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I am trying to find a DCP formulation for the following convex objective function (using CVXPY):

Let $x$ be the $N$-dimensional vector variable on which we optimize on, $c$ be a known scalar value such that $0 < c \le 1$ and and $L$ be a known $N$-dimensional vector with $L_i > 0 \: \forall i=1,..., N$. The (convex) objective is formulated as:

$$ f(x) = \sum_{i=1}^N\left(c^{\max(x_i,0)/L_i}\cdot L_i - \ln(c)\cdot\max(x_i,0)\right) $$

Single out any $i=1,...,N$, the issue I am facing is that each convex element of the sum above is given by the sum of a nonconvex and a convex function:

$$ c^{\max(x_i,0)/L}\cdot L_i - \ln(c)\cdot\max(x_i,0) $$

The expected behavior of the above sum is to be equal to: \begin{align} c^{\max(x_i,0)/L}\cdot L_i - \ln(c)\cdot\max(x_i,0)=\begin{cases} c^{x_i/L}\cdot L_i - \ln(c)\cdot x&\quad x_i \ge 0 \\ L_i &\quad x_i < 0 \end{cases} \end{align}

But using standard atoms it is not trivial to provide a DCP certificate for the above due to non-convexity of the first term.

Do you have any ideas/suggestions on how one could rewrite the problem and make it DCP compliant?

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  • $\begingroup$ Which of x,L,c are optimization variables, and which are input data? $\endgroup$ Oct 29, 2021 at 11:40
  • $\begingroup$ @MarkL.Stone added more color in the corpus of the question. Answering your question: x is the variable on which we optimize on. L and c are known scalar vector and value, respectively. $\endgroup$
    – LowOdds
    Oct 29, 2021 at 12:34
  • $\begingroup$ I think this is convex, but haven't proven it. The scalar case should suffice. In any event, this could be a candidate for @ErlingMOSEK 's challenge twitter.com/J_P_Vielma/status/1014513695605608448 as a (possible example of) convex expression which can't be modeled by (extreme) DCP. But, maybe one of the clever MOSEK people can come up with a DCP formulation (perhaps some way of effectively combining the two terms) $\endgroup$ Oct 29, 2021 at 15:20
  • $\begingroup$ Do you have a convexity proof? if so, it might be helpful to show it. If the proof is "constructive" (in a DCP building block kind of way), that would be even better. $\endgroup$ Oct 29, 2021 at 15:37

1 Answer 1

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For all $c>0$ the function of one variable $$g(x)=c^{x/L}\cdot L - \ln(c)\cdot x=\exp(\ln(c)x/L)\cdot L-\ln(c)\cdot x$$ is convex increasing on $x\geq 0$ (derivative check) and DCP representable (verbatim). So we can model your extension to all real numbers via

$$u\geq \max(x,0),$$ $$t\geq g(u).$$

Now if $x\geq 0$ then we have $t\geq g(x)$ and if $x\leq 0$ then we have $t\geq g(0)=L$.

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