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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 12, 2013 12:38 PM
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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:
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> > 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.
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> Thanks for all the replies. It is surprising that the function value goes to zero for n -> infinity.
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> Let x(i) = value of items i, i = 2,3,... 0<=x(i) <= 1
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> x(1) = 1 value of items 1.
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> Item 1 is the best, in terms of value. All items cost $1 per unit of investment.
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> We have a total of $1 to invest and let y(i) be invested in item i.
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> y(1) = 1 all other y(j) = 0 is optimal.
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> But suppose one wants to diversify and an intuitive approach is to say that
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> y(i) is proportional to x(i).
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> Then the allocations of $1 to the different items are
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> 1 /( 1 + sum {x(i)} for item 1
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> and x(i) / (1 + sum{x(i)} for all other items
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> Total value received from a $1 investment
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> = 1 . 1 /(1 + sum(x(i)) + sum( x(i).x(i)) / [1+ sum(x(i))]
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> = [1 + sum{x(i)^2}] / [1 + sum{x(i)}]
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> I.e., the function under question.
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> (1) Is there a contradiction here - all items generate positive value but the total goes to zero as n goes to infinity.
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> (2) Is there a better way to diversify?
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> 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)).



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