Abstract

The Routing Open Shop Problem deals with \(n \) jobs located in the nodes of an edge-weighted graph \(G=(V,E) \) and \(m \) machines that are initially in a special node calleddepot. The machines must process all jobs in arbitrary order so that each machine processes at most one job at any one time and each job is processed by at most one machine at any one time. The goal is to minimize the makespan; i.e., the time when the last machine returns to the depot. This problem is known to be NP-hard even for the two machines and the graph containing only two nodes. In this article we consider the particular case of the problem with a \(2\)-node graph, unit processing time of each job, and unit travel time between every two nodes. The conjecture is made that the problem is polynomially solvable in this case; i.e., the makespan depends only on the number of machines and the loads of the nodes and can be calculated in time \(O(\log mn) \). We provide some new bounds on the makespan in the case of \(m = n \) depending on the loads distribution.

中文翻译：

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关于具有单位处理时间的两个节点上的路由开店问题

摘要

路由开放车间问题处理\（n \）个作业，这些作业位于边缘加权图\（G =（V，E）\）和\（m \）机器的节点中，这些机器最初位于称为仓库的特殊节点中。机器必须以任意顺序处理所有作业，以便每台机器在任何时间最多处理一个作业，并且每一次最多由一台机器处理每个作业。目标是最大程度地缩短制造时间。即最后一台机器返回仓库的时间。即使对于两台机器以及仅包含两个节点的图形，此问题也是已知的NP难题。在本文中，我们考虑带有\（2 \）的问题的特殊情况-节点图，每个作业的单位处理时间以及每两个节点之间的单位行进时间。据推测，在这种情况下该问题可以多项式解决。即，制造期仅取决于机器的数量和节点的负载，并且可以在时间\（O（\ log mn）\）中进行计算。根据负载分布，在\（m = n \）的情况下，我们为制造期提供了一些新的边界。