so I'm struggling to find how the pivot points were found in the following optimization problem: $$Minimize\ z = a + b + c$$ $$Subject\ to:$$ $$a - b - c ≤ 0$$ $$a + b + c ≥ 4$$ $$a + b - c = 2$$ $$a, b ≥ 0$$
I got the problem into this form: $$Maximize\ z* + a + b + c’ – c’’ = 0$$ Such that $c = c' - c''$ and $z*=-z$ $$Subject\ to:$$ $$a – b – c’ – c’’ + S_1 = 0$$ $$a + b + c’ – c’’ – L_1 + A_1 = 4$$ $$a + b – c’ – c’’ + A_2 = 2$$ $$a, b, c’, c’’, S_1, L_1, A_1, A_2 ≥ 0$$ With the first tableu being:
\begin{array}{rrrrrrrrr|r} & a & b & c' & c'' & S_1 & L_1 & A_1 & A_2 & \\ \hline & 1 &-1 &-1 &-1 & 1 & 0 & 0 & 0 & 0 \\ & 1 & 1 & 1 &-1 & 0 & -1 & 1 & 0 & 4 \\ & 1 & 1 &-1 &-1 & 0 & 0 & 0 & 1 & 2 \\ \hline & 1 & 1 & 1 &-1 & 0 & 0 & 0 & 0 & 0 \end{array} The problem is, from here, I got no success (using c'' as pivot column), so I checked a few simplex calculators to see what I was doing wrong, and they seemed to get to the same sort of setup as me, and with the same tableu. However, from there, they took column a as pivot column, which is the most positive number? They then got the ratios, and took row 1 as pivot row (ratio of 0), which I guess makes sense?
After creating the second tableu, they took column b as the pivot coulumn (again most positive z number(tied) at 2), BUT, then they used row 3, which had a ratio of 1, meanwhile row 1 still had a ratio of 0.
Same sort of scenario with the third and fourth* tableus, achieving a final answer of $$a=2, b=1, c=1(1-0)$$ $$min z = 4$$
As far as I can tell the answer is right, but I cannot for the life of me figure out how they got the pivot points.
Any help is greatly appreciated.
Edit: Also, in the final tableu, which is used as the solution, the z row still contains positive and negative numbers, as well as zeros.
Edit 2: Thanks to this question, as far I can tell the zero was not selected in the second and third tableus because the pivot in that case would be nonpositive.
Edit 3: Second, third, and fourth (*whoops there's four) tableus: \begin{array}{rrrrrrrrr|r} & a & b & c' & c'' & S_1 & L_1 & A_1 & A_2 & \\ \hline & 1 &-1 &-1 &-1 & 1 & 0 & 0 & 0 & 0 \\ & 0 & 2 & 2 & 0 &-1 & -1 & 1 & 0 & 4 \\ & 0 & 2 & 0 & 0 &-1 & 0 & 0 & 1 & 2 \\ \hline & 0 & 2 & 2 & 0 &-1 & 0 & 0 & 0 & 0 \end{array}
\begin{array}{rrrrrrrrr|r} & a & b & c' & c'' & S_1 & L_1 & A_1 & A_2 & \\ \hline & 1 & 0 &-1 &-1 & 1/2 & 0 & 0 & 1/2 & 1 \\ & 0 & 0 & 2 & 0 & 0 & -1 & 1 &-1 & 2 \\ & 0 & 1 & 0 & 0 &-1/2 & 0 & 0 & 1/2 & 1 \\ \hline & 0 & 0 & 2 & 0 & 0 & 0 & 0 &-1 &-2 \end{array}
Final: \begin{array}{rrrrrrrrr|r} & a & b & c' & c'' & S_1 & L_1 & A_1 & A_2 & \\ \hline & 1 & 0 & 0 &-1 & 1/2 &-1/2 & 1/2 & 0 & 2 \\ & 0 & 0 & 1 & 0 & 0 &-1/2 & 1/2 &-1/2 & 1 \\ & 0 & 1 & 0 & 0 &-1/2 & 0 & 0 & 1/2 & 1 \\ \hline & 0 & 0 & 0 & 0 & 0 & 1 &-1 & 0 & -4 \end{array}
So, to reiterate, what would cause someone to use a positive z-row entry to determine the pivot column, and what determines whether you've reached the optimal solution? As I thought it was the z-row containing no negative enties. Thanks ^^