Online Bottleneck Matching Problem with Two Heterogeneous Sensors in a Metric Space

Abstract

In this paper, we consider the online matching problem with two heterogeneous sensors $s_1$ and $s_2$ in a metric space $(X,d)$. If a request r is assigned to sensor $s_1$, the service cost of r is the distance $d(r,s_1)$. Otherwise, r is assigned to sensor $s_2$, and the service cost of r is $\frac{d(r,s_2)}{w}$, where $w\ge 1$ is the weight of sensor $s_2$. The goal is to minimize the maximum matching cost, we design an optimal online algorithm with a competitive ratio of $1+w+\frac{1}{w}$ for $1\le w\le 1.839$, and an optimal online algorithm with a competitive ratio of $\frac{w+1+\sqrt{w^2+6w+1}}{2}$ for $w>1.839$.

1. Introduction

In a metric space $(X,d)$, X is a set of points and $d(·,·)$ is a distance function. $S=\{s_1,s_2,\dots ,s_n\}\subseteq$ is a set of sensors, and $R=\{r_1,r_2,\dots ,r_n\}\subseteq X$ is a set of requests. Each request arrives one by one in an online fashion. When a request $r_j\in R$ arrives, it must be immediately and irrevocably matched to some unmatched sensor $s_i$. The cost of matching $r_j$ to $s_i$ is the distance $d(r_j,s_i)$.

For a minimization problem and an input instance I, let $C^A\left(I\right)$ ($C^A$ for short) and $C^{OPT}\left(I\right)$ ($C^{OPT}$ for short) be the costs of the feasible solution obtained by an online algorithm A and an optimal off-line algorithm, respectively. An online algorithm A is $\rho$-competitive (or the competitive ratio of A is at most $\rho$) if $C^A\le \rho C_{OPT}$ for any input instance I. For an online problem, if there is no algorithm with competitive ratio less than r, then r is a lower bound of the problem. If there is an algorithm whose competitive ratio matches the lower bound, this algorithm is called an optimal online algorithm.

Online problem and algorithm can be abstracted as a request-answer game between the algorithm designer and adversary. Each time the adversary gives a request, the algorithm decides how to respond based on the previous responses of both parties and the current request, and the adversary gives the next request based on the responses of both parties so far. The goal of the algorithm is to make it perform as well as possible, while the goal of the adversary is exactly the opposite.

Online-matching based models have many applications, e.g., when assigning cars to parking spots, advertisers to ad slots, skis to skiers at a rental station, drivers and riders in Uber. The classical Online Minimum Matching (OMM) is to find a matching M such that the total cost of matching all requests is minimized. Kalyanasundaram and Pruhs [1] and Khuller et al. [2] independently introduced the OMM problem and proved that the Permutation is an optimal online algorithm with a competitive ratio of $2m-1$. Meanwhile, Kalyanasundaram and Pruhs [1] also proved that the greedy algorithm is $(2^m-1)$-competitive, and this bound is tight. Meyerson et al. [3] presented a randomized online algorithm with competitive ratio of $O\left({log}^{3}m\right)$. Bansal et al. [4] proposed an $O\left({log}^{2}m\right)$-competitive randomized algorithm.

If the metric space $(X,d)$ is a line, Gupta and Lewi [5] gave an $O(logn)$-competitive randomized algorithm for the OMM problem. Fuchs et al. [6] gave a lower bound 9.001 for the OMM problem. Antoniadis et al. [7] designed a deterministic online algorithm with competitive ratio $O(n^{log(3+ϵ)-1}/ϵ)$ for any $ϵ>0$. Nayyar and Raghvendra [8] proved that the competitive ratio of the deterministic online algorithm proposed in [9] is $O\left({log}^{2}n\right)$, which is improved to $O(logn)$ [10] for the OMM problem. Recently, Peserico and Scquizzato [11] proved that the competitive ratio of any randomized online algorithm for the OMM problem exceeds $\sqrt{{log}_2(n+1)}/15$.

