Re: [Help-glpk] Save solution in GLPK

[Cedric[FR]]

I don't understand the zero bound.

My new column has a lower bound equal at zero. 
So, It's always feasible.

And, the status is non-basis.

I tested my example without the routine lpx_std_basis

Now, the result is good:

      0:   objval =   0.000000000e+00   infeas =   1.000000000e+00 (0)
     15:   objval =   3.497400000e+04   infeas =   0.000000000e+00 (0)
*    15:   objval =   3.497400000e+04   infeas =   0.000000000e+00 (0)
*    29:   objval =   2.652371454e+02   infeas =   0.000000000e+00 (0)
OPTIMAL SOLUTION FOUND
*    29:   objval =   2.652371454e+02   infeas =   0.000000000e+00 (0)
*    31:   objval =   6.586143114e+01   infeas =   0.000000000e+00 (0)
OPTIMAL SOLUTION FOUND

But, At the 55 st problem, I obtain:

spx_simplex: initial basis is invalid




Andrew Makhorin wrote:
> 
>>> At the second simplex, GLPK starts with an objval=0 instead of the
>>> precedent solution.
>> 
>>> Why?
>> 
>> If you add columns to the problem object, they become non-basic,
>> so the optimal basis must remain at least primal feasible, however,
>> it is primal infeasible (as seen from the solver output). Which
>> changes in the problem object did you make before re-optimiziation?
>> Looks like you changed bounds of some basic rows/columns.
> 
> Make sure that after addition of new columns to the problem object
> they are non-basic and on their *zero* bounds. If a new column has
> a non-zero bound, adding it (along with the column of the constraint
> matrix) may make the basis primal infeasible.
> 
> 
> 
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