6 Consider the following LP problem.
| Minimise | \(2 a - 4 b + 5 c - 30\), |
| subject to | \(3 a + 2 b - c \geqslant 10\), |
| \(- 2 a + 4 c \leqslant 35\), |
| \(4 a - b \leqslant 20\), |
| and | \(a \leqslant 6 , b \leqslant 8 , c \leqslant 10\). |
- Since \(a \leqslant 6\) it follows that \(6 - a \geqslant 0\), and similarly for \(b\) and \(c\). Let \(6 - a = x\) (so that \(a\) is replaced by \(6 - x ) , 8 - b = y\) and \(10 - c = z\) to show that the problem can be expressed as
$$\begin{array} { l l }
\text { Maximise } & 2 x - 4 y + 5 z ,
\text { subject to } & 3 x + 2 y - z \leqslant 14 ,
& 2 x - 4 z \leqslant 7 ,
& - 4 x + y \leqslant 4 ,
\text { and } & x \geqslant 0 , y \geqslant 0 , z \geqslant 0 .
\end{array}$$ - Represent the problem as an initial Simplex tableau. Perform two iterations of the Simplex algorithm, showing how each row was obtained. Hence write down the values of \(a , b\) and \(c\) after two iterations. Find the value of the objective for the original problem at this stage.
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