The OMM problem is closely related to the Online Bottleneck Matching (OBM) problem which is to find a matching M such that the maximum cost is minimized. Kalyanasundaram and Pruhs [1] also introduced the OBM problem and proved that the Permutation algorithm is $(2m-1)$-competitive, and gave a lower bound of $m+1$. Idury and Schaffer [12] gave a lower bound approximately $1.44m$ for the OBM problem even if the metric space is a real line. Anthony and Chung [13] proved that the greedy algorithm achieves a competitive ratio of $m{2}^{m-1}$.

A generalized version of the OMM problem, which is called online b-matching [1], online transportation [1], the fire station problem [14], the school assignment problem [14], or online facility assignment [15], is also considered, where each sensor can be matched multiple times. Recently, Itoh et al. [16] presented several lower bounds on the competitive ratio for this problem with different number of sensors. Xiao and Li [17] considered the semi-matching problem with two heterogeneous sensors, and proposed two optimal online algorithms.

In this paper, we describe our problems and some preliminaries in Section 2. In Section 3, we consider an OBM problem with two heterogeneous sensors. We gives two optimal online algorithms, and the competitive ratios are shown in Figure 1. Finally, we make a summary in Section 4.

2. Preliminaries

Let $(X,d)$ be a metric space, where X is a set of points and $d(·,·)$ is a distance function. We are given a set of two heterogeneous sensors $S=\{s_1,s_2\}$ and two requests $r_1$, $r_2$ in X, where each sensor $s_i$ is characterized by the position $p\left(s_i\right)\in X$ and the weight $w_i>0$, $i=1,2$. A request $r_j$ is also characterized by the position $p\left(r_j\right)\in X$, $j=1,2$.

In the online fashion, the positions of the sensors are known in advance. The requests arrive one-by-one and the request $r_1$ must be assigned to some unmatched sensor before the request $r_2$ arrives. A sensor $s_i$ is called available if the $s_i$ is not matched by one request. If $r_j$ is matched with the sensor $s_i$, the cost of pair $(r_j,s_i)$ is

$$\begin{array}{c}\hfill c(r_j,s_i)=d(r_j,s_i)/w_i\end{array}$$

For convenience, let ${\sigma }^{-1}\left(i\right)$ be the request matched by sensor $s_i$ for $i=1,2$. Without loss of generality, we assume that $w_1=1$ and $w_2=w\ge 1$.

The online bottleneck matching problem with two heterogeneous sensors ($OBM\left(2\right)$) is to find a matching $\sigma$ such that the maximum cost

$$\begin{array}{c}\hfill max\{\underset{j:r_j={\sigma }^{-1}\left(1\right)}{max}d(r_j,s_1),\underset{j:r_j={\sigma }^{-1}\left(2\right)}{max}d(r_j,s_2)/w\}\end{array}$$

is minimized. Clearly, if $w=1$, this problem is exactly the problem considered in [18] and has an optimal online algorithm with competitive ratio 3.

3. The OBM(2) Problem

In this section, for different ranges of w, we use the adversary approach to obtain the worst case competitive ratios of the OBM(2) problem. Let $\beta \approx 1.839$ be the real root of the equation $x^3-x^2-x-1=0$, implying that

$$\begin{array}{c}\hfill {\beta }^3-{\beta }^2-\beta -1=0.\end{array}$$

We propose two optimal online algorithms for $1<w\le \beta$ and $w>\beta$ in next two subsections respectively.

3.1. The Case $1\le w\le \beta$

Theorem 1.

When $1\le w\le \beta$, any online algorithm A for OBM(2) has a competitive ratio at least $1+w+\frac{1}{w}$, even if $(X,d)$ is a line.

Proof. 

As shown in Figure 2, let $p\left(s_1\right)=0$, $p\left(s_2\right)=1$, and $x=\frac{1}{1+w}$. The first request arrives at position $p\left(r_1\right)=p\left(s_2\right)-x$. If $r_1$ is matched with sensor $s_2$, the last request $r_2$ arrives at position $p\left(r_2\right)=p\left(s_2\right)+(1-x)w$. Therefore, $C^A\ge d(r_2,s_1)=1+(1-x)w$, and $C^{OPT}=d(r_1,s_1)=\frac{d(r_2,s_2)}{w}=(1-x)$, implying that

$$\begin{array}{c}\hfill \frac{C^A}{C^{OPT}}\ge \frac{1+(1-x)w}{1-x}=\frac{1}{1-x}+w=1+w+\frac{1}{w}.\end{array}$$

