0
$\begingroup$

Starting with a sum such as $\sum_{i} b_{i}$, where $b_{i} > 0$ are real numbers for all $i$ under consideration, I have a corresponding vector of real numbers $a_{i} > 0$ for all $i$. I want to form the sum $\sum_{i} f(a_{i})b_{i}$ and then scale it with a constant $\gamma > 0$ such that

$\gamma\sum_{i} f(a_{i})b_{i} = \sum_{i} b_{i}$, (*)

under the conditions that

  1. $\gamma f(x) = 0.7$ for all $\gamma x \leq 0.7$
  2. $0.7 \leq \gamma f(x) \leq 2$ for all $0.7 \leq \gamma x \leq 2$
  3. $\gamma f(x) = 2$ for all $\gamma x > 2$.

One solution I thought about is letting $f$ be defined by

$f(x, \gamma) = min(\frac{2}{\gamma}, x) + \frac{0.7}{\gamma} - min(\frac{0.7}{\gamma}, x)$,

which I believe would satisfy all the above constraints. I have not been able to find a closed form solution to (*) using such an $f$ (it seems to me it is not possible). I have solved this for $\gamma$ numerically using the bisection method, since

$g(\gamma) = \gamma\sum_{i} f(a_{i}, \gamma)b_{i} - \sum_{i} b_{i}$ is a continuous increasing function with one root.

I have two questions:

  1. Can anyone think of any ways that might solve the problem using a different $f$, that might result in a closed form solution?
  2. Are there any approaches that might result in a closed form solution to the problem using the $f$ I have tried?

I am not sure what are appropriate tags for this problem. Thank you in advance to anyone that took the time to read this post.

Edit: I should also mention that preferably I would like all factors $\gamma f(a_{i})$ that are not set to either 0.7 or 2 to be adjusted by the same constant.

$\endgroup$

0

You must log in to answer this question.

Browse other questions tagged .