t=1}^T\sum _{k=1}^{2}H(C_{kt}^{\ast })$$

$$21\sum _{k=1}^2[H(C_{kt}^{\ast })+|y_k^t-y_k^{t-1}|]+{\Phi }_t(x^t)-{\Phi }_{t-1}(x^{t-1})$$

$$\sum _{k=1}^2|z_k-x_k^{t-1}|\le 2\sum _{k=1}^2|y_k^t-y_k^{t-1}|-{\Phi }_t(z)+{\Phi }_{t-1}(x^{t-1})$$

$$\sum _{k=1}^2[H(C_{kt})+|x_k^t-z_k|]\le 21\sum _{k=1}^{2}H(C_{kt}^{\ast })-{\Phi }_t(x^t)+{\Phi }_t(z)$$

$$\sum _{t=1}^T\sum _{k\in F}{S}_k^t=\frac{1}{N}\sum _{t=1}^T\sum _{j=1}^{K\cdot N}d(Y_j^{t-1},Y_j^t)$$

$$\begin{array}[]{ll@{}ll}\text{minimize}&\sum\limits_{t=1}^{T}\sum\limits_{k=1}^{K}S_{k}^{t}\\ &\\ \text{s.t.}&\sum\limits_{k\in F}f^{t}_{kj}=\frac{1}{N}&\forall j\in V_{t}^{+},t\in\{1,T\}\\ &\sum\limits_{j\in V_{t}^{+}}f_{kj}^{t}=1&\forall k\in F,t\in\{1,T\}\\ &S_{k}^{t}=\sum\limits_{j,l\in V}d(j,l)S_{kjl}^{t}&\forall k\in F,t\in\{1,T\}\\ &\sum\limits_{j\in V_{t-1}^{+}}S_{kjl}^{t}=f_{kl}^{t}&\forall k\in F,l\in V_{t}^{+},t\in\{1,T\}\\ &\sum\limits_{l\in V_{t}^{+}}S_{kjl}^{t}=f_{kj}^{t-1}&\forall k\in F,j\in V_{t-1}^{+},t\in\{1,T\}\\ \end{array}$$

$$\begin{array}[]{ll@{}ll}\text{minimize}&\sum\limits_{t=1}^{T}\sum\limits_{j\in V_{t-1}^{+},l\in V_{t}^{+}}d(j,l)F_{jl}^{t}\\ &\\ \text{s.t.}&\sum\limits_{l\in V_{t}^{+}}F^{t}_{jl}=\frac{1}{N}&\forall j\in V_{t-1}^{+},t\in\{1,T\}\\ &\sum\limits_{j\in V_{t-1}^{+}}F^{t}_{jl}=\frac{1}{N}&\forall l\in V_{t}^{+},t\in\{1,T\}\\ \end{array}$$

$$\frac{1}{N}\sum _{p=1}^{N}ConCos{t}_i^t(So{l}_p)=\sum _{i\in C}\sum _{j\in V}d(\text{Loc}(i,t),j)x_{ij}^t$$

Full text

1 Introduction

Facility Location is a classical problem that has been widely studied in both combinatorial optimization and operations research, due to its many practical applications. It provides a simple and natural model for industrial planning, network design, machine learning, data clustering and computer vision Drezner and Hamacher [12], Lazic [19], Caragiannis et al. [7], Betzler et al. [5]. In its basic form, $K$-Facility Location instances are defined by the locations of $n$ agents in a metric space. The goal is to find $K$ facility locations so as to minimize the sum of distances of the agents to their nearest facility.

In many natural location and network design settings, agent locations are not known in advance. Motivated by this fact, Meyerson [21] introduced online facility location problems, where agents arrive one-by-one and must be irrevocably assigned to a facility upon arrival. Moreover, the fast increasing volume of available data and the requirement for responsive services has led to new, online clustering algorithms Liberty et al. [20], balancing the quality of the clusters with their rate of change over time. In practical settings related to online data clustering, new data points arrive, and the decision of clustering some data points together should not be regarded as irrevocable (see e.g., Fotakis [15] and the references therein).

More recently, understanding the dynamics of temporally evolving social or infrastructure networks has been the central question in many applied areas such as viral marketing, urban planning etc. Dynamic facility location proposed by Eisenstat et al. [13] has been a new tool to analyze temporal aspects of such networks. In this time dependent variant of facility location, agents may change their location over time and we look for the best tradeoff between the optimal connections of agents to facilities and the stability of solutions between consecutive timesteps. The stability of the solutions is modeled by introducing an additional moving cost (or switching cost), which has a different definition depending on the particular setting.

Model and Motivation. In this work, we study the multistage $K$-facility reallocation problem on the real line, introduced by de Keijzer and Wojtczak [10]. In $K$-facility reallocation, $K$ facilities are initially located at $(x_{1}^{0},\ldots,x_{K}^{0})$ on the real line. Facilities are meant to serve $n$ agents for the next $T$ days. At each day, each agent connects to the facility closest to its location and incurs a connection cost equal to this distance. The locations of the agents may change every day, thus we have to move facilities accordingly in order to reduce the connection cost. Naturally, moving a facility is not for free, but comes with the price of the distance that the facility was moved. Our goal is to specify the exact positions of the facilities at each day so that the total connection cost plus the total moving cost is minimized over all $T$ days. In the online version of the problem, the positions of the agents at each stage $t$ are revealed only after determining the locations of the facilities at stage $t-1$.

For a motivating example, consider a company willing to advertise its products. To this end, it organizes $K$ advertising campaigns at different locations of a large city for the next $T$ days. Based on planned events, weather forecasts, etc., the company estimates a population distribution over the locations of the city for each day. Then, the company decides to compute the best possible campaign reallocation with $K$ campaigns over all days (see also [10] for more examples).

de Keijzer and Wojtczak [10] fully characterized the optimal offline and online algorithms for the special case of a single facility and presented a dynamic programming algorithm for $K\geq 1$ facilities with running time exponential in $K$. Despite the practical significance and the interesting theoretical properties of $K$-facility reallocation, its computational complexity and its competitive ratio (for the online variant) are hardly understood.

Contribution. In this work, we resolve the computational complexity of $K$-facility reallocation on the real line and take a first step towards a full understanding of the competitive ratio for the online variant. More specifically, in Section 3, we present an optimal algorithm with running time polynomial in the combinatorial parameters of $K$-facility reallocation (i.e., $n$, $T$ and $K$). This substantially improves on the complexity of the algorithm, presented in [10], that is exponential in $K$. Our algorithm solves a Linear Programming relaxation and then rounds the fractional solution to determine the positions of the facilities. The main technical contribution is showing that a simple rounding scheme yields an integral solution that has the exact same cost as the fractional one.

Our second main result concerns the online version of the problem with $K=2$ facilities. We start with the observation that online $K$-facility reallocation problem with $K\geq 2$ facilities is a natural and interesting generalization of the classical $K$-server problem, which has been a driving force in the development of online algorithms for decades. The key difference is that, in the $K$-server problem, there is a single agent that changes her location at each stage and a single facility has to be relocated to this new location at each stage. Therefore, the total connection cost is by definition [math], and we seek to minimize the total moving cost.

From a technical viewpoint, the $K$-facility reallocation problem poses a new challenge, since it is much harder to track the movements of the optimal algorithm as the agents keep coming. It is not evident at all whether techniques and ideas from the $K$-server problem can be applied to the $K$-facility reallocation problem, especially for more general metric spaces. As a first step towards this direction, we design a constant-competitive algorithm, when $K=2$. Our algorithm appears in Section 4 and is inspired by the double coverage algorithm proposed for the $K$-server problem Koutsoupias [18].

Related Work. We can cast the $K$-facility reallocation problem as a clustering problem on a temporally evolving metric. From this point of view, $K$-facility reallocation problem is a dynamic $K$-median problem. A closely related problem is the dynamic facility location problem, Eisenstat et al. [13], An et al. [1]. Other examples in this setting are the dynamic sum radii clustering Blanchard and Schabanel [6] and multi-stage optimization problems on matroids and graphs Gupta et al. [17].