If $r_1$ is matched with sensor $s_1$, the last request $r_2$ arrives at position $p\left(r_2\right)=p\left(s_1\right)-\frac{x}{w}$. Therefore, $C^A\ge \frac{d(r_2,s_2)}{w}=\frac{1}{w}+\frac{x}{w^2}$, $C^{OPT}=\frac{d(r_1,s_2)}{w}=d(r_2,s_1)=\frac{x}{w}$, implying that

$$\begin{array}{c}\hfill \frac{C^A}{C^{OPT}}\ge \frac{\frac{1}{w}+\frac{x}{w^2}}{\frac{x}{w}}=\frac{1}{x}+\frac{1}{w}=1+w+\frac{1}{w}.\end{array}$$

According to the above analysis, there is no online algorithm which competitive ratio strict smaller than $1+w+\frac{1}{w}$.    □

When $1\le w\le \beta$, our Algorithm 1 is described as follows.

Algorithm 1:A1

Theorem 2.

The competitive ratio of Algorithm 1 is at most $1+w+\frac{1}{w}$.

Proof. 

Let ${\sigma }^{*}$ be the off-line optimal solution, and $\sigma$ be the feasible solution produced by the Algorithm 1. Let $r_j$ be the request attaching the maximum in the $\sigma$, and $r_k$ be the request attaching the maximum in the ${\sigma }^{*}$, $j,k\in \{1,2\}$. For $j=1,2$, if ${\sigma }^{*}\left(r_j\right)=\sigma \left(r_j\right)$, implying that Algorithm 1 produces an optimal solution. Otherwise,

$$\begin{array}{c}\hfill {\sigma }^{*}\left(r_j\right)\ne \sigma \left(r_j\right),\forall j=1,2,\end{array}$$

(1)

we distinguish the following four cases.

Case 1.$\sigma \left(r_j\right)=s_1$ and ${\sigma }^{*}\left(r_k\right)=s_1$ ($\sigma \left(r_k\right)=s_2$ and ${\sigma }^{*}\left(r_j\right)=s_2)$.

In this case, we have $C^{A1}=d(r_j,s_1)$ and $C^{OPT}=d(r_k,s_1)$. By (1), we have $j\ne k$. If $j<k$, by the choice of Algorithm 1, we have $d(r_j,s_1)\le wd(r_j,s_2)$. Since $w\le \beta$, we have

$$\begin{array}{c}\hfill w^2\le 1+w+\frac{1}{w},\end{array}$$

(2)

implying that

$$\begin{array}{c}\hfill C^{A1}=d(r_j,s_1)\le wd(r_j,s_2)=wd(r_j,{\sigma }^{*}\left(r_j\right))\le w^2{C}^{OPT}\le (1+w+\frac{1}{w})C^{OPT}.\end{array}$$

If $j>k$, by the choice of Algorithm 1, we have $d(r_k,s_1)>wd(r_k,s_2)$. Hence,

$$\begin{array}{ccc}\hfill C^{A1}& =& d(r_j,s_1)\hfill \\ & \le & d(r_j,s_2)+d(s_1,s_2)\hfill \\ & \le & d(r_j,s_2)+d(r_k,s_1)+d(r_k,s_2)\hfill \\ & \le & d(r_j,{\sigma }^{*}\left(r_j\right))+(1+\frac{1}{w})d(r_k,s_1)\hfill \\ & =& d(r_j,{\sigma }^{*}\left(r_j\right))+(1+\frac{1}{w})d(r_k,{\sigma }^{*}\left(r_k\right))\hfill \\ & \le & w{C}^{OPT}+(1+\frac{1}{w})C^{OPT}\hfill \\ & =& (1+w+\frac{1}{w})C^{OPT}.\hfill \end{array}$$

Case 2.$\sigma \left(r_j\right)=s_1$ and ${\sigma }^{*}\left(r_k\right)=s_2$.

In this case, we have $C^{A1}=d(r_j,s_1)$ and $C^{OPT}=\frac{d(r_k,{\sigma }^{*}\left(r_k\right))}{w}=\frac{d(r_k,s_2)}{w}$. By (1) and $j,k=1,2$, we have $j=k$.

