Parallel Machines (Q||ΣCi , Q|pmtn|ΣCi , R||ΣCi , R|pmtn|Cmax, R|pmtn|Lmax) Lesson 8

Q||ΣCi t Assume that i1, i2,..., ir is the sequence of jobs to be processed on machine Mj. t Mj i1 i2 i3 ● ● ● ir pi1/sj (pi1+ pi2+pi3 +...+pir)/sj (pi1+ pi2) /sj (pi1+ pi2+pi3)/sj The contribution of these jobs is This implies that in an optimal schedule the jobs on machine Mj are sequenced according to nondecreasing processing requirements pi.

Optimal Strategy Let t1, t2,..., tn be a nondecreasing sequence of the n smallest numbers in the set If then schedule job i on Mj as the k-th last job because we assume p1 ≥ p2 ≥ ... ≥ pn.

Q||ΣCi Consider an instance I with 2 machines and 10 jobs. Let s1= 2 and s2= 5. Let p1= 10, p2= 9, p3= 8, p4= 7, p5= 6, p6= 5, p7= 4, p8= 3, p9= 2, and p10= 1.

Q||ΣCi Consider an instance I with 2 machines and 10 jobs. Let s1= 2 and s2= 5. Let p1= 10, p2= 9, p3= 8, p4= 7, p5= 6, p6= 5, p7= 4, p8= 3, p9= 2, and p10= 1. First, we find the 10 smallest numbers:

Q||ΣCi Consider an instance I with 2 machines and 10 jobs. Let s1= 2 and s2= 5. Let p1= 10, p2= 9, p3= 8, p4= 7, p5= 6, p6= 5, p7= 4, p8= 3, p9= 2, and p10= 1. First, we find the 10 smallest numbers: Second, we find the assignment:

Q||ΣCi Consider an instance I with 2 machines and 10 jobs. Let s1= 2 and s2= 5. Let p1= 10, p2= 9, p3= 8, p4= 7, p5= 6, p6= 5, p7= 4, p8= 3, p9= 2, and p10= 1. First, we find the 10 smallest numbers: Second, we find the assignment: M1 10 7 3 M2 t 1 2 3.2 4.6 6.5 8.4

Q||ΣCi Consider an instance I with 2 machines and 10 jobs. Let s1= 2 and s2= 5. Let p1= 10, p2= 9, p3= 8, p4= 7, p5= 6, p6= 5, p7= 4, p8= 3, p9= 2, and p10= 1. First, we find the 10 smallest numbers: Second, we find the assignment: M1 10 7 3 M2 9 8 6 5 4 2 1 t 1 2 3.2 4.6 6.5 8.4

Optimal Strategy Let t1, t2,..., tn be a nondecreasing sequence of the n smallest numbers in the set If then schedule job i on Mj as the k-th last job because we assume p1 ≥ p2 ≥ ... ≥ pn.

Optimal Strategy Let t1, t2,..., tn be a nondecreasing sequence of the n smallest numbers in the set If then schedule job i on Mj as the k-th last job because we assume p1 ≥ p2 ≥ ... ≥ pn. Optimality of this strategy is a consequence of the following result about Monge array.

Monge Array

Q|pmtn|ΣCi To solve this problem we apply an adapted version of the SPT-rule. Order the jobs according to nondecreasing processing requirements, and schedule each successive job preemptively so as to minimize it completion time. In other words, we schedule job n on the fastest machine M1 until it is completed at time t1 = pn/s1. Then we schedule job n – 1 first on machine M2 for t1 time units and then on machine M1 from time t1 to time t2 ≥ t1 until it is completed. Job n – 2 is scheduled on M3 for t1 time units, on M2 for t2 – t1 time units, and on machine M1 from time t2 to time t3 ≥ t2 until it is completed, etc.

Q|pmtn|ΣCi Consider an instance I with 2 machines and 10 jobs. Let s1= 2 and s2= 5. Let p1= 10, p2= 9, p3= 8, p4= 7, p5= 6, p6= 5, p7= 4, p8= 3, p9= 2, and p10= 1. M1 9 8 M2 10 9 8 t 0.2 0.52 0.992 A simple interchange argument may be used to prove the correctness of the algorithm.