In Friggstad and Salavatipour [16], a mobile facility location problem was introduced, which can be seen as a one stage version of our problem. They showed that even this version of the problem is $NP$-hard in general metric spaces using an approximation preserving reduction to $K$-median problem.

Online facility location problems and variants have been extensively studied in the literature, see Fotakis [15] for a survey. Divéki and Imreh [11] studied an online model, where facilities can be moved with zero cost. As we have mentioned before, the online variant of the $K$-facility reallocation problem is a generalization of the $K$-server problem, which is one of the most natural online problems. Koutsoupias [18] showed a $(2K-1)$-competitive algorithm for the $K$-server problem for every metric space, which is also $K$-competitive, in case the metric is the real line Bartal and Koutsoupias [4]. Other variants of the $K$-server problem include the $(H,K)$-server problem Bansal et al. [3, 2], the infinite server problem Coester et al. [8] and the $K$-taxi problemFiat et al. [14], Coester and Koutsoupias [9].

2 Problem Definition and Preliminaries

Definition 1** ($K$-Facility Reallocation Problem)**

We are given a tuple $(x^{0},C)$ as input. The $K$ dimensional vector $x^{0}=(x^{0}_{1},\ldots,x^{0}_{K})$ describes the initial positions of the facilities. The positions of the agents over time are described by $C=(C_{1},\ldots,C_{T})$. The position of agent $i$ at stage $t$ is $\alpha_{i}^{t}$ and $C_{t}=(\alpha_{1}^{t},\ldots,\alpha_{n}^{t})$ describes the positions of the agents at stage $t$.

Definition 2

A solution of K-Facility Reallocation Problem is a sequence $x=(x^{1},\ldots,x^{T})$. Each $x^{t}=(x_{1}^{t},\ldots,x_{K}^{t})$ is a $K$ dimensional vector that gives the positions of the facilities at stage $t$ and $x_{k}^{t}$ is the position of facility $k$ at stage $t$. The cost of the solution $x$ is

[TABLE]

Given an instance $(x^{0},C)$ of the problem, the goal is to find a solution $x$ that minimizes the $Cost(x)$. The term $\sum_{t=1}^{T}\sum_{k=1}^{K}|x_{k}^{t}-x_{k}^{t-1}|$ describes the cost for moving the facilities from place to place and we refer to it as moving cost, while the term $\sum_{t=1}^{T}\sum_{i=1}^{n}\min_{1\leq k\leq K}|\alpha_{i}^{t}-x_{k}^{t}|$ describes the connection cost of the agents and we refer to it as connection cost.

In the online setting, we study the special case of $2$-facility reallocation problem. We evaluate the performance of our algorithm using competitive analysis; an algorithm is $c$-competitive if for every request sequence, its online performance is at most $c$ times worse (up to a small additive constant) than the optimal offline algorithm, which knows the entire sequence in advance.

3 Polynomial Time Algorithm

Our approach is a typical LP based algorithm that consists of two basic steps.

• •

Step 1: Expressing the $K$-Facility Reallocation Problem as an Integer Linear Program.

• •

Step 2: Solving fractionally the Integer Linear Program and rounding the fractional solution to an integral one.

3.1 Formulating the Integer Linear Program

A first difficulty in expressing the $K$-Facility Reallocation Problem as an Integer Linear Program is that the positions on the real line are infinite. We remove this obstacle with help of the following lemma proved in [10].

Lemma 3.1

Let $(x_{0},C)$ an instance of the $K$-facility reallocation problem. There exists an optimal solution $x^{*}$ such that for all stages $t\in\{1,T\}$ and $k\in\{1,K\}$,

[TABLE]

According to Lemma 3.1, there exists an optimal solution that locates the facilities only at positions where either an agent has appeared or a facility was initially lying. Lemma 3.1 provides an exhaustive search algorithm for the problem and is also the basis for the Dynamic Programming approach in [10]. We use Lemma 3.1 to formulate our Integer Linear Program.

The set of positions $Pos=C_{1}\cup\ldots\cup C_{T}\cup x^{0}$ can be represented equivalently by a path $P=(V,E)$. In this path, the $j$-th node corresponds to the $j$-th leftmost position of $Pos$ and the distance between two consecutive nodes on the path equals the distance of the respective positions on the real line. Now, the facility reallocation problem takes the following discretized form: We have a path $P=(V,E)$ that is constructed by the specific instance $(x^{0},C)$. Each facility $k$ is initially located at a node $j\in V$ and at each stage $t$, each agent $i$ is also located at a node of $P$. The goal is to move the facilities from node to node such that the connection cost of the agents plus the moving cost of the facilities is minimized.

To formulate this discretized version as an Integer Linear Program, we introduce some additional notation. Let $d(j,l)$ be the distance of the nodes $j,l\in V$ in $P$, $F$ be the set of facilities and $C$ be the set of agents. For each $i\in C$, $\text{Loc}(i,t)$ is the node where agent $i$ is located at stage $t$. We also define the following $\{0,1\}$-indicator variables for all $t\in\{1,T\}$: $x_{ij}^{t}=1$ if, at stage $t$, agent $i$ connects to a facility located at node $j$, $f_{kj}^{t}=1$ if, at stage $t$, facility $k$ is located at node $j$, $S_{kjl}^{t}=1$ if facility $k$ was at node $j$ at stage $t-1$ and moved to node $l$ at stage $t$. Now, the problem can be formulated as the Integer Linear Program depicted in Figure1.

The first three constraints correspond to the fact that at every stage $t$, each agent $i$ must be connected to a node $j$ where at least one facility $k$ is located. The constraint $\sum_{j\in V}f_{kj}^{t}=1$ enforces each facility $k$ to be located at exactly one node $j$. The constraint $S_{k}^{t}=\sum_{j,l\in V}d(j,l)S_{kjl}^{t}$ describes the cost for moving facility $k$ from node $j$ to node $l$. The final two constraints ensure that facility $k$ moved from node $j$ to node $l$ at stage $t$ if and only if facility $k$ was at node $j$ at stage $t-1$ and was at node $l$ at stage $t$ ($S_{kjl}^{t}=1$ iff $f_{kl}^{t}=1$ and $f_{kj}^{t-1}=1$).

We remark that the values of $f_{kj}^{0}$ are determined by the initial positions of the facilities, which are given by the instance of the problem. The notation $x_{ij}^{t}$ should not be confused with $x_{k}^{t}$, which is the position of facility $k$ at stage $t$ on the real line .

3.2 Rounding the Fractional Solution

Our algorithm, described in Algorithm 1, is a simple rounding scheme of the optimal fractional solution of the Integer Linear Program of Figure 1. This simple scheme produces an integral solution that has the exact same cost with an optimal fractional solution.

Theorem 3.1

Let $x$ denote the solution produced by Algorithm 1. Then

[TABLE]

where $x_{ij}^{t},S_{k}^{t}$ denote the values of these variables in the optimal fractional solution of the Integer Linear Program (1).

Theorem 3.1 is the main result of this section and it implies the optimality of our algorithm. We remind that by Lemma 3.1, there is an optimal solution that locates facilities only in positions $C_{1}\cup\ldots\cup C_{T}\cup x^{0}$. This solution corresponds to an integral solution of our Integer Linear Program, meaning that $Cost(x^{*})$ is greater than or equal to the cost of the optimal fractional solution, which by Lemma 3.1 equals $Cost(x)$. We dedicate the rest of the section to prove Theorem 3.1. The proof is conducted in two steps and each step is exhibited in sections 3.3 and 3.4 respectively.

In section 3.3, we present a very simple rounding scheme in the case, where the values of the variables of the optimal fractional solution satisfy the following assumption.

Assumption 1

Let $f_{jk}^{t}$ and $c_{j}^{t}$ be either $1/N$ or [math], for some positive integer $N$.

