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From: Luis Felipe on 28 May 2010 17:11
Hi,

Have you ever seen the following optimization problem?:

n
max prod (p_i)^a_i
i=1

n
subject to sum p_i = 1
i=1

where a_i is a positive constant for i=1,...,n

I would like to know if that optimization problem has any application
(or has been used to solve anything). Thank you for your comments.
From: christian.bau on 28 May 2010 17:31
On May 28, 10:11 pm, Luis Felipe <luispip...(a)gmail.com> wrote:
> Hi,
>
> Have you ever seen the following optimization problem?:
>
>              n
> max   prod (p_i)^a_i
>            i=1
>
>                                   n
> subject to                sum  p_i  =  1
>                                  i=1
>
> where   a_i   is a positive constant for i=1,...,n
>
> I would like to know if that optimization problem has any application
> (or has been used to solve anything). Thank you for your comments.

I'd try to figure out what is max (x^n) * (y^m) under the condition x
+ y = c. I bet that will give you a ratio x/y that depends on n and
m.

Let's see:
x^n * (c - x)^m = max
d/dx (x^n * (c-x)^m) = 0

n x^(n-1) * (c-x)^m - m (c-x)^(m-1) * x^n = 0
n x^(n-1) * (c-x)^m = m (c-x)^(m-1) * x^n
n (c-x) = m * x
n*c = (m + n)*x
x = c * n / (m + n)
y = c * m / (m + n)
x / n = y / m

I'd say calculate S = sum of the a_i, then let p_i = a_i / S.
If I made any mistakes in the calculation then they should be easy to
fix.






From: Rob Johnson on 28 May 2010 17:40
In article <c10669bc-97ef-4970-bcdd-5efe0022512f(a)v37g2000vbv.googlegroups.com>,
Luis Felipe <luispipe16(a)gmail.com> wrote:
>Have you ever seen the following optimization problem?:
>
> n
>max prod (p_i)^a_i
> i=1
>
> n
>subject to sum p_i = 1
> i=1
>
>where a_i is a positive constant for i=1,...,n
>
>I would like to know if that optimization problem has any application
>(or has been used to solve anything). Thank you for your comments.

The first condition can be written as

n
---
> a_i log(p_i) [1]
---
i=1

is maximized, subject to the condition

n
---
> p_i = 1 [2]
---
i=1

Let Dp_i be the variation of p_i. To maximize [1], we need

n
--- a_i
> --- Dp_i = 0 [3]
--- p_i
i=1

Restriction [2] implies that

n
---
> Dp_i = 0 [4]
---
i=1

Thus, we need to find the p_i that, for every Dp_i that satisfies [4],
also satisfies [3]. According to standard orthogonalization results,
(<http://www.whim.org/nebula/math/orthovar.html>), [3] and [4] imply
that there is a constant k so that

a_i
--- = k [5]
p_i

Thus, [2] and [5] imply

n
---
k = > a_i [6]
---
i=1

and [5] and [6] imply that

n
/ ---
p_j = a_j / > a_i [7]
/ ---
i=1

Rob Johnson <rob(a)trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font
From: Tony on 28 May 2010 17:52
On May 28, 11:11 pm, Luis Felipe <luispip...(a)gmail.com> wrote:
> Hi,
>
> Have you ever seen the following optimization problem?:
>
>              n
> max   prod (p_i)^a_i
>            i=1
>
>                                   n
> subject to                sum  p_i  =  1
>                                  i=1
>
> where   a_i   is a positive constant for i=1,...,n
>
> I would like to know if that optimization problem has any application
> (or has been used to solve anything). Thank you for your comments.

This is a simple exercise in Lagrange multipliers. Let P =
prod(p_i^a_i). To maximise P subject to sum(p_i) = 1, we must make the
gradient of P perpendicular to the plane defined by sum(p_i) = 1, i.e.
parallel to the vector (1,1,...,1).
We have, for each j, partial dP/dp_j = P a_j/p_j, and we want this
parallel to (1,1,...,1). So a_j/p_j is the same for all j! Let p_j = c
a_j, so c sum(a_i) = sum(p_i) = 1. Thus c = 1/sum(a_i), giving

p_j = a_j / sum(a_i)

Looking again at your post, I see that you didn't actually ask for a
solution. But I've written it out now, so here it is.
From: Tony on 28 May 2010 17:57
Well, at least we all came up with the same answer :-)
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