Outline

1. Outline • Questions? • Exam Results • Go over Exam • New homework • Lecture • Efficiency • Smith’s algorithm • Wassenhove and Gelders

2. Min Fbar subject to Tmax =0 • This algorithm is due to Smith (1956) and builds the schedule from the back as does Lawler’s • Tmax = 0, of course, must be possible • Again we build the schedule from the back: • a. Look at unscheduled jobs, • Find all the jobs that are due at or after Tau • Select the one with the longest processing time • (arbitrary for ties) and place it last • b. Reduce Tau and the set of unscheduled jobs and repeat

3. Efficiency • What is meant by efficiency? (for nonequivalent measures) • Given two measures R1 and R2, an efficient schedule is one when we can find no better schedule such that • R1 <= R1’ and R2 <= R2’ • In particular, we will consider the measures Fbar and Tmax and find the schedule with the best combination of the two measures under a given relationship between the two measures, e.g., 2Tmax + (Fbar)^2 • This relationship, of course, must make sense for our particular situation

4. Efficiency (continued) • Rephrasing: • S produces: Tmax and Fbar • S’ produces: T’max and Fbar’ • Then S is better than, or dominates S’ if • Tmax <= T’max • Fbar <= Fbar’ • The comparisons must be strict in at least one • Tmax <= T’max or Tmax < T’max orTmax < T’max • Fbar < Fbar’ Fbar <= Fbar’ Fbar < Fbar’

5. Efficiency (continued) • Smith’s algorithm has been extended by Wassenhove and Gelders: • In Smith algorithm we arbitrarily break ties when two jobs have due dates greater than Tau and have equal processing times. • In the extension, the job with the later due date is selected

6. Efficiency (continued) • Instead of minimizing Fbar subject to Tmax =0, which frequently is not achievable, let’s minimize it subject to Tmax<= delta • Our task will be to find all the schedules for all the possible values of delta. (We won’t actually have to do this) • We begin with stating that this is no different than increasing all the due dates by delta and using Smith’s algorithm • We want to find all the efficient schedules • We need to assume that all p and d are integral (always possible - using shorter time measurements)

7. Efficiency (continued) • What is the tardiest that a job can be? • When its due date is 0 and it is processed last • Therefore Tmax <= the sum of the processing times • First let us consider a small problem of four jobs. There are 24 possible schedules and we could try each one, finding both Fbar and Tmax and find the best one -- not the best approach, but let us look at how we would find the efficient ones from the 24 that we had generated

8. Efficiency (continued) Fbar • Each point represents the result of a schedule. There are fewer than 24 points because several schedules yield identical results. The corners of the shaded area are the dominant schedules Tmax

9. Efficiency (continued) • Better in both worse in Fbar worse in both worse in Tmax

10. Efficiency (continued) • Outlining the algorithm: • 1. Find delta = sum of the processing times • 2. Add delta to all the due dates times • 3. Apply Smith’s algorithm to get S1 with Fbar1 and Tmax1 • 4. Set delta to Tmax - 1 • 5. Repeat from step 2 until there is no schedule with Tmax <=delta

11. W&G