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Re: Re: Finding the optimum by repeteadly zooming on the solution space (or something like that)
*To*: mathgroup at smc.vnet.net
*Subject*: [mg92597] Re: [mg92575] Re: Finding the optimum by repeteadly zooming on the solution space (or something like that)
*From*: "Mauricio Esteban Cuak" <cuak2000 at gmail.com>
*Date*: Tue, 7 Oct 2008 07:06:30 -0400 (EDT)
*References*: <200810060814.EAA28778@smc.vnet.net>
I appreciate both your help!
I was trying it the other way because I'm more interested in the speed of
the solution than in the precision. Ultimately, what I need to do is to find
the optimum "a" for at least thousands of values of r.
Say, for a couple of values:
Table[NMaximize[{obj, cpoAg1 == 0 && cpoAg2 == 0 && x > 0.1 && y > 0.1}, {a,
x, y}], {r, 0.4, 1, 0.2}]]
It seems that my approach doesn't improve the speed for most values of r, so
I'll try your suggestions.
If someone can suggest me a way of speeding things up, I'd really appreciate
it.
Kind Regards,
cd
2008/10/6 Bill Rowe <readnews at sbcglobal.net>
> On 10/5/08 at 6:05 AM, cuak2000 at gmail.com (Mauricio Esteban Cuak)
> wrote:
>
> >Hello everyone.I know somebody has probably writ mathematica code on
> >this problem. A general answer would probably benefit more people,
> >but I'll do the specific example, 'cause I'm not sure how to explain
> >it in another way.
>
> >I have these restrictions:
>
> >cpoAg1= ( -x + (70*a*(x^0.4 + y^0.4)^0.75)/x^0.6);
>
> >(*and*)
>
> >cpoAg2= ((70*(1 - a)*(x^0.4 + y^0.4)^0.75)/y^0.6 - 2*y);
>
> >(* And I need to find the "a" that will maximise this following
> >function. I don't need and exact solution, but something that is
> >sufficiently near *)
>
> >obj = -x^2/2 + 100*(x^0.4 + y^0.4)^1.75 - y^2;
>
> >(* "A" goes between 0 and 1 so I discretise to obtain the {x,y} that
> >maximise every "a"
>
> >the Table command gives me the list of rules which I can replace on
> >"obj" later to find the maximum. No problem here*)
>
> >rules = Table[
> >FindRoot[{cpoAg1 == 0, cpoAg2 == 0}, {{x, 0.1}, {y, 0.1}}], {a, 0.=
> 1,
> >1, 0.1}
> >];
>
> >(*the maximum is found with this following function *)
>
> >Max[Thread[ReplaceAll[obj, rules]]];
>
> >So far so good...I could discretise "a" as thinly as I want, but I
> >can't afford the luxury of being that inneficient, 'cause I've got
> >to do other things later with this part of the program.
>
> I don't understand why you would take this approach. Why not use
> the built-in function NMaximize, That is:
>
> In[20]:= NMaximize[{obj,
> cpoAg1 == 0 && cpoAg2 == 0 && x > 0.1 && y > 0.1}, {a, x, y}]
>
> Out[20]= {2084.72,{a->0.527688,x->22.6727,y->13.7173}}
>
>
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