Parallel Machine Models (Deterministic)
5.1 The Makespan without Preemptions
First, the problemP m||Cmaxis considered. This problemis of interest because minimizing the makespan has the effect of balancing the load over the various machines, which is an important objective in practice.
It is easy to see that P2 || Cmax is NP-hard in the ordinary sense as it is equivalent toPARTITION(see Appendix D). During the last couple of decades many heuristics have been developed for P m || Cmax. One such heuristic is described below.
The Longest Processing Time first (LPT) rule assigns att= 0 themlongest jobs to them machines. After that, whenever a machine is freed the longest job among those not yet processed is put on the machine. This heuristic tries to place the shorter jobs more towards the end of the schedule, where they can be used for balancing the loads.
In the next theoreman upper bound is presented for Cmax(LP T)
Cmax(OP T),
whereCmax(LP T) denotes the makespan of the LPT schedule andCmax(OP T) denotes the makespan of the (possibly unknown) optimal schedule. This type
of worst case analysis is of interest as it gives an indication of how well the heuristic is guaranteed to performas well as the type of instances for which the heuristic performs badly.
Theorem 5.1.1. ForP m||Cmax Cmax(LP T) Cmax(OP T) ≤4
3 − 1 3m.
Proof. By contradiction. Assume that there exists one or more counterexamples with the ratiostrictlylarger than 4/3−1/3m. If more than one such counterex- ample exist, there must exist an example with the smallest number of jobs.
Consider this “smallest” counterexample and assume it has n jobs. This smallest counterexample has a useful property: under LPT the shortest job is the last job to start its processing and also the last job to finish its process- ing. That this is true can be seen as follows: first, under LPT by definition the shortest job is the last to start its processing. Also, if this job is not the last to complete its processing, the deletion of this smallest job will result in a counterexample with fewer jobs (the Cmax(LP T) remains the same while the Cmax(OP T) may remain the same or may decrease). So for the smallest coun- terexample the starting time of the shortest job under LPT isCmax(LP T)−pn. Since at this point in time all other machines are still busy it follows that
Cmax(LP T)−pn≤ n−1
j=1 pj
m .
The right hand side is an upper bound on the starting time of the shortest job.
This upper bound is achieved when scheduling the firstn−1 jobs according to LPT results in each machine having exactly the same amount of processing to do. Now
Cmax(LP T)≤pn+ n−1
j=1pj
m =pn(1− 1 m) +
n j=1pj
m .
Since
Cmax(OP T)≥ n
j=1pj
m
the following series of inequalities holds for the counterexample:
4 3 − 1
3m < Cmax(LP T)
Cmax(OP T) ≤pn(1−1/m) +n
j=1pj/m Cmax(OP T)
= pn(1−1/m) Cmax(OP T) +
n
j=1pj/m
Cmax(OP T) ≤pn(1−1/m) Cmax(OP T)+ 1.
114 5Parallel Machine Models (Deterministic) Thus
4 3− 1
3m <pn(1−1/m) Cmax(OP T)+ 1 and
Cmax(OP T)<3pn.
Note that this last inequality is astrict inequality. This implies that for the smallest counterexample the optimal schedule may result in at most two jobs on each machine. It can be shown that if an optimal schedule is a schedule with at most two jobs on each machine then the LPT schedule is optimal and the ratio of the two makespans is equal to one (see Exercise 5.11.b). This contradiction
completes the proof of the theorem.
Example 5.1.2 (A Worst Case Example of LPT)
Consider 4 parallel machines and 9 jobs, whose processing times are given in the table below:
jobs 1 2 3 4 5 6 7 8 9 pj 7 7 6 6 5 5 4 4 4
Scheduling the jobs according to LPT results in a makespan of 15. It can be shown easily that for this set of jobs a schedule can be found with a makespan of 12 (see Figure 5.1). This particular instance is thus a worst case
when there are 4 machines in parallel. ||
What would the worst case be, if instead of LPT an arbitrary priority rule is used? Consider the case where at timet= 0 the jobs are put in an arbitrary list. Whenever a machine is freed the job that ranks, among the remaining jobs, highest on the list is put on the machine. It can be shown that the worst case of this arbitrary list rule satisfies the inequality
Cmax(LIST)
Cmax(OP T) ≤2− 1 m.
(This result can be shown via arguments that are similar to the proof of Theo- rem5.6.1 in the section on online scheduling.)
However, there are also several other heuristics for theP m||Cmax problem that are more sophisticated than LPT and that have tighter worst-case bounds.
These heuristics are beyond the scope of this book.
Consider now the same problem with the jobs subject to precedence con- straints, i.e.,P m|prec|Cmax. Froma complexity point of view this problem has to be at least as hard as the problemwithout precedence constraints. To obtain some insights into the effects of precedence constraints, a number of special cases have to be considered. The special case with a single machine
0 4 8 12 16 4
3
5 6 2
1
Cmax(LPT) = 15
7
8 9
t
0 4 8 12
8 7
3
9 4 2
1
Cmax(OPT) = 12
5 6
t
Fig. 5.1Worst case example of LPT
is clearly trivial. It is enough to keep the machine continuously busy and the makespan will be equal to the sum of the processing times. Consider the special case where there are an unlimited number of machines in parallel, or where the number of machines is at least as large as the number of jobs, i.e.,m≥n.
This problemmay be denoted byP∞ |prec|Cmax. This is a classical problem in the field of project planning and its study has led to the development of the well-knownCritical Path Method (CPM) and Project Evaluation and Re- view Technique (PERT). The optimal schedule and the minimum makespan are determined through a very simple algorithm.