1

Online solver gave me that this problem was unbounded. $$Maximise − 7 + x − 2y + 3z − w_3$$ subject to $$w_1 = 3 − 2x − y + z$$ $$w_2 = 2 + 8x + 2y − 5z$$ $$w_0 = 7 − x + 2y − 3z + w_3$$ $$x, y, z, w_0, w_1, w_2, w_3 \ge 0$$

I can increase $x$ by a maximum of $\frac{2}{8}$ and $w_1$ is my leaving variable.

Maximise $$\frac{-11}{2} − \frac{w_1}{2} − \frac{5y}{2} + \frac{7z}{2} + w_3$$

$$x = \frac{3}{2} − \frac{w_1}{2} − \frac{y}{2} + \frac{z}{2}$$ $$w_2 = 2 + 8x + 2y − 5z$$ $$w_0 = 7 − x + 2y − 3z + w_3$$ $$x, y, z, w_0, w_1, w_2, w_3 \ge 0$$

So is the giveaway that I can set $z$ to whatever value and then adjust all others to satisfy the non-negativity constraints?

stackdsewew
  • 1,047

1 Answers1

1

You are correct that $w_1$ will be leaving variable, but $\frac{2}{8}$ is not the correct bound for $x$.

You should be looking only at nonpositive coefficients with $x$, notably $-2x$ and $-x$, and you choose minumum from $\frac{-p_i}{q_{ij}}$ where $p$ is vector of real numbers, here $(3,2,7)$ and $q_{ij}$ are coefficients of non-basic variables.

Here we are looking at $x$ collumn, so you go through $q_{i1}$, which is $( -2, 8 -1)$ and pick minimum $\frac{-p_i}{q_{ij}}$, which in this case is $\frac{3}{2}$ in first row.

Rupert Ehringer
  • 193