Although Assumption 1 is very restrictive and its not generally satisfied, it is the key step for proving the optimality guarantee of the rounding scheme presented in Algorithm 1. Then, in section 3.4 we use the rounding scheme of section 3.3 to prove Theorem 3.1. In the upcoming sections, $c_{j}^{t},x_{ij}^{t},f_{kj}^{t},S_{kjl}^{t},S_{k}^{t}$ will denote the values of these variables in the optimal fractional solution of the ILP (1).

3.3 Rounding Semi-Integral solutions

Throughout this section, we suppose that Assumption 1 is satisfied; $f_{kj}^{t}$ and $c_{j}^{t}$ are either $1/N$ or [math] , for some positive integer $N$. If the optimal fractional solution meets these requirements, then the integral solution presented in Lemma 3.2 has the same overall cost. The goal of the section is to prove Lemma 3.2.

Definition 3

$V_{t}^{+}$* denotes the set of nodes of $P$ with a positive amount of facility ($c_{j}^{t}$) at stage $t$,*

[TABLE]

We remind that since $c_{j}^{t}=1/N$ or [math], $|V_{t}^{+}|=K\cdot N$. We also consider the nodes in $V_{t}^{+}=\{Y_{1}^{t},\ldots,Y_{K\cdot N}\}$ to be ordered from left to right.

Lemma 3.2

Let $Sol$ be the integral solution that at each stage $t$ places the $m$-th facility at the $(m-1)N+1$ node of $V_{t}^{+}$ i.e. $Y_{(m-1)N+1}^{t}$. Then, $Sol$ has the same cost as the optimal fractional solution.

The term m**-th facility** refers to the ordering of the facilities on the real line according to their initial positions $\{x_{1}^{0},\ldots,x_{K}^{0}\}$. The proof of Lemma 3.2 is quite technically complicated, however it is based on two intuitive observations about the optimal fractional solution.

Observation 1

The set of nodes at each agent $i$ connects at stage $t$ are consecutive nodes of $V_{t}^{+}$. More precisely, there exists a set $\{Y_{\ell}^{t},\ldots,Y_{\ell+N-1}^{t}\}\subseteq V_{t}^{+}$ such that

[TABLE]

Proof:

Let an agent $i$ that at some stage $t$ has $x_{iY_{j}^{t}}^{t}>0,x_{iY_{\ell}^{t}}^{t}<1/N$ and $x_{iY_{h}^{t}}^{t}>0$ for some $j<\ell<h$. Assume that $\text{Loc}(i,t)\leq Y_{\ell}^{t}$ and to simplify notation consider $x_{\ell}=x_{iY_{\ell}^{t}}^{t},x_{h}=x_{iY_{h}^{t}}^{t}$. Now, increase $x_{\ell}$ by $\epsilon$ and decrease $x_{h}$ by $\epsilon$, where $\epsilon=\min(1/N-x_{\ell},x_{h})$. Then, the cost of the solution is decreased by $(d(\text{Loc}(i,t),h)-d(\text{Loc}(i,t),\ell))\epsilon>0$, thus contradicting the optimality of the solution. The same argument holds if $\text{Loc}(i,t)\geq Y_{\ell}^{t}$. The proof follows since $\sum_{j\in V}x_{ij}^{t}=1$. $\square$

Observation 2

Under Assumption 1, the $m$-th facility places amount of facility $f_{mj}^{t}=1/N$ from the $(m-1)N+1$ to the $mN$ node of $V_{t}^{+}$ i.e. to nodes $\{Y^{t}_{(m-1)N+1},\ldots,Y_{mN}^{t}\}$.

Observation 2 serves in understanding the structure of the optimal fractional solution under Assumption 1. However, it will be not used in this form in the rest of the section. We use Lemma 3.3 instead, which is roughly a different wording of Observation 2 and its proof can be found in subsection A.1 of the Appendix.

Lemma 3.3

Let $S^{t}_{k}$ the fractional moving cost of facility $k$ at stage $t$. Then

[TABLE]

Observations 1, and Lemma 3.3 (Observation 2) are the key points in proving Lemma 3.2.

Definition 4

Let $Sol_{p}$ the integral solution that places at stage $t$ the $m$-th facility at the $(m-1)N+p$ node of $V_{t}^{+}$ i.e. $Y_{(m-1)N+p}^{t}$.

Notice that the integral solution $Sol$ referred in Lemma 3.2 corresponds to $Sol_{1}$. The proof of Lemma 3.2 follows directly by Lemma 3.4 and Lemma 3.5 that conclude this section.

Lemma 3.4

Let $S_{k}^{t}$ be the moving cost of facility $k$ at stage $t$ in the optimal fractional solution and MovingCost$(Sol_{p})$ the total moving cost of the facilities in the integral solution $Sol_{p}$. Then,

[TABLE]

Proof:

By the definition of the solutions $Sol_{p}$ we have that:

[TABLE]

The last equality comes from Lemma 3.3. $\square$

Lemma 3.4 states that if we pick uniformly at random one of the $N$ integral solutions $\{Sol_{p}\}_{p=1}^{N}$, then the expected moving cost that we will pay is equal to the moving cost paid by the optimal fractional solution. Interestingly, the same holds for the expected connection cost. This is formally stated in Lemma 3.5 and it is where Observation 1 comes into play.

Lemma 3.5

Let $ConCost^{t}_{i}(Sol_{p})$ denote the connection cost of agent $i$ at stage $t$ in $Sol_{p}$. Then,

[TABLE]

As already mentioned, the proof of Lemma 3.5 crucially makes use of Observation 1 and is presented in the subsection A.1 of the Appendix. Combining Lemma 3.4 and 3.5 we get that if we pick an integral solution $Sol_{p}$ uniformly at random, the average total cost that we pay is $Z_{LP}^{*}$, where $Z_{LP}^{*}$ is the optimal fractional cost. More precisely,

[TABLE]

Since $Sol_{p}\geq Z_{LP}^{*}$, we have that $Sol_{1}=\dots=Sol_{N}=Z_{LP}^{*}$ and this proves Lemma 3.2.

3.4 Rounding the General Case

In this section, we use Lemma 3.2 to prove Theorem 3.1. As already discussed, Assumption 1 is not satisfied in general by the fractional solution of the linear program (1). Each $S_{kj\ell}^{t}$ will be either [math] or $A_{kj\ell}^{t}/N_{kj\ell}^{t}$, for positive some integers $A_{kj\ell}^{t},N_{kj\ell}^{t}$. However, each positive $f_{kj}^{t}$ will have the form $B_{kj}^{t}/N$, where $N=\Pi_{S_{kj\ell}^{t}>0}N_{kj\ell}^{t}$. This is due to the constraint $f_{kj}^{t}=\sum_{j\in V}S_{kj\ell}^{t}$.

Consider the path $P^{\prime}=(V^{\prime},E^{\prime})$ constructed from path $P=(V,E)$ as follows: Each node $j\in V$ is split into $KN$ copies $\{j_{1},\ldots,j_{KN}\}$ with zero distance between them. Consider also the LP (1), when the underlying path is $P^{\prime}=(V^{\prime},E^{\prime})$ and at each stage $t$, each agent $i$ is located to a node of $V^{\prime}$ that is a copy of $i$’s original location, $Loc^{\prime}(i,t)=\ell\in V^{\prime}$, where $\ell\in\text{Copies}(\text{Loc}(i,t))$. Although these are two different LP’s, they are closely related since a solution for the one can be converted to a solution for the other with the exact same cost. This is due to the fact that for all $j,h\in V$, $d(j,h)=d(j^{\prime},h^{\prime})$, where $j^{\prime}\in\text{Copies}(j)$ and $h^{\prime}\in\text{Copies}(h)$.

The reason that we defined $P^{\prime}$ and the second LP is the following: Given an optimal fractional solution of the LP defined for $P$, we will construct a fractional solution for the LP defined for $P^{\prime}$ with the exact same cost, which additionally satisfies Assumption 1. Then, using Lemma 3.2 we can obtain an integral solution for $P^{\prime}$ with the same cost. This integral solution for $P^{\prime}$ can be easily converted to an integral solution for $P$. We finally show that these steps are done *all at once *by the rounding scheme of Algorithm 1 and this concludes the proof of Theorem 3.1.

Given the fractional positions $\{f_{kj}^{t}\}_{t\geq 1}$ of the optimal solution of the LP formulated for $P=(V,E)$, we construct the fractional positions of the facilities in $P^{\prime}=(V^{\prime},E^{\prime})$ as follows: If $f_{kj}^{t}=B_{kj}^{t}/N$, then facility $k$ puts a $1/N$ amount of facility in $B_{kj}^{t}$ nodes of the set $\text{Copies}(j)=\{j_{1},\ldots,j_{KN}\}$ that have a [math] amount of facility. The latter is possible since there are exactly $KN$ copies of each $j\in V$ and $c_{j}^{t}\leq K$ (that is the reason we required $KN$ copies of each node). The values of the rest of the variables are defined in the proof of Lemma 3.7 that is presented in the end of the section. The key point is that the produced solution $\{f_{k\ell}^{{}^{\prime}t},c_{j}^{{}^{\prime}t},S_{kj\ell}^{{}^{\prime}t},x_{ij}^{{}^{\prime}t},S_{k}^{{}^{\prime}t}\}$ will satisfy the following properties (see Lemma 3.7):

• •

its cost equals $Z_{LP}^{*}$

• •

$f^{{}^{\prime}t}_{k\ell}=1/N$ or [math], for each $\ell\in V^{\prime}$

• •

$c^{{}^{\prime}t}_{\ell}=1/N$ or $0,$ for each $\ell\in V^{\prime}$

• •

$c^{t}_{j}=\sum\limits_{\ell\in\text{Copies}(j)}c^{{}^{\prime}t}_{\ell}$, for each $j\in V$

Clearly, this solution satisfies Assumption 1 and thus Lemma 3.2 can be applied. This implies that the integral solution for $P^{\prime}$ that places the $m$-th facility to the $(m-1)N+1$ node of $V^{{}^{\prime}+}_{t}$ ($Y^{{}^{\prime}t}_{(m-1)N+1}\in V^{\prime}$) has cost $Z_{LP}^{*}$. So the integral solution for $P$ that places the $m$-th facility to the node $j_{m}^{t}\in V$, such that $Y^{{}^{\prime}t}_{(m-1)N+1}\in\text{Copies}(j_{m}^{t})$, has again cost $Z_{LP}^{*}$.

A naive way to determine the nodes $j_{m}^{t}$ is to calculate $N$, construct $P^{\prime}$ and its fractional solution, find the nodes $Y^{{}^{\prime}t}_{(m-1)N+1}$ and determine the nodes $j_{m}^{t}$ of $P$. Obviously, this rounding scheme requires exponential time. Fortunately, Lemma 3.6 provides a linear time rounding scheme to determine the node $j_{m}^{t}$ given the optimal fractional solution of $P=(V,E)$. This concludes the proof of Theorem 3.1.

Lemma 3.6

The $(m-1)N+1$ node of $V_{t}^{\prime+}$ is a copy of the node $j_{m}^{t}\in V$ if and only if

[TABLE]

.

Proof:

Let $(m-1)N+1$ node of $V_{t}^{\prime+}$ be a copy of the node $j_{m}^{t}\in V_{t}^{+}$. Then

[TABLE]

[TABLE]

The above equations hold because of the property $c_{\ell}^{t}=\sum_{\ell^{\prime}\in\text{Copies}(\ell)}c_{\ell^{\prime}}^{{}^{\prime}t}$ and that $c_{\ell^{\prime}}^{{}^{\prime}t}$ is either [math] or $1/N$.

Now, let $\sum_{\ell=1}^{j_{m}^{t}-1}c_{\ell}^{t}\leq m-1<\sum_{\ell=1}^{j_{m}^{t}}c_{\ell}^{t}$ and assume that $(m-1)N+1$-th node of $V_{t}^{+}$ is a copy of $j\in V$. If $j<j_{m}^{t}$, then $\sum_{\ell=1}^{j}c_{\ell}^{t}>m-1$ and if $j>j_{m}^{t}$, then $\sum_{\ell=1}^{j_{m}^{t}}<m-1$. As a result, $j=j_{m}^{t}$. $\square$

Remark 1

We remark that the nodes $j_{m}^{t}$ can be determined with an even simpler way than that presented in Algorithm 1. That is $j_{m}^{t}$ is the most left node such that $f_{mj}^{t}>0$. However, this rounding strategy requires some additional analysis.

Lemma 3.7

Let $\{f_{kj}^{t},c_{j}^{t},S_{kjl}^{t},x_{ij}^{t}\}_{t\geq 1}$ the optimal fractional solution for the LP $(1)$ with underlying path $P$. Then, there exists a solution $\{f_{kj}^{{}^{\prime}t},c_{j}^{{}^{\prime}t},S_{kjl}^{{}^{\prime}t},x_{ij}^{{}^{\prime}t},S_{k}^{{}^{\prime}t}\}_{t\geq 1}$ of the LP $(1)$ with underlying path $P^{\prime}$ such that

Its cost is $Z_{LP}^{*}$. 2. 2.

$f^{{}^{\prime}t}_{k\ell}=1/N$* or [math], for each $\ell\in V^{\prime}$* 3. 3.

$c^{{}^{\prime}t}_{\ell}=1/N$* or [math], for each $\ell\in V^{\prime}$* 4. 4.

$c^{t}_{j}=\sum\limits_{\ell\in\text{Copies}(j)}c^{{}^{\prime}t}_{\ell}$, for each $j\in V$

Proof:

First, we set values to the variables $f_{kj}^{{}^{\prime}t}$. Initially, all $f_{kj}^{{}^{\prime}t}=0$. We know that if $f_{kj}^{t}>0$, then it equals $B_{kj}^{t}/N$, for some positive integer $B_{kj}^{t}$. For each such $f_{kj}^{t}$, we find $u_{1},\ldots,u_{B_{kj}^{t}}\in\text{Copies}(j)$ with $f_{ku_{h}}^{{}^{\prime}t}=0$. Then, we set $f_{ku_{h}}^{{}^{\prime}t}=1/N$ for $h=\{1,B_{kj}^{t}\}$. Since there are $KN$ copies of each node $j\in V$ and $\sum_{j\in V}f_{kj}^{t}\leq K$, we can always find sufficient copies of $j$ with $f_{ku}^{{}^{\prime}t}=0$. When this step is terminated, we are sure that conditions $2,3,4$ are satisfied.

We continue with the variables $S_{kj\ell}^{{}^{\prime}t}$. Initially, all $S_{kj\ell}^{{}^{\prime}t}=0$. Then, each positive $S_{kj\ell}^{t}$ has the form $B_{kj\ell}^{t}/N$. Let $B=B_{kj\ell}^{t}$ to simplify notation. We now find $B$ copies of $u_{1},\ldots,u_{B}$ of $j$ and $v_{1},\ldots,v_{B}$ of $\ell$ so that

• •

$f_{ku_{1}}^{{}^{\prime}t}=\dots=f_{ku_{B}}^{{}^{\prime}t}=f_{kv_{1}}^{{}^{\prime}t}=\dots=f_{kv_{B}}^{{}^{\prime}t}=1/N$

• •

$S_{ku_{1}h}^{{}^{\prime}t}=\dots=S_{ku_{B}h}^{{}^{\prime}t}=S_{khv_{1}}^{{}^{\prime}t}=\dots=S_{khv_{B}}^{{}^{\prime}t}=0$ for all $h\in V^{\prime}$

We then set $S_{ku_{1}v_{1}}^{{}^{\prime}t}=\dots=S_{ku_{B}v_{B}}^{{}^{\prime}t}=1/N$. Again, since $\sum_{\ell\in V}S_{kj\ell}^{t}=f_{kj}^{t}$ and $\sum_{j\in V}S_{kj\ell}^{t}=f_{k\ell}^{t}$ we can always find $B_{kj\ell}^{t}$ pairs of copies of $j$ and $\ell$ that satisfy the above requirements. We can now prove that the movement cost of each facility $k$ is the same in both solutions.

