Single-machine Sequencing
2.5 Flexibility in the Basic Model
Again, the MDD rule is weaker than such rules as SPT and SWPT. It tells us that if we examined an optimal sequence, we would find that each pair of jobs is sequenced consistently with MDD; however, starting at time zero and sequen- cing the jobs by MDD may not produce an optimal schedule. To put it another way, the MDD rule represents a necessary condition for optimality, but it is not a sufficient condition.
We conclude our treatment of theT-problem with some specialized results concerning optimal sequences:
•
If the EDD sequence produces no more than one tardy job, it yields the min- imum value ofT.•
If all jobs have the same due date, thenTis minimized by SPT sequencing.•
If it is impossible for any job to be on time in any sequence, thenTis mini- mized by SPT sequencing.•
If SPT sequencing yields no jobs on time, then it minimizesT.The weighted version of the total tardiness problem is even more difficult to solve than theT-problem, which itself is NP-hard, and we postpone its discus- sion until we examine more general methods of solution in Chapter 3.
possibly subject to some constraint that represents a proxy for the negotiation process.
Suppose that the due date can be selected at the job’s release date (rj). The selection of the due date represents a target for the flow allowance or the amount of time that the job will spend in the system. We might select due dates according to one of the following rules:
CON:constantflow allowance:dj=rj+γ SLK:equal slackflow allowance:dj=rj+pj+β TWK:total workflow allowance:dj= rj+αpj
where each rule contains a single tightness parameter (γ,β, orα) that must be specified. For equal release dates, however, any one of these due date rules will result in agreeable due dates and processing times. By Theorem 2.8, it follows that SPT (which will be equivalent to EDD) minimizes total tardiness.
When due dates are completely discretionary, it is not difficult to minimize total tardiness: for any schedule we could select the due dates to be loose enough that no job would be late. However, in an environment where due dates can be selected, it seems reasonable to seek the tightest due dates possible. Tight due dates correspond to short flow allowances and thus represent commitments to customers that orders will be filled promptly. Of course, such commitments would be meaningless if there were no hope that they could be met. Therefore, we impose the constraint that no job is allowed to be tardy, and we examine how to set the due dates so they are as tight as possible.
To measure the tightness of a set of due dates, we use the sum of the due dates or
D=
n
j= 1
dj
The problem becomes one of minimizing D, subject to the requirement thatCj≤dj.
In principle, we can easily find an optimal solution to this problem. For any schedule, the tightest possible set of due dates is obviously given by dj=Cj. Therefore,Dcan be minimized by minimizing the sum of the completion times or, equivalently, total flowtime. Since we know by Theorem 2.3 that this is accomplished by SPT, our solution can be found by constructing an SPT sched- ule of the jobs, computing the completion time of each job in this schedule, and setting the due date of each job equal to its completion time. This optimal solu- tion requires that the due date of each job depends on specific information about every other job in the schedule, which we refer to as a comprehensive information base. A more practical approach is to rely on such rules as CON, SLK, and TWK, in which the selection of a due date depends only on information about the job itself (its release date and its processing time) and
2.5 Flexibility in the Basic Model 31
on a tightness parameter. Next, we might ask whether one of those three limited information rules is best.
It is possible to show that for any set ofnjobs, CON due dates are dominated by either SLK or TWK due dates. That is,Dwill never be larger under SLK or TWK than it is under CON.
∎Example 2.3 Consider a problem containingn= 3 jobs, as described in the table below, withrj= 0 for all jobs.
Jobj 1 2 3
pj 1 2 16
Suppose our problem consisted of just the first two jobs. Then the tightness parameters would be selected as follows:
CON: γ= 3 For whichD= 6 SLK: β= 1 For whichD= 5 TWK: α= 1.5 For whichD= 4.5
In this case, the optimal (full information) value isD= 4. When our problem consists of all three jobs, the results are as follows:
CON: γ= 19 For whichD= 57
SLK: β= 3 For whichD= 28
TWK: α= 1.5 For whichD= 28.5
Here, the optimal (full information) value isD= 23. Our two examples demon- strate that either TWK or SLK can be the best of the three rules. The examples also illustrate the fact that CON is always dominated. A computational study (see Baker and Bertrand, 1981) suggests that TWK tends to be the best rule most of the time and that its advantage grows with larger problem sizes and with var- iability among processing times. Therefore, in practice, a good approach is to use TWK and adjustαby trial and error to maintain the shop due date perfor- mance on target.