If $j=1$, by the choice of Algorithm 1, we have $d(r_1,s_1)\le wd(r_1,s_2)$. Therefore, by (2),

$$\begin{array}{c}\hfill C^{A1}=d(r_j,s_1)\le wd(r_j,s_2)=w^2{C}^{OPT}\le (1+w+\frac{1}{w})C^{OPT}.\end{array}$$

If $j=2$, by the choice of Algorithm 1, we have $d(r_1,s_1)>wd(r_1,s_2)$. Hence,

$$\begin{array}{ccc}\hfill C^{A1}& =& d(r_2,s_1)\hfill \\ & \le & d(r_2,s_2)+d(s_1,s_2)\hfill \\ & \le & d(r_2,s_2)+d(r_1,s_1)+d(r_1,s_2)\hfill \\ & \le & d(r_2,{\sigma }^{*}\left(r_2\right))+(1+\frac{1}{w})d(r_1,s_1)\hfill \\ & =& d(r_2,{\sigma }^{*}\left(r_2\right))+(1+\frac{1}{w})d(r_1,{\sigma }^{*}\left(r_1\right))\hfill \\ & \le & w{C}^{OPT}+(1+\frac{1}{w})C^{OPT}\hfill \\ & =& (1+w+\frac{1}{w})C^{OPT}.\hfill \end{array}$$

Case 3.$\sigma \left(r_j\right)=s_2$ and ${\sigma }^{*}\left(r_k\right)=s_2$ ($\sigma \left(r_k\right)=s_1$ and ${\sigma }^{*}\left(r_j\right)=s_1)$.

In this case, we have $C^{A1}=\frac{d(r_j,s_2)}{w}$ and $C^{OPT}=\frac{d(r_k,s_2)}{w}=\frac{d(r_k,{\sigma }^{*}\left(r_k\right))}{w}$. By (1), we have $j\ne k$. If $j<k$, by the choice of Algorithm 1, we have $d(r_j,s_1)>wd(r_j,s_2)$. Hence,

$$\begin{array}{c}\hfill C^{A1}=\frac{d(r_j,s_2)}{w}<\frac{d(r_j,s_1)}{w^2}=\frac{d(r_j,{\sigma }^{*}\left(r_j\right))}{w^2}\le \frac{1}{w^2}\frac{d(r_k,{\sigma }^{*}\left(r_k\right))}{w}\le \frac{1}{w^2}{C}^{OPT}\le C^{OPT}.\end{array}$$

If $j>k$, by the choice of Algorithm 1, we have $d(r_k,s_1)\le wd(r_k,s_2)$. Hence,

$$\begin{array}{ccc}\hfill C^{A1}& =& \frac{d(r_j,s_2)}{w}\hfill \\ & \le & \frac{1}{w}(d(r_j,s_1)+d(s_1,s_2))\hfill \\ & \le & \frac{1}{w}(d(r_j,s_1)+d(r_k,s_1)+d(r_k,s_2))\hfill \\ & \le & \frac{1}{w}(d(r_j,{\sigma }^{*}\left(r_j\right))+(1+w)d(r_k,s_2))\hfill \\ & =& \frac{1}{w}(d(r_j,{\sigma }^{*}\left(r_j\right))+(1+w)d(r_k,{\sigma }^{*}\left(r_k\right)))\hfill \\ & \le & \frac{1}{w}(C^{OPT}+(1+w)w{C}^{OPT})\hfill \\ & =& (1+w+\frac{1}{w})C^{OPT}.\hfill \end{array}$$

Case 4.$\sigma \left(r_j\right)=s_2$ and ${\sigma }^{*}\left(r_k\right)=s_1$.

In this case, we have $C^{A1}=\frac{d(r_j,s_2)}{w}$ and $C^{OPT}=d(r_k,s_1)=d(r_k,{\sigma }^{*}\left(r_k\right))$. By (1), we have $j=k$.

