![]() ![]() ![]() ![]() I can not decide which of these is going to be optimal. In each case we will have take in to account the chosen jobs being compatible, needless to say. If I go for alternate fashion, I choose the first possible job in the first resource, then second possible job in the second resource and so on. If I go for sequential manner, I schedule all the possible jobs in the first resource and then do the same for second resource with the jobs yet to be scheduled. My problem is, how do I proceed after that? Do I choose the resources in sequential or in an alternate fashion? ![]() How do I schedule my jobs now? As my idea goes, again you start by sorting the requests in order of finish time. Now, let us say we have two resources instead of one. A formal explanation is given by a Charging argument. Continue until the set of candidate intervals is empty. Remove x, and all intervals intersecting x, and all intervals in the same group of x, from the set of candidate intervals. In order to schedule the $n$ job requests over one resource, you sort the requests in order of finish time, choose the request with earliest finish time, choose the next compatible one, and so on. Select the interval, x, with the earliest finishing time. Consider the interval scheduling problem, see also here. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |