Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math.independent

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
analyst41@hotmail.com

Posts: 117
Registered: 4/27/05
Re: An optimization problem
Posted: Sep 14, 2013 9:44 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On Thursday, September 12, 2013 12:38:04 PM UTC-4, Ray Vickson wrote:
> On Wednesday, September 11, 2013 7:09:48 PM UTC-7, anal...@hotmail.com wrote:
>

> > On Saturday, September 7, 2013 9:00:16 AM UTC-4, anal...@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.
>
> >
>
> >
>
> >
>
> > Thanks for all the replies. It is surprising that the function value goes to zero for n -> infinity.
>
> >
>
> >
>
> >
>
> > Let x(i) = value of items i, i = 2,3,... 0<=x(i) <= 1
>
> >
>
> >
>
> >
>
> > x(1) = 1 value of items 1.
>
> >
>
> >
>
> >
>
> > Item 1 is the best, in terms of value. All items cost $1 per unit of investment.
>
> >
>
> >
>
> >
>
> > We have a total of $1 to invest and let y(i) be invested in item i.
>
> >
>
> >
>
> >
>
> > y(1) = 1 all other y(j) = 0 is optimal.
>
> >
>
> >
>
> >
>
> > But suppose one wants to diversify and an intuitive approach is to say that
>
> >
>
> >
>
> >
>
> > y(i) is proportional to x(i).
>
> >
>
> >
>
> >
>
> > Then the allocations of $1 to the different items are
>
> >
>
> >
>
> >
>
> > 1 /( 1 + sum {x(i)} for item 1
>
> >
>
> >
>
> >
>
> > and x(i) / (1 + sum{x(i)} for all other items
>
> >
>
> >
>
> >
>
> > Total value received from a $1 investment
>
> >
>
> >
>
> >
>
> > = 1 . 1 /(1 + sum(x(i)) + sum( x(i).x(i)) / [1+ sum(x(i))]
>
> >
>
> >
>
> >
>
> > = [1 + sum{x(i)^2}] / [1 + sum{x(i)}]
>
> >
>
> >
>
> >
>
> > I.e., the function under question.
>
> >
>
> >
>
> >
>
> > (1) Is there a contradiction here - all items generate positive value but the total goes to zero as n goes to infinity.
>
> >
>
> >
>
> >
>
> > (2) Is there a better way to diversify?
>
> >
>
> >
>
> >
>
> > Thanks again.
>
>
>
> I think your model is in error: if the i'th investment quantity y(i) is proportional to x(i) and the total investment = 1, then y(i) = x(i)/sum(x(j)), NOT x(i)/[1+ sum(x(j). So, your investment payoff should be sum(x(i)^2)/sum(x(i)).


Just a notation-confusion.

In the investment problem x(1) = 1 and the sum is over i = 2 to i = infinity.

The first item is the best investment and its expected benefit normalized to 1(thats why 0<=x(j) <=1 for j = 2,3,...).



Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.