Hi Ron,
1)
A pivot step should do two things:
a. Improve the value of the objective function.
For this, we decide which non-basic variable $x_e$ enters by picking one with $c_e \ge 0$.
Then, increasing the value of $x_e$ by $\epsilon$ changes the objective function by $c_e\epsilon\ge 0$.
b. Maintain feasibility.
For this, we decide which basic variable $x_l$ leaves by picking the one for which $a_{le} > 0$ and the ratio $b_l/a_{le}$ is minimal.
Explanation:
Recall that for each basic variable we have $x_i = b_i - \sum_{j\in N} a_{ij} x_j$. So the effect of increasing the value of $x_e$ by $\epsilon$ is to change the value of $x_i$ by $-a_{ie} \epsilon$. We don't want $x_i$ to become negative, and thus we need $b_i-a_{ie}\epsilon$ to be non-negative. If $a_{ie} \le 0$, there's no limitation, but for positive $a_{ie}$ this places an upper bound on $\epsilon$.
2)
The initialization begins with a pivot step that ensures feasibility (i.e., makes all $b_i$ non-negative), so you're actually asking whether it's possible to stop after this pivot step.
No.
We have to finish optimizing the auxiliary program, since only when $x_0=0$ the slack form is usable in the original problem.