Speed-robust scheduling: sand, bricks, and rocks

Abstract

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 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 ≈ 1.58. The particular machine environment in which all machines have either speed 0 or 1 was studied before by Stein and Zhong (ACM Trans Algorithms 16(1):1-20, 2020. https:// doi.org/10.1145/3340320). For this setting, we provide an algorithm for scheduling infinitesimal jobs with an optimal robustness factor of 1+ √ 2 2 ≈ 1.207. It lays the foundation for an algorithm matching the lower bound of 4 3 for equal-size jobs.

Scheduling with Speed Predictions

“…Our main result is a deterministic algorithm for minimizing makespan in the scheduling with speed predictions (SSP) model that is 1+α consistent and 2+2/α robust, for any α ∈ (0, 1) (Theorem 2). When the predictions are accurate, the 1 + α consistency outperforms the best-known approximation for speed-robust scheduling without predictions of 2 − 1/m [11], while maintaining a 2 + 2/α robustness guarantee that holds even when the predictions are arbitrarily wrong. To obtain a polynomial time algorithm, the consistency and robustness both increase by a 1 + ϵ factor, for any constant ϵ ∈ (0, 1), due to the PTAS for makespan minimization on related machines that we use as a subroutine [14].…”

“…Recall that in the setting without predictions, the best known algorithm is (2 − 1/m)-robust (and thus also (2 − 1/m)-consistent) [11]. Since we have shown that algorithms with near-optimal consistency must have unbounded robustness, a main question is thus whether it is even possible to achieve a consistency that improves over (2 − 1/m) while also obtaining bounded robustness.…”

“…A recent line of work considers scheduling problems where there is uncertainty surrounding the available machines (e.g. [1,12,26,11]). In particular, we emphasize scheduling with an unknown number of parallel machines, introduced in [26] where, given a set of jobs, there is a first partitioning stage where they must be partitioned into bags without knowing the number of machines available and then, in a second scheduling stage, the algorithm learns the number of machines and the bags must be scheduled on the machines without being split up.…”

“…In particular, we emphasize scheduling with an unknown number of parallel machines, introduced in [26] where, given a set of jobs, there is a first partitioning stage where they must be partitioned into bags without knowing the number of machines available and then, in a second scheduling stage, the algorithm learns the number of machines and the bags must be scheduled on the machines without being split up. This problem was generalized to speed-robust scheduling [11] where there are m machines, but speeds of the machines are unknown in the partitioning stage and are revealed in the scheduling stage 1 . We will use the speed robust scheduling model in the rest of this paper, as it captures applications where partial packing decisions have to be made with only partial information about the machines.…”

“…Without predictions, [11] achieves a (2 − 1/m)-approximation. Thus, if we do not trust the predictions, we can ignore them and use this algorithm to achieve a 2 − 1/m consistent and 2 − 1/m robust algorithm.…”