Find the optimal solution without going through the ERO's

Question

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All I got is that

$$12y_1 + 7y_2 + 10y_3 = 2(0) + 4(10.4) + 3(0) + 1(0.4)$$

and

$y_2 = 0$ because $x_6$ is in basis.

How do I find $y_1$ and $y_3$ without going through the simplex method?

I took the dual and that doesn't seem to help.

I know that $y_1 = z_5 - c_5$ and $y_3 = z_7 - c_7$, but how do I find $z_i - c_i$?

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From Chapter 2 here.

Answer 1

Consider the dual:

$$\min z' = 12y_1 +7y_2 + 10y_3$$

s.t.

$$\begin{bmatrix} 3 & 1 & 2\\ 1 & 3 & 1\\ 1 & 2 & 3\\ 4 & 3 & -1 \end{bmatrix}\begin{bmatrix} y_1\\ y_2\\ y_3 \end{bmatrix} \ge \begin{bmatrix} 2\\ 4\\ 3\\ 1 \end{bmatrix}$$

Since the dual of the dual is the primal, the dual of the dual variables are the primal variables.

Using complementary slackness on the dual, we have:

$$x_1 \mathcal{S}_1 = 0$$

$$x_2 \mathcal{S}_2 = 0$$

$$x_3 \mathcal{S}_3 = 0$$

$$x_4 \mathcal{S}_4 = 0$$

$$\because x_2, x_4 > 0, \mathcal{S}_2 = \mathcal{S}_4 = 0$$

Thus,

$$\begin{bmatrix} 1 & 3 & 1\\ 4 & 3 & -1 \end{bmatrix}\begin{bmatrix} y_1\\ y_2\\ y_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$$

$$\to \begin{bmatrix} 1 & 3 & 1\\ 4 & 3 & -1 \end{bmatrix}\begin{bmatrix} y_1\\ 0\\ y_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$$

$$\to \begin{bmatrix} 1 & 1\\ 4 & -1 \end{bmatrix}\begin{bmatrix} y_1\\ y_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$$

$$\to y_1 = 1, y_3 = 3$$

Answer 2

HINT: With these values of $x_1,x_2,x_3,x_4$, it is clear that the basic variables are $x_2,x_4$ and $x_6$ (the slack variable for the second constraint).

Therefore, it is easy to find the optimal tableau, from which you can directly read the values of the dual variables (do you see where?).