Two-phase simplex

Question

Question:

Consider the following LP:
$\min z=4 x_{1}+2 x_{2}+4 x_{3}$
s.t.
$x_{2}+x_{3} \leq 2$
$x_{1}-x_{2}-2 x_{3} \leq-1$
$x_{2}+x_{3}=2$v $x_{1}, x_{2}, x_{3} \geq 0$
Solve this problem using the (TWO-PHASE) SIMPLEX algorithm, and state the optimal values of basic variables, non-basic variables, and $z$. Also indicate which of the five possible (special) cases applies / apply to this problem (e.g. a unique solution, an unbounded problem, etc.) and why.

Solution:

The LP is converted into the standard form.
$\min z=4 x_{1}+2 x_{2}+4 x_{3}$
s.t.
$x_{2}+x_{3}+s_{1}=2,[1]$
$-x_{1}+x_{2}+2 x_{3}-e_{2}+a_{2}=1,[2]$
$x_{2}+x_{3}+a_{3}=2,[3]$
$x_{1}, x_{2}, x_{3}, s_{1}, e_{2}, a_{2}, a_{3} \geq 0$
We hence use the 2-phase simplex. The problem of the first phase is to minimise $z^{\prime}=a_{2}+a_{3},[0]$, s.t. $[1],[2],[3]:$
Table 3: Phase 1 - First iteration
$$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \text { BV } & x_{1} & x_{2} & x_{3} & s_{1} & e_{2} & a_{2} & a_{3} & \text { rhs } & \text { ratio } \\ \hline[0] & -1 & 2 & 3 & 0 & -1 & 0 & 0 & 3 & / \\ {[1]} & 0 & 1 & 1 & 1 & 0 & 0 & 0 & 2 & 2 / 1 \\ {[2]} & -1 & 1 & 2 & 0 & -1 & 1 & 0 & 1 & 1 / 2^{*} \\ {[3]} & 0 & 1 & 1 & 0 & 0 & 0 & 1 & 2 & 2 / 1 \\ \hline \end{array} $$ Introduce $x_{3}$ and $a_{2}$ will become non-basic:
Table 4: Phase 1 - Second iteration $$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \text { BV } & x_{1} & x_{2} & x_{3} & s_{1} & e_{2} & a_{2} & a_{3} & \text { rhs } & \text { ratio } \\ \hline[0] & 1 / 2 & 1 / 2 & 0 & 0 & 1 / 2 & / & 0 & 3 / 2 & / \\ {[1]} & 1 / 2 & 1 / 2 & 0 & 1 & 1 / 2 & / & 0 & 3 / 2 & 3^{*} \\ {[2]} & -1 / 2 & 1 / 2 & 1 & 0 & -1 / 2 & / & 0 & 1 / 2 & / \\ {[3]} & 1 / 2 & 1 / 2 & 0 & 0 & 1 / 2 & / & 1 & 3 / 2 & 3^{*} \\ \hline \end{array} $$ Introducing $x_{2}$ will make $x_{3}$ non-basic, but introducing $x_{1}$ will make $a_{3}$ nonbasic, and this seems therefore more desirable:

Original Image

The upper part of the image attached is the question and the lower part is the solution.
I did the exact same table as table4 in the picture, yet I put the ratio of $[2]$ as $0.5/0.5=1$ instead. Could someone please tell me why it is not one but $n/a$ ?

Answer 1

I believe Table 4 shows the quotients in the case where $x_1$ enters the basis. In the text they essentially say that we could let $x_2$ enter, in which case the quotient would be $\frac{1/2}{1/2}=1$ as you say, but we prefer to let $x_1$ enter as that allows us to get rid of an auxiliary variable (in this case $a_3$).