tag:blogger.com,1999:blog-593563533834706486.post4796073745868292805..comments2024-03-28T10:35:10.453-04:00Comments on Yet Another Math Programming Consultant: Another difficult MIPErwin Kalvelagenhttp://www.blogger.com/profile/09496091402502236997noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-593563533834706486.post-88358895688136182102009-01-16T09:04:00.000-05:002009-01-16T09:04:00.000-05:00In my case the problem is highly combinatorial, wi...In my case the problem is highly combinatorial, with lots of symmetry, and a poorly defined cost structure to distinguish solutions. I got better performance by removing some of the symmetry. About your case: there is very little I can say in general, without studying the actual model.Erwin Kalvelagenhttps://www.blogger.com/profile/09496091402502236997noreply@blogger.comtag:blogger.com,1999:blog-593563533834706486.post-56494887391098301602009-01-16T08:55:00.000-05:002009-01-16T08:55:00.000-05:00what I mean by expensive is the struture of the co...what I mean by expensive is the struture of the constraints not the numbers. The constraints may be too loose to narrow down the solutions especially in some continuous-type representation rather than discrete models.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-593563533834706486.post-56935975730108156182009-01-15T11:11:00.000-05:002009-01-15T11:11:00.000-05:00For a MIP more equations is actually often better,...For a MIP more equations is actually often better, so constraints being 'too expensive' is probably not an issue.Erwin Kalvelagenhttps://www.blogger.com/profile/09496091402502236997noreply@blogger.comtag:blogger.com,1999:blog-593563533834706486.post-20200263863744038102009-01-15T10:17:00.000-05:002009-01-15T10:17:00.000-05:00I have a very similiar situation. any thought why ...I have a very similiar situation. any thought why they are so difficult to converge ? I am guessing some constraints are too expensive....Anonymousnoreply@blogger.com