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Re: [Help-glpk] Using max on an array of decision variables

From: Mark Gritter
Subject: Re: [Help-glpk] Using max on an array of decision variables
Date: Fri, 7 Sep 2007 17:44:52 -0500

What you can do is use an auxiliary variable to hold the maximum.
Let's call that m.  Then you need a new set of constraints saying that
"m" has to be greater than or equal to all the completion times.

s.t. maxim{s in SERVERS} : m >= CompletionTime[s]

Your objective should then be to minimize m.  The minimization ensures
that m is the maximum (and no larger) of the completion times.

If you don't use CompletionTime[s] for anything else you can eliminate
it and just write your constraints in the form m >= {forumula for
completion time for s}.


On 9/7/07, glpkuser <address@hidden> wrote:
> As part of my decision variables, I will have an array of numbers, where each
> element is the execution time to complete tasks at a server. The LP is to
> spread some load across the servers such that the maximum completion time
> across all tasks is minimised. I cannot figure out how to express this
> objective.
> I have something like this:
> set SERVERS;
> var CompletionTimes{s in SERVERS} >= 0;
> s.t. compl{s in SERVERS} : CompletionTime[s] = .......;
> minimize z: max(CompletionTimes{s in SERVERS}); # Doesn't work
> set SERVERS := 1 2 3 4 5;
> --
> View this message in context: 
> Sent from the Gnu - GLPK - Help mailing list archive at
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> Help-glpk mailing list
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