Re: [Help-glpk] Slow performance on "Select minimum" task

[Jan van Rijn]

Dear Michael,

Thanks a lot, this seems like a very elegant formulation -- the performance seems much better than my initial performance.

I need to add two things:

- It is my conjecture that there should be an additional constraint, i.e., y[r,c] > 0 (Otherwise non-selected rows could participate towards a lower score)

- M[r,c] should contain positive values (which guarantees that y[r,c] == 1 iff x[r] - SUM x[s] == 1)

Thanks again. This really helped a lot :)

Best regards,

Jan

2018-06-05 11:05 GMT-04:00 Michael Hennebry <address@hidden>:

On Tue, 29 May 2018, Jan van Rijn wrote:

Yes, it is indeed a matrix of floats.

2018-05-29 15:45 GMT-04:00 Kevin Martin <address@hidden>:

I have re-read your problem and I may have misinterpreted it. I thought
your matrix was binary, as in each element was in {0,1}, the sum condition
was supposed to be i:M(j,i)=0, which would be the sum of all the row
inclusion (x_j) variables where the Matrix element for the column was 0.
The idea being that if none of the rows corresponding to a are 0 selected,
the minimum of the column must be 1.

As I re-read your original email, I now think that each element may be
fractional somewhere in the closed interval [0,1]. If this is the case, I
think the problem may be quite hard, for general M, I can?t think of an
obvious way to formulate it better.

A similar mechanism still works:

y[r,c] >= x[r] - SUM x[s]
                 s in Q[r,c]

x is binary
y need not be specified binary
Q[r,c] = { s : M[s,c]< M[r,c] }
The objective is SUM y[r,c]*M[r,c]
                 r,c

The definition of Q assumes no ties.
Ties may be broken arbitrarily, but must be consistent.
You may turn M[s,c]< M[r,c] into a lexigraphic comparison
of (M[s,c], s) and (M[r,c], r) .

--
Michael   address@hiddenedu
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