$$maximize\quad 2x_1-6x_2$$ \begin{align}s.t \quad x_1+x_2+x_3 \ge 2 \\ 2x_1-x_2+x_3 \le 1\end{align}

Becomes $$maximize\quad 2x_1-6x_2$$ \begin{align}s.t\quad x_1+x_2+x_3-s_1=2 \\ 2x_1+x_2-x_3+s_2=1\end{align}

Now I add artificial variables.

$$maximize\quad -a_0$$ \begin{align}s.t\quad x_1+x_2+x_3-s_1+a_0=2 \\ 2x_1+x_2-x_3+s_2=1\end{align}

rearranging, $$maximize \quad x_1+x_2+x_3-2-s_1$$ \begin{align}a_0=-x_1-x_2-x_3+2+s_1\\ s_2=1-2x_1-x_2+x_3\end{align}

I will increase $x_2$ by 1 and $s_2$ will be my leaving variable.

This is where I get stuck.

I know the process for how to solve it, I just don't know where the variables go and how to frame it.

  • $\begingroup$ What is the method you are following? $\endgroup$ – Colliot Apr 1 '17 at 11:47
  • $\begingroup$ I don't know the name but this method turns it into an auxiliary problem. $\endgroup$ – user430765 Apr 1 '17 at 11:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.