[petsc-users] question

[Gideon Simpson]

Suppose I’m specifying the Jacobian.

-gideon

> On Oct 24, 2016, at 4:02 PM, Kong, Fande <fande.kong at inl.gov> 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 <jychang48 at gmail.com <mailto:jychang48 at gmail.com>> wrote:
> Sorry forgot to hit reply all
> 
> On Monday, October 24, 2016, Justin Chang <jychang48 at gmail.com <mailto:jychang48 at 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.simpson at 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
> 
> 
> 

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