only_fj! version requests f and j separately at the same x
rleegates opened this issue · comments
As the title suggests, using method newton, no linesearch. I could search the code for where this occurs, if needed. Since in my case requesting only f and then only j at the same x from fj is less efficient than requesting both simultaneously, this generates significant overhead. With fj! generating some debugging output, I get:
Iter f(x) inf-norm Step 2-norm
------ -------------- --------------
┌ Warning: Requested both stiffness and RHS.
└ @ Solvers ...
0 3.751768e+03 NaN
┌ Warning: Requested only RHS. norm(x) = 190.759526036454
└ @ Solvers ...
1 3.523550e+00 6.360172e-02
┌ Warning: Requested only stiffness. norm(x) = 190.759526036454
└ @ Solvers ...
┌ Warning: Requested only RHS. norm(x) = 190.72736454878574
└ @ Solvers ...
2 2.828951e-02 1.358687e-03
┌ Warning: Requested only stiffness. norm(x) = 190.72736454878574
└ @ Solvers ...
┌ Warning: Requested only RHS. norm(x) = 190.727259268085
└ @ Solvers ...
3 2.609981e-07 1.455108e-08
┌ Warning: Requested only stiffness. norm(x) = 190.727259268085
└ @ Solvers ...
┌ Warning: Requested only RHS. norm(x) = 190.72726075663638
└ @ Solvers ...
4 1.096478e-09 2.914448e-12Is there some reason why this is the intended behavior?
That's not the intended behavior, no. I can try to look into it.
But some code to go by would be nice :)
Okay, I got it.
...imho, it's the static line-search.
It's because the value is already checked in the line search, yes. So the cheapest thing is to only calculate the Jacobian next time around (I mean given the current setup). I've changed it here #243 you can try that branch