<p>To maximize the expression, given $x_i$, should set $y_i = m$ for the value of $i$ maximizing $x_i - \alpha$. </p>

<p>This implies you should set $x_2, \dots, x_n = \alpha$ and $x_1 = 1 - (n-1) \alpha$.</p>

<p>(I assume here that $0^0 = 1$; otherwise you must set $x_i = \alpha + \epsilon$, and the resulting function does not realize its supremum)</p>

CC BY-SA 3.0
2011-12-01T00:05:29.277Z
Thanks. But $\alpha$ is a real number here.
user7430
CC BY-SA 3.0
2011-11-30T21:41:40.900Z
2011-12-01T02:07:14.017Z
2011-12-01T02:07:14.017Z
David Harris
<p>Please let me know whether there are closed-form optimal results (or upper bound) for the following optimization problem:</p>

<p>$$\max (\prod_{1\leq i\leq n}(x_i)^{y_i}-\prod_{1\leq i\leq n}(x_i-\alpha)^{y_i})$$</p>

<p>with the following  constraints:</p>

<p>$$x_i\geq \alpha; \sum_{1\leq i\leq n} x_i=1$$</p>

<p>$$y_i\in\{0,1,\dots,m\}; \sum_{1\leq i\leq n} y_i=m$$</p>

<p>Only $x_i,y_i$ are variables. $\alpha$ is known and  $0\leq \alpha \leq1/n$.</p>

<p>Thank you very much.</p>

CC BY-SA 3.0
2011-11-30T21:58:19.500Z
Can you motivate?
Mahdi Cheraghchi
CC BY-SA 3.0
2011-11-30T23:21:07.763Z
Is this homework?
Jeffε
CC BY-SA 3.0
2011-12-01T00:42:12.113Z
It's not homework. I'm trying to bound the optimality of piecewise linear approximation of a nonconvex function.
user7430
<upper-bounds><convex-optimization>
Upper bound of an optimization problem
user7430
# Upper bound of an optimization problem

(Asked by: username_0 on 2011-11-30)

Please let me know whether there are closed-form optimal results (or upper bound) for the following optimization problem:

$$\max (\prod\_{1\leq i\leq n}(x\_i)^{y\_i}-\prod\_{1\leq i\leq n}(x\_i-\alpha)^{y\_i})$$

with the following constraints:

$$x\_i\geq \alpha; \sum\_{1\leq i\leq n} x\_i=1$$

$$y\_i\in\{0,1,\dots,m\}; \sum\_{1\leq i\leq n} y\_i=m$$

Only $x\_i,y\_i$ are variables. $\alpha$ is known and $0\leq \alpha \leq1/n$.

Thank you very much.

**Tags:** upper-bounds, convex-optimization

> **Comments:**
> **username_1**: Can you motivate?  
> **username_2**: Is this homework?  
> **username_0**: It's not homework. I'm trying to bound the optimality of piecewise linear approximation of a nonconvex function.  

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## 1 answers

### Answer by username_3 on 2011-11-30. Score: 1

To maximize the expression, given $x\_i$, should set $y\_i = m$ for the value of $i$ maximizing $x\_i - \alpha$.

This implies you should set $x\_2, \dots, x\_n = \alpha$ and $x\_1 = 1 - (n-1) \alpha$.

(I assume here that $0^0 = 1$; otherwise you must set $x\_i = \alpha + \epsilon$, and the resulting function does not realize its supremum)

> **Comments:**
> **username_0**: Thanks. But $\alpha$ is a real number here.  

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