<p>I'm sorry I don't fully understand the question, but if the point is to find the maximum, considering the local maximum is the global, then any gradient based method is going to take you directly to it. For instance artificial neural networks (gradient backpropagation).</p>

CC BY-SA 3.0
2011-09-18T22:52:38.827Z
2011-09-18T22:52:38.827Z
Trylks
<p>Let the objective be to maximize the sum of $f_i(x_i)$ where all $f_i$ are strictly increasing convex functions. Maximizing a convex function is hard as a local maximum might not be a global one. However for the special case described above, since $f_i$ are strictly increasing, from what I understand, the local maximum should be the global one. </p>

<p>If so, are there any special techniques that can optimize the objective (or maybe a piecewise approximation of it) subject to linear inequality constraints quickly?</p>

<p>Thanks</p>

CC BY-SA 3.0
2011-09-19T10:00:38.540Z
If all $f_i$ are strictly convex then a local maximum must be the global maximum.  If all $f_i$ are strictly increasing then the maximum is unbounded.  To maximize you would have to set $x_i=\infty$ for all $x_i$.
James King
CC BY-SA 3.0
2011-09-19T21:07:37.110Z
@James - The OP asked for maximizing w/ respect to linear inequality constraints.
Opt
CC BY-SA 3.0
2012-01-23T08:18:37.900Z
Use the Lagrange multiplier, you'll solve a similar problem with the same results.
luke14free
CC BY-SA 3.0
2012-01-25T01:25:40.137Z
Local maxima are not necessarily the global maximum in your situation.  For example, consider maximizing 2x^2+y^2 subject to x+y≤1, x≥0, y≥0.  Point (1,0) is a local maximum but not the global maximum.
Tsuyoshi Ito
<ds.algorithms><optimization>
Maximizing strictly increasing convex function
Opt
# Maximizing strictly increasing convex function

(Asked by: username_0 on 2011-09-18)

Let the objective be to maximize the sum of $f\_i(x\_i)$ where all $f\_i$ are strictly increasing convex functions. Maximizing a convex function is hard as a local maximum might not be a global one. However for the special case described above, since $f\_i$ are strictly increasing, from what I understand, the local maximum should be the global one.

If so, are there any special techniques that can optimize the objective (or maybe a piecewise approximation of it) subject to linear inequality constraints quickly?

Thanks

**Tags:** ds.algorithms, optimization
> **Comments:**
> **username_1**: If all $f_i$ are strictly convex then a local maximum must be the global maximum.  If all $f_i$ are strictly increasing then the maximum is unbounded.  To maximize you would have to set $x_i=\infty$ for all $x_i$.  
> **username_0**: @James - The OP asked for maximizing w/ respect to linear inequality constraints.  
> **username_2**: Use the Lagrange multiplier, you'll solve a similar problem with the same results.  
> **username_3**: Local maxima are not necessarily the global maximum in your situation.  For example, consider maximizing 2x^2+y^2 subject to x+y≤1, x≥0, y≥0.  Point (1,0) is a local maximum but not the global maximum.  

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## 1 answers

### Answer by username_4 on 2011-09-18. Score: 1

I'm sorry I don't fully understand the question, but if the point is to find the maximum, considering the local maximum is the global, then any gradient based method is going to take you directly to it. For instance artificial neural networks (gradient backpropagation).


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