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

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 analyst41@hotmail.com Posts: 139 Registered: 4/27/05
Re: An optimization problem
Posted: Sep 11, 2013 10:09 PM
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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)}
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> 0 <=xi <=1 for all i.
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> The maximum function value of 1 occurs at either all x's = 0 or all x's = 1.
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> Can an explicit formula be given for the minimum?
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> 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.

Date Subject Author
9/7/13 analyst41@hotmail.com
9/7/13 RGVickson@shaw.ca
9/7/13 quasi
9/7/13 RGVickson@shaw.ca
9/8/13 quasi
9/8/13 quasi
9/7/13 RGVickson@shaw.ca
9/7/13 William Elliot
9/8/13 quasi
9/8/13 Peter Percival
9/8/13 quasi
9/8/13 Peter Percival
9/8/13 quasi
9/8/13 Peter Percival
9/8/13 Timothy Murphy
9/9/13 AP
9/9/13 RGVickson@shaw.ca
9/11/13 analyst41@hotmail.com
9/12/13 RGVickson@shaw.ca
9/14/13 analyst41@hotmail.com

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