In page 3 of this paper on linear programming, they define $L(d,n)$ to be the set of linear program with $d$ variables and $n$ constraints. They also define a linear objective function $\phi$.
For a linear programming problem $O \in L(d,n)$, $P$ is the polyhedron corresponding to $O$, and $G(P)$ is a directed graph where for any two vertices $u, v \in P$ s.t. $\phi(u) > \phi(v)$, there exists corresponding vertices $u',v' \in G(P)$ s.t. an edge exists directed from $u$ to $v$ ($G(P)$ includes $\infty$). The height of $G(P)$ is the maximum length of the minimal path from a vertex $v \in G(P)$ to some $w \in G(P)$ which maximizes the objective function $\phi$ if it exists. The diameter of $G(P)$ is the maximal length over the shortest paths between any two vertices $u, v \in G(P)$.
The claim I don't understand is that if we let $\delta(d,n)$ denote the maximum diameter over all $O \in L(d,n)$ and $H(d,n)$ denote the maximum height of $G(P)$ over all $O \in L(d,n)$, then $\delta(d,n) \leq H(d,n)$.
The reason I don't understand this is that if the diameter is maximal, then for $O \in L(d,n)$ where there is a path from $u$ to $v$ that is a height of $G(P)$, then wouldn't the diameter of $G(P)$ be at least as big as this height?
I appreciate any feedback. It is my first time on stack overflow, so if there is something I am not doing properly, please alert me. Thanks.
