[Clp] Help: Incorrect solution from CBC

[Greg Gruber]

C190 is an indicator variable -- if it is 1 then 4000000 <= C225 <= 1E15.
If C190 is 0, then C225 must equal 0.

I cannot change the limits on C225.  Should I adjust the integer tolerance
for C190?

I tried setting integerT from 1E-6 to 1E-10 and it had no effect.

Thanks again for the help!

Greg,
>
> The problems seem to stem from the range of numbers in model e.g. a pair
> of constraints -
>
>   -1E+15 C190 +1 C225 <= +0
>
> and
>
>   -4000000 C190 +1 C225 >= +0
>
> where C190 is binary
>
> I can see this causing problems (for both codes) - if C190 is 1.0e-9 is
> this integer feasible?
>
> Solving two ways I get
>
> ma.sol:Optimal - objective value 1664550445.89107466
> mb.sol:Optimal - objective value 1676264519.23880672
>
> while changing e15 to e7 (or e6 - which will fix C190 to an exact zero)
> I get
>
> mc.sol:Optimal - objective value 1647162379.14471364
>
> If you want you can send me the glpsol solution and I can see if I can
> duplicate that solution.
>
> John Forrest
>
>
>
>
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