- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Report Inappropriate Content
I am using the out-of-core `pardiso` with 3,267,775,116 non-zero entries.
Are there any estimates how much RAM phase 11 should take? I understand that this is the METIS symbolic factorization. Is there a way to get finer control for this stage?
I have tried to use a user-defined permutation (setting `iparm[4]=1`, and the permutation arrya `perm = (MKL_INT) 1..N`). This was to no avail.
Currently, I am getting 120GB of RAM used, which utterly defeats the purpose of the out-of-core part for the rest of the code.
Thank you.
Link Copied
- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Report Inappropriate Content
I would also like to add that the stage in question (11) is also the slowest one. Even with `iparm[1]` = 3 (which I assume should employ an openMP routine of some sort), there is only one thread being used, which hangs for over 30 minutes (100% user time, no system time).
I could not find any documentation on what phase `11` actually does - other than "analysis" (https://software.intel.com/en-us/node/521690).
Again, thanks for the support!
- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Report Inappropriate Content
Hi Ivan,
Currently MKL pardiso doesn't support ability to estimate size of RAm needed for reordering step. However i am bit surprised that reordering step spent most time of whole solution. Can i ask you to sent me the matrix to check what's happened and why reordering step provide poor performance? Probably such behaviour could be resolved.
Thanks,
Alex
- Subscribe to RSS Feed
- Mark Topic as New
- Mark Topic as Read
- Float this Topic for Current User
- Bookmark
- Subscribe
- Printer Friendly Page