Scheduling jobs on identical parallel machines
- 1. Greedy Algorithm & Local Search Scheduling Jobs On Identical Parallel Machines Prepared By Mir Md. Asif Hossain Student Id : 0805070
- 2. Problem Overview ‘n’ Different Jobs ‘j’=1 to n With processing time P j =1 to ‘n’ JOB 1 JOB 2 JOB 3 JOB ‘N’ All Jobs Are Present At Time 0
- 3. Problem Overview ‘m’ Identical Parallel Machine ‘i’=1 to m M1 M2 M4 Mm
- 4. Problem Overview Aim is to complete all jobs as soon as possible JOB 1 Schedule JOB 3 JOB 2 JOB 5 JOB 3 JOB 4 JOB 1 JOB 5 JOB 6 JOB 2 JOB 6 NP HARD JOB 4
- 5. Problem Overview Goal If job ‘J’ completes at a time Cj then minimize Cmax = max(j=1,...,n Cj) Cj is often called the makespan or length of the schedule
- 6. Local Search Algorithm For Scheduling Jobs At Identical Parallel Machine 1.Start with any schedule 2. Consider job ‘l’ that finishes last 3. If there any machine to which it can be reassigned such that this job finishes earlier . Reassigned the Job. 4. Repeat step until no other Job left to reassigned.
- 7. Example Consider a scenario with 7 jobs and 4 parallel machine JOBS JOB 1 JOB 2 JOB 3 JOB 4 JOB 5 JOB 6 JOB 7 Machines
- 8. Example 1.Start with any schedule 2. Consider job ‘l’ that finishes last 3. If there any machine to which it can be reassigned such that this job finishes earlier . Reassigned the Job. 4. Repeat step until no other Job left to reassigned.
- 9. Example 1.Start with any schedule 2. Consider job ‘l’ that finishes last 3. If there any machine to which it can be reassigned such that this job finishes earlier . Reassigned the Job. 4. Repeat until no other Job left to reassigned. JOB 1 JOB 3 JOB 5 JOB 4 JOB 6 JOB 7 JOB 2
- 10. Example 1.Start with any schedule 2. Consider job ‘l’ that finishes last 3. If there any machine to which it can be reassigned such that this job finishes earlier . Reassigned the Job. 4. Repeat steps until no other Job left to reassigned. JOB 1 JOB 3 JOB 5 JOB 4 JOB 6 JOB 7 JOB 2
- 11. Example 1.Start with any schedule 2. Consider job ‘l’ that finishes last 3. If there any machine to which it can be reassigned such that this job finishes earlier . Reassigned the Job. 4. Repeat steps until no other Job left to reassigned. JOB 1 JOB 3 JOB 5 JOB 4 JOB 6 JOB 7 JOB 2
- 12. Example 1.Start with any schedule 2. Consider job ‘l’ that finishes last 3. If there is any machine to which it can be reassigned such that this job finishes earlier . Reassigned the Job. 4. Repeat step 3 until no other Job left to reassigned. JOB 1 JOB 3 JOB 5 JOB 4 JOB 6 JOB 7 JOB 2
- 13. Example 1.Start with any schedule 2. Consider job ‘l’ that finishes last 3. If there is any machine to which it can be reassigned such that this job finishes earlier . Reassigned the Job. 4. Repeat step 3 until no other Job left to reassigned. JOB 1 JOB 3 JOB 5 JOB 4 JOB 6 JOB 7 JOB 2
- 14. Example 1.Start with any schedule 2. Consider job ‘l’ that finishes last 3. If there is any machine to which it can be reassigned such that this job finishes earlier . Reassigned the Job. 4. Repeat step 3 until no other Job left to reassigned. JOB 1 JOB 3 JOB 5 JOB 4 JOB 6 JOB 7 JOB 2
- 15. Example 1.Start with any schedule 2. Consider job ‘l’ that finishes last 3. If there is any machine to which it can be reassigned such that this job finishes earlier . Reassigned the Job. 4. Repeat step 3 until no other Job left to reassigned. JOB 1 JOB 3 JOB 5 JOB 4 JOB 6 JOB 7 JOB 2
- 16. Example 1.Start with any schedule 2. Consider job ‘l’ that finishes last 3. If there is any machine to which it can be reassigned such that this job finishes earlier . Reassigned the Job. 4. Repeat step 3 until no other Job left to reassigned. JOB 1 JOB 3 JOB 5 JOB 4 JOB 6 JOB 7 JOB 2
- 17. Example 1.Start with any schedule 2. Consider job ‘l’ that finishes last 3. If there is any machine to which it can be reassigned such that this job finishes earlier . Reassigned the Job. 4. Repeat steps until no other Job left to reassigned. JOB 1 JOB 3 JOB 6 JOB 5 JOB 4 JOB 7 JOB 2
- 18. Example 1.Start with any schedule 2. Consider job ‘l’ that finishes last 3. If there is any machine to which it can be reassigned such that this job finishes earlier . Reassigned the Job. 4. Repeat steps until no other Job left to reassigned. JOB 1 JOB 3 JOB 6 JOB 7 JOB 5 JOB 4 JOB 2
- 19. Example 1.Start with any schedule 2. Consider job ‘l’ that finishes last 3. If there is any machine to which it can be reassigned such that this job finishes earlier . Reassigned the Job. 4. Repeat steps until no other Job left to reassigned. JOB 1 JOB 3 JOB 6 JOB 7 JOB 5 JOB 4 JOB 2
- 20. Example 1.Start with any schedule 2. Consider job ‘l’ that finishes last 3. If there is any machine to which it can be reassigned such that this job finishes earlier . Reassigned the Job. 4. Repeat steps until no other Job left to reassigned. JOB 1 JOB 3 JOB 6 JOB 7 JOB 5 JOB 4 JOB 2
