Approximation Algorithms for scheduling

Transcription

1 Approximation Algorithms for scheduling Ahmed Abu Safia I.D.: , McGill University, 2004 (COMP 760)

2 Approximation Algorithms for scheduling Leslie A. Hall The first Chapter of the book entitled Approximation Algorithms for NP-Hard problems edited by Dorit S. Hochbaum Introduction Sequencing with release dates to minimize Lateness Jackson s Rule A simple 3/2 approximation algorithms Identical Parallel Machines Unrelated Parallel Machines Conclusion ג Introduction The problem: We have a set of n jobs J 1,..J n and m identical machines. M 1,..., M m. Each job J j must be processed without interruption for a time p j > 0 on one of the m machines, each of which can process at most one job at a time. List scheduling (LS) Whenever a machine becomes available the next job on the list is assigned to begin processing on that machine. claim: The total makespan (the last job s finish time) of this schedule is at most 2 - (1/m) times the makespan of the optimal schedule. proof Let us denote the starting time and completion time of job J j in the heuristic schedule as s j and C j, respectively, for j = 1,...,n. see (fig-1) If we focus on the last job to complete processing (call it J k ), then no machines can be idle at any time prior to s k, since otherwise job J k would have been processed there. C max p k C max (1/m) n j=1 C LS max = C k = s k + p k (1/m) j k p j + p k p j 2

3 n = (1/m) p j + (1 1/m)p k j=1 C max + (1 1/m)C max = (2 1/m)C max Sequencing with release dates to minimize Lateness Remember: Once a job begins processing on a machine it must complete its processing without interruption, which contrasts with preemptive scheduling models. Problems are described by three fields: The first: represents the machine environment, the second describes any special conditions or constraints in the model, and the third describes the objective function criterion. P = p j here we consider a single machine, each job J j has a delivery time q j, as well as processing time p j > 0. We define the lateness of a job as the delivery completion time. L j = s j + p j + q j Our goal is to find a schadule that minimizes the maximaum lateness over all jobs, L max = max 1 j n L j On a single machine we can see that an optimal ordering of the jobs to minimize L max is given by ordering the jobs according to non-increasing delivery time. Lets introduce, job release dates into the model, each job J j can not begin its processing before its release date r j. This model is an example of 1 r j L max. This problem is strongly NP-hard, yet one of the easiest. We will use the following lower bounds of the single-machine problem: Jackson s Rule L max P L max r j + p j + q j, j = 1,..., n Lets looks at a generalized version of List Scheduling Theorem: L LS max < 2L max and this bound is tight Proof: Let J k be the job whose lateness attains that of the schedule, i.e.,l LS max = C k +q k. Since there is no idel time on the machine between time r k and s k, we have: L LS max = s k + p k + q k < (r k + P ) + p k + q k = r k + p k + q k + P 2L max Jackson s Rule apply list scheduling to a list ordered by non-increasing delivery times. Corollary: L J max < 2L max and this bound is tight Let s examine this heuristic in more details. We have ane instance of 1 r j L max and a schedule given by Jackson s rule. Define: critical job J c : one whose lateness attains that of the schedule 3

4 critical sequence: Those jobs tracing backward from J c in J a : the first job in. S: the set of jobs in the critical sequence. L J max = L c = r a + P (S) + q c Observe that for every job J j in the critical sequence, r a r j. For any set of jobs S L max min J j S r j + P (S) + min J j S q j if q c q j for all J j S then the schedule is optimal. L J max L max + q c How does Jackson s Rule go wrong?? If S contains a job J b with q b < q c, then the sequence might not be optimal. Let J b be the last job in S with q b < q c, we call it an interference job, then: L J max L max + p b proof: consider a sub set S S of those jobs processed after J b. By the way J b was chosen, q j q c for all jobs J j S, implies q j q b for all jobs J j S, implies that no job J j S could have been available at time s b, else it would have been taken priority over J b. Thus, r j > s b for all J j S. So, L max min r J j S j + P (S ) + min q J j S j > s b + P (S ) + q k and L J max = s b + p b + P (S ) + q k < L max + p b A simple 3/2 approximation algorithms Due to Nowicki and Smutnicki. consider an instance of 1 r j L max. Let job J d, if it exists, be the unique job with p d > P/2. In no such a job exists, then the schedule given by Jackson s satisfies L J max L max + P/2 (3/2)L max Define: A = {J j : j d, r j q j, j = 1,..., n} When J d exists, clearly P (A) + P (B) < P/2 Algorithm NS B = {J j : j d, r j > q j, j = 1,..., n} step-1: Construct a schedule by Jackson s rule, and determine a critical job J c and a critical sequence. If there exists no interference job J b, then stop and return this schedule. step-2: If min{p b, q c } P/2, then stop and return the schedule of step-1. Otherwise order the jobs of A according to nondecreasing release dates and of B according to non-increasing delivery times. Construct a schedule given by the ordered set A, followed by J b, followed by the ordered set B. Return the better of this schedule and that constructed in step-1. Identical Parallel Machines Theorem: List scheduling is a 2-approximation algorithm for the scheduling problem P r j, prec L max. proof is left to you as an exercise 4

5 Unrelated Parallel Machines Due to Lenstra, Shmoys, and Tardos. Based on a 2-relaxed decision procedure; that is given a target length of T, the procedure with either correctly deduce that no schedule exits, or it will construct a schedule with make span at most 2T. We will binary search on T. Given T, construct the following: for each job J j : for each machine M i : M(j) := {i : p ij T } J(i) := {j : p ij T } Now consider the following linear feasibility system x ij = 1, j = 1,..., n; i M(j) j J(i) p ij x ij T, i = 1,..., m; x ij 0, i = 1,..., m, j = 1,..., n. A basic feasible solution has at most n+m variables are positive and at most m jobs split (fractional) assignment. As before construct it in two phases: first assign each un-split job to its prober machine and then we focus on assigning the jobs with fractional assignment. If we construct a matching of the fractional jobs to the machines in such a way that each job gets matched to a machine it is already partially assigned to, then we will have constructed an assignment of the fractional jobs that is guaranteed to have length at most T. Fortunately, the feasibility system explained is an example of a generalized assignment problem, whose basic feasible solutions are well known to have a very special form. Construct a graph whose nodes are the jobs and machines of the instances, and whose edges (i,j) correspond to those variables x ij > 0. If the solution x is basic, then the graph constructed will consist of a forest of trees and 1-trees., in which job nodes and machine nodes alternate. delete the job nodes with integral assignment from the graph. (see fig-2), Our task is to construct a matching in this graph, in which every job node gets matched. This matching will corresponds to the desired assignment. HOW Conclusion The results presented here are the most basic models in scheduling. There is a massive literature and if you are interested then there is a list provided by the author and can be found at the end of the chapter. 5