The speed-robust scheduling problem is a two-stage problem where given $m$ machines, jobs must be grouped into at most $m$ bags while the processing speeds of the given $m$ machines are unknown. After the speeds are revealed, the grouped jobs must be assigned to the machines without being separated. To evaluate the performance of algorithms, we determine upper bounds on the worst-case ratio of the algorithm's makespan and the optimal makespan given full information. We refer to this ratio as the robustness factor. We give an algorithm with a robustness factor $2-1/m$ for the most general setting and improve this to $1.8$ for equal-size jobs. For the special case of infinitesimal jobs, we give an algorithm with an optimal robustness factor equal to $e/(e-1) \approx 1.58$. The particular machine environment in which all machines have either speed $0$ or $1$ was studied before by Stein and Zhong (SODA 2019). For this setting, we provide an algorithm for scheduling infinitesimal jobs with an optimal robustness factor of $(1+\sqrt{2})/2 \approx 1.207$. It lays the foundation for an algorithm matching the lower bound of $4/3$ for equal-size jobs.
翻译:速度- robust 排程问题是一个两阶段的问题, 给出 $ $ 的机器, 工作必须分组成最多 $ $ 美元 的包件, 而给定的 $ 美元 机器的处理速度未知 。 在显示速度后, 分组的工作必须分给机器, 而不分离 。 要评估算法的性能, 我们确定算法最坏比例的上限, 算法最差比例的缩放和最优比例的完整信息 。 我们将此比率称为稳健系数 2019 。 我们给出一个算法, 最通用设置的算法为 2-1 / / 美元 美元, 并且将这个算法改进为1.8 美元 。 对于无限工作的特殊情况, 我们给出一个最稳健的算法, 等于 $/ (e-1) \ approx 1. 2.58 。 。 所有机器都具有速度 $ 或 1 美元 美元 美元 美元 和 Zhong (SO 2019 ) 。 对于此设置, 我们给出一个算算算算法 最稳妥的算法, $ 1xxxxxx 4xxx 基底平 。 。