Multiple sink location problem in path networks with a combinational objective

Abstract

In this paper, we consider the k-sink location problem in a path network with the goal of optimizing a combinational function of the maximum completion time and the total completion time. Let $P=\left( V,E\right)$ be an undirected path network with n vertices. Each vertex has a positive weight, indicating the initial amount of supplies, and each edge has a positive length and a uniform capacity, which is the maximum amount of supplies that can enter the edge per unit time. Our goal is to identify k sink locations on the path P so that all supplies will be successfully evacuated and the given objective function is optimized. This paper presents two efficient polynomial time algorithms, which achieve $O\left( n \right)$ for $k=1$ and $O\left( n^6 \right)$ for general k, respectively.

Appendices

Proof of Lemma 2

Proof

Compute $f_1^L\left( v_i\right)$ in increasing order of i. By definition, $f_1^L\left( v_1\right) =0$. According to Eq. (2), $f_1^L\left( v_2\right) =\left( v_2-v_1\right) \tau + w_1/c$. For $2 < i \le n$, we have

$$\begin{aligned} f_1^L\left( v_i\right) =\max \left\{ f_1^L\left( v_{i-1}\right) +\left( v_i - v_{i-1} \right) \tau , ~~\left( v_i - v_{i-1} \right) \tau + \frac{\sum _{1 \le j \le i-1} w_j}{c} \right\} . \end{aligned}$$

Therefore, $f_1^L\left( v_i\right)$ for all $v_i$ can be computed in $O\left( n\right)$ time. Because of the symmetry of $f_1^L\left( v_i\right)$ and $f_1^R\left( v_i\right)$, $f_1^R\left( v_i\right)$ for all $v_i$ can also be computed similarly in reverse order of i. The calculation is as follows:

$$\begin{aligned} \begin{aligned}&f_1^R\left( v_n\right) = 0, \\&f_1^R\left( v_{n-1}\right) = \left( v_n-v_{n-1}\right) \tau + w_n/c, \\&f_1^R\left( v_i\right) = \max \left\{ f_1^R\left( v_{i+1}\right) +\left( v_{i+1} - v_{i} \right) \tau , ~~\left( v_{i+1} - v_{i} \right) \tau + \frac{\sum _{i+1 \le j \le n} w_j}{c} \right\} ,\\&(1 \le i < n).\\ \end{aligned} \end{aligned}$$

$\square$

Proof of Lemma 3

Proof

We focus on the calculation of $f_2^L \left( v_i\right)$, without loss of generality, because $f_2^R \left( v_i\right)$ can be computed in a similar manner. Compute $f_2^L\left( v_i\right)$ in increasing order of i. By definition, $f_2^L\left( v_1\right) =0$. According to Eq. (4), $f_2^L\left( v_2\right) =\left( v_2-v_1\right) w_1 \tau + {w_1}^2/\left( 2c\right)$. Below, we show how to compute $f_2^L\left( v_{i+1}\right)$ in constant time, given $f_2^L\left( v_{j}\right)$, $j \le i$.

Different sink points usually mean different congestion situations, that is, different $P^\prime$. For convenience, let $P_i^\prime$ be the new path, where $v_i$ is the sink point. Therefore, we need to obtain n new paths. We show how to transform P to $P^\prime$ in $O\left( n\right)$ time. The transformation of the left side of the sink point $v_i$ is taken as an example, and the right side can be treated in the same manner. In the following, $P_i^\prime$ is always used to mean the left side of $P_i^\prime$. To determine whether a vertex $v_{i-1}$ will congest at vertex $v_{i}$, we actually need to know whether $\left( v_{i}-v_{i-1} \right) \tau$ is greater than $w_{i}/c$. $\left( v_{i}-v_{i-1} \right) \tau \ge w_{i}/c$ means that $v_{i-1}$ does not need to wait at $v_{i}$. In this case, the two vertices can keep their respective weights. $\left( v_{i}-v_{i-1} \right) \tau < w_{i}/c$ means that $v_{i-1}$ should wait at $v_{i}$, that is, congestion occurs. In this case, $v_{i-1}$ and $v_{i}$ are merged into a single new vertex, whose weight is $\left( w_i+w_{i-1}\right)$. If the sink point is $v_i$, $v_{i-1}$ will never congest at $v_i$. In $O\left( n\right)$ time, we can determine whether $v_{i-1}$ will congest at $v_{i}$, for all $2\le i \le n$.

Without loss of generality, suppose that

$$\begin{aligned} f_2^L\left( v_{i}\right) =\sum _{1\le j \le i-1} \left[ \tau \left( v_i-v_j\right) w_j+ \left. {{w_j}^2 } \big / {2c} \right. \right] . \end{aligned}$$

If $v_{i-1}$ will not congest at $v_i$, we have the following recursive formula:

$$\begin{aligned} \begin{aligned} f_2^L\left( v_{i+1}\right)&=\sum _{1\le j \le i} \left[ \tau \left( v_{i+1}-v_j\right) w_j + \frac{\left( w_{j}\right) ^2}{2c}\right] \\&= f_2^L\left( v_{i}\right) + \tau \left( v_{i+1} -v_{i}\right) \cdot \left( \sum _{1\le j \le i} w_j\right) + \frac{\left( w_{i}\right) ^2}{2c}. \end{aligned} \end{aligned}$$

Otherwise, if $v_{i-1}$ will congest at $v_{i}$, then $v_{i-1}$ and $v_i$ should be merged as a new vertex. In this case, we also have the following recursive formula,

$$\begin{aligned} \begin{aligned} f_2^L\left( v_{i+1}\right)&=\sum _{1\le j \le {i-2}} \left[ \tau \left( v_{i+1}-v_j\right) w_j + \frac{\left( w_{j}\right) ^2}{2c}\right] \\&\quad +\tau \left( v_{i+1}-v_i\right) \left( w_{i-1}+w_{i}\right) + \frac{\left( w_{i-1}+w_{i}\right) ^2 }{2c}\\&= f_2^L\left( v_{i}\right) + \tau \left( v_{i+1} -v_{i}\right) \cdot \left( \sum _{1\le j \le i} w_j\right) + \frac{\left( w_{i-1}+w_{i }\right) ^2}{2c} \\&\quad -\tau \left( v_{i}-v_{i-1}\right) w_{i-1} - \frac{\left( w_{i-1}\right) ^2}{2c}. \end{aligned} \end{aligned}$$

Therefore, we can obtain all $f_2^L\left( v_i\right)$ within $O\left( n\right)$ time. The lemma follows. $\square$