Lateness Criteria

Single-machine Sequencing

2.4 Problems with Due Dates: Elementary Results

2.4.1 Lateness Criteria

Recall that job lateness is defined asLj=Cj−dj, or the discrepancy between the due date of a job and its completion time. A somewhat remarkable result is that minimum total lateness is achieved by SPT.

∎Theorem 2.5 Total lateness is minimized by SPT sequencing.

Proof. By definition, L=

n

j= 1

Lj=

n

j= 1

Cj−dj =

n

j= 1

Fj−dj =

n

j= 1

Fj– ⁿ

j= 1

dj=F− ⁿ

j= 1

dj

The last term is the sum of the given due dates and is therefore a constant.

Because Ldiffers fromF by a constant that is independent of sequence, the sequence that minimizes L must be the sequence that minimizesF, and this

sequence is given by SPT. □

This result is somewhat remarkable because a sequencing rule that ignores due date information is optimal for a due date-oriented criterion. However, another interpretation of the result might be thatLis only superficially a due date-oriented performance measure.

2 Single-machine Sequencing 22

Instead of using SPT, an intuitive approach to meeting due dates might well be to sequence the jobs according to some measure of due date urgency. One obvious measure of urgency for a given job is the time until its due date. Sequen- cing the jobs byearliest due date(EDD) cannot guarantee, however, thatLwill be minimized, because only SPT guarantees that. Instead we can show that EDD sequencing minimizes the maximum lateness in the schedule.

∎Theorem 2.6 Maximum lateness and maximum tardiness are minimized by earliest due date (EDD) sequencing (d_[1]≤d_[2]≤ ≤d_[n]).

Proof. We again employ the method of adjacent pairwise interchange (see Figure 2.4). Consider a sequence S that is not the EDD sequence. That is, somewhere in S there must exist a pair of adjacent jobs, i and j, with j followingi, such thatdi>dj. Now construct a new sequence,S, in which jobs i and jare interchanged and all other jobs complete at the same time as in S. Then

Li S =p B +pi–di Lj S =p B +pj–dj

L_j S =p B +p_i+p_j–d_j L_i S =p B +p_j+p_i–d_i from which it follows thatLj(S) >Li(S) andLj(S) >Lj(S). Hence,

Lj S > max Li S ,Lj S

LetLAB= max{Lk|k Aork B}, and notice thatLABis the same under bothS andS. Then

Lmax S = max LAB,Li S ,Lj S ≥max LAB,Li S ,Lj S =Lmax S In other words, the interchange of jobsiandjdoes not increase the value ofL_max and may actually reduce (improve) it. Therefore, an optimal sequence can be constructed as follows:

1) Begin with an arbitrary non-EDD sequence.

2) Find a pair of adjacent jobsiandj, withjfollowingi, such thatdi>dj. 3) Interchange jobsiandj.

4) Return to Step 2 iteratively until an EDD sequence is constructed. At each iteration, L_max either remains the same or is reduced. Because an EDD sequence can be reached from any other sequence in this manner, there can be no other sequence with a value ofL_maxlower than that corresponding to EDD sequencing.

Again, ties do not disturb the logic. A similar argument establishes that EDD minimizesT_max, beginning with the inequality

Tmax S = max 0,Lmax S ≥max 0,Lmax S =Tmax S □ 2.4 Problems with Due Dates: Elementary Results 23

A second measure of urgency for a given job is the time until its due date less the time required to process it. This urgency measure is calledslack time, and, at timet, the slack time of jobjis represented as (dj−t−pj). In particular, among jobs with identical due dates, the longest is most urgent. Slack time may appear to be a more sophisticated quantification of urgency than the due date alone.

Nevertheless, there is little to be said for optimality of minimum slack time (MST) sequencing in the single-machine problem. Its only general property involves a mirror image of Theorem 2.6, which is of questionable usefulness in this situation.

∎Theorem 2.7 Among schedules with no idle time, the minimum job late- ness is maximized by minimum slack time (MST) sequencing (d_[1]−p_[1]≤ d_[2]−p_[2]≤ ≤d_[n]−p_[n]).

Proof. The proof is a mirror image of the proof of Theorem 2.6 and utilizes an adjacent pairwise interchange argument. Observe that L_min is not a regular measure of performance–hence the need, in Theorem 2.7, to restrict consid-

eration to schedules without inserted idle time. □

An important variation of the basic model involves the designation of both a primary and a secondary measure of performance. The primary measure is the dominant criterion, but if there are alternative optima with respect to the pri- mary measure, we then want to identify the best sequence among those alter- natives with respect to a secondary measure.

For example, suppose that a tardiness-based measure (such asT_max) is the pri- mary measure and that several sequences are considered“perfect”because they contain no tardy jobs. Furthermore, suppose thatFis the secondary measure.

Then, to construct a perfect sequence that minimizesF, we can employ a result known asSmith’s rule:

Jobimay be assigned the last position in sequence only if (S1)di≥ ⁿ

j= 1

pjand

(S2)pi≥pkamong all jobsksuch thatdk≥ ⁿ

j= 1

pj

This rule should seem quite logical, for if some other job were to come last in sequence, then there would be room for improvement. If (S1) is violated, then total tardiness can be reduced by shifting some job that satisfies (S1) to the last position. If (S1) holds but (S2) is violated, thenF can be reduced, without increasing tardiness, by interchanging the last job with a job that satisfies (S2). Once Smith’s rule has identified the last amongn jobs, there remain (n−1) jobs to which the rule can be applied. If we continue in this

2 Single-machine Sequencing 24

fashion, the rule eventually constructs an optimal sequence, working backward.

∎ Example 2.1 Consider a problem containing n = 5 jobs, as described in the table.

Jobj 1 2 3 4 5

pj 1 2 3 4 5

dj 9 13 11 15 10

It is not hard to verify (using EDD) that a perfect sequence exists. The only job that satisfies (S1) is job 4, which is placed last. At the next stage, jobs 2 and 3 both satisfy (S1), and job 3 is chosen to be fourth, according to (S2). Next, jobs 1, 2, and 5 all satisfy (S1), and job 5 is chosen to be third. Finally, job 2 is chosen to be second, leaving job 1 to be first. In this manner, Smith’s rule generates a perfect sequence withF= 38. In contrast, EDD yieldsF= 42.