Re: [Help-glpk] Problem : basis

[Cedric[FR]]

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.
> 
> 
> 
> 
> 
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> 
> 

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