1 IOE/MFG 543 Chapter 5: Parallel machine models (Sections )

2 Section 5.3: Total completion time Pm||  C j On a single machine On a single machine C j =  k=1  p (k) C j =  k=1  p (k) where p (j) is the processing time of the jth job processed on the machine Then Then  C j =np (1) +(n-1)p (2) +...+p (n) => Shortest Processing Time first rule minimizes  C j => Shortest Processing Time first rule minimizes  C j j

3 Total completion time Pm||  C j (2) The same argument can be extended to parallel machines The same argument can be extended to parallel machines Theorem Theorem –The SPT rule is optimal for Pm||  C j In fact, a number of optimal schedules exist In fact, a number of optimal schedules exist

4 Weighted total completion time Pm||  w j C j WSPT rule is not always optimal WSPT rule is not always optimal In practice it usually does pretty well In practice it usually does pretty well Worst case bound Worst case bound  w j C j (WSPT) < 1 (1+√2) ≈ 1.2  w j C j (OPT) 2

5 Other completion time models Precedence constraints Pm|prec|  C j Precedence constraints Pm|prec|  C j –Strongly NP-hard Non-identical machines Rm||  C j Non-identical machines Rm||  C j –Can be formulated as an integer program –The solution of the corresponding linear program gives an optimal schedule => Can be solved in polynomial time

6 Section 5.4: Preemptions Pm|prmp|  C j The SPT non-preemptive rule is still optimal The SPT non-preemptive rule is still optimal –A special case of a more general result for Qm|prmp|  C j Recall: v i is the speed on machine i in Qm models Recall: v i is the speed on machine i in Qm models –p j can be interpreted as the required work to complete job j (processing time = p j /v i )

7 Qm|prmp|  C j SRPT-FM rule: SRPT-FM rule: –Process the jobs such that the job with the shortest remaining processing time is put on the fastest machine. The job with the second shortest remaining processing time is put on the second fastest machine, etc. –Whenever the fastest machine completes a job, all the remaining jobs are moved up on the machines Theorem Theorem –The SRPT-FM rule is optimal for Qm|prmp|  C j

8 Example Qm|prmp|  C j 4 machines 4 machines 7 jobs 7 jobs Use the SRPT-FM to solve Qm|prmp|  C j Use the SRPT-FM to solve Qm|prmp|  C j machine i 1234 vivivivi4221 job j pjpjpjpj

9 Section 5.5 Due date related objectives Single machine problems that can be solved “easily” Single machine problems that can be solved “easily” –e.g., 1|prec|L max, 1|r j,prmp|L max Single machine tardiness problems (1||  T j ) are NP-hard Single machine tardiness problems (1||  T j ) are NP-hard If all due dates are 0 If all due dates are 0 –Then Pm||L max is equivalent to Pm||C max => Pm||L max is NP-hard

10 Qm|prmp|L max One of few parallel machine problems that can be solved in polynomial time One of few parallel machine problems that can be solved in polynomial time First question: Is there a schedule such that L max ≤ z ? First question: Is there a schedule such that L max ≤ z ? –This can be formulated (backwards) as Qm|r j,prmp|C max –Let d j =d j +z be a hard deadline

11 Qm|prmp|L max (2) Solve the problem backwards Solve the problem backwards –Find the job k with the latest deadline –Let r k =0 and r j =d k -d j Solve Qm|r j,prmp|C max by applying LRPT-FM Solve Qm|r j,prmp|C max by applying LRPT-FM –The backward schedule is optimal for Qm|prmp|L max

12 Example P2|prmp|L max 4 jobs 4 jobs Let z=0 Let z=0 Solve P2|r j,prmp|C max by LRPT rule Solve P2|r j,prmp|C max by LRPT rule job j 1234 pjpjpjpj3338 djdjdjdj pjpjpjpj3338 rjrjrjrj5410

13 Section 5.6 Discussion Parallel machine models are much harder than single machine models! Parallel machine models are much harder than single machine models! Later in the course we will study heuristics that can be used to obtain “good” schedules Later in the course we will study heuristics that can be used to obtain “good” schedules