I have the following linear program:
Maximize $15x + 2y + z$ subject to $$x \le 10 \\ x + y \le 17 \\ 2x + 3 \le 25 \\ y + z \ge 11$$
I created the following Simplex Tableau:
x y z s1 s2 s3 s4 p
s1 1 0 0 1 0 0 0 0 10
s2 1 1 0 0 1 0 0 0 17
s3 2 0 3 0 0 1 0 0 25
s4 0 1 1 0 0 0 -1 0 11
p -15 -2 -1 0 0 0 0 1 0
According to what I understand from the Simplex Algorithm, the first pivot column is the most negative element in the last row (-15 in this case, so first column) and the first pivot row is going to be the row where dividing the rightmost column value by the value in the pivot column is at a minimum, so for this example a minimum of 10/1, 17/1, 25/2, 11/0 = 10/1 which is the first row.
However, when I run the program through a Linear Program Solver, the first pivot is actually in the second column (where -2 is the value in the last row) and second to last row. Why is this the case?
The reason your software chooses the $C^\pi_y = -2$ is due the behaviors of the artificial variable that exists in the fourth constraint. For example, we'll use the Big M method to standardize the problem of the form:
$$max \text{ }\text{ } w -15x -2y -z +Ma_4 = 0$$ Subject to, $$x + s_1 = 10$$ $$x+y+s_2 = 17$$ $$2x + 3z + s_3 = 25$$ $$y + z - e_4 + a_4 = 11$$ $$x,y,z, s_1, s_2, s_3, e_4, a_4 \ge 0$$
From this, we'll get the following tableau:
Notice that we are missing a basic variable in the fourth row, and the closest value we have in the fourth row to becoming a basic variable is the artificial variable $a_4$, so we take row 4 times $-M$ and add it to the objective row to get the following tableau:
From here, since M is the largest possible number we can pick (or what the computer can handle), then the $y$ column, whose $C^\pi_y = -M-2$ is the most negative, is the pivoting column. Thus why your software chose to pivot the $y$ column.