I have the following LP
Solve the following problem by the simplex tableau method starting with the basic feasible solution corresponding to the vertex (x1, x2) = (4, 0).
$$ \text{maximize }z = -x_1 + 2x_ 2 \\ \text{subject to: } \\ 3x_1 + 4x_2 = 12 \\ 2x_ 1-x_ 2 \leqslant 12 $$
It is clear to me how to initialize when an initial solution is not given, but how could I build my table once it is given?
I just solved it. Since we have the initial vertex, we should before adjust the problem:
$$ 3x_1+4x_2=12\\ 2x_1-x_2+x_3=12\\ $$
Once it is adjusted, we should use the vertex and check how the restrictions work with them:
$$ 3\cdot(4)+4\cdot(0)=12\\ 2\cdot(4)-1\cdot(0)+x_3=12 $$
Wich makes $x_3=4$. So now, we can adapt our initial solution to $(4,0,4)$ and now we have an initial basis composed by $(x_1,x_3)$ and finally, we can build the tableau as:
| $x_1$ | $x_2$ | $x_3$ | ||
|---|---|---|---|---|
| z | -1 | 2 | 0 | |
| $x_1$ | 3 | 4 | 0 | 12 |
| $x_3$ | 2 | -1 | 1 | 12 |