[CIG-SHORT] Change in dt means new Jacobian

[Matthew Knepley]

On Sun, Sep 2, 2012 at 12:24 PM, Birendra jha <bjha7333 at yahoo.com> wrote:

> Dear developers
>
> I am trying to run my Pylith simulation faster. Right now it takes about 3
> min for Linear solve (using default KSP solver, about 960 iterations) and 3
> min for integrating residual. I have about 190000 hex cells (mesh is very
> refined in the center and becomes coarse outward). No faults. Elastic
> materials. I used non-uniform user specified timesteps.
>
> I thought of following options:
>

(0) Is this an optimized build?


> (1) Try different solver. I am thinking of using field split
> preconditioner with multigrid, as in tet4/step02.cfg.
>

If there are no faults, FieldSplit makes no sense. Just use ML. It should
take < 10 iterations.


> (2) Switch from vtk to hdf5 output
>

Definitely, also back off the frequency of output.


> (3) Run in parallel. I am looking into running it using sge. Or just
> running it on one multicore machine.
>

Do bother with this until you figure out how to run in serial.


> (4) Increase linear solve tolerance from 1E-12 to 1E-11.
>

I would try and gauge what tolerance you need by doing some manufactured
solutions.

     Matt

(5) Fix timestep.
>
> Are these good options to consider?
>
> I have a question on (5). Manual says
> Warning: Changing the time step requires recomputing the Jacobian of the
> system, which can greatly increase the runtime if the time-step size
> changes frequently.
>
> Where does this happen in the program, i.e a change in dt setting
> needNewJacobian=true? I ask this because I could not find message on the
> stdout about reforming Jacobian when I am using non-uniform dt (it is
> possible that I missed it, the timesteps remains constant for a while then
> changes then becomes constant again).
>
> Thanks and regards
> Birendra
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