[TABLE]

The second equality follows from the fact that $h,h^{\prime}$ are copies of $j,\ell$ respectively and thus $d(h,h^{\prime})=d(j,\ell)$.

Finally, set values to the variables $x_{ij}^{{}^{\prime}t}$ for each $j\in V^{\prime}$. Again, each positive $x_{ij}^{t}$ equals $B_{ij}^{t}/N$, for some positive integer. We take $B_{ij}^{t}$ copies of $j$, $u_{1},\dots,u_{B_{ij}^{t}}$ and set $x_{iu_{1}}^{{}^{\prime}t}=\dots=x_{iu_{B_{ij}^{t}}}^{{}^{\prime}t}=1/N$. The connection cost of each agent $i$ remains the same since

[TABLE]

The third equality holds since $\text{Loc}^{\prime}(i,t)\in\text{Copies}(\text{Loc}(i,t))$. $\square$

4 A Constant-Competitive Algorithm for the Online 2-Facility Reallocation

Problem

In this section, we present an algorithm for the online 2-facility reallocation problem and we discuss the core ideas that prove its performance guarantee. The online algorithm, denoted as Algorithm 2, consists of two major steps.

In Step $1$, facilities are initially moved towards the positions of the agents. We remark that in Step $1$, the final positions of the facilities at stage $t$ are not yet determined. The purpose of this step is to bring at least one facility close to the agents. This initial moving consists of three cases (see Figure 2), depending only on the relative positions of the facilities at stage $t-1$ and the agents at stage $t$.

In Step $2$, our algorithm determines the final positions of the facilities $x_{1}^{t},x_{2}^{t}$. Notice that after Step $1$, at least one of the facilities is inside the interval $[\alpha_{1}^{t},\alpha_{n}^{t}]$, meaning that at least one of the facilities is close to the agents. As a result, our algorithm may need to decide between moving the second facility close to the agents or just letting the agents connect to the facility that is already close to them. Obviously, the first choice may lead to small connection cost, but large moving cost, while the second has the exact opposite effect. Roughly speaking, Algorithm 2 does the following: If the connection cost of the agents, when placing just one facility optimally, is not much greater than the cost for moving the second facility inside $[\alpha_{1}^{t},\alpha_{n}^{t}]$, then Algorithm 2 puts the first facility to the position that minimizes the connection cost, if one facility is used. Otherwise, it puts the facilities to the positions that minimize the connection cost, if two facilities are used. The above cases are depicted in Figure 3. We formalize how this choice is performed, introducing some additional notation.

Definition 5

**

• •

$C_{t}=\{\alpha_{1}^{t},\ldots,\alpha_{n}^{t}\}$* denotes the positions of the agents at stage $t$ ordered from left to right.*

• •

If $C$ is a set of positions with $|C|=2k,k\in\mathbb{N}_{>0}$, then $M_{C}$ denotes the median interval of the set, which is the interval $[\alpha_{n/2},\alpha_{n/2+1}]$. If $|C|=2k+1,k\in\mathbb{N}_{0}$, then $M_{C}$ is a single point.

• •

$H(C)$* denotes the optimal connection cost for the set $C$ when all agents of $C$ connect to just one facility. That is $H(C)=\sum_{\alpha\in C}|\alpha-M_{C}|.$ We also define $H(\emptyset)=0$.*

• •

$C_{1t}^{*}$* (resp. $C_{2t}^{*}$) denotes the positions of the agents that connect to facility $1$ (resp. $2$) at stage $t$ in the optimal solution $x^{*}$. $C_{1t}$ (resp. $C_{2t}$) denotes the positions of the agents that connect to facility $1$ (resp. $2$) at stage $t$ in the solution produced by Algorithm 2.*

We first mention that Algorithm 2 seems much more complicated than it really is (the first two cases are symmetric both in Step 1 and Step 2). In fact, only the last two cases are difficult to handle and we explain them subsequently. The performance guarantee of Algorithm 2 is formally stated in Theorem 4.1.

Theorem 4.1

Let $x=\{x_{1}^{t},x_{2}^{t}\}_{t\geq 1}$ the solution produced by Algorithm 2 and $x^{*}$ the optimal solution. Then,

[TABLE]

where $x_{1}^{0},x_{2}^{0}$ are the initial positions of the facilities.

The rest of the section is dedicated to provide a proof sketch (some proofs are included in subsection A.2 of the Appendix) of Theorem 4.1. Although it is possible to improve the competitive ratio of Algorithm 2 by a much more technically involved analysis, we stress here that it is not possible to turn the result into any constant factor. The reason is that the 2-facility reallocation problem on the line is a generalization of 2-server problem on the line, which has a lower bound of 2 on the competitive ratio of any online algorithm. Before proceeding, we present Lemma 4.1 that is a key component in the subsequent analysis and that reveals the real difficulty of the online $2$-facility reallocation problem.

Lemma 4.1

Let the optimal solution $x^{*}$ and $C_{1t}^{*},C_{2t}^{*}$ the set of agents that connect respectively to facility $1$ and $2$ at stage $t$. Let the solution $y^{t}=(y_{1}^{t},y_{2}^{t})$ defined as follows:

[TABLE]

Then, the following inequality holds.

[TABLE]

Proof:

Since $\sum_{t=1}^{T}\sum_{k=1}^{2}H(C_{kt}^{*})=\sum_{t=1}^{T}\sum_{k=1}^{2}\sum_{a\in C_{kt}^{*}}|x_{k}^{*t}-a|,$ we only have to prove that

[TABLE]

From the triangle inequality, we have that

[TABLE]

The right hand side of the inequality is maximized, when $y_{k}^{t}\neq x_{k}^{*t}$ and $y_{k}^{t-1}\neq x_{k}^{*t-1}$ for $k=1,2$, namely when $C_{kt}^{*},C_{k(t-1)}^{*}\neq\emptyset$. Since $y_{k}^{t}$ (resp. $y_{k}^{t-1}$) is the median agent (lies in the median interval of $C_{kt}^{*}$ in the case $|C_{kt}^{*}|=2k$) of $C_{kt}^{*}$ (resp.$C_{k(t-1)}^{*}$) in this case, $x_{k}^{*t}$ (resp. $x_{k}^{*t-1}$) has to connect $y_{k}^{t}$ (resp. $y_{k}^{t-1}$). Thus, $\sum_{t=1}^{T}\sum_{k=1}^{2}|y_{k}^{t}-x_{k}^{*t}|+|y_{k}^{t-1}-x_{k}^{*t-1}|$ can be upper bounded by the optimal connection cost, which is

[TABLE]

$\square$

Lemma 4.1 indicates that the real difficulty of the problem is not determining the exact positions of the facilities in the optimal solution, but to determine the service clusters that the optimal solution forms. In fact, if we knew the clusters $C_{1t}^{*},C_{2t}^{*}$, then Lemma 4.1 provides us with a $3$-approximation algorithm. Obviously, this information cannot be acquired in the online setting, since $C_{1t}^{*},C_{2t}^{*}$ depend on the future positions of the agents that we do not know. We prove that Algorithm 2 has an approximation guarantee of $21$ with respect to the solution $y$, that directly translates to an approximation guarantee of $63$ with respect to $Cost(x^{*})$. The latter is formally stated in Lemma 4.2 and is the main result of this section.

