[Cbc] Initial integer feasible solution

[Pratim Vakish]

Thanks Bob.

However, I think that I am doing what you suggest and it is not working.
Here is what I do.
I set my integer variables to a given value as follows:

let {i in T} x[i]:= xinit[i];

The solution defined by the xinit values is integer feasible.
However, CBC does not use this integer solution as the incumbent one in the tree, since CBC takes a lot of time and nodes in the tree before coming up with a first integer feasible solution.

So could you tell me why I am having this issue and how to resolve it?


Many thanks,

Pratim



From: 4er at iems.northwestern.edu
To: pratim_usc at hotmail.com; cbc at list.coin-or.org
Subject: RE: [Cbc] Initial integer feasible solution
Date: Thu, 2 Oct 2008 23:51:09 -0500






















Just assign the incumbent values to the AMPL variables, and AMPL
will automatically pass those values to the solver.  Then the solver is
free to do whatever it can with them.

 

Bob Fourer

4er at ampl.com

 

 











From:
cbc-bounces at list.coin-or.org [mailto:cbc-bounces at list.coin-or.org] On Behalf Of Pratim Vakish

Sent: Thursday, October 02, 2008
8:12 PM

To: cbc at list.coin-or.org

Subject: [Cbc] Initial integer
feasible solution



 

Hello,



I am solving large-scale MIP problems using the CBC solver and the AMPL
interface.



For some problems, I know an integer feasible solution of good
"quality" and I would like CBC to use it as first incumbent solution
in the branch-and-bound tree.

I think that the CBC option CbcModel::setBestSolution()
allows to do that.



I use AMPL and I would like to know what is the AMPL command to use this
option?

How can I tell CBC (with AMPL) which solution I would like to use as first
incumbent solution?



Thanks very much,

 

Pratim

 

 









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