2.5.2 Job Selection Decisions
In the basic sequencing model, the workload is given, and due dates are given, and the scheduling task is to find the best sequence. When due dates are deci- sions, as in the previous section, we face a variation of the basic sequencing model that allows for additional flexibility. When, instead, the workload is a decision, we face a different kind of flexibility. Specifically, in thejob selection model, we must decide which of the available jobs to accept (and which to
2 Single-machine Sequencing 32
reject). As in the basic model, processing times and other parameters are given, along with a penalty (or opportunity cost) for each job rejected. For the jobs selected, the scheduling task remains one of finding the best sequence.
In the job selection model, the objective thus has two components: a time-based performance measure, such as total flowtime, and an economic measure, the total penalty. Because rejected jobs are simply removed from the scheduling problem, a suitable representation for the economic measure is a lump-sum penaltyej> 0, incurred if jobjis rejected. This penalty could represent a contractual payment or an opportunity cost, such as lost profit or the inefficiency cost of assigning the job to some other resource. A more direct way of representing the economic measure, however, is to associate revenue with each job that is processed.
Consider the performance measure for accepted jobs. Standard scheduling objectives, such as makespan (Cmax), weighted flowtime (Fw), and maximum tardiness (Tmax), are not intrinsically economic criteria. In addition, they meas- ure time intervals, whereas the rejection penalty is a lump sum. Therefore, the most consistent scheduling objectives are also lump-sum measures, such as the maximum number of on-time jobs or the minimum weighted number of tardy jobs. Nevertheless, the scheduling and economic components may not be com- mensurable, so in some cases we may prefer to adopt a weighting factor to com- bine them into a single objective function.
Perhaps the simplest nontrivial model involves lump-sum revenues and costs.
Suppose jobjis characterized by a processing timepj, a due datedj, and a tar- diness penaltywj, which is incurred if the job is accepted and completed after its due date. If jobjis accepted, then it generates revenue of vj; otherwise, it is rejected. No direct cost is associated with a job’s rejection. (All parameters are assumed to be positive.) For the solution, we define the following sets:
A= the set of accepted jobs R= the set of rejected jobs Z= the set of late jobs Y= the set of on-time jobs
We use lowercase to denote the number of jobs in each set:a= the number of accepted jobs, etc. Thusn=a+randa=y+z=n−r. Finally, for a given job set S, we can write the objective function as follows:
F S =
A
vj–
Z
wj
The expressionF(S) is meant to convey that, conceptually, the first step is to select the jobs inA, after which the next step is to sequence them optimally.
Suppose the revenues and costs are identical:vj=vand wj=w. That means the net revenue for an accepted job completed late isv−w. Ifv−w> 0, then acceptance is preferred to rejection, even if tardiness occurs; it is optimal to accept all jobs and then maximize the net benefit,vn−wz(S). This criterion cor- responds to minimizing the number of tardy jobsz(S), which is achieved with
2.5 Flexibility in the Basic Model 33
Algorithm 2.1. Thus, this particular case reduces to a well-known scheduling problem without rejection that can be solved in polynomial time.
Ifv−w≤0, then rejection is no worse than acceptance with tardy completion. In this case, it is optimal to complete all accepted jobs on time and reject the rest. The objective function becomesav=vy(S). We can maximize this quantity by maxi- mizing the number of on-time jobs, which is also accomplished by Algorithm 2.1.
In the identical parameter case, the problem reduces to finding two sets– an on-time set and its complement. Furthermore, the impact on the objec- tive function of completing a job on time versus placing it in the comple- mentary set is the same for all jobs. However, when the parameters are job dependent, we cannot rely on Algorithm 2.1 to provide an optimal solu- tion. Moreover, the general form of the problem involves finding three sets –an on-time set of accepted jobs, a tardy set of accepted jobs, and the set of rejected jobs. Given that a job will not be accepted and completed on time, the choice between rejecting the job and accepting it but completing it late is dictated by the value ofvj−wj. Moreover, that preference can be determined at the outset. Thus, let cj= max{0,vj−wj}. Whencj= 0, if jobjis not com- pleted on time, it should be rejected. Similarly, when cj > 0, if job j is not completed on time, it should still be accepted. Even when these observations are applied, the general problem remains challenging. To obtain optimal solutions, we would need to implement more powerful techniques, such as those described in Chapter 3.