Accept or Reject: Can we get the work done in time? Marjan van den Akker Joint work with Han Hoogeveen

Outline of the talk Problem description Review Moore-Hogdson Stochastic processing times –consequences –four classes of instances Summary

Problem setting n jobs become available at time 0 Known processing time Known due date Known reward for completing in time (currently 1) Decision to make now: accept or reject minimize number of tardy jobs on a single machine

Moore-Hodgson 1.Number the jobs in EDD order 2.Let S denote the EDD schedule 3.Find the first job not on time in S (suppose this is job j) 4.Remove from S the largest available job from jobs 1,…,j 5.Continue with Step 3 for this new schedule S until all jobs are on time

Solving the problem from scratch Observations First the on time jobs On time jobs in EDD order Forget about the late jobs Knowing the on time set is sufficient

Dominance rule Let E 1 and E 2 be two subsets of jobs 1,…,j All jobs in E 1 and E 2 are on time (feasible) Cardinality of E 1 and E 2 is equal The total processing time of the jobs in E 2 is more than the total processing time of the jobs in E 1 Then subset E 2 can be discarded.

Proof (sketch) Take an optimal schedule starting with E 2 (remainder: jobs from j+1, …, n) E2E2 remainder E1E1 time 0

Use dynamic programming Find E j * (k): feasible subset of jobs 1,…,j with cardinality k and minimum total processing time Use state variables f j (k) equal to p(E j * (k)) Define z j as maximum number of on time jobs from jobs 1, …, j Initialization: f j (k)=0 for j=k=0 (and +  otherwise)

Recurrence relation Put f j+1 (0)=0 f j+1 (k)=min{f j (k),f j (k-1)+p j+1 } (k=1,…,z j ) If f j (z_j)+p j+1  d j+1 then z j+1 =z j +1 and f j+1 (z j+1 )=f j (z j )+p j+1 ; otherwise, z j+1 =z j.

Relation with Moore-Hodgson The set E j *(k-1) can be computed from E j * (k) by removing the largest job! Recurrence relation f j+1 (k)=min{f j (k),f j (k-1)+p j+1 } Equal to: remove the largest job

Moore-Hodgson 1.Number the jobs in EDD order 2.Compute the values f j (z j ): –If f j (z_j)+p j+1  d j+1 then z j+1 =z j +1 and f j+1 (z j+1 )=f j (z j )+p j+1 i.e. J j+1 is added –else z j+1 = z j and f j+1 (z j+1 ) = min{f j (z j ),f j (z j -1)+p j+1 } i.e. largest job is removed

Stochastic processing times Completion times are uncertain Decision about accept or reject must be made before running the schedule When do you consider a job on time?

On time stochastically Work with a sequence of on time jobs (instead of a set of completion times) Add a job to this sequence and compute the probability that it is ready on time If this probability is large enough (at least equal to the minimum success probability msp) then accept it as on time

Classes of processing times Gamma distribution Negative binomial distribution Equally disturbed processing times p_j Normal distribution Jobs must be independent

Class 1: Gamma distribution Parameters a_j and b (common) If x_1 and x_2 follow the gamma distribution and are independent, then x_1+x_2 is gamma distributed with parameters a_1+a_2 and b.

More gamma Define S as the set of the job j and all its predecessors in the schedule Define p(S) as the sum of all processing times of jobs in S Then C_j=p(S) follows a gamma distribution with parameters a(S) and b.

Even more gamma Denote the msp of job j by y_j Job j is on time if the probability that C_j is no more than d_j is at least y_j C_j depends on a(S) only Given d_j and y_j, you can compute the maximum value of a(S) such that P(C j <= d j ) is at least y j : call it D_j

Last of Gamma Treat D_j as ordinary due dates Treat a_j as ordinary deterministic processing times Then the dominance rule still holds You can use Moore-Hodgson! Negative binomial distribution: similar

More complicated problems p j Var dj dj msp j Job Job Normally distributed processing times Optimum: first job 2 and then job 1 From now on: equal msp values  EDD-order

Equal disturbances On time probability of job j depends on: –Number of predecessors (on time jobs before j) –Total processing time of its predecessors Dominance rule: given the cardinality of the on time set, take the one with minimum total processing time Use dynamic programming with state variables f j (k) that indicate the minimum total processing possible (as before) Hence: Moore-Hodgson’s solves it!

Normal distribution (1) Parameters: expected processing time of job j and variance of job j Reminder: expected value and variances of X 1 +X 2 are equal to the respective sums Necessary for computing the on time probability of job j: –Total processing time of predecessors –Total variance of predecessors

Normal distribution (2) Dominance rule: if cardinality and total processing time are equal, then take the set with minimum total variance (msp > 0.5) Use state variables f j (k,P): –k is cardinality of on time set –P is total processing time of on time set –f j (k,P) is minimum variance possible

Normal distribution: details Running time pseudo-polynomial Problem is NP-hard Role of total variance and total processing time in the dominance rule and in the DP is interchangeable

What to remember (optional) Moore-Hodgson = Dynamic Programming DP is applicable in a stochastic environment –Stochastic on time: work with the minimum success probability –EDD sequence optimal for the on time set?? Weighted case can be solved in a similar way

Yes: single machine, minimize number of tardy jobs Solvable in O(n log n) time by Moore-Hodgson Known problem?

Miscellaneous remarks DP computes more state variables than necessary DP can be used for the weighted case: –Use f j (W) with W is the total weight of the on time set (instead of cardinality of the on time set) DP can be used for more problems (to be shown next)

Negative binomial distribution Parameters s_j and p (common for all jobs) If independent, then C_j=p(S) follows a negative binomial distribution with parameters s(S) and p Same as gamma distribution