DOLFIN

about the use of LU solver

Asked by Dupront

Hello,

I am using the following command that is called repeatedly in a time loop:

"""
solve( A1, u1.vector(), b1, solver_parameters={"symmetric":True, "linear_solver":"mumps", "reuse_factorization":True})
"""

The matrix A1 is supposed to be constant and I want the factorization to be performed
only one time.

Everything is working but it seems to me that the "reuze_factorization" option has not
been taken into account.
If I print the parameters (info(parameters,True)), I read that the value that this parameter
has is "false". Also the time spent in the method solve is always the same at each time step.
I was expecting a real decrease knowing that the factorization is quite expensive.

Could someboby help me with that ?
Thanks

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eduardo rodriguez (edu-ampu4) said : 		#1

I'm solving a similar problem and I impose the condition before the loop:

solver=LUSolver(A1)
solver.parameters['reuse_factorization']=True

and then inside the loop:

solver.solve(u1.vector(),b1)

I suppose that you can modify the others parameters before the loop too

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Dupront (michel-dupront) said : 		#2

Thank you very much.

Unfortunately it is not working in my script.
I am still trying to make something work ...

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Jan Blechta (blechta) said : 		#3

info(parameters, True) prints this: parameters['lu_solver']['reuse_factorization'].

But changing solver.parameters['reuse_factorization'] does not change global parameter parameters['lu_solver']['reuse_factorization']. Neither do passing {"reuse_factorization": True} to solver.solve() method.

Conversely a change of parameters['lu_solver']['reuse_factorization'] prior to creating LUSolver instance will be inherited in solver.parameters.

Can you help with this problem?

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