NO - Summary of Chapter 3 (Part II)

Given the base $\overline{\overline{B}}$, we can set non-basic variables ${\overline{x}}_N=0$ and compute

$${\overline{x}}_B={\overline{\overline{B}}}^{-1}\cdot \overline{b}$$

If ${\overline{x}}_B<0$ the basis is unfeasible. Else if ${\overline{x}}_B\ge 0$ the basis is feasible. → So we have that a BFS (Basic Feasible Solution) is obtained setting ${\overline{x}}_B={\overline{\overline{B}}}^{-1}\cdot \overline{b}$

NOTE: It can be proven that each vertex corresponds to a BFS