the step-size, i.e., $$x_{\jmath_2^k} = \max_{\bar{\c}_{j^k}<0}
\Big\{
\min_{i \in  \{ 1,\ldots,m \}} \bar{b}_i / \bar{a}_{j^k,i},
\hspace{0.1in} \mbox{subject to}
\hspace{0.1in}    \bar{a}_{j^k,i}>0
\Big\}.
\label{iniVar2}$$ This strategy will be justified again in the proof of Theorem \[thm2\] and in the discussion of Remark \[longestStep\]. If ${\jmath_2^k}={\jmath_1^k}$ (which means that the most negative rule will generate the longest step), then we take the second entering variable $x_{\jmath_2^k}$ which has the second largest step-size.

Now we discuss how to choose the leaving variables. To make our notation simple, we drop the iteration index $k$ if it does not cause confusion. Let $\bar{\A}_{(\jmath_1,\jmath_2)} = 
\A_B^{-1} \A_{(\jmath_1,\jmath_2)}$ where $\A_{(\jmath_1,\jmath_2)}$ is composed of the $\jmath_1$ and $\jmath_2$ columns of $\A_N$, and $\bar{\c}_{(\jmath_1,\jmath_2)} < \0$ be the two corresponding elements in $\bar{\c}_N$. For two entering indices $(\jmath_1,\jmath_2) \in N^k$ such that that $\x_{(\jmath_1,\jmath_2)}^{\Tr} =[x_{\jmath_1},x_{\jmath_2}] \ge \0$, we need $$\x^{k+1}=\x_{B^k}=\A_{B^k}^{-1}\b
-\A_{B^k}^{-1}\A_{N^k}\x_{N^k} 
= \bar{\b} - \bar{\A}_{(\jmath_1,\jmath_2)}
\x_{(\jmath_1,\jmath_2)} \ge \0.
\label{updatedXB}$$ Therefore, the problem of finding a good new vertex is reduced to minimize the following linear programming problem. $$\begin{aligned}
\begin{array}{cl}
\min  &  \bar{\c}_{(\jmath_1,\jmath_2)}^{\T}
\x_{(\jmath_1,\jmath_2)}, \\ 
\mbox{\rm subject to} 
&  \bar{\A}_{(\jmath_1,\jmath_2)}
\x_{(\jmath_1,\jmath_2)} \le \bar{\b}, 
\hspace{0.1in} \x_{(\jmath_1,\jmath_2)} \ge \0.
\end{array}
\label{twoDim}\end{aligned}$$ Here the third merit criterion is introduced, which is to determine the 