Consider the many shared resource scheduling problem where jobs have to be scheduled on identical parallel machines with the goal of minimizing the makespan. However, each job needs exactly one additional shared resource in order to be executed and hence prevents the execution of jobs that need the same resource while being processed. Previously a $(2m/(m+1))$-approximation was the best known result for this problem. Furthermore, a $6/5$-approximation for the case with only two machines was known as well as a PTAS for the case with a constant number of machines. We present a simple and fast 5/3-approximation and a much more involved but still reasonable 1.5-approximation. Furthermore, we provide a PTAS for the case with only a constant number of machines, which is arguably simpler and faster than the previously known one, as well as a PTAS with resource augmentation for the general case. The approximation schemes make use of the N-fold integer programming machinery, which has found more and more applications in the field of scheduling recently. It is plausible that the latter results can be improved and extended to more general cases. Lastly, we give a $5/4 - \varepsilon$ inapproximability result for the natural problem extension where each job may need up to a constant number (in particular $3$) of different resources.
翻译:考虑许多共享资源调度问题,即工作必须安排在相同的平行机器上,以尽量减少混凝土;然而,每个工作都需要完全增加一个共享资源,以便执行,从而防止执行在处理过程中需要同样资源的工作。以前,美元(200万美元/(m+1)美元)的调整是这一问题最已知的结果。此外,只使用两台机器的案件需要6/5美元,而使用固定数量的机器的案件需要PTAS。我们提出了一个简单快捷的5/3比和更多参与但仍然合理的1.5比共享资源的资源。此外,我们只为案件提供固定数量的机器的PTAS,这比以前已知的机器简单和更快,以及增加一般案件资源的一个PTAS。近似的做法是使用N倍整齐的编程机制,最近,在排程领域发现越来越多的应用程序。我们有理由认为,后一种结果可以改进,而后一种结果可以扩大到更普通的1美元。最后,我们提供一个固定数量的5美元,一个固定的3美元。