Maximizing the minimum load: The cost of selfishness

Xujin Chen, Leah Epstein, Elena Kleiman, Rob Van Stee

Research output: Contribution to journalArticlepeer-review


We consider a scheduling problem on m machines, where each job is controlled by a selfish agent. Each agent is only interested in minimizing its own cost, defined as the total load of the machine that its job is assigned to. We consider the objective of maximizing the minimum load (the value of the cover) over the machines. Unlike the regular makespan minimization problem, which was extensively studied in a game-theoretic context, this problem has not been considered in this setting before. We study the price of anarchy (poa) and the price of stability (pos). These measures are unbounded already for two uniformly related machines [11], and therefore we focus on identical machines. We show that the pos is 1, and derive tight bounds on the pure poa for m≤7 and on the overall pure poa, showing that its value is exactly 1.7. To achieve the upper bound of 1.7, we make an unusual use of weighting functions. Finally, we show that the mixed poa grows exponentially with m for this problem. In addition, we consider a similar setting of selfish jobs with a different objective of minimizing the maximum ratio between the loads of any pair of machines in the schedule. We show that under this objective the pos is 1 and the pure poa is 2, for any m≥2.

Original languageEnglish
Pages (from-to)9-19
Number of pages11
JournalTheoretical Computer Science
StatePublished - 22 Apr 2013

Bibliographical note

Funding Information:
The first author’s research was supported in part by the National Science Foundation of China under grant 11222109. The fourth author’s research was supported by the German Research Foundation (DFG).


  • Envy-ratio
  • Machine covering
  • Price of anarchy
  • Scheduling

ASJC Scopus subject areas

  • Theoretical Computer Science
  • General Computer Science


Dive into the research topics of 'Maximizing the minimum load: The cost of selfishness'. Together they form a unique fingerprint.

Cite this