TITLE: simplex method : Entering Variable
QUESTION [5 upvotes]: In the Simplex method, a variable that enters the basis, cannot depart the basis in the very next iteration. Please explain..why so ?

REPLY [6 votes]: This isn't true.  For a counterexample, consider 
$$\max \text{ } Z = x_1 + 2x_2$$ 
subject to
$$x_1 + 3x_2 + s = 3,$$
$$x_1, x_2, s \geq 0.$$
where $s$ is the slack variable.
The initial basis is $\{s\}$.  Using Dantzig's rule for selecting the entering basic variable, we would pick $x_2$, as it gives the largest per-unit increase.  Since $x_2$ enters, $s$ must leave.  Our new dictionary looks like 
$$Z = 2 + \frac{1}{3}x_1 - \frac{2}{3}s,  $$
$$x_2 = 1 - \frac{1}{3} x_1 - \frac{1}{3} s.$$ 
Thus we can increase $Z$ by increasing $x_1$.  Let $x_1$ enter the basis; then $x_2$ must leave, yielding the optimal dictionary:
$$Z = 3 - x_2 - s,$$
$$x_1 = 3 - 3x_2 - s.$$
The point is that $x_2$ entered the basis in the first iteration and left in the second, providing a counterexample to your statement.