On Dec 28, 2020, at 9:31 PM, Salazar De Troya, Miguel
<[email protected]<mailto:[email protected]>> wrote:
Hello,
Thanks for your response, Hong. I see that all cost functionals are evaluated
in a single backward run.
All gradients, not necessarily the cost functionals.
However, I want to do it separately. I want to isolate the evaluation of the
gradients for each cost functional.
What is the motivation of doing multiple TSAdjointSolve() calls in your case?
Note that evaluating the gradients in one call is more efficient because you do
not have to load the same checkpoints multiple times.
Can you please elaborate on how to reuse the trajectory for multiple calls?
Specifically, how to set the trajectory back to the end so I can call
TSAdjoint() again?
This is the last thing you want to do. Before each adjoint run, you can reset
TS into the same state as when the forward run ends by specifying the final
time, the step size and the step number. You will be limited to use disk
(default option) for checkpointing. Here is an example modified from ex20adj.c:
diff --git a/src/ts/tutorials/ex20adj.c b/src/ts/tutorials/ex20adj.c
index 8ca9e0b7ba..e185bc4721 100644
--- a/src/ts/tutorials/ex20adj.c
+++ b/src/ts/tutorials/ex20adj.c
@@ -277,6 +277,10 @@ int main(int argc,char **argv)
ierr = TSGetSolveTime(ts,&user.ftime);CHKERRQ(ierr);
ierr = TSGetStepNumber(ts,&user.steps);CHKERRQ(ierr);
+ for (PetscInt iter=1; iter<3; iter++) {
+ ierr = TSSetTime(ts,user.ftime);CHKERRQ(ierr);
+ ierr = TSSetTimeStep(ts,0.001);CHKERRQ(ierr);
+ ierr = TSSetStepNumber(ts,user.steps);CHKERRQ(ierr);
/* - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Adjoint model starts here
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - */
@@ -321,7 +325,7 @@ int main(int argc,char **argv)
ierr = VecRestoreArray(user.mup[1],&x_ptr);CHKERRQ(ierr);
ierr = VecRestoreArray(user.lambda[1],&y_ptr);CHKERRQ(ierr);
ierr = PetscPrintf(PETSC_COMM_WORLD,"\n sensivitity wrt parameters:
d[z(tf)]/d[mu]\n%g\n",(double)PetscRealPart(derp));CHKERRQ(ierr);
-
+ }
/* - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Free work space. All PETSc objects should be destroyed when they
are no longer needed.
Hong (Mr.)
Miguel
From: "Zhang, Hong" <[email protected]<mailto:[email protected]>>
Date: Monday, December 28, 2020 at 6:16 PM
To: "Salazar De Troya, Miguel"
<[email protected]<mailto:[email protected]>>
Cc: "Salazar De Troya, Miguel via petsc-users"
<[email protected]<mailto:[email protected]>>
Subject: Re: [petsc-users] Calculating adjoint of more than one cost function
separately
On Dec 27, 2020, at 5:01 PM, Salazar De Troya, Miguel via petsc-users
<[email protected]<mailto:[email protected]>> wrote:
Hello,
I am interested in calculating the gradients of an optimization problem with
one goal and one constraint functions which need TSAdjoint for their adjoints.
I’d like to call each of their adjoints in different calls, but it does not
seem to be possible without making compromises.
If you are calculating the derivatives to the same set of parameters, the
adjoints of all cost functionals can be done with a single backward run.
For instance, one could set TSCreateQuadratureTS() and TSSetCostGradients()
with different quadratures (and their gradients) for each adjoint call (one at
a time). This would evaluate the cost functions in the backwards run though,
whereas one typically computes the cost functions in a different routine than
the adjoint call (like in line searches evaluations)
The second argument of TSCreateQuadratureTS() allows you to choose if the
quadrature is evaluated in the forward run or in the backward run. The choice
typically depends on the optimization algorithms. Some optimization algorithms
may expect users to provide an objective function and its gradient as a bundle;
in this case, the choice does not make a difference. Some other algorithms may
occasionally evaluate the objective function without evaluating its gradient,
then evaluating the quadrature in the forward run is definitely a better choice.
One could also set TSCreateQuadratureTS() with the goal and the constraint
functions to be evaluated at the forward run (as typically done when computing
the cost function). The problem would be that the adjoint call now requires two
sets of gradients for TSSetCostGradients() and their adjoint are calculated
together, costing twice if your routines for the cost and the constraint
gradients are separated.
You can put the two sets of gradients in vector arrays and pass them to
TSSetCostGradients() together. Only one call to TSAdjointSolve() is needed. See
the example src/ts/tutorials/ex20adj.c, where we have two independent cost
functionals, and their adjoints correspond to lambda[0]/mup[0] and
lambda[1]/mup[1] respectively. After performing a TSAdjontSolve, you will get
the gradients for both cost functionals.
The only solution I can think of is to set TSCreateQuadratureTS() with both the
goal and constraint functions in the forward run. Then, in the adjoint calls,
reset TSCreateQuadratureTS() with just the cost function I am interested in
(either the goal or the constraint) and set just a single TSSetCostGradients().
Will this work? Are there better alternatives?
TSCreateQuadratureTS() is needed only when you have integral terms in the cost
functionals. It has nothing to do with the procedure to compute the adjoints
for multiple cost functionals simultaneously. Do you have integrals in both the
goal and the constraint? If so, you can create one quadrature TS and evaluate
both integrals together. For example, you may have r[0] (the first element of
the output vector in your cost integrand) for the goal and r[1] for the
constraint. Just be careful that the adjoint variables (array lambda[]/mup[])
should be organized in the same order.
Even if successful, there is the problem that the trajectory goes back to the
beginning when we perform a TSAdjointSolve() call. Subsequent calls to
TSAdjointSolve() (for instance for another cost function) are invalid because
the trajectory is not set at the end of the simulation. One needs to call the
forward problem to bring it back to the end. Is there a quick way to set the
trajectory state to the last time step without having to run the forward
problem? I am attaching an example to illustrate this issue. One can uncomment
lines 120-122 to obtain the right value of the derivative.
Most likely you need only one call to TSAdjointSolve(). Reusing the trajectory
for multiple calls is also doable. But I doubt you would need it.
Hong (Mr.)
Thanks
Miguel
Miguel A. Salazar de Troya
Postdoctoral Researcher, Lawrence Livermore National Laboratory
B141
Rm: 1085-5
Ph: 1(925) 422-6411
<simple-ode.py>
