Re: [Bug-glpk] Error delete column

[Andrew Makhorin]

> That means that the basis matrix, i.e. a square matrix built from
> columns of the constraint matrix corresponding to basic variables, is
> singular, so the current basis defined by statuses of rows and columns
> is invalid in the sense that it does not define any basic solution.

> So I must do:

> glpkParamPtr.presolve=GLP_OFF;

>   int ret=glp_simplex(glpkProbPtr,&glpkParamPtr);

>   if (ret == GLP_EBADB){
>     ret = lpx_warm_up(glpkProbPtr);
>     if (ret == LPX_E_BADB){
>       lpx_std_basis(glpkProbPtr);
>       if(glp_simplex(glpkProbPtr,&glpkParamPtr)!=0)
>         cout<<"could not solve LP"<<endl;
>     }
>   } 
>   else if(ret == GLP_ESING){
>     ret = lpx_warm_up(glpkProbPtr);
>     if (ret == LPX_E_SING){
>       lpx_std_basis(glpkProbPtr);
>     }
>     if(glp_simplex(glpkProbPtr,&glpkParamPtr)!=0)
>       cout<<"could not solve LP"<<endl;
>   }

> Because, the LP solver use the precedent solution as initial solution
> before to resolve the LP.

If you did not remove active rows and basic columns and not change
constraint coefficients, the basis must remain valid.

> So, I must find a valid initial solution if a basis matrix is singular
> to obtain a invertible basis matrix

Yes. However, it is important to understand why the basis became
invalid.