[petsc-users] solving Ax=b with constant A

[Matthew Knepley]

On Mon, May 12, 2014 at 12:27 PM, <likunt at caltech.edu> wrote:

> Dear Petsc developers,
>
> I am solving a linear system Ax=b, while A is constant and b is changing
> in each time step. Here is the code I wrote:
>
> /****************************************************************
> ...compute matrix A...
> KSPCreate(PETSC_COMM_WORLD, &ksp);  CHKERRQ(ierr);
> KSPSetOperators(ksp, A, A, SAME_PRECONDITIONER);  CHKERRQ(ierr);
> KSPSetTolerances(ksp, 1.e-5, 1.E-50, PETSC_DEFAULT, PETSC_DEFAULT);
> KSPSetFromOptions(ksp);
> for(int step=0; step<STEP; step++)
> {
>     ... compute vector b ...
>     KSPSolve(ksp, b, x);
> }
> *****************************************************************/
>
> I tested a system with size 1725*1725, on 4 processors, it takes 0.06s.
> Would you please let me know if there is a way to improve its efficiency?
>

It would be amazing if we could do that given the description. First, we do
not know
exactly what solver is being used (-ksp_view), but lets assume its
GMRES/ILU(0)
which is the default. Second, we have no idea what the convergence was like
(-ksp_monitor_true_residual -ksp_converged_reason), so we do not know what
the bottleneck is, and have no performance monitoring (-log_summary).
Lastly,
even if we had that we have no idea what the operator is so that we could
make
intelligent suggestions for other preconditioners.

   Matt


> Thanks.
>



-- 
What most experimenters take for granted before they begin their
experiments is infinitely more interesting than any results to which their
experiments lead.
-- Norbert Wiener
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