- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Report Inappropriate Content
Dear All
I am testing the example given in the MKL manual for pardiso with partial solution in Fortran.
As stated for iparm(31), when its value is 3, selected components of the solution vectors will be computed. However, I tried many times, with different perm values, iparm(31) values(=1, 2 or 3), and paramater locations, it always give the whole solution.
Since for my real problem(with thousands of DOF), only a few elements in rhs are nonzero and only a few selected components of the solution are needed. I have to realize the partial solution function to cut the time cost.
I hope you can help me. Thanks a lot.
The attachment is my fortran code.
Jim Cheng
Link Copied
- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Report Inappropriate Content
Hello Jim,
1) (minor) First, looking at your code I see
iparm(7) = 16 ! default logical fortran unit number for output
doesn't agree with Intel MKL documentation (iparm[7] is the output iparm parameter which can give the number of iterative refinement steps performed). Hence, the question is whether you're actually intend to use Intel MKL PARDISO or some other solver with a similar interface?
2) Second, you also have
iparm(8) = 9 ! numbers of iterative refinement steps
but the partial solve by its nature does not support iterative refinement. So, I believe due to this setting PARDISO actually performed a full solve.
Could you please switch this iparm off (set iparm(8)=1) and see what happens?
Hope this helps!
Best,
Kirill

- Subscribe to RSS Feed
- Mark Topic as New
- Mark Topic as Read
- Float this Topic for Current User
- Bookmark
- Subscribe
- Printer Friendly Page