I have a polyhedral set for constraining $x$: \begin{align} \mathcal{P} = \{x \in \mathbb{R}^n_{+} : \ Dx \leq d \} \end{align} where $D \in \mathbb{R}^{m \times n}, d \in \mathbb{R}^m$. I find the Chebyshev center of this polyhedron, by solving: \begin{array}{ll} \max &\, r\\ \text{s.t.} & D_i^\top x_c + r \|D_i^\top\| \leq d_i \ \text{for }i =1,\ldots,m \end{array} Now I have this $x_c$, Chebyshev center of $\mathcal{P}.$ I split this polyhedron by two parts. Let $r \in \mathbb{R}^n$ be a random vector. I can divide $\mathcal{P}$ by two halves by adding $r^\top x_c \leq r^\top x$ or $r^\top x_c \geq r^\top x$ to have $\mathcal{P}^\geq, \mathcal{P}^{\leq}$.
Now, in this way, there are definitely new redundant constraints. I have seen some ways to find these redundant constraints, but my case is more specific since I divide based on the center. Therefore, I am wondering if there is an easy way to address this issue...