If $j=1$, by the choice of Algorithm 1, we have $d(r_j,s_1)>wd(r_j,s_2)$. Therefore,

$$\begin{array}{c}\hfill C^{A1}=\frac{d(r_j,s_2)}{w}<\frac{d(r_j,s_1)}{w^2}=\frac{d(r_j,{\sigma }^{*}\left(r_j\right))}{w^2}\le \frac{1}{w^2}{C}^{OPT}\le C^{OPT}.\end{array}$$

If $j=2$, by the choice of Algorithm 1, we have $d(r_1,s_1)\le wd(r_1,s_2)$. Hence,

$$\begin{array}{ccc}\hfill C^{A1}& =& \frac{d(r_j,s_2)}{w}\hfill \\ & \le & \frac{1}{w}(d(r_j,s_1)+d(s_1,s_2))\hfill \\ & \le & \frac{1}{w}(d(r_j,s_1)+d(r_1,s_1)+d(r_1,s_2))\hfill \\ & \le & \frac{1}{w}(d(r_j,s_1)+(1+w)d(r_1,s_2))\hfill \\ & =& \frac{1}{w}(d(r_j,{\sigma }^{*}\left(r_j\right))+(1+w)d(r_1,{\sigma }^{*}\left(r_1\right)))\hfill \\ & \le & \frac{1}{w}(C^{OPT}+(1+w)w{C}^{OPT})\hfill \\ & =& (1+w+\frac{1}{w})C^{OPT}.\hfill \end{array}$$

Therefore, the theorem holds.    □

3.2. The Case $w>\beta$

Theorem 3.

When $w>\beta$, any online algorithm A for OBM(2) has a competitive ratio at least $\frac{w+1+\sqrt{w^2+6w+1}}{2}$, even if $(X,d)$ is a line.

Proof. 

Let $x=\frac{3w+1-\sqrt{w^2+6w+1}}{4w}$. As shown in Figure 3, assume that $p\left(s_1\right)=0$ and $p\left(s_2\right)=1$. The first request arrives at position $p\left(r_1\right)=p\left(s_2\right)-x$. If $r_1$ is matched with sensor $s_2$, the last request $r_2$ arrives at position $p\left(r_2\right)=p\left(s_2\right)+(1-x)w$, implying $C^A\ge d(r_2,s_1)=1+(1-x)w$ and $C^{OPT}=d(r_1,s_1)=\frac{d(r_2,s_2)}{w}=1-x$. Thus,

$$\begin{array}{c}\hfill \frac{C^A}{C^{OPT}}\ge \frac{1+(1-x)w}{1-x}=\frac{1}{1-x}+w=\frac{w+1+\sqrt{w^2+6w+1}}{2}.\end{array}$$

If $r_1$ is matched with sensor $s_1$, the last request $r_2$ arrives at position $p\left(r_2\right)=p\left(s_1\right)-\frac{x}{w}$, implying that $C^A\ge d(r_1,s_1)=1-x$, and $C^{OPT}=\frac{d(r_1,s_2)}{w}=d(r_2,s_1)=\frac{x}{w}$. Thus,

$$\begin{array}{c}\hfill \frac{C^A}{C^{OPT}}\ge \frac{1-x}{\frac{x}{w}}=\frac{w+1+\sqrt{w^2+6w+1}}{2}.\end{array}$$

Therefore, the theorem holds.    □

When $w>\beta$, for convenience, let

$$\begin{array}{c}\hfill w_0=\frac{w+1+\sqrt{w^2+6w+1}}{2w}.\end{array}$$

Our Algorithm 2 is described as follows.

Algorithm 2:A2

Theorem 4.

The competitive ratio of Algorithm 2 is at most $\frac{w+1+\sqrt{w^2+6w+1}}{2}$.

Proof. 

As in the proof of Theorem 2, let $r_j$ be the request attaching the maximum in the feasible solution $\sigma$ produced by the Algorithm 2, and $r_k$ be the request attaching the maximum in the optimal solution ${\sigma }^{*}$, $j,k=1,2.$ We assume that

$$\begin{array}{c}\hfill \sigma \left(r_j\right)\ne {\sigma }^{*}\left(r_j\right),\forall j=1,2,\end{array}$$

(3)

and distinguish the following four cases.

Case 1.$\sigma \left(r_j\right)=s_1$ and ${\sigma }^{*}\left(r_k\right)=s_1$ ($\sigma \left(r_k\right)=s_2$ and ${\sigma }^{*}\left(r_j\right)=s_2)$.

