2
$\begingroup$

Given the following sequences:

let value = $(b_0^{p_0})(b_1^{p_1})\cdots(b_n^{p_n})$

let productOfExponents = $p_0 \cdot p_1 \cdots p_n$

Where $p_i\geq 0$ and $p_i$ an element of $\mathbb{N}$ for all i
And $b_i < b_{i+1}$ and $b_i$ an element of $\mathbb{N}$ for all i

What is an efficient algorithm to minimize value given productOfExponents must be greater than anArbitraryNaturalNumber?

Update

The $b_i$ values are fixed and non-controllable. Only $p_i$ values can be modified.

1 Answers 1

1

Let $P$ be the given "anArbitraryNaturalNumber". To get a hold on the problem I propose to consider the $p_i$ as continuous variables and put $p_i\log b_i=:x_i>0$ $\ (1\leq i\leq n)$. Then we have to minimize $\log v=\sum_{i=1}^n x_i$ under the condition \prod_{i=1}^n x_i= P' where P':= P\ \prod_{i=1}^n\log b_i\ . Now it is well known that for a given value of the product $\prod_{i=1}^n x_i$ the sum $\sum_{i=1}^n x_i$ is minimal when all $x_i$ are equal. This implies that we should chose $p_i:={\lambda \over \log b_i}\quad(1\leq i\leq n)$ for a certain $\lambda$, and this $\lambda$ is determined by the condition $\prod_{i=1}^n p_i=P$. In terms of the original problem this means that all powers $b_i^{p_i}$ should have the same value $e^\lambda$.

Now this is only the beginning ...

  • 0
    My apologies - the $b_i$ values are fixed. only $p_i$ can be modified. I've updated my question. Thanks for taking the time to answer the question in it's previous incarnation!2012-01-19