I am new to CVX and am trying to simulate this convex problem I found in a paper.

$$\min_{\gamma,\mathbf{mu},\mathbf{G},\mathbf{\Omega},t} \text{Tr}(\mathbf{G}\mathbf{C}\mathbf{G}^H)+t \\ s.t. -t-\frac{\gamma}{2}\sum\limits_{i=1}^n\log(1-\frac{2\mu_i}{\gamma})-\gamma \log(\delta)\leq0\\\mu_1\geq\mu_2\geq\ldots\geq\mu_n\geq0\\2\mu_1<\gamma\\ \text{S}_k(C^{1/2} \Lambda C^{1/2})\leq \sum\limits_{i=1}^{k} \mu_i \quad \forall k = 0,1,\dots,n-1\\ \text{Tr}(\Omega C)\leq \sum\limits_{i=1}^{n}\mu_i\\\Omega\geq \Lambda$$

where $$\Lambda = (\mathbf{G}\mathbf{H}-\mathbf{I})^H(\mathbf{G}\mathbf{H}-\mathbf{I})$$ and $S_k(\mathbf{A})$ is the sum of the largest $k$ eigenvalues of $\mathbf{A}$. In the problem, the scalar $\delta$, matrices $\mathbf{H}$ and $\mathbf{C}$ are assumed known. $\mathbf{C}$ is symmetric positive definite. I am finding it difficult to reformulate the first constraint with the log summation.

Any help will be greatly appreciated.

  • $\begingroup$ It is a nonconvex problem so you will never be able to implement this using CVX (the name CVX tells you why). I suspect you've flipped a bunch of signs. $\endgroup$ Nov 26 '14 at 20:38
  • $\begingroup$ Are you sure? The paper I am reading clearly says that it is convex. The paper can be found at bit.ly/1ycHCVa $\endgroup$ Nov 26 '14 at 20:41
  • $\begingroup$ As I said, you must have written down the wrong model. $t=-\infty$ is optimal, and $\Sigma =0$ is trivially feasible and there is no reason to pick any other choice, making the two last constraints redundant. $\endgroup$ Nov 26 '14 at 20:49
  • $\begingroup$ I meant $\Omega=0$ $\endgroup$ Nov 26 '14 at 20:55
  • 1
    $\begingroup$ The log stuff is convex, as it is a negated perspective of a logarithm, alternatively it is a difference of a kullback-leibler term and an entropy. If you wish, you can open a thread on the YALMIP google groups forum and I will show you an implementation using YALMIP + the solver scs which solves the problem exactly using semidefinite+exponential cones. $\endgroup$ Nov 28 '14 at 7:13

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