In this case, we have $C^{A2}=d(r_j,s_1)$ and $C^{OPT}=d(r_k,{\sigma }^{*}\left(r_k\right))=d(r_k,s_1)$. By (3), we have $j\ne k$. If $j<k$, by the choice of Algorithm 2, we have $d(r_j,s_1)\le w_{0}d(r_j,s_2)$. Therefore,

$$\begin{array}{c}\hfill C^{A2}=d(r_j,s_1)\le w_{0}d(r_j,s_2)=w_{0}d(r_j,{\sigma }^{*}\left(r_j\right))\le w_{0}w{C}^{OPT}=\frac{w+1+\sqrt{w^2+6w+1}}{2}{C}^{OPT}.\end{array}$$

If $j>k$, by the choice of Algorithm 2, we have $d(r_k,s_1)>w_{0}d(r_k,s_2)$. Hence,

$$\begin{array}{ccc}\hfill C^{A2}& =& d(r_j,s_1)\hfill \\ & \le & d(r_j,s_2)+d(s_1,s_2)\hfill \\ & \le & d(r_j,s_2)+d(r_k,s_1)+d(r_k,s_2)\hfill \\ & \le & d(r_j,{\sigma }^{*}\left(r_j\right))+(1+\frac{1}{w_0})d(r_k,s_1)\hfill \\ & =& d(r_j,{\sigma }^{*}\left(r_j\right))+(1+\frac{1}{w_0})d(r_k,{\sigma }^{*}\left(r_k\right))\hfill \\ & \le & w{C}^{OPT}+(1+\frac{1}{w_0})C^{OPT}\hfill \\ & =& (1+w+\frac{1}{w_0})C^{OPT}\hfill \\ & =& \frac{w+1+\sqrt{w^2+6w+1}}{2}{C}^{OPT}.\hfill \end{array}$$

Case 2.$\sigma \left(r_j\right)=s_1$ and ${\sigma }^{*}\left(r_k\right)=s_2$.

In this case, we have $C^{A2}=d(r_j,s_1)$ and $C^{OPT}=\frac{d(r_k,{\sigma }^{*}\left(r_k\right))}{w}=\frac{d(r_k,s_2)}{w}$. By (3) and $j,k=1,2$, we have $j=k$. If $j=1$, by the choice of Algorithm 2, we have $d(r_j,s_1)\le w_{0}d(r_j,s_2)$. Therefore,

$$\begin{array}{c}\hfill C^{A2}=d(r_j,s_1)\le w_{0}d(r_j,s_2)=w_{0}w{C}^{OPT}=\frac{w+1+\sqrt{w^2+6w+1}}{2}{C}^{OPT}.\end{array}$$

If $j=2$, by the choice of Algorithm 2, we have $d(r_1,s_1)>w_{0}d(r_1,s_2)$. Hence,

$$\begin{array}{ccc}\hfill C^{A2}& =& d(r_j,s_1)\hfill \\ & \le & d(r_j,s_2)+d(s_1,s_2)\hfill \\ & \le & d(r_j,s_2)+d(r_1,s_1)+d(r_1,s_2)\hfill \\ & \le & d(r_j,{\sigma }^{*}\left(r_j\right))+(1+\frac{1}{w_0})d(r_1,s_1)\hfill \\ & =& d(r_j,{\sigma }^{*}\left(r_j\right))+(1+\frac{1}{w_0})d(r_1,{\sigma }^{*}\left(r_1\right))\hfill \\ & \le & w{C}^{OPT}+(1+\frac{1}{w_0})C^{OPT}\hfill \\ & =& (1+w+\frac{1}{w_0})C^{OPT}\hfill \\ & =& \frac{w+1+\sqrt{w^2+6w+1}}{2}{C}^{OPT}.\hfill \end{array}$$

Case 3.$\sigma \left(r_j\right)=s_2$ and ${\sigma }^{*}\left(r_k\right)=s_2$ ($\sigma \left(r_k\right)=s_1$ and ${\sigma }^{*}\left(r_j\right)=s_1)$.

