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OUTLINE

OUTLINE. Questions? Did you hear or notice anything since last class that has to do something with this class? New Homework More algorithms Review Moore’s Lawler’s 2 machine flow shops Quiz. Basic Solutions (continued). n/1//N T is minimized by Moore’s algorithm

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OUTLINE

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  1. OUTLINE • Questions? • Did you hear or notice anything since last class that has to do something with this class? • New Homework • More algorithms • Review Moore’s • Lawler’s • 2 machine flow shops • Quiz

  2. Basic Solutions (continued) • n/1//NT is minimized by Moore’s algorithm • 1. Order the jobs in EDD order • 2. If no jobs are tardy, the schedule is optimal • 3. Find the first tardy job(kth job) • 4. From 1<= j<=k, select the job with the longest processing time and place it last • 5. Repeat step 3 • If we give the jobs weights (importance), we move the one with the least weight to the end of the schedule

  3. Precedence constraints • A precedence constraint requires that one job be completed before another job, but not necessarily immediately before it • We cannot do the second job without the first • We reduce the number of potential solutions

  4. Lawler’s Algorithm • Applies to any measure that is a linear function of Completion time • Allows different cost functions for each job as long as it is non-decreasing in C (Completion time) • Minimizes the maximum of the measures • The simplest example is Lmax • It is a function of C • It is the same for each job • It is non-decreasing in C • Li = f(Ci) = Ci - di

  5. Lawler’s Algorithm (continued) • The algorithm is based on selecting those jobs that may be processed last (are not predecessors to other jobs) and placing them last in the sequence, the one with the least cost being placed at the end • The schedule is built from its end forward • 1. Find the subset V that contains all the jobs that may be placed last • 2. From V select the job with the least cost when placed last and schedule it last • 3. Reduce the problem by the placed job and repeat from step 1

  6. Lawler’s Algorithm (continued) • Note that when we apply this to the Lmax problem without precedence constraints, all jobs may be placed last and the one with the largest due date will incur the least cost and will be placed last. We thus build an EDD schedule, but from the back. • When we add precedence constraints, the set of jobs that may go last is reduced and changes from step to step as we build the schedule from the back

  7. Lawler’s Algorithm - example one

  8. Lawler’s Algorithm - example two

  9. Flow shops • n/m/F/Regular • ri = 0 • 1. We need to consider only schedules that have the same ordering on the first two machines (French 67, Pinedo 133) • 2. Therefore n/2/F/Regular only has to consider permutation schedules (French 67) • n/m/F/Fmax • 1. We need to consider only schedules with the same sequence on the last two machines • 2. Therefore n/3/F/Fmax only has to consider permutation schedules

  10. Johnson’s algorithm • We combine the two results from the last slide: • n/2/F/Fmax • This algorithm builds the schedule from both ends M1 M2

  11. Johnson’s algorithm (1954) • It takes advantage of the facts that : • 1. The sooner the second machine can start processing, the sooner we can finish • 2. The less time the first machine is idle, the sooner we finish

  12. Johnson’s algorithm - example

  13. Johnson’s algorithm for three machines • Johnson’s algorithm forms the basis of a large collection of heuristic approaches to scheduling. • Most are based on this extension to three machines: • We form two virtual machines from the three by • 1. The processing times on the first virtual machine are the sum of the processing times on the first and second machine • 2. The processing time on the second virtual machine are the sum of the processing times on the second and third machine • Then we apply Johnson’s algorithm to the two virtual machines and find our permutation

  14. Johnson’s algorithm for three machines (continued) • There is a restriction to this procedure • The processing times on the middle machine must be less than or equal to either the minimum of the processing times on machine 1 or 3 (but not necessarily both) • When you draw the Gantt chart for the example on the next page, you can see how the middle machine does not affect the outcome because of the above restriction

  15. Johnson’s algorithm for three machines - example

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