Suppose I’m specifying the Jacobian. -gideon
> On Oct 24, 2016, at 4:02 PM, Kong, Fande <[email protected]> wrote: > > 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 <[email protected] > <mailto:[email protected]>> wrote: > Sorry forgot to hit reply all > > On Monday, October 24, 2016, Justin Chang <[email protected] > <mailto:[email protected]>> 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 <[email protected] <>> > 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 > > >
