On the hardness of the classical job shop problem

In a classical job shop problem, n jobs have to be processed onm machines, where the machine orders of the jobs are given. Computationalexperiments show that there are huge differences in the hardness of the job shop problem tominimize makespan depending on the given machine orders. We study a partial order“\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document} $$\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{ \succ } $$ \end{document}” on the set of sequences, i.e., feasiblecombinations of job orders and machine orders, with the property thatB\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document} $$\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{ \succ } $$ \end{document}B implies thatthe makespan of the semiactive schedule corresponding to sequence B isless than or equal to the makespan of any schedule corresponding to B.The minimal sequences according to this partial order are called irreducible.We present a polynomial algorithm to decide whether B\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document} $$\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{ \succ } $$ \end{document}B holds andwe develop a new enumeration algorithm for irreducible sequences. To explain the differencesin the hardness of job shop problems, we study the relation between the hardness of a jobshop problem and the number of irreducible sequences corresponding to the given machine orders.