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I am maximizing a linear functional subject to an integrates to one constraint. More explicitly, my problem is $$\begin{align} &\max_{x \in \mathbb{R}^n}\quad c \cdot x\\ &\text{subject to} \int_{a}^{b} \prod_{i=1}^{n} e^{x_i \cdot y^i} dy <=d \end{align}$$ where $a, b, d \in \mathbb{R}$, $c \in \mathbb{R}^n$, and $n \in \mathbb{N} $ are constants.

I have plugged this into the convex solver SCS, relaxing my integral constraint to a crude Riemann-sum type estimate. Because the sum of exponential constraints is so large in this expression, it takes a very long time to solve. This is my first experience working numerically with a constraint like this, and I would like to increase the speed and/or accuracy of this solution. Is there any more efficient, or computationally tractable, method to implement such an integration constraint like the one above?

I suppose I am hoping for some "convergence in epigraph" type result for approximating this constraint. Is this something that is realistic to hope for?

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  • $\begingroup$ I don't see the polynomial $p(x)$ in the formulation of your problem. Am I missing something? $\endgroup$
    – nicoguaro
    May 5 '15 at 17:58
  • $\begingroup$ Whoops! I removed the $p(x)$ term in the expression to make it more explicit, but forgot to remove it from the description. Fixed. $\endgroup$ May 5 '15 at 18:41
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    $\begingroup$ Something isn't right here. Is $x$ a vector or scalar? If it's a vector, does $\max cx$ mean the maximum component, and if it's a scalar, what does $x_i$ mean? Additionally, $y$ appears to be a scalar since you are integrating it over an interval of it, so do you mean $y^i$ as exponentiation? Once you clear these things up, I think you can simplify the constraint by applying replacing the product of exponentials with an exponential of a sum. After that you may be able to explicitly integrate the constraint over $y$ and then take log of both sides to make the values reasonable. $\endgroup$
    – Bill Barth
    May 6 '15 at 0:31
  • $\begingroup$ Hi Bill. I edited the problem for clarification. Thanks for your comments. $\endgroup$ May 8 '15 at 0:38

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