Unrelated Machines We have n independent jobs i = 1,..., n to be processed on m machines. The processing time of job i on machine Mj is pij (i = 1,..., n; j = 1,..., m). This model is a generalization of the uniform machine model we get by setting pij = pi/sj .

No preemption. The unrelated machines model is a generalization of the uniform machine model. Remind that, problem Q||ΣCi is the only problem which was polynomially solvable in the case of non-preemptable jobs with arbitrary processing times. We show that a polynomial algorithm also exists for the corresponding problem with unrelated machines.

R||ΣCi We reduce this problem to assignment problem. Again, if i1, i2,..., ir is the sequence of jobs processed on machine Mj. Mj i1 i2 i3 ● ● ● ir t pi1j (pi1j+ pi2j+pi3j +...+pirj) (pi1j+ pi2j) (pi1j+ pi2j+pi3j) Then the contribution of these jobs in the objective function is

Assignment t Mj i1 i2 i3 ● ● ● ir pi1j (pi1j+ pi2j+pi3j +...+pirj) (pi1j+ pi2j) (pi1j+ pi2j+pi3j) We define a position of a job on a machine by considering the job processed last on the first position, the job processed second from last on the second position, etc. To solve problem R||ΣCi we have to assign the jobs i to positions k on machines j. The cost of assigning job i to (k, j) is kpij.

Property of Assignment Problem Note that an optimal solution of this assignment problem has the following property: if some job i is assigned to position k > 1 on machine j, then there is also a job assigned to position k – 1 on machine j. Otherwise, scheduling job i in position k – 1 would improve the total assignment cost. A solution of the Assignment Problem always yields an optimal solution of our scheduling problem.

R|pmtn|Cmax We solve problem R|pmtn|Cmax in two steps. We formulate a linear program to calculate for each job i and each machine j the amount of time tij machine j works on job i in an optimal schedule. We construct a corresponding schedule.

R|pmtn|Cmax Given nm positive integers pij, which represents the total processing times of job i on machine Mj. Let tij be the processing times of that part of job i which is processed on Mj . Then tij/ pij is the fraction of time that job i spends on machine j, and must hold in order for job i to be completed.

LP the total time job i spends on all machines. the total time machine Mj spends processing jobs.

Schedule An optimal solution of LP determines for each job i and each machine j the amount of time tij machine j works on job i in an optimal schedule. M1 1 2 5 M2 1 2 3 M3 3 4 M4 1 3 5 Cmax

Schedule An optimal solution of LP determines for each job i and each machine j the amount of time tij machine j works on job i in an optimal schedule. Also we have

LP the total time job i spends on all machines. the total time machine Mj spends processing jobs.

Schedule An optimal solution of LP determines for each job i and each machine j the amount of time tij machine j works on job i in an optimal schedule. Also we have To problem of finding a corresponding schedule is equivalent to the problem of finding a solution to the O|pmtn|Cmax problem.

O|pmtn|Cmax An optimal schedule of the O|pmtn|Cmax problem achieves the bound An optimal schedule of the O|pmtn|Cmax problem can be constructed in polynomial time by using the reduction to Bipartite Matching problem.

R|pmtn|Lmax Assume that the jobs are numbered in nondecreasing due date order, i.e. d1  d2  ...  dn. Let be the total amount of time that machine Mj spends on job i within the time period I1=[0, d1 +Lmax]. Furthermore, for k = 2,..., n let be the total amount of time that machine Mj spends on job i within the time period Ik=[dk–1 +Lmax, dk +Lmax]. t d1 d2 d3 d4 t d1+Lmax d2+Lmax d3+Lmax d4+Lmax

LP the total time job i spends on all machines in Ik. the total time machine Mj spends processing jobs in Ik.

Schedule Given an optimal solution of this LP, an Lmax-optimal schedule can be obtained by constructing for each of the time periods Ik (k = 1,..., n) a corresponding schedule using the data given by the matrix For each interval Ik we get an instance of O|pmtn|Cmax problem and problem R|pmtn|Lmax is polynomially solvable.

Exercise Formulate LP for R| pmtn, ri |Lmax .