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Topic: An optimization problem
Replies: 19   Last Post: Sep 14, 2013 9:44 AM

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RGVickson@shaw.ca

Posts: 1,655
Registered: 12/1/07
Re: An optimization problem
Posted: Sep 9, 2013 4:32 AM
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On Sunday, September 8, 2013 11:51:53 PM UTC-7, AP wrote:
> On Sat, 7 Sep 2013 06:00:16 -0700 (PDT), analyst41@hotmail.com wrote:
>
>
>

> >
>
> >consider f(x1,x2,...xn) = {1 + sum(xi)^2)} / {1 +sum{x(i)}
>
> >
>
> >0 <=xi <=1 for all i.
>
> >
>
> >The maximum function value of 1 occurs at either all x's = 0 or all x's = 1.
>
> >
>
> >Can an explicit formula be given for the minimum?
>
> >
>
> >Thanks.
>
> if minimum (or maximum) for x1,x2,...xn then
>
> df/dx_i(x1,x2,...xn)=0
>
>
>
> so 2x_i(1+sumx_i)-(1+sumx_i^2)=0 and x_i=a with
>
> 2a=(1+na^2)/(1+na)
>
> na^2+2a-1=0
>
> and because 0<=a<=1, a=(sqrt(n+1)-1)/n
>
>
>
> but one must see if there is min or max
>
>
>
> f(a,a..,a)=(1+na^2)/(1+na)=2a
>
>
>
> the sign of f(x1,x2,...xn)-f(a,a,..,a)
>
> is the sign of
>
> 1+sumx_i^2-2a(1+sumx_i)
>
> =sum(x_i-a)^2-na^2+1-2a
>
> =sum(x_i-a)^2>=0
>
>
>
> and f has a mini for (a,a...a)


There is no need to check: for a given value of the sum(x_i) the function is strictly convex in (x_1,x_2,...,x_n), so a point that satisfies the Lagrangian conditions is a *global* minimum. That is: it is globally minimal to take all x_i = a and to then minimize over z.



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