Re: [Help-glpk] Assertion failed

[Andrew Makhorin]

> I have a problem with running glpk-4.10:
> 
> I run the following MIP instance using command (glpsol --cpxlp
> ~/tmp_prob.lp):
> 
> Minimize
>  obj: x_1
> 
> Subject To
>  r_1: - 1003 x_1 + x_2 >= 0
>  r_2: - 1999 x_2 + x_3 >= 0
>  r_3: - 4000 x_3 + x_4 >= 0
>  r_4: + 8000 x_4 + x_5 <= 0
>  r_5: + 2 x_5 + x_6 >= 0
>  r_6: - x_1 <= -1002
>  r_7: - x_2 <= -1005006
>  r_8: - x_3 <= 138476654
>  r_9: - x_4 <= -144165184
>  r_10: - x_6 <= -1
> 
> Bounds
>  x_1 free
>  x_2 free
>  x_3 free
>  x_4 free
>  x_5 free
>  x_6 free
> 
> Generals
>  x_1
>  x_2
>  x_3
>  x_4
>  x_5
>  x_6
> 
> End
> 
> and I get the following output (error):
> 
> lpx_read_cpxlp: 10 rows, 6 columns, 15 non-zeros
> lpx_read_cpxlp: 6 integer columns, none of which are binary
> lpx_read_cpxlp: 32 lines were read
> lpx_simplex: original LP has 10 rows, 6 columns, 15 non-zeros
> lpx_simplex: presolved LP has 5 rows, 6 columns, 10 non-zeros
> lpx_adv_basis: size of triangular part = 5
>       0:   objval =   1.002000000e+03   infeas =   1.000000000e-00 (0)
> Assertion failed: spx->p != 0; file glpspx2.c; line 668
> 
> It seems like an error in glpk or is there anything I should do to be able
> to
> solve this problem?

The error is caused by extremely bad properties of your problem.
Using free integral variables is a not good idea. Besides, as it
follows, for example, from constraint r_9, x4 being declared as
integral variable must be greater than 144165184; it is very
problematic for any solver based on finite precision arithmetic
to satisfy such condition.

Try to reformulate your problem. In particular, if a value of some
variable is expected to be greater than 100 in optimal solution,
there is no reason in most cases to consider it as integral.

Andrew Makhorin