- 21. Example 1.Start with any schedule 2. Consider job ‘l’ that finishes last 3. If there is any machine to which it can be reassigned such that this job finishes earlier . Reassigned the Job. 4. Repeat step 3 until no other Job left to reassigned. JOB 1 JOB 3 JOB 6 JOB 7 JOB 5 JOB 4 JOB 2
- 22. Example 1.Start with any schedule 2. Consider job ‘l’ that finishes last 3. If there is any machine to which it can be reassigned such that this job finishes earlier . Reassigned the Job. 4. Repeat steps until no other Job left to reassigned. JOB 1 JOB 3 JOB 6 JOB 7 JOB 5 JOB 4 JOB 2
- 23. Example 1.Start with any schedule 2. Consider job ‘l’ that finishes last 3. If there is any machine to which it can be reassigned such that this job finishes earlier . Reassigned the Job. 4. Repeat steps until no other Job left to reassigned. JOB 1 JOB 2 JOB 3 JOB 5 JOB 6 JOB 7 JOB 4
- 24. Algorithm Analysis We first provide some natural lower bounds on the length of an optimal schedule Cmax Total work done Average work in each machine is P/m So
- 25. Algorithm Analysis Consider solution of local search algorithm. Let ℓ be a job that completes last in this final schedule; the completion time of job ℓ, Cℓ Partition Total Time into two parts Sℓ = Cℓ − pℓ pℓ
- 26. Algorithm Analysis Total work done in 1st part m*Sℓ <= total work So From 2.4 Hence
- 27. Theorem 2.5: The local search procedure for scheduling jobs on identical parallel machines is a 2-approximation algorithm Proof From inequality 2.5 we can write that hence the total length of the schedule produced is at most By applying lower bounds 2.3 and 2.4 we get that schedule length is at most 1/m is significant only when m is small.
- 28. Greedy Algorithm For Scheduling Jobs At Identical Parallel Machine Whenever a machine is Idle If there is at least a job in the list then One of the remaining jobs is assigned to start processing on that machine Often Called As List Scheduling Algorithm
- 29. Example Consider a scenario with 7 jobs and 4 parallel machine JOBS JOB 1,3 JOB 2,2 JOB 3,5 JOB 4,2 JOB 5,3 JOB 6,2 JOB 7,4 Machines
- 30. Example Consider a scenario with 7 jobs and 4 parallel machine JOBS JOB 1,3 JOB 2,2 JOB 3,5 JOB 4,2 JOB 5,3 JOB 6,2 JOB 7,4 Machines JOB 1,3
- 31. Example Consider a scenario with 7 jobs and 4 parallel machine JOBS Machines JOB 1,3 JOB 2,2 JOB 3,5 JOB 4,2 JOB 5,3 JOB 6,2 JOB 7,4 JOB 2,2
- 32. Example Consider a scenario with 7 jobs and 4 parallel machine JOBS Machines JOB 1,3 JOB 3,5 JOB 4,2 JOB 5,3 JOB 6,2 JOB 7,4 JOB 2,2 JOB 3,5
- 33. Example Consider a scenario with 7 jobs and 4 parallel machine JOBS Machines JOB 1,3 JOB 2,2 JOB 4,2 JOB 5,3 JOB 6,2 JOB 7,4 JOB 3,5 JOB 4,2
- 34. Example Consider a scenario with 7 jobs and 4 parallel machine JOBS Machines JOB 1,3 JOB 2,2 JOB 5,3 JOB 3,5 JOB 5,3 JOB 6,2 JOB 7,4 JOB 4,2
- 35. Example Consider a scenario with 7 jobs and 4 parallel machine JOBS Machines JOB 1,3 JOB 2,2 JOB 5,3 JOB 3,5 JOB 6,2 JOB 7,4 JOB 4,2 JOB 6,2
- 36. Example Consider a scenario with 7 jobs and 4 parallel machine JOBS Cmax = 3+4 = 7 Machines JOB 1,3 JOB 2,2 JOB 7,4 JOB 5,3 JOB 3,5 JOB 4,2 JOB 7,4 JOB 6,2
- 37. Theorem 2.6: The list scheduling algorithm for the problem of minimizing the makespan on m identical parallel machines is a 2-approximation algorithm. Proof It can be proofed that list scheduling can be reduced to local search algorithm. To see this, If we uses this schedule as the starting point for the local search procedure, that algorithm would immediately declare that the solution cannot be improved.
- 38. Improving Our Greedy Approach Different lists yield different results Add additional greedy rule to sort the jobs in the list in non decreasing order This greedy algorithm is called the longest processing time rule, or LPT.
- 39. Example Consider a scenario with 7 jobs and 4 parallel machine JOBS JOB 3,5 JOB 7,4 JOB 5,3 JOB 1,3 JOB 4,2 JOB 2,2 JOB 6,2 Machines
- 40. Example Consider a scenario with 7 jobs and 4 parallel machine JOBS JOB 3,5 JOB 7,4 JOB 5,3 Machines JOB 3,5 JOB 7,4 JOB 6,2 JOB 1,3 JOB 4,2 JOB 2,2 JOB 6,2 JOB 5,3 JOB 2,2 JOB 1,3 JOB 4,2
- 41. Theorem 2.7: The longest processing time rule is a 4/3-approximation algorithm for scheduling jobs to minimize the makespan on identical parallel machines. Any counterexample for which the last job ℓ to complete is not the smallest can yield a smaller counterexample, simply by omitting all of the jobs ℓ +1, . . . , n; the length of the schedule produced is the same, and the optimal value of the reduced input can be no larger. Hence the reduced input is also a counterexample.