Lemma 4.2

Let $x=\{x_{1}^{t},x_{2}^{t}\}_{t\geq 1}$ be the solution produced by Algorithm 2. Then, the cost paid by solution $x$ at stage $t$, $\sum_{k=1}^{2}|x_{k}^{t}-x_{k}^{t-1}|+\sum_{i=1}^{n}\min_{k=1,2}|x_{k}^{t}-\alpha_{i}^{t}|$, is at most

[TABLE]

where* $\Phi_{t}(x_{1},x_{2})=2(|x_{1}-y_{1}^{t}|+|x_{2}-y_{2}^{t}|)+|x_{1}-x_{2}|$.*

Lemma 4.2 directly implies Theorem 4.1 by applying a telescopic sum over all $t$ and then applying Lemma 4.1. Notice that the additive term $|x_{1}^{0}-x_{2}^{0}|$ in Theorem 4.1 depends only on the initial positions of the facilities and follows from the fact that $\Phi_{0}(x^{0})=|x_{1}^{0}-x_{2}^{0}|$. Since the additive term is a constant independent from the request sequence (the client positions $C_{t}$), it is common to define the competitive ratio of an online algorithm as in Section 2. In the rest of the section, we present the proof ideas of Lemma 4.2, which come together with explaining Steps $1$ and $2$ of our algorithm. Let us start with explaining Step $1$. First, note that since $x_{1}^{0}\leq x_{2}^{0}$, then $x_{1}^{t}\leq x_{2}^{t}$ by our algorithm construction. Now, assume that $x_{2}^{t-1}\leq\alpha_{1}^{t}$ (second case). Before deciding the exact positions of the facilities, we can safely move facility $2$ to the right until reaching $\alpha_{1}^{t}$. The term safely means that this moving cost is roughly upper bounded by the moving cost $\sum_{k=1}^{2}|y_{k}^{t}-y_{k}^{t-1}|$. This safe moving applies to all three cases of Step $1$ in Algorithm 2 and is formally stated in Lemma 4.3.

Lemma 4.3

Let $z=(z_{1},z_{2})$ denote the values of the variables $z_{1},z_{2}$ after Step 1 of Algorithm 2. Then,

[TABLE]

The proof of Lemma 4.3 can be found in subsection A.2 of the Appendix. Lemma 4.3 reveals the basic idea of Step $1$ performed by the online algorithm. We remind that Step $1$ is performed by Algorithm 2 if both facilities are outside the interval $C_{t}$ at the beginning of stage $t$. Therefore, it distinguishes between the three cases depicted in Figure 2 (We show 2 cases since the case with both facilities on the right of the agents is symmetric to the first). Moving with the same speed towards the interval $[a_{1}^{t},a_{n}^{t}]$ results to the same moving cost for both facilities; both facilities will move the distance of the facility which is closest to its closest agent.

According to the geometry of the agents’ positions, we can identify a safe move whose cost is also paid by solution $y$ for moving the facilities. Moreover, the proof of Lemma 4.3 reveals why we compare our algorithm with the solution $y$ and not directly with $x^{*}$. All these safe moves are based on the fact that either $y_{1}^{t}$ or $y_{2}^{t}$ lies in the $C_{t}=[\alpha_{1}^{t},\alpha_{n}^{t}]$ (the latter does not necessarily hold for $x^{*}$). Finally, the potential function $\Phi_{t}(x_{1},x_{2})$ is crucial, since it permits safe moves, when all agents are on the right/left of the facilities (first/second case) as well as when they are contained in the interval $[x_{1}^{t-1},x_{2}^{t-1}]$ (third case). This idea was first developed for the $K$-server problem Koutsoupias [18].

Up next, we analyze the ideas of Step $2$. We now need to bound the connection cost plus some additional moving cost from the point where the safe move stopped. The following lemma formalizes the guarantees provided by Algorithm 2 after the execution of Step $2$. The full proof of Lemma 4.4 can be found in subsection A.2 of the Appendix.

Lemma 4.4

Let $x^{t}=(x_{1}^{t},x_{2}^{t})$ denote the locations of facilities at stage $t$ after the execution of Step 2. Then,**

[TABLE]

When Algorithm 2 performs Step $2$, we know that at least one facility lies inside the interval $C_{t}$. This facility will definitely connect some agents of $C_{t}$, since we can charge it a small moving cost even if it connects all agents. Thus, the algorithm needs only to decide whether it will connect agents by using only one facility or by using both facilities. The decision depends on the distance between the facility, which is outside of $C_{t}$ (in case there is one), and the closest agent to this facility. If this distance is ”small” (resp. if the facility is already inside the interval), Algorithm 2 will connect agents to both facilities minimizing the connection cost using two facilities. This will guarantee that the moving cost and connection cost incurred are relatively small compared to the cost of solution $y$.

Now, if the facility, which is outside the interval $C_{t}$, is ”far” from its closest agent, Algorithm 2 moves this facility towards $C_{t}$ by a distance, depending on the optimal connection cost using one facility, and serves all agents using the facility, which is inside $C_{t}$. Then, we can prove that this move is sufficient to bound the total cost of the algorithm compared to the cost of solution $y$, even if $y$ has arbitrarily smaller connection cost (if it uses both facilities to serve the agents). The choices of Algorithm 2 are depicted in Figure 3. We provide more detailed Figures based on the analysis of Algorithm 2 in subsection A.2 of the Appendix.

5 Open Problems

Regarding the offline variant of the $K$-facility reallocation problem, it would be interesting to consider the problem in general metric spaces. Since $K$-facility reallocation is essentially a dynamic $K$-median problem, a main open problem is to design approximation algorithms for this problem as well as to find lower bounds on the approximation ratio of any offline algorithm in general metric spaces. Turning to the online variant, the main question arising is to design an online algorithm for online $K$-facility reallocation problem on the line. This variant with any number of facilities seems much more intriguing. It would also be interesting to consider randomized algorithms for both the online and the offline variant.

Appendix A Appendix

A.1 Omitted proofs of

Section 3.4

Lemma 3.3 Let $S^{t}_{k}$ the fractional switching cost of facility $k$ at stage $t$. Then,

[TABLE]

Proof:

By Assumption 1, $c_{j}^{t}=1/N$ if $j\in V_{t}^{+}=\{Y_{1}^{t},\ldots,Y_{KN}^{t}\}$ and [math] otherwise. Notice that the connection cost of the optimal fractional solution only depends on the variables $c_{j}^{t}$. As a result, $f_{kj}^{t},S_{k}^{t},S_{kjl}^{t}$ must be the optimal solution of the following linear program.

[TABLE]

Instead of proving that the minimum cost of the above linear program is $\frac{1}{N}\sum\limits_{t=1}^{T}\sum\limits_{j=1}^{K\cdot N}d(Y_{j}^{t-1},Y_{j}^{t})$, we prove this for the following more convenient relaxation of the above LP.

[TABLE]

It is easy to prove that the LP (1) is a relaxation of the first by setting $F_{jl}^{t}=\sum_{k\in F}S_{kjl}^{t}$. Moreover, the above LP describes a flow problem between the nodes $V_{t}^{+}$, where $F_{jl}^{t}$ is the amount of flow going from node $j\in V_{t-1}^{+}$ to node $l\in V_{t}^{+}$ (see Figure 4).

