32 - Section3.2TheMaximumLateness Preparedby: IEE 533 1...

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IEE 533 1 Section 3.2  The Maximum Lateness Prepared by: Pornsarun Wirojanagud
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IEE 533 2 Maximum Lateness 1|prec|h max   where h max  = max (h 1 (C 1 ), …, h n (C n )) with  h j , j = 1,…,n is nondecreasing cost functions The objective is a due date-related as the functions h j   may take any one of the due date-related penalty  functions
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IEE 533 3 Due Date-Related Penalty Functions T j C j d j Tardiness L j C j d j Lateness C j d j U j 1 Unit Penalty
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IEE 533 4 Maximum Lateness The completion of the last job occurs at the  makespan C max  =  Σ p j , which is independent of the  schedule  Let J = set of jobs already scheduled which are processed during the time interval J c  = set of jobs still to be scheduled J’ = set of jobs in J c  that can be scheduled  immediately before set J (set of schedulable jobs) The following backward algorithm yields optimal  solution - max J j j max C , p C
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IEE 533 5 Algorithm 3.2.1 (Minimizing Maximum Cost) Step 1 Set J = 0, J c  = {1,…,n} and J’ the set of all jobs with no  successors. Step 2 Let j* be such that Add j* to J Delete j* from J c   Modify J’ to represent the new set of schedulable jobs Step 3   If J c  = 0 STOP, otherwise go to step 2. = c c J k k j ' J j J j j * j p h min p h
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IEE 533 6 Algorithm 3.2.1 (Minimizing Maximum Cost) The worst case computation time required by this  algorithm can be established as follow There are n steps needed to schedule the n jobs. At each step at most n jobs have to be considered. The overall running time of the algorithm is bounded by  O(n 2 ).
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IEE 533 7 Theorem 3.2.2 Proof: By contradiction At a given iteration job j**, selected from J’, does not have the  minimum completion cost among the jobs in J’ If job j** is selected, the minimum cost job j* must then be  scheduled at a later iteration That implies that job j* has to appear in the sequence before  job j** A number of jobs may even appear between jobs j* and j** c J j j * j p h Algorithm 3.2.1 yields optimal schedule for 1|prec|h max
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IEE 533 8 Theorem 3.2.2 (cont.) j* j** h j (C j ) C j* ,C j** h j** h j*
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IEE 533 9 Theorem 3.2.2 (cont.) To show that this sequence cannot be optimal Take job j* and insert it in the schedule immediately  following job j** All jobs in the original schedule between jobs j* and j**, 
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This note was uploaded on 10/07/2011 for the course IEE 533 taught by Professor Johnfowler during the Spring '10 term at ASU.

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32 - Section3.2TheMaximumLateness Preparedby: IEE 533 1...

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