Using the two-phase simplex method I am asked to

Minimise: z= 3x2-x3+8 
Subject to:

I am not sure how to go about this question as I have only used this method with maximsie problems. My initial thought was that the problem is already in canonical form, so there is no need to add slack, surplus or artificial variables and the problem can be solved using the simplex method? However I have tried this and it doesnt work. The answers to this question are x1=5, x2=0, x3=5 and z=3. If anyone could show how they get to this stage I would really appreciate it. Ps. sorry for the poor markdown I am still new e.g. x2 means 'x small 2' not 'x squared'.


You have to use artificial variables ($a_i$), because of the equality signs.

$\begin{array}{|c|c|c|c|c|c|} \hline x_1 &x_2&x_3&a_1&a_2&RHS \\ \hline 0&3&-1&0&0&-8 \\ \hline 1&1&\color{red}{ \boxed{ 1}} &1&0&10 \\ \hline 2&3&1&0&1&15 \\ \hline \end{array}$

$\color{red}{ \boxed{ 1}}$ is the first pivot element. Both artificial variables have to be kicked out of the base.

$\begin{array}{|c|c|c|c|c|c|} \hline x_1 &x_2&x_3&a_1&a_2&RHS \\ \hline 1&4&0&1&0&2 \\ \hline 1&1&1 &1&0&10 \\ \hline \color{red}{ \boxed{ 1}}&2&0&-1&1&5 \\ \hline \end{array}$

$\begin{array}{|c|c|c|c|c|c|} \hline x_1 &x_2&x_3&a_1&a_2&RHS \\ \hline 0&2&0&2&-1&-3 \\ \hline 0&-1&1 &2&-1&5 \\ \hline 1&2&0&-1&1&5 \\ \hline \end{array}$

Et voila the optimal solution is $(x_1,x_2,x_3,z)=(5,0,5,3)$

  • $\begingroup$ Firstly thank you very much for the response you have certainly made things clearer, however i am still confused about a couple of things. Why did you not add a new objective function in regards to the artificial variables i.e. w=a1+a2? And why did you not stop at the second stage, as all the coefficients of the z function were positive ? Thanks again! $\endgroup$ – BigAl1992 Sep 17 '15 at 13:17
  • $\begingroup$ @AlexdeTurris You can use the new objective function, to make it easier to choose the right column. But in this case it was obvious to me to choose the third column first, because it had a negative coefficient at the objective function (z). And then I had to choose the first column. The reason is, if I would have chosen the second column, the objective function (z) would has got negative coefficients again. Therefore I have chosen the first column. Yes, you are almost right. You are finished, if all artificial variables are out of the base and the coefficients of z are all non-negative. $\endgroup$ – callculus Sep 17 '15 at 14:13
  • $\begingroup$ Brilliant thank you ! $\endgroup$ – BigAl1992 Sep 17 '15 at 14:29

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