I understand the column status. For row, if I understood well, the row is active if Ax=b for the problem Ax<=b with x auxiliary variables. So, structural variables = 0
my algorithm is: ->solve problem ->delete columns and rows ->add columns ->resolve problem But, if I delete column or rows, I must to have a basis solution to resolve my problem. So I must resolve my problem after fix the column status my algorithm become: ->solve problem ->change status ->resolve problem ->delete columns and rows ->add columns ->resolve problem Andrew Makhorin wrote: > >> Yes, I delete Rows and Cols > > If the current basis is valid (whether optimal or not) and if you > delete active rows (i.e. the rows for which glp_get_row_stat returns > GLP_NL, GLP_NU, GLP_NF, or GLP_NS) and/or basic columns (i.e. the > columns for which glp_get_col_stat returns GLP_BS), the basis becomes > invalid. To keep it valid you either have not to delete such rows or > columns or have to change the statuses of remaining rows and columns > appropriately. > >> For Row, must I use glp_set_row_stat(lp , i , GLP_NL)? > >> For Column, must I use glp_set_col_stat(lp , i , GLP_NL)? > > Do you understand what is the row/column status? > >> How can I fix my column at 0? > > glp_set_col_bnds(lp, j, GLP_FX, 0.0, 0.0); > >> must I resolve my problem before to add column or row? > >> Or > >> must I only start again the routine lpx_std_basis(lp)? > > lpx_std_basis makes all auxiliary variables basic and all structural > variables non-basic, i.e. using it you lose all the current basis > information. > > > > > > _______________________________________________ > Help-glpk mailing list > [email protected] > http://lists.gnu.org/mailman/listinfo/help-glpk > > -- View this message in context: http://www.nabble.com/Problem-%3A-basis-tf3882882.html#a11022242 Sent from the Gnu - GLPK - Help mailing list archive at Nabble.com. _______________________________________________ Help-glpk mailing list [email protected] http://lists.gnu.org/mailman/listinfo/help-glpk
