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Problem 10

# You are given the final simplex tableau for the dual problem. Give the solution to the primal problem and the solution to the associated dual problem. Problem: Minimize $$\quad C=2 x+3 y$$ subject to \begin{aligned} x+4 y & \geq 8 \\ x+y & \geq 5 \\ 2 x+y & \geq 7 \\\ x \geq 0, y & \geq 0 \end{aligned} Final tablcau: $$\begin{array}{cccccc|c} u & v & w & x & y & P & \text { Constant } \\ \hline 0 & 1 & \frac{7}{3} & \frac{4}{3} & -\frac{1}{3} & 0 & \frac{5}{5} \\ 1 & 0 & -\frac{1}{3} & -\frac{1}{3} & \frac{1}{3} & 0 & \frac{1}{3} \\ \hline 0 & 0 & 2 & 4 & 1 & 1 & 11 \end{array}$$

Expert verified
The solution to the primal problem is $$(x, y) = (\frac{1}{3}, \frac{5}{3})$$ with the minimum value of the objective function $$C = \frac{17}{3}$$. The solution to the dual problem is $$(u, v, w) = (1, 0, \frac{7}{3})$$ with the maximum value of the objective function $$W = \frac{65}{3}$$.
See the step by step solution

## Step 1: Identify the primal and dual problems

The given problem is a minimization problem; the primal problem is: Minimize $$C=2x + 3y$$ subject to: $$x + 4y \geq 8$$ $$x + y \geq 5$$ $$2x + y \geq 7$$ $$x, y \geq 0$$ Now, let's write the dual problem: Maximize $$W=8u + 5v + 7w$$ subject to: $$u + v + 2w \leq 2$$ $$4u + v + w \leq 3$$ $$u, v, w \geq 0$$

## Step 2: Identify the optimal solution for both primal and dual problems from the final simplex tableau

The final simplex tableau is: $$\begin{array}{cccccc|c} u & v & w & x & y & P & \text { Constant } \\ \hline 0 & 1 & \frac{7}{3} & \frac{4}{3} & -\frac{1}{3} & 0 & \frac{5}{3} \\ 1 & 0 & -\frac{1}{3} & -\frac{1}{3} & \frac{1}{3} & 0 & \frac{1}{3} \\ \hline 0 & 0 & 2 & 4 & 1 & 1 & 11 \end{array}$$ From the tableau, we can find the optimal solution for both primal and dual problems: For the primal problem: $$x = \frac{1}{3}$$ $$y=\frac{5}{3}$$ $$C_{min} = 2(\frac{1}{3}) + 3(\frac{5}{3}) = \frac{2}{3} + 5 = \frac{17}{3}$$ The solution to the primal problem is $$(x, y) = (\frac{1}{3}, \frac{5}{3})$$ and the minimum value of C is $$\frac{17}{3}$$. For the dual problem: $$u=1$$ $$v=0$$ $$w=\frac{7}{3}$$ $$W_{max} = 8(1) + 5(0) + 7(\frac{7}{3}) = 8 + \frac{49}{3} = \frac{65}{3}$$ The solution to the dual problem is $$(u, v, w) = (1, 0, \frac{7}{3})$$ and the maximum value of W is $$\frac{65}{3}$$.

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