# How to calculate $\max\{\frac{1}{2}\sum_{i=1}^5(1-X_{i,i+1}) \mid X \text{ PSD}, X_{ii}=1 \forall i\in [5]\}$ analytically.

I'm trying to calculate $$sdp(C_5) =\max\{\frac{1}{2}\sum_{i=1}^5(1-X_{i,i+1}) \mid X \text{ PSD}, X_{ii}=1 \forall i\in [5]\}$$ analytically.

• As $$\begin{pmatrix}1 & -\frac{1}{2} & 0&0&-\frac{1}{2}\\ -\frac{1}{2}&1 & -\frac{1}{2}&0&0\\ 0&-\frac{1}{2}&1 & -\frac{1}{2}&0\\ 0&0&-\frac{1}{2}&1 & -\frac{1}{2}\\ -\frac{1}{2}&0&0 & -\frac{1}{2}&1\\ \end{pmatrix}$$ is feasible, we know $$sdp(C_5) \geq \frac{15}{4}$$.
• Now, every principal submatrix of $$X$$ is necesairly also SDP, in order for $$X$$ to be SDP, so we need $$\begin{pmatrix}1&x\\ x&1\end{pmatrix}$$ to be SDP also. we know the eigenvalues are of the form $$\lambda = 1 \pm \sqrt{x^2}$$, so we know $$-1\leq x\leq 1$$ and therefore $$sdp(C_5) \leq 5$$.
Is it even possible to calculate this analytically? Because i've tried finding the minimal $$y$$ for which $$\begin{pmatrix}1 & -1 & z\\ -1 & 1&y\\z&y&1\end{pmatrix}$$ is PSD, without any clear luck.