Routing and Scheduling Problems with Two Agents on a Line-Shaped Network

Abstract

We consider routing and scheduling problems with two agents on a line-shaped network in this paper. There are two agents and each agent has some jobs which are located in the network. Let $L =(V,E)$ be a line-shaped network, where $V =\{v_{0}\}\bigcup V^{A}\bigcup V^{B}$ is the set of $n+1$ vertices and E is a set of edges. A job v is located at some vertex v, which is also denoted as v. The travel time d(u, v) is associated with each edge $\{u,v\}\in E$. The vehicle starts from an initial vertex $v_{0}\in V$ and visits all jobs for their processing. The objective is to find a route of the vehicle that minimizes the completion time of agent A under the constraint condition that the completion time of agent B is no more than the threshold value Q. We express this problem as $line-1|C_{max}^{B}\le Q|C_{max}^{A}$. For the problem without release time, an O(n) time algorithm is provided. For the problem with release time, we show that this problem is NP-hard even though there is only one job in agent B and the jobs in agent A have no release time. Finally we give a (3, 3)-approximation algorithm for the before general problem.

This research was supported by the National Natural Science Foundation of China under Grant 11871213.

Appendix

For the problem $line-1|r(v_{j})=0,C_{max}^{B}\le Q|C_{max}^{A}$, when the rightmost endpoint belongs to agent A, we will give the following Algorithm C. This algorithm is actually similar to Algorithm A. Let $M=\{v_{1}^{A},...,v_{j}^{A}\}$ be the vertex set of agent A located before $v_{n_{B}}^{B}$ on the line-shaped network L.

$\mathbf{Algorithm} C$

Step1. According to the following candidate schedules to construct the schedule $\sigma ^{A}$.

For $h=0,1,...,j-1$,

$\sigma _{h}^{1A}$: first the vehicle processes the jobs in $V^{B}\bigcup \{v_{1}^{A},...,v_{h}^{A}\}$ from the depot $v_{0}$ to $v_{n_{B}}^{B}$. Next, the vehicle goes to $v_{n_{A}}^{A}$ without processing any job, and finally the vehicle returns to vertex $v_{h+1}^{A}$ to process the remaining jobs in $J^{A}$ one by one.

$\sigma _{h}^{2A}$: first the vehicle processes the jobs in $V^{B}\bigcup \{v_{1}^{A},...,v_{h}^{A}\}$ from the depot $v_{0}$ to $v_{n_{B}}^{B}$. Next, the vehicle goes to $v_{h+1}^{A}$ without processing any job, and finally the vehicle processes the remaining jobs in $V^{A}$ from $v_{h+1}^{A}$ to $v_{n_{A}}^{A}$ in turn.

For $h=j$,

$\sigma _{j}^{A}$: the vehicle processes all the jobs one by one $v_{0}$ to $v_{n_{A}}^{A}$.

Step2. Choose the best feasible schedule from $\{\sigma _{h}^{1A}, h=0,1,...,j-1\}$, $\{\sigma _{h}^{2A}, h=0,1,...,j-1\}$ and $\sigma _{j}^{A}$ as $\sigma ^{A}$ with the makespan no more than Q for agent B.

Step3. By the following candidate schedules to construct the schedule $\sigma ^{B}$.

For $l=0,1,...,n_{B}-1$,

$\sigma _{l}^{B}$: first the vehicle processes the jobs in $V^{A}\bigcup \{v_{1}^{B},...,v_{l}^{B}\}$ from the depot $v_{0}$ to $v_{n_{A}}^{A}$. Next, the vehicle returns to vertex $v_{l+1}^{B}$ to process the remaining jobs in $V^{B}$ in turn.

Step4. Choose the best feasible schedule from $\{\sigma _{l}^{B}, l=0,1,...,n_{B}-1\}$ as $\sigma ^{B}$ under the threshold value constraint.

Step5. If $C_{max}^A(\sigma ^{B})\le C_{max}^A(\sigma ^{A})$, select $\sigma ^{B}$ as the optimal schedule $\sigma$. Otherwise, choose $\sigma ^{A}$ as the optimal schedule $\sigma$.

According to different value ranges of Q, we can give following four cases. (1) $2L^{A}-d(v_{0},v_{1}^{B})+H\le Q$; (2)$L^{A}+d(v_{n_{A}}^{A},v_{n_{B}}^{B})+H\le Q< 2L^{A}-d(v_{0},v_{1}^{B})+H$; (3) $L^{B}+H^{B}\le Q< L^{A}+d(v_{n_{A}}^{A},v_{n_{B}}^{B})+H$; (4) $Q< L^{B}+H^{B}$. By the similar method to Theorem 2.1, we can prove that the problem $line-1|r(v_{j})=0,C_{max}^{B}\le Q|C_{max}^{A}$ is solvable in O(n) time, when the rightmost endpoint belongs to agent A.