In this case, we have $C^{A2}=\frac{d(r_j,s_2)}{w}$ and $C^{OPT}=\frac{d(r_k,s_2)}{w}=\frac{d(r_k,{\sigma }^{*}\left(r_k\right))}{w}$. By (3), we have $j\ne k$. If $j<k$, by the choice of Algorithm 2, we have $d(r_j,s_1)>w_{0}d(r_j,s_2)$. Hence,

$$\begin{array}{c}\hfill C^{A2}=\frac{d(r_j,s_2)}{w}<\frac{d(r_j,s_1)}{w{w}_0}=\frac{d(r_j,{\sigma }^{*}\left(r_j\right))}{w{w}_0}\le \frac{1}{w{w}_0}{C}^{OPT}\le C^{OPT}.\end{array}$$

If $j>k$, by the choice of Algorithm 2, we have $d(r_k,s_1)\le w_{0}d(r_k,s_2)$. Hence,

$$\begin{array}{ccc}\hfill C^{A2}& =& \frac{d(r_j,s_2)}{w}\hfill \\ & \le & \frac{1}{w}(d(r_j,s_1)+d(s_1,s_2))\hfill \\ & \le & \frac{1}{w}(d(r_j,s_1)+d(r_k,s_1)+d(r_k,s_2))\hfill \\ & \le & \frac{1}{w}\left(d(r_j,{\sigma }^{*}\left(r_j\right)+(1+w_0)d(r_k,s_2))\right)\hfill \\ & =& \frac{1}{w}(d(r_j,{\sigma }^{*}\left(r_j\right))+(1+w_0)d(r_k,{\sigma }^{*}\left(r_k\right)))\hfill \\ & \le & \frac{1}{w}(C^{OPT}+(1+w_0)w{C}^{OPT})\hfill \\ & =& (1+w_0+\frac{1}{w})C^{OPT}\hfill \\ & \le & \frac{w+1+\sqrt{w^2+6w+1}}{2}{C}^{OPT},\hfill \end{array}$$

where the last inequality follows from that $w>\beta$.

Case 4.$\sigma \left(r_j\right)=s_2$ and ${\sigma }^{*}\left(r_k\right)=s_1$.

In this case, we have $C^{A2}=\frac{d(r_j,s_2)}{w}$ and $C^{OPT}=d(r_k,s_1)$. By (3), we have $j=k$.

If $j=1$, by the choice of Algorithm 2, we have $d(r_j,s_1)>w_{0}d(r_j,s_2)$. Therefore,

$$\begin{array}{c}\hfill C^{A2}=\frac{d(r_j,s_2)}{w}<\frac{d(r_j,s_1)}{w{w}_0}=\frac{d(r_j,{\sigma }^{*}\left(r_j\right))}{w{w}_0}\le \frac{1}{w{w}_0}{C}^{OPT}\le C^{OPT}.\end{array}$$

If $j=2$, by the choice of Algorithm 2, we have $d(r_1,s_1)\le w_{0}d(r_1,s_2)$. Hence,

$$\begin{array}{ccc}\hfill C^{A2}& =& \frac{d(r_j,s_2)}{w}\hfill \\ & \le & \frac{1}{w}(d(r_j,s_1)+d(s_1,s_2))\hfill \\ & \le & \frac{1}{w}(d(r_j,s_1)+d(r_1,s_1)+d(r_1,s_2))\hfill \\ & \le & \frac{1}{w}(d(r_j,s_1)+(1+w_0)d(r_1,s_2))\hfill \\ & =& \frac{1}{w}(d(r_j,{\sigma }^{*}\left(r_j\right))+(1+w_0)d(r_1,{\sigma }^{*}\left(r_1\right)))\hfill \\ & \le & \frac{1}{w}(C^{OPT}+(1+w_0)w{C}^{OPT})\hfill \\ & =& (1+w_0+\frac{1}{w})C^{OPT}\hfill \\ & \le & \frac{w+1+\sqrt{w^2+6w+1}}{2}{C}^{OPT},\hfill \end{array}$$

where the last inequality follows from $w>\beta$.

Therefore, the theorem holds. □

4. Discussion

In this paper, we studied an online matching problem with two heterogeneous sensors in a metric space. For the bottleneck objective, we proposed two optimal online algorithms for different values of w. It is interesting to study the online matching problem with more than three sensors.