I need to solve the following problem:

\[\left\{\begin{array}{lll} {\displaystyle \inf_{(\beta,\lambda)\in\mathbb{R}^{2},s\in\mathbb{R}^{N}} } & {\displaystyle \lambda \varepsilon +\frac{1}{N}\sum_{i=1}^{N}s_{i}} &\\ \mbox{subject to} &\beta^{2}+4a_{i}\lambda\beta+4a_{i}^{2}\lambda-4\lambda s_{i}+4s_{i}\leq 0 & \forall i\leq N \\ &\lambda >1& \\ & \beta\geq 0.& \end{array}\right.\tag{\(\bigstar\)}\]

where \(\varepsilon, a_{i}\in \mathbb{R}\) are fixed for all \(i=1,2,\ldots,N\).

My intention is to use CVX (see this link), but I have problems, as it is stated ( \(\bigstar\) ) the solver does not interpret it.

Remark: I do not know if the following is useful, but note that the constrain \[\beta^{2}+4a_{i}\lambda\beta+4a_{i}^{2}\lambda-4\lambda s_{i}+4s_{i}\leq 0 \quad \forall i\leq N\] can be changed by \[\frac{\beta^{2}}{4(\lambda-1)}+\frac{\lambda}{\lambda-1}a_{i}(\beta+a_{i})-s_{i}\leq 0 \quad \forall i\leq N\] I showed that the function \(f:\mathbb{R}^{3}\rightarrow \mathbb{R}\) given by \(f(\beta,\lambda,s):=\frac{\beta^{2}}{4(\lambda-1)}+\frac{\lambda}{\lambda-1}a(\beta+a)-s\) is convex in \(\mathbb{R}\times\mathbb{R}_{\geq 1}\times\mathbb{R}\) for any \(a\in \mathbb{R}\).

My attempt: CVX can solve geometric programs (GP), the problem is that ( \(\bigstar\) ) is not a geometric program because \( a_{i} \) can be negative, my idea was to express ( \(\bigstar\) ) as a geometric program but what arrived was the following equivalent formulation:

\(\left\{ \begin{array}{lll} {\displaystyle \inf_{\beta,\lambda,s_{i},r_{i}} } & r_{0} & \\ \mbox{subject to} & \varepsilon\lambda r_{0}^{-1}+\frac{1}{N}\sum_{i=1}^{N}s_{i}r_{i}\leq 1 &\\ &\left.\begin{array}{l}\beta^{2}r_{i}^{-1}+4a_{i}\lambda \beta r_{i}^{-1}+4a_{i}^{2}\lambda r_{i}^{-1}+4s_{i}r_{i}^{-1}+r_{i}^{-1}\leq 1 \\ 4\lambda s_{i}r_{i}^{-1}+r_{i}^{-1}\geq 1 \end{array}\right\} &\forall i\in I_{1} \\ &\left.\begin{array}{l}\beta^{2}r_{i}^{-1}+4a_{i}^{2}\lambda r_{i}^{-1}+4s_{i}r_{i}^{-1}+r_{i}^{-1}\leq 1 \\ 4\lambda s_{i}r_{i}^{-1}+4(-a_{i})\lambda\beta r_{i}^{-1}+r_{i}^{-1}\geq 1 \end{array}\right\} &\forall i\in I_{2} \\ & \lambda >1 & \\ & \beta\geq 0. \end{array}\right.\tag{\(\blacktriangle\)}\)

where \( I_{1}:=\left\{i \:|\: 0<i\leq N \mbox{ such that } a_{i}\geq 0 \right\} \) and \( I_{2}:=\left\{i \:|\: 0<i\leq N \mbox{ such that } a_{i}< 0 \right\} \).

The problem of ( \(\blacktriangle\) ) is that it is not yet a geometric program, the difficulty is in the constrains \(\geq 1\). The only advantage that has ( \(\blacktriangle\) ) is that all constrains are in terms of posynomials, really do not know if this is an advantage because CVX has not yet interpreted it, it shows an error in the restrictions \(\geq 1\).

Since my attempt has not had the desired results I hope that someone will read this problem, my can help with a way to solve it or a way of expressing it in such a way that CVX can solve it.

asked 11 Dec '17, 20:11

Diego%20Fonseca's gravatar image

Diego Fonseca
accept rate: 0%

edited 12 Dec '17, 18:14


Given your revised problem statement, I have deleted my previous answer. However, given that your revised formulation is not convex over its entire "natural" domain (i.e., is non-convex for lambda < 1), I think it unlikely you will ever find a formulation which is accepted by CVX. CVX can not accept all convex problems, and problems which are convex only over a portion of the natural domain, even if there are other constraints constraining it to be in the convex region, usually (although there are exceptions) can not be formulated in CVX.

(12 Dec '17, 19:11) Mark L Stone

Constraining a non-monomial posynomial to be >= a monomial (for your special case, the constant 1), is non-convex. That is why CVX rejects it.

(12 Dec '17, 19:12) Mark L Stone

After reading http://cvxr.com/cvx/doc/ and http://ask.cvxr.com/t/why-isnt-cvx-accepting-my-model-read-this-first/570 , you can feel free to post an appropriate question at http://ask.cvxr.com/ , which is a specialized forum for CVX.

(12 Dec '17, 19:15) Mark L Stone
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Asked: 11 Dec '17, 20:11

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Last updated: 12 Dec '17, 19:17

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