If you are using the matrix-free method, the number of function evaluations is way more than the number of Newton iterations.
Fande, On Mon, Oct 24, 2016 at 2:01 PM, Justin Chang <jychan...@gmail.com> wrote: > Sorry forgot to hit reply all > > On Monday, October 24, 2016, Justin Chang <jychan...@gmail.com> wrote: > >> It depends on your SNES solver. A SNES iteration could involve more than >> one function evaluation (e.g., line searching). Also, -snes_monitor may say >> 3 iterations whereas -snes_view might indicate 4 function evaluations which >> could suggest that the first call was for computing the initial residual. >> >> On Mon, Oct 24, 2016 at 2:22 PM, Gideon Simpson <gideon.simp...@gmail.com >> > wrote: >> >>> I notice that if I use -snes_view, >>> >>> I see lines like: >>> total number of linear solver iterations=20 >>> total number of function evaluations=5 >>> Just to clarify, the number of "function evaluations" corresponds to the >>> number of Newton (or Newton like) steps, and the total "number of linear >>> solver iterations” is the total number of iterations needed to solve the >>> linear problem at each Newton iteration. Is that correct? So in the >>> above, there are 5 steps of Newton and a total of 20 iterations of the >>> linear solver across all 5 Newton steps. >>> >>> -gideon >>> >>> >>
