Competitive analysis of preemptive single-machine scheduling

Abstract

We give an online algorithm for minimizing the total weighted completion time on a single machine where preemption of jobs is allowed and prove that its competitive ratio is at most 1.57.

Introduction

Many scheduling problems with the objective of minimizing the average (weighted) completion time can be approximated efficiently to within any constant arbitrarily close to 1, i.e. they have polynomial time approximation schemes (PTAS’s) [1]. However, in the online setting where we assume jobs arrive over time and the schedule is constructed without knowledge of jobs arriving in the future, the best known approximation ratio is often much worse than what we get from offline computing. For some problems, lower bounds show that no algorithm that works online can have a small approximation ratio, but for other problems these bounds are missing and they may have simple, efficient online algorithms with ratios close to 1. In this paper, we study the single-machine scheduling problem with preemption of jobs and the average weighted completion time objective. This problem does have a PTAS and the best known lower bound on the competitive ratio (online approximation ratio) is only 1.073 [5]. We give a simple, $O(nlogn)$-time online algorithm and prove that its competitive ratio is at most 1.57. (A slightly different proof is given in the author’s Ph.D. Thesis [11].) Our proof is based on the elegant technique of modified instances.

An instance $I$ of our problem is given by a set $J\subset \mathbb{N}$ and for each $j\in J$ a positive processing time $p_j$, non-negative weight $w_j$ and non-negative release time $r_j$. We assume all numbers are rational and write ${\rho }_j=w_j/p_j$. A schedule determines for each moment in time which job is processed on the machine. Each job $j$ needs to be processed for exactly $p_j$ time units and must not start before time $r_j$. Preemption is allowed, i.e., we may split the $p_j$ units over several intervals. Given a feasible schedule $\sigma$, we denote the completion time of job $j$ by $C_j$ and denote the weighted sum of completion times by $Z\left(\sigma \right)$. We denote the minimum weighted sum of completion times over all feasible schedules for $I$ by $\text{<mi mathsize="small" is="true">OPT</mi>}\left(I\right)$.

In the non-preemptive setting no deterministic online algorithm can be better than 2-competitive [7]. A 2-competitive algorithm was given by Anderson and Potts [3]. A simplified proof by Queyranne [9] shows that the ratio holds with respect to the preemptive problem as well. Goemans, Wein and Williamson (cited as a personal communication in [10]) noted that scheduling at any point in time the job with highest ratio of weight over processing time gives a 2-competitive algorithm. This is sometimes called the preemptive weighted shortest processing time rule (WSPT). Megow [8] considers the shortest weighted remaining processing time rule (SWRPT) in which at any moment the job with largest ratio of weight over remaining processing time is processed. She shows that the ratio of this algorithm is at most 2 but not better than 1.21 and conjectures that it is much smaller than 2. In contrast to the non-preemptive problem, the best lower bound on the ratio for the preemptive problem is only 1.073 [5]. A randomized online algorithm with ratio 4/3 was given by Schulz and Skutella [10]. For an overview of approximation ratios for average completion time scheduling see [2], [4], [12].

The mean busy time of a job in a schedule is the average time at which it is processed. Formally, let ${\delta }_j\left(t\right)$ be the indicator function of job $j$ in a given schedule, i.e., ${\delta }_j\left(t\right)=1$ if job $j$ is processed at time $t$ and ${\delta }_j\left(t\right)=0$ otherwise. Then, the mean busy time $M_j$ of job $j$ is defined by

$$M_j=\frac{1}{p_j}{\int }_0^{\infty }{\delta }_j\left(t\right)t\cdot dt\text{.}$$

Observe that for any job $M_j\le C_j-p_j/2$. Equality holds if and only if job $j$ is not preempted. Given an instance $I$, we say that functions ${\delta }_j\left(t\right):[0,\infty )\to [0,1]$ for all $j$ define a relaxed schedule if

• ${\sum }_j{\delta }_j\left(t\right)\le 1$ for all $t\ge 0$, and

• ${\int }_{r_j}^{\infty }{\delta }_j\left(t\right)\cdot dt=p_j$ for all $j\in J$.

Intuitively, the machine can process more than one job at a time, although at a lower speed. We restrict to functions that are piecewise constant with only a polynomial number of discontinuities. Consequently, each job finishes in finite time. Consider the problem of minimizing the mean busy time over all relaxed schedules, i.e., minimize ${\sum }_j{w}_j{M}_j={\sum }_j\frac{w_j}{p_j}{\int }_0^{\infty }{\delta }_j\left(t\right)t\cdot dt$ subject to (i) and (ii). For a given instance $I$, we denote this problem by $M\left(I\right)$ and its optimal value by ${\text{<mi mathsize="small" is="true">OPT</mi>}}^M\left(I\right)$. It is easy to show that the optimal solution is found by scheduling jobs preemptively in order of non-increasing ratio of $w_j/p_j$ (the preemptive WSPT rule) [6]. If $\prec$ is a complete order of the jobs such that $w_j/p_j\ge w_k/p_k$ if $j\prec k$, then we call the schedule that processes jobs preemptively in order of $\prec$ the WSPT schedule w.r.t. $\prec$. For a given (relaxed) schedule $\Delta$ (defined by some functions ${\delta }_j\left(t\right)$), we denote its weighted mean busy time by $Z^M\left(\Delta \right)$.