1 / 18

Lateness Models

Lateness Models. Contents 1. Lawler’s algorithm which gives an optimal schedule with the minimum cost h max when the jobs are subject to precedence relationship 1 | prec | h max 2. A branch-and-bound algorithm for the scheduling problems with the objective to minimise lateness

dudleyr
Télécharger la présentation

Lateness Models

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lateness Models • Contents • 1. Lawler’s algorithm which gives an optimal schedule with the minimum cost hmax • when the jobs are subject to precedence relationship • 1 | prec | hmax • 2. A branch-and-bound algorithm for the scheduling problems with the objective to minimise lateness • 1 | rj | Lmax • Literature: • Scheduling, Theory, Algorithms, and Systems, Michael Pinedo, Prentice Hall, 1995, Chapter 3.2 or new: Second Addition, 2002, Chapter 3.

  2. Lawler’s Algorithm • Backward algorithm which gives an optimal schedule for • 1 | prec | hmaxhmax= max ( h1(C1), ... ,hn(Cn) ) • hjare nondecreasing cost functions • Notation • makespan Cmax =  pj completion of the last job • J set of jobs already scheduled • they have to be processed during the time interval • JC complement ofset J, set of jobs still to be scheduled • J'  JC set of jobs that can be scheduled immediately before set J (schedulable jobs)

  3. Lawler’s Algorithm for 1 | | hmax Step 1. J =  JC = {1,...,n} k = n Step 2. Let j* be such that Place j* in J in the k-th order position Delete j* from JC Step 3. If JC =  then Stop else k = k - 1 go to Step 2

  4. ... 3 5 10 Example (no precedence relationships between jobs) J = JC={1, 2, 3} jobs still to be scheduled Cmax = 10 h1(10) = 11 h2(10) =12 h3(10) =10 Job 3 is scheduled last and has to be processed in [5, 10].

  5. 1 2 3 5 10 2 1 3 5 10 J = {3} JC={1, 2} jobs still to be scheduled Cmax = 5 h1(5) = 6 h2(5) = 6 Either job 1 or job 2 may be processed before job 3. or Two schedules are optimal: 1, 2, 3 and 2, 1, 3

  6. Lawler’sAlgorithm for 1 | prec | hmax Step 1. J = , JC = {1,...,n} J' the set of all jobs with no successors k = n Step 2. Let j* be such that Place j* in J in the k-th order position Delete j* from JC Modify J' to represent the set of jobs which can be scheduledimmediately before set J. Step 3. If JC =  then Stop else k = k - 1 go to Step 2

  7. ... 3 5 10 2 3 2 5 10 1 2 3 5 10 Example. What will happen in the previous example if theprecedence 1  2 has to be taken into account? J =  JC={1, 2, 3} still to be scheduled J'={2, 3} have no successors Cmax = 10 h2(10) = 12 h3(10) = 10 J = {3} JC={1, 2} still to be scheduled J'={2} can be scheduled immediately before J Cmax = 5 h2(5) = 6 J = {3, 2} JC={1} J'={1} h1(2) = 3 Optimal schedule: 1, 2, 3, hmax= 10 2

  8. 1 || Lmax is the special case of the 1 | prec | hmax where hj = Cj - dj algorithm results in the schedule that orders jobs in increasing orderof their due dates - earliest due date first rule (EDD) 1 | rj | Lmax is NP hard , branch-and-bound is used 1 | rj , prec | Lmax similar branch-and-bound

  9. S S1 S2 Sn . . . S12 S13 . . . . . . • Branch-and-bound algorithm • Search space can grow very large as the number of variables in the problem increases! • Branch-and-bound is a heuristic that works on the idea of successive partitioning of the search space. S = S1 S2... Sn S1 S2...  Sn = 

  10. S S1 S2 Sn . . . S12 S13 . . . • We need some means for obtaining a lower bound on the cost for any particular solution (the task is to minimise the cost). fboundf(x), xS1 fbound f(x), xS2 there is no need to exploreS2

  11. Branch-and-bound algorithm Step 1 Initialise P =  Si(determine the partitions) Initialise fbound Step 2 Remove best partition Si from P Reduce or subdivide Si into Sij Update fbound P = PSij For all SijP do if lower bound of f(Sij) > fbound then remove Sij from P Step 3 If not termination condition then go to Step 2

  12. *,*,*,* 1,*,*,* 2,*,*,* n,*,*,* . . . 1,2,*,* 1,3,*,* . . . . . . Branch-and-bound algorithm for1 | rj | Lmax • Solution space contains n! schedules (n is number of jobs). • Total enumeration is not viable !

  13. *,*,*,* 1,*,*,* 2,*,*,* n,*,*,* . . . 1,2,*,* 1,3,*,* . . . . . . 1st level 2nd level Branching rule: k-1 level, j1, ... , jk-1 are scheduled, jk need to be considered if no job still to be scheduled can notbe processed before the release time of jk that is: J set of jobs not yet scheduled t is time when jk-1 is completed

  14. Lower bound: • Preemptive earliest due date (EDD) rule is optimal for 1 | rjprmp | Lmax • A preemptive schedule will have a maximum lateness not greater than a non-preemtive schedule. • If a preemptive EDD rule gives a nonpreemptive schedule then all nodes with a larger lower bound can be disregarded.

  15. 1 2 12 3 7 2 1 0 5 9 2 1 2 0 3 7 9 Example. • Non-preemptive schedules L1=3 L2=6 Lmax=6 0 L1=5 L2=-1 Lmax=5 • Preemptive schedule obtained using EDD L1=3 L2=3 Lmax=3 • the lowest Lmax !

  16. Example *,*,*,* L.B. = 5 L.B. = 7 1,*,*,* 2,*,*,* 3,*,*,* 4,*,*,* job 1 could be processedbefore job 4 1,2,*,* 1,3,*,* job 2 could be processedbefore job 3 L.B. = 5 L.B. = 6 1,3,4,2

  17. *, *, *, * 1,*,*,* 1 [0, 4] L1=-4 3 [4, 5] 4 [5, 10] L4=0 3 [10, 15] L3=4 2 [15, 17] L2=5 2,*,*,* 2 [1, 3] L2=-9 1 [3, 7] L1=-1 4 [7, 12] L4=2 3 [12, 18] L3=7 4,*,*,* either job 1 or 2can be processedbefore 4 ! 3,*,*,* job 2can be processedbefore 3 ! 1,2,*,* 1 [0, 4] L1=-4 2 [4, 6] L2=-6 4 [6, 11] L4=1 3 [11, 17] L3=6 1,3,*,* 1 [0, 4] L1=-4 3 [4, 10] L3=-1 4 [10, 15] L4=5 2 [15, 17] L3=5 Schedule: 1, 3, 4, 2,

  18. Summary • 1 | prec | hmax , hmax=max( h1(C1), ... ,hn(Cn) ), Lawler’s algorithm • 1 || Lmax EDD rule • 1 | rj | Lmax is NP hard , branch-and-bound is used • 1 | rj , prec | Lmax similar branch-and-bound • 1 | rj, prmp | Lmax preemptive EDD rule

More Related