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Suppose I've a simple linear programming problem

Maximize some arbitrary objective function subject to the following constraints:

$x_1+2x_2 \leq 6$

$x_1-x_2 \leq 4$

$x_2 \leq 2$

$x_1,x_2 \geq 0$

It was simple enough to graph and the extreme points are at:

$(0,0), (0,2), (4,0), (2,2), (\frac{14}{3}, \frac{2}{3})$

For each extreme point how do I determine which are the basic and non-basic variables?

I think there should be 2 basic and 3 non-basic, right? I also think for the point $(0,0)$ 2 slack variables are the basic variables? But there are 3 in this problem.

I'm not sure how to know exactly which without an objective function. Any ideas?

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You will have as many basic variables as constraints, i.e., $3$ here. Typically, a basic variable has a positive value.

If you consider the point $(0,0)$, it means that $x_1=x_2=0$, and that the slack variables $e_1,e_2,e_3$ are positive (for the constraints to hold). So the basic variables are $e_1,e_2,e_3$.

For point $(0,2)$, $x_1=0$ and $x_2>0$, so $x_2$ is a basic variables. You need two more. You can either find them algebraically by plugging the values of $x_1$ and $x_2$ in the constraints: you will find that $e_3=0$, and that $e_1,e_2>0$, so $e_1,e_2$ are your remaining basic variables. You can also work graphically. $(0,2)$ is at the intersection between $x_2=2$ and $x_1=0$, in other words at this point only the third constraint is active, which means that first and second constraints are inactive, i.e., $e_1,e_2 >0$.

Can you do the same for the other points?

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  • $\begingroup$ Got it! I'd known this before but forgot :D Thanks !! $\endgroup$ – Gregory Peck Mar 2 '17 at 22:43

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