[petsc-users] PETSc/SLEPc: Memory consumption, particularly during solver initialization/solve

Ale Foggia amfoggia at gmail.com
Wed Oct 10 02:38:23 CDT 2018


Jed, Jose and Matthew,
I've finally managed to make massif (it gives pretty detailed information,
I like it) work in the correct way in the cluster and I'm able to track
down the memory consumption, and what's more important (for me), I think
now I'm able to make a more accurate prediction of the memory I need for a
particular size of the problem. Thank you very much for all your answers
and suggestions.

El vie., 5 oct. 2018 a las 9:38, Jose E. Roman (<jroman at dsic.upv.es>)
escribió:

>
>
> > El 4 oct 2018, a las 19:54, Ale Foggia <amfoggia at gmail.com> escribió:
> >
> > Jose:
> > - By each step I mean each of the step of the the program in order to
> diagonalize the matrix. For me, those are: creation of basis, preallocation
> of matrix, setting values of matrix, initializing solver,
> solving/diagonalizing and cleaning. I'm only diagonalizing once.
> >
> > - Regarding the information provided by -log_view, it's confusing for
> me: for example, it reports the creation of Vecs scattered across the
> various stages that I've set up (with PetscLogStageRegister and
> PetscLogStagePush/Pop), but almost all the deletions are presented in the
> "Main Stage". What does that "Main Stage" consider? Why are more deletions
> in there that creations? It's nor completely for me clear how things are
> presented there.
>
> I guess deletions should match creations. Seems to be related to using
> stages. Maybe someone from PETSc can give an explanation, but looking at a
> PETSc example that uses stages (e.g. dm/impls/plex/examples/tests/ex1.c) it
> seems that some destructions are counted in the main stage while the
> creation is counted in another stage - I guess it depends on the points
> where the stages are defined. The sum of creations matches the sum of
> destroys.
>
> >
> > - Thanks for the suggestion about the solver. Does "faster convergence"
> for Krylov-Schur mean less memory and less computation, or just less
> computation?
> >
>
> It should be about the same memory with less iterations.
>
> Jose
>
>
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