
Re: An optimization problem
Posted:
Sep 12, 2013 12:38 PM


On Wednesday, September 11, 2013 7:09:48 PM UTC7, anal...@hotmail.com wrote: > On Saturday, September 7, 2013 9:00:16 AM UTC4, 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)).

