In 2-phases simplex method what kind of operations must be done to get the canonical form tableau?

In this step(phase 2 of 2-phases method) after the remotion of artificial variables columns of auxiliary problem and copy of coefficents of objective functions, I don't know what are the operations performed to get canonical form tableau (as you see in the picture)

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the objective function was

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  • $\begingroup$ Canonical form, I'm assuming, is $\min c^{T}x$ subject to $Ax \leq b$, $x \geq 0$?. It varies from text-to-text. $\endgroup$ – clocktower Feb 3 '16 at 22:44
  • $\begingroup$ I call this standard form of objective function, before start the simplex method I use the standard form but this isn't a problem. I mean the canonical form of the latest step tableau (in phase 2 of 2-phases simplex method). The step that transform the first table in the second table of my picture. $\endgroup$ – AndreaF Feb 3 '16 at 22:52
  • $\begingroup$ If I'm understanding your confusion. For two-phase simplex, you're solving an auxiliary problem to get the second tableau. There's no straightforward transformation. Unfortunately, I've not encountered two-phase with equality constraints. $\endgroup$ – clocktower Feb 3 '16 at 22:54
  • $\begingroup$ Exactly, I solve an auxiliary problem. my question is about the operation performed after the remotion of the artificial variables used in the auxiliary problem in phase 2 since I have to get a solution of the original problem after the first phase iterations. $\endgroup$ – AndreaF Feb 3 '16 at 22:59

To get the matrix back in canonical form, you simply need to make sure that any basic variable has a 0 coefficient in the objective function row. So looking at the matrix

$$\begin{bmatrix} 1&0&-1&-1&0&2 \\ 0&0&-4&-4&1&10 \\ 0&1&3&1&0&4 \\ -2&-1&2&0&0&0 \\ \end{bmatrix}$$

You simply need to add 2 times the first row to the last (objective row) and 1 times the third row to the last (objective row). This will make sure there is a coefficient of 0 in the objective row for the first two columns, since they are the columns that contain the basic variables.

Doing this, you should get that after adding $R4 + 2R1$ you get $$\begin{bmatrix} 1&0&-1&-1&0&2 \\ 0&0&-4&-4&1&10 \\ 0&1&3&1&0&4 \\ -2&-1&0&-2&0&4 \\ \end{bmatrix}$$ This makes sure that the basic variable $x_1$ has a 0 coefficient in the objective function row. Doing the same for the basic variable $x_2$, adding $R4+R3$, you should get:

$$\begin{bmatrix} 1&0&-1&-1&0&2 \\ 0&0&-4&-4&1&10 \\ 0&1&3&1&0&4 \\ -2&0&-3&-1&0&8 \\ \end{bmatrix}$$

Which is the matrix in canonical form.

  • $\begingroup$ Thanks for the answer. So not in all situation I have to add 2 time the 1st and 1 time the 3rd. I have to choose random rows to sum to latest to get 0 in all basic variables. Right? Is there any other criterion to choose the rows that I have to sum or could I choose what row want to get 0? $\endgroup$ – AndreaF Feb 4 '16 at 0:15

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