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