Simplex tableau, constraints, and slack variables

[Guest]

I'm doing some revision of the Simplex Algorithm, I've got so far with it but then come up with stuff that I think contradicts each other, so was wondering if someone could tell me where I'm going wrong?

Here are my constraints (modified so they have slack variables, etc):

. . .x + y + z + r = 3

. . .2x + y + 2z + s = 5

. . .x + 2y + z + t = 6

. . .f(x, y, z) - x - 2y - 3z = 0

After writing out the tableau and boiling it down I get and end result of:

. . .y + 2r - s = 1

. . .x + z - r + s = 3

. . .-3r + s + t = 2

. . .f(x, y, z) + 2x + r + s = 8

What I don't get is, if the equation is maxed when r, s, and t are all 0, then 0 = 2 from the third equation. All I can think is I've gone wrong somewhere. I can put my method up if you'd like.
Any help appreciated.
Thanks guys
Josh

 

[JakeD]

The solution is y = 1, z = 2 and t = 2 (not t = 0). This interactive simplex method solver will give a detailed walk-through of a solution that I found very helpful.