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Total Completion Time Scheduling Under Scenarios

Thomas Bosman, Martijn van Ee, Ekin Ergen, Csanad Imreh, Alberto Marchetti-Spaccamela, Martin Skutella, Leen Stougie

TL;DR

An almost complete picture of the evolving complexity landscape is painted, drawing the line between easy and hard in this classical problem under uncertainty, in which the uncertainty is modeled by a set of scenarios.

Abstract

Scheduling jobs with given processing times on identical parallel machines so as to minimize their total completion time is one of the most basic scheduling problems. We study interesting generalizations of this classical problem involving scenarios. In our model, a scenario is defined as a subset of a predefined and fully specified set of jobs. The aim is to find an assignment of the whole set of jobs to identical parallel machines such that the schedule, obtained for the given scenarios by simply skipping the jobs not in the scenario, optimizes a function of the total completion times over all scenarios. While the underlying scheduling problem without scenarios can be solved efficiently by a simple greedy procedure (SPT rule), scenarios, in general, make the problem NP-hard. We paint an almost complete picture of the evolving complexity landscape, drawing the line between easy and hard. One of our main algorithmic contributions relies on a deep structural result on the maximum imbalance of an optimal schedule, based on a subtle connection to Hilbert bases of a related convex cone.

Total Completion Time Scheduling Under Scenarios

TL;DR

An almost complete picture of the evolving complexity landscape is painted, drawing the line between easy and hard in this classical problem under uncertainty, in which the uncertainty is modeled by a set of scenarios.

Abstract

Scheduling jobs with given processing times on identical parallel machines so as to minimize their total completion time is one of the most basic scheduling problems. We study interesting generalizations of this classical problem involving scenarios. In our model, a scenario is defined as a subset of a predefined and fully specified set of jobs. The aim is to find an assignment of the whole set of jobs to identical parallel machines such that the schedule, obtained for the given scenarios by simply skipping the jobs not in the scenario, optimizes a function of the total completion times over all scenarios. While the underlying scheduling problem without scenarios can be solved efficiently by a simple greedy procedure (SPT rule), scenarios, in general, make the problem NP-hard. We paint an almost complete picture of the evolving complexity landscape, drawing the line between easy and hard. One of our main algorithmic contributions relies on a deep structural result on the maximum imbalance of an optimal schedule, based on a subtle connection to Hilbert bases of a related convex cone.
Paper Structure (12 sections, 7 theorems, 7 equations, 1 figure, 2 tables)

This paper contains 12 sections, 7 theorems, 7 equations, 1 figure, 2 tables.

Table of Contents

  1. Introduction
  2. Preliminaries
  3. Our Contribution
  4. Scheduling Under Arbitrary $K$ is Hard, but $K=2$ is Easy
  5. NP-hardness for unbounded number of scenarios
  6. Computing an ideal schedule for two scenarios
  7. The MinMax Version
  8. Constant number of machines
  9. Any number of machines
  10. The MinAvg Version
  11. Constant number of machines
  12. Any number of machines

Key Result

proposition thmcounterproposition

For two machines and all jobs having unit lengths and weights, it is NP-hard to approximate MinMaxSTC within a factor $2-\varepsilon$ and MinAvgSTC within ratio 1.011. The latter even holds if all scenarios contain only two jobs.

Figures (1)

  • Figure 1: Left: original schedule. Right: equivalent "weight-schedule". In both cases the objective value is equal to the total area of the rectangles.

Theorems & Definitions (9)

  • proposition thmcounterproposition
  • proposition thmcounterproposition
  • theorem 1.2
  • proof
  • theorem 1.3
  • theorem 1.4
  • theorem 1.5
  • theorem 1.6
  • proof