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I thought of writing this question Minimum for this function in a different way, if it helps.

I want to minimize $$\sum_{i=1}^n a_ix_i + \nu \sum_{i=1}^n b_i 2^{x_i} ,$$
where $a_i \in [0,1]$, $b_i \in (0,\infty)$, and $x_i \in [x_{\min},0)$, and
$$ \sum_{i=1}^n 2^{x_i} = 1 .$$

$x_{\min}$, $\nu$, $a_i$ and $b_i$ are constants.

I guess the tricky part is to minimize the function while ensuring all $2^{x_i}$ sum up to $1$ for these $n$ variables.

Thank you for your patience and help.

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2 Answers 2

up vote 2 down vote accepted

First let $x = (x_1, \dots, x_n)$ and write $f(x) = \sum_i^n a_i x_i + \nu\sum_i^n b_i 2^{x_i}$ and $g(x) = \sum_i^n 2^{x_i} - 1$. Now the method of Lagrange's multipliers tells you that you have a solution $x'$ iff it is an extremal point of the function $f(x) + \lambda g(x)$ with $\lambda$ a parameter to be determined. So you get equations $a_i + (\nu b_i + \lambda) 2^{x'_i} \log{2} = 0$ for all $1 \leq i \leq n$. It's now easy to express $x'_i$ as a function of lambda and plugging this into the remaining constraint $g(x') = 0$ to find the lambda and complete the solution. Hope this helps.

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I agree but is this (math.stackexchange.com/questions/9105/…) not going to happen here. I will get a polynomial equation for $\lambda$, which will be not solvable if degree is greater than 5. –  SkypeMeSM Nov 7 '10 at 16:54
2  
It will be solvable numerically. –  Yuval Filmus Nov 7 '10 at 16:54
1  
It will not be solvable in radicals (in general). Is this a problem? Another thing is that x' need not be a minimum but only a stationary point, so you will need to further inspect the stationary points. The minimum can also be attained on the boundary of the domain D(f), g=0 which means that some x_k might be equal to x_{min} but that only reduces the problem to having less variables. –  Marek Nov 7 '10 at 17:26
    
Ok. I understand what you are saying. I guess I got my answer here. :) Thank you all. –  SkypeMeSM Nov 7 '10 at 17:29

I don't think rewriting this way helps. In the first place, how would you show that if $y$ is the optimal solution of $\min_y f(y)$, then $x = 2^y$ is the optimal solution to $\min_x f(\log x)$?

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This is not a good answer. –  The Chaz 2.0 Dec 13 '11 at 5:23

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