Worst-case analysis of the LPT algorithm for single processor scheduling with time restrictions

We consider the following scheduling problem. We are given a set S of jobs which are to be scheduled sequentially on a single processor. Each job has an associated processing time which is required for its processing. Given a particular permutation of the jobs in S, the jobs are processed in that order with each job started as soon as possible, subject only to the following constraint: For a fixed integer B≥2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$B \ge 2$$\end{document}, no unit time interval [x,x+1)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$[x, x+1)$$\end{document} is allowed to intersect more than B jobs for any real x. There are several real world situations for which this restriction is natural. For example, suppose in addition to the jobs being executed sequentially on a single main processor, each job also requires the use of one of B identical subprocessors during its execution. Each time a job is completed, the subprocessor it was using requires one unit of time to reset itself. In this way, it is never possible for more than B jobs to be worked on during any unit interval. In Braun et al. (J Sched 17: 399–403, 2014a) it is shown that this problem is NP-hard when the value B is variable and a classical worst-case analysis of List Scheduling for this situation has been carried out. We prove a tighter bound for List Scheduling for B≥3\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$B\ge 3$$\end{document} and we analyze the worst-case behavior of the makespan τLPT(S)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\tau _\mathrm{LPT}(S)$$\end{document} of LPT (longest processing time first) schedules (where we rearrange the set S of jobs into non-increasing order) in relation to the makespan τo(S)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\tau _o(S)$$\end{document} of optimal schedules. We show that LPT ordered jobs can be processed within a factor of 2-2/B\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$2-2/B$$\end{document} of the optimum (plus 1) and that this factor is best possible.