Speed-robust scheduling revisited

TL;DR

The paper investigates speed-robust scheduling on uniformly related machines, proposing a two-stage bagging framework in which a first-stage partition into $b$ bags precedes a second-stage allocation once machine speeds are revealed. It delivers new bounds across job-type regimes: a $1.6$-robust algorithm for bricks with $b=m$, tight up to a $1.58$ lower bound; tight upper and lower bounds for infinitesimal (sand) jobs when $b\ge m$ with the factor $\overline{\rho}(m,b)=\frac{m^b}{m^b-(m-1)^b}$; and a $q$-pebble result achieving robustness near the infinitesimal limit, with a constructive coin-based approach for bricks that also integrates fractional and integral analyses. The key technical contributions include a geometric-sand construction using $U=m^b$, $L=m^b-(m-1)^b$, $t_j$ sequences, and a greedy second stage, plus a novel integrality-driven bag-size algorithm for bricks that attains the 1.6 bound. Collectively, the results advance understanding of robustness in two-stage scheduling under unknown machine environments and provide practically efficient algorithms with provable guarantees for several natural job regimes. The work also highlights a path toward improving the general $b=m$ bound and motivates further exploration of the rocks and mixed regimes.

Abstract

Speed-robust scheduling is the following two-stage problem of scheduling $n$ jobs on $m$ uniformly related machines. In the first stage, the algorithm receives the value of $m$ and the processing times of $n$ jobs; it has to partition the jobs into $b$ groups called bags. In the second stage, the machine speeds are revealed and the bags are assigned to the machines, i.e., the algorithm produces a schedule where all the jobs in the same bag are assigned to the same machine. The objective is to minimize the makespan (the length of the schedule). The algorithm is compared to the optimal schedule and it is called $ρ$-robust, if its makespan is always at most $ρ$ times the optimal one. Our main result is an improved bound for equal-size jobs for $b=m$. We give an upper bound of $1.6$. This improves previous bound of $1.8$ and it is almost tight in the light of previous lower bound of $1.58$. Second, for infinitesimally small jobs, we give tight upper and lower bounds for the case when $b\geq m$. This generalizes and simplifies the previous bounds for $b=m$. Finally, we introduce a new special case with relatively small jobs for which we give an algorithm whose robustness is close to that of infinitesimal jobs and thus gives better than $2$-robust for a large class of inputs.

Figures (8)

• Figure 3.1: An example of bag sizes chosen for $m = 2$ and $b = 4$.
• Figure 5.1: Graphical representation of the first three chosen bags for $n = 13$, $m = 10$. The dots represent coins and the boxes represent chosen bags. The number of coins inside a box represent the cost of the bag. Vertical lines emphasize the multiples of $m$, which determine the bag costs.
• Figure 5.2: Graphical representation of the first chosen bag of size $\lceil\lambda\rceil$.
• Figure 5.3: Tabular and graphical representation of the execution of \ref{['alg-bag-size']} for $n = 45$, $m = 9$ and $\rho = 1.6$. The numbers above bags represent their sizes. The sum of bag sizes is actually $46 > n = 45$, to solve this, we can for example replace one bag of size 3 with a bag of size 2.
• Figure 5.4: Fractional solution for $n = 45$, $m = 9$ and $\rho = 1.6$ produced by \ref{['alg-fractional']}. Notice that we always use only one bag size (cost) between consecutive multiples of $m$. Compare this to Figure \ref{['fig_45_9']} where bag cost $5$ "overflows" the line at $4m$ coins.

Theorems & Definitions (13)

• Theorem 1
• Lemma 2
• Theorem 3
• Theorem 4
• Theorem 5
• Theorem 6
• Definition 7
• Definition 9
• Lemma 12
• Lemma 13