We are ready for the final step of our proof. First, observe that $F^{t}_{Y_{j}^{t-1}Y_{j}^{t}}$ is feasible solution for the above LP since $|V_{t-1}^{+}|=|V_{t}^{+}|=K\cdot N$. If we prove that this assignment minimizes the objective, then we are done. Assume that in the optimal solution, $F^{t}_{Y_{1}^{t-1}Y_{1}^{t}}<1/N$. Since $\sum\limits_{l\in V_{t}^{+}}F^{t}_{Y_{1}^{t-1}l}=\frac{1}{N}$, there exists $Y_{j}^{t}$ such that $F^{t}_{Y_{1}^{t-1}Y_{j}^{t}}>0$. Similarly, by using the second constraint we obtain that $F^{t}_{Y_{j^{\prime}}^{t-1}Y_{1}^{t}}>0$. Let $\epsilon=\min(F^{t}_{Y_{1}^{t-1}Y_{j}^{t}},F^{t}_{Y_{j^{\prime}}^{t-1}Y_{1}^{t}})$. Observe that if we increase $F^{t}_{Y_{1}^{t-1}Y_{1}^{t}}$, $F^{t}_{Y_{j^{\prime}}^{t-1}Y_{j}^{t}}$ by $\epsilon$ and decrease $F^{t}_{Y_{1}^{t-1}Y_{j}^{t}}$, $F^{t}_{Y_{j^{\prime}}^{t-1}Y_{1}^{t}}$ by $\epsilon$, we obtain another feasible solution. The cost difference of the two solutions is $D=\epsilon(d(Y_{1}^{t-1},Y_{j}^{t})+d(Y_{j^{\prime}}^{t-1},Y_{1}^{t})-d(Y_{1}^{t-1},Y_{1}^{t})-d(Y_{j^{\prime}}^{t-1},Y_{j}^{t}))$. If we prove that $D$ is no negative, we are done. We show the latter using the fact that $Y_{1}^{t-1}\leq Y_{j^{\prime}}^{t-1}$ and $Y_{1}^{t}\leq Y_{j}^{t}$. More precisely,

• •

If $Y_{1}^{t-1}\leq Y_{1}^{t}$ then $D\geq 0$ since $Y_{1}^{t}\leq Y_{j}^{t}$.

• •

If $Y_{1}^{t-1}\geq Y_{1}^{t}$ then $D\geq 0$ since $Y_{1}^{t-1}\leq Y_{j^{\prime}}^{t-1}$.

Until now, we have shown that in the optimal solution, the node $Y_{1}^{t-1}$ sends all of her flow to the node $Y_{1}^{t}$. Meaning that $Y_{1}^{t}$ does not receive flow by any other node apart from $Y_{1}^{t-1}$. By repeating the same argument, it follows that in the optimal solution each node $Y_{j}^{t-1}$ sends all of her flow to $Y_{j}^{t}$. $\square$

Lemma 3.5 Let ConCost${}^{t}_{i}(Sol_{p})$ denote the connection cost of agent $i$ at stage $t$ in $Sol_{p}$ of Definition 4. Then

[TABLE]

Proof:

We will prove that $\frac{1}{N}\sum_{p=1}^{N}\text{ConCost}_{i}^{t}(Sol_{p})$ equals $\sum_{j\in V}d(\text{Loc}(i,t),j)x_{ij}^{t}$. We remind that by Assumption 1, $c_{j}$ is $1/N$ if $j\in V_{t}^{+}$ and [math] otherwise. As a result, in the optimal fractional solution, each agent $i$ finds the $N$ closest to $\text{Loc}(i,t)$ nodes of $V_{t}^{+}$ and receives a $1/N$ amount of service from each one of them. Let us call this set $N_{i}^{t}$. By Observation 1, the nodes in $N_{i}^{t}$ must be consecutive nodes of $V_{t}^{+}$ i.e. $N_{i}^{t}=\{Y_{l}^{t},\dots,Y_{l+N-1}^{t}\}$ and

[TABLE]

Since $Sol_{p}$ puts facilities in the positions $\{Y_{(m-1)\cdot N+p}^{t}\}_{m=1}^{K}$, there exists a unique node $Y_{l(p)}^{t}\in N_{i}^{t}$ in which $Sol_{p}$ puts a facility. $Y_{l(p)}^{t}$ is the closest node to $\text{Loc}(i,t)$ from all the nodes in which $Sol_{p}$ puts a facility. As a result, $ConCost_{i}^{t}(Sol_{p})=d(\text{Loc}(i),Y_{l(p)}^{t})$. Now, summing over $p$ we get,

[TABLE]

$\square$

A.2 Omitted proofs of Section 4

Lemma 4.3 Let $z=(z_{1},z_{2})$ denote the values of the variables $z_{1},z_{2}$ after Step 1 of Algorithm 2. Then,

[TABLE]

Proof:

Assume that $x_{2}^{t-1}\leq\alpha_{1}^{t}$, then Algorithm 2 will first move facility $2$ to $\alpha_{1}^{t}$ ($z_{1}=x_{1}^{t-1},z_{2}=\alpha_{1}^{t}$), paying moving cost equal to $|\alpha_{1}^{t}-x_{2}^{t-1}|$. This moving cost can be bounded with the use of the potential function $\Phi$. More specifically, we have that $\Phi_{t}(z)-\Phi_{t}(x^{t-1})+\Phi_{t}(x^{t-1})-\Phi_{t-1}(x^{t-1})$

[TABLE]

In the considered case $z_{1}=x_{1}^{t-1},z_{2}=\alpha_{1}^{t}$, the difference $\Phi_{t}(z)-\Phi_{t}(x^{t-1})$ in the potential function equals the quantity $2(|y_{2}^{t}-\alpha_{1}^{t}|-|y_{2}^{t}-x_{2}^{t-1}|)+|x_{1}^{t-1}-\alpha_{1}^{t}|-|x_{1}^{t-1}-x_{2}^{t-1}|$. By the definition of solution $y$ in Lemma 4.2, either $y_{1}^{t}$ or $y_{2}^{t}$ lies in the interval $[\alpha_{1}^{t},\alpha_{n}^{t}]$. Since either $y_{1}^{t}$ or $y_{2}^{t}$ lies in the interval $[a_{1}^{t},a_{2}^{t}]$ and $y_{1}^{t}\leq y_{2}^{t}$, we have that $a_{1}^{t}\leq y_{2}^{t}$. Meaning that $z_{2}$ is closer to $y_{2}^{t}$ than $x_{2}^{t-1}$ and consequently $2(|y_{2}^{t}-a_{1}^{t}|-|y_{2}^{t}-x_{2}^{t-1}|)=-2|x_{2}^{t-1}-a_{1}^{t}|$. Therefore, $\Phi_{t}(z)-\Phi_{t}(x^{t-1})=-2|x_{2}^{t-1}-a_{1}^{t}|+|x_{1}^{t-1}-\alpha_{1}^{t}|-|x_{1}^{t-1}-x_{2}^{t-1}|=-|a_{1}^{t}-x_{2}^{t-1}|=-|z_{2}-x_{2}^{t-1}|$, which completes the proof of Lemma 4.3 for this case of Step 1.

Notice that inequality (2) holds for all three cases of Step 1. Thus, one just need to prove that $\Phi_{t}(z)-\Phi_{t}(x^{t-1})\leq-\sum_{k=1}^{2}|z_{k}-x_{k}^{t-1}|$ for the other two cases. We prove it for the third case of Step 1, since the second case ($x_{1}^{t-1}\geq\alpha_{n}^{t}$) is just symmetric to the first case.

In the third case of Step 1, we have that $x_{1}^{t-1}<a_{1}^{t}$, $x_{2}^{t-1}>a_{n}^{t},z_{1}=x_{1}^{t-1}+\min(|x_{1}^{t-1}-a_{1}^{t}|,|x_{2}^{t-1}-a_{n}^{t}|)$ and $z_{2}=x_{2}^{t-1}-\min(|x_{1}^{t-1}-a_{1}^{t}|,|x_{2}^{t-1}-a_{n}^{t}|)$. The difference $\Phi_{t}(z)-\Phi_{t}(x^{t-1})$ in the potential function equals the quantity $2(|z_{1}-y_{1}^{t}|-|x_{1}^{t-1}-y_{1}^{t}|+|z_{2}-y_{2}^{t}|-|x_{2}^{t-1}-y_{2}^{t}|)+|z_{1}-z_{2}|-|x_{1}^{t-1}-x_{2}^{t-1}|$. Now, $|z_{1}-z_{2}|$ - $|x_{1}^{t-1}-x_{2}^{t-1}|=-2\min(|x_{1}^{t-1}-\alpha_{1}^{t}|,|x_{2}^{t-1}-\alpha_{n}^{t}|)=-\sum_{k=1}^{2}|z_{k}-x^{t-1}_{k}|$. Assume that $y_{1}^{t}\in[a_{1}^{t},a_{n}^{t}]$, then $\sum_{k=1}^{2}(|z_{k}-y_{k}^{t}|-|x_{k}^{t-1}-y_{k}^{t}|)\leq 0$ since $|z_{1}-y_{1}^{t}|-|x_{1}^{t-1}-y_{1}^{t}|=-\min(|x_{1}^{t-1}-\alpha_{1}^{t}|,|x_{2}^{t-1}-\alpha_{n}^{t}|)$ and $|z_{2}-y_{2}^{t}|-|x_{2}^{t-1}-y_{2}^{t}|\leq\min(|x_{1}^{t-1}-\alpha_{1}^{t}|,|x_{2}^{t-1}-\alpha_{n}^{t}|)$. As a result, inequality (2) holds. Using the same argument in case $y_{2}^{t}\in[a_{1}^{t},a_{n}^{t}]$ completes the proof. $\square$

