[SciPy-user] [OpenOpt] evaluation of f(x) and df(x)

[dmitrey]

Hi Emanuele,

if df(x1) is obtained via finite-difference calculations then f(x1) is 
stored and compared during next call to f / df, and vice versa: if f(x1) 
is called then the value obtained is stored and compared during next 
call to f and/or finite-difference df.

At least it is intended so, I can take more precise look if you have 
noticed it doesn't work properly.

Regards, D.

Emanuele Olivetti wrote:
> Dear All and Dmitrey,
>
> in my code the evaluation of f(x) and df(x) shares many
> intermediate steps. I'd like to re-use what is computed
> inside f(x) to evaluate df(x) more efficiently, during f(x)
> optimization. Then is it _always_ true that, when OpenOpt
> evaluates df(x) at a certain x=x^*, f(x) too was previously
> evaluated at x=x^*? And in case f(x) was evaluated multiple
> times before evaluating df(x), is it true that the last x at
> which f(x) was evaluated (before computing df(x=x^*))
> was x=x^*?
>
> If these assumptions holds (as it seems from preliminary
> tests on NLP using ralg), the extra code to take advantage
> of this fact is extremely simple.
>
> Best,
>
> Emanuele
>
> P.S.: if the previous assumptions are false in general, I'd
> like to know it they are true at least for the NLP case.
>
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