Answer to Question #235309 in Operations Research for mona

Question

Consider the following LPP

Maximize Z = 2x₁ + 3x₂

Subject to

x₁ + 3x₂ ≤ 6

3x₁ + 2x₂ ≤ 6

x₁, x₂ ≥ 0

a. Determine all the basic solutions of the problem, and classify them as feasible and

infeasible.

b. Carry out the full tableau implementation of the simplex method, starting with the basic

feasible solution (x₁, x₂) = (0, 0)

Answer 1

$a. \text{ Solving the system of linear equations given in the constraints simultaneously, we }\\ \text{have