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

This preview shows pages 1–10. Sign up to view the full content.

IEE 533 Scheduling 1 Section 3.2  The Maximum Lateness Prepared by: Pornsarun Wirojanagud

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
IEE 533 Scheduling 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
IEE 533 Scheduling 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
IEE 533 Scheduling 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
IEE 533 Scheduling 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
IEE 533 Scheduling 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 ).
IEE 533 Scheduling 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
IEE 533 Scheduling 8 Theorem 3.2.2 (cont.) j* j** h j (C j ) C j* ,C j** h j** h j*
IEE 533 Scheduling 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**

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern