values of the two entering variables to minimize the objective function under the constraints of (\[twoDim\]). These constraints make sure that the updated $\x_B \ge \0$ and the leaving variables are zeros. As this problem has only two variables, the solution is slightly more complicate than the selection of the single leaving variable in existing pivot rules, but is still simple and straightforward. Divide $\bar{\A}_{(\jmath_1,\jmath_2)}$ into two parts: $\bar{\A}_1$ has the rows with at least one positive element, and $\bar{\A}_2 \le \0$ has the rows with all elements smaller than or equal to zero . Partition $\bar{\b}=\A_B^{-1} \b$ into the corresponding $\bar{\b}_1$ and $\bar{\b}_2$. Since elements in $\bar{\A}_2$ are smaller than to equal to zero, in view of (\[updatedXB\]) or (\[twoDim\]), introducing positive entering variables will keep the corresponding elements in $\x_B$ to be positive. For $\bar{\A}_1$, in view of (\[updatedXB\]) or (\[twoDim\]), introducing positive variables may change the sign of $\x_B$. If the number of rows in $\bar{\A}_1$ is greater than or equal to $2$, for any two independent rows $(i_1,i_2)$ of $\bar{\A}_1$, solving $$\bar{\A}_1(i_1,i_2) \left[ \begin{array}{c}
x_{\jmath_1} \\ x_{\jmath_2} \end{array} \right]
= \bar{\b}_1(i_1,i_2)
\label{subEq}$$ will give a possible vertex in the convex polygon defined in (\[twoDim\]). Therefore $\x_2 