Lemma 4.4 Let $x^{t}=(x_{1}^{t},x_{2}^{t})$ denote the locations of facilities at stage $t$ after the execution of Step 2. Then,

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Proof:

Observe that by Algorithm 2, either $a_{1}^{t}\leq z_{1}\leq a_{n}^{t}$ or $a_{1}^{t}\leq z_{2}\leq a_{n}^{t}$. As a result, we need to prove the claim for the following 4 cases:

• •

$a_{1}\leq z_{1}\leq a_{n}$ and $z_{2}-a_{n}\geq 3H(C_{t})$

• •

$a_{1}\leq z_{1}\leq a_{n}$ and $z_{2}-a_{n}<3H(C_{t})$

• •

$a_{1}\leq z_{2}\leq a_{n}$ and $a_{1}-z_{1}\geq 3H(C_{t})$

• •

$a_{1}\leq z_{2}\leq a_{n}$ and $a_{1}-z_{1}<3H(C_{t})$

We will prove just the first and the second case since the third is symmetric to the first and the forth is symmetric to the second.

In case $a_{1}\leq z_{1}\leq a_{n}$ and $z_{2}-a_{n}\geq 3H(C_{t})$, Algorithm 2 puts facility $1$ in the median of $C_{t}$, namely $x_{1}^{t}=M_{C_{t}}$ (or $x_{1}^{t}\in M_{C_{t}}$ in case the number of agents is even), and moves facility $2$ to the left by a distance of $3H(C_{t})$.

First note that $\sum_{k=1}^{2}H(C_{kt})\leq H(C_{t})$ since $x_{1}^{t}\in M_{C_{t}}$. Then $|x_{1}^{t}-z_{1}|\leq|a_{1}^{t}-a_{n}^{t}|\leq H(C_{t})$ because both $x_{1}^{t}$ and $z_{1}$ lie in the interval $[a_{1}^{t},a_{n}^{t}]$ and $|x_{2}^{t}-z_{2}|=3H(C_{t})$ by Algorithm 2. Therefore, we have that $\sum_{k=1}^{2}H(C_{kt})+|x_{k}^{t}-z_{k}|\leq 5H(C_{t})$.

By the geometry of this case and the aforementioned bounds,

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Since $\sum_{k=1}^{2}\left[H(C_{kt})+|x_{k}^{t}-z_{k}|\right]\leq 5H(C_{t})$, the challenge is bounding $\sum_{k=1}^{2}\left(|x_{k}^{t}-y_{k}^{t}|-|z_{k}-y_{k}^{t}|\right)$ by the connection cost $\sum_{k=1}^{2}H(C_{kt}^{*})$. In case $\sum_{k=1}^{2}H(C_{kt}^{*})=H(C_{t})$ meaning that $y^{t}$ connects all agents to just one facility things are quite easy, at least intuitively. The real difficulty arises when $C_{1t}\neq\emptyset$ and $C_{2t}\neq\emptyset$, where $\sum_{k=1}^{2}H(C_{kt}^{*})$ can be arbitrarily smaller than $H(C_{t})$. As we will see in this case $x^{t}$ gets closer to $y^{t}$ and the term $\sum_{k=1}^{2}\left(|x_{k}^{t}-y_{k}^{t}|-|z_{k}-y_{k}^{t}|\right)$ becomes negative.

We start with the most challenging case, where $C_{1t}^{*}\neq\emptyset$ and $C_{2t}^{*}\neq\emptyset$. We remind that our goal is showing that $x^{t}$ gets closer to $y^{t}$. Since $C_{2t}^{*}\neq\emptyset$ and $y_{2}^{t}\in M_{C_{2t}^{*}}$ we get that $y_{2}^{t}\leq a_{n}^{t}$ and as a result $|x_{2}^{t}-y_{2}^{t}|-|z_{2}-y_{2}^{t}|=|x_{2}^{t}-z_{2}|-3H(C_{t}).$

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Now, assume that $C_{1t}^{*}=\emptyset$ or $C_{2t}^{*}=\emptyset$ meaning that $\sum_{k=1}^{2}H(C_{kt}^{*})=H(C_{t})$. As a result, bounding everything by $H(C_{t})$ serves our purpose. More formally,

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The forth inequality follows from the fact that $\sum_{k=1}^{2}H(C_{kt})+|x_{k}^{t}-z_{k}|\leq 5H(C_{t})$.

We now need to treat the second case where $a_{1}\leq z_{1}\leq a_{n}$ and $z_{2}-a_{n}<3H(C_{t})$. Since Algorithm 2 computes the optimal clustering $(C_{1t},C_{2t})$ and puts $x_{1}^{t}$ in the interval $M_{C_{1t}}$ and $x_{2}^{t}$ in the interval $M_{C_{2t}}$, we are ensured that the connection cost of our solution is less than the connection cost of $y^{t}$, $\sum_{k=1}^{2}H(C_{kt})\leq\sum_{k=1}^{2}H(C_{kt}^{*})$, so we are mostly concerned in bounding $\sum_{k=1}^{2}|x_{k}^{t}-z_{k}|$.

The easy case is when $\sum_{k=1}^{2}H(C_{kt}^{*})=H(C_{t})$. A small difference with the previous case is that we don’t know how $|x_{2}^{t}-z_{2}|$ is. However, $z_{1},x_{1}^{t},x_{2}^{t}\in[a_{1}^{t},\ldots a_{n}^{t}]$ and $|x_{2}^{t}-z_{2}|=|x_{2}^{t}-a_{n}^{t}|+|a_{n}^{t}-z_{2}|$. Thus, $|x_{1}^{t}-z_{1}|+|x_{2}^{t}-a_{n}^{t}|\leq H(C_{t})$, $|a_{n}^{t}-z_{2}|\leq 3H(C_{t})$ and therefore $\sum_{k=1}^{2}[H(C_{kt})+|x_{k}^{t}-z_{k}|]\leq 5H(C_{t})$. So we can again bound everything by $H(C_{t}).$

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Things become more complicated, when the connection cost $\sum_{k=1}^{2}H(C_{kt}^{*})$ is relatively small ($C_{1t}^{*}\neq\emptyset$ and $C_{2t}^{*}\neq\emptyset$), where bounding everything by $H(C_{t})$ does not work. However, the solutions $x^{t}$ and $y^{t}$ will be relatively close in this case. More formally,

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We need to upper bound the distance $\sum_{k=1}^{2}|x_{k}^{t}-y_{k}^{t}|$. Observe that in the solution $x^{t}$, the agent at position $a_{1}^{t}$ connects to the left facility (facility 1) and the agent at position $a_{n}^{t}$ connects to the right facility (facility 2), $|x_{1}^{t}-a_{1}^{t}|+|x_{2}^{t}-a_{n}^{t}|\leq\sum_{k=1}^{2}H(C_{kt}).$ Since $C_{1t}^{*}\neq\emptyset$ and $C_{2t}^{*}\neq\emptyset$, the same holds for the solution $y^{t}$. As a result,

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$\square$