[Math] Choosing pivot row in Simplex – slack variables allowed

linear algebralinear programmingsimplex

I have a question concerning the Simplex method to solve linear optimization problems.
I have the following problem:

$$ f(x,y,z) = x+2y+3z$$
Constraints:
$$x+y+z \leq 3$$
$$2x+2y+z \geq 4$$

So my first tableau is (a and b are the slack variables)

 x      y      z      a      b      v
 1      1      1      1      0      3  ->  -3/3
-2     -2    (-1)      0      1    -4  ->  -3/4
-1     -2    (-3)     0      0

As far as I understood, I choose the column by searching the smallest number in the last row (which represents the function) which is here -3. Then I have to divide this number by each value in the column v. The smallest result there shows me which row is the pivot row so I have the element. (I put them in braces in the tableau)

So my next tableau is

 x      y      z      a      b      v
 1      1      1      1      0      3
 2      2      1      0      -1     4 
 5      4      0      0      -3

So, I would see that -3 is the smallest number in the last row, so the column containing the values of slack variable b is the pivot column.
But in the solution to this problem, they now used 4 (so the y column) and I do not understand why?

Best Answer

The error is in the ratio computations. Although choosing the z-column is correct, the ratios are based off of the z- and v-values in each row. Thus, your ratios should be: Column 1: v/z = 3/1 = 3 Column 2: v/z = -4/-1 = 4

Since Column 1 has the smaller ratio, it will be the pivot point. The rest should fall into suit.

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