Re: [Help-glpk] Problem : basis

[Cedric[FR]]

If I changed column/row status, I didn't must to delete column/row? 


Cedric[FR] wrote:
> 
> 
> 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
>> address@hidden
>> http://lists.gnu.org/mailman/listinfo/help-glpk
>> 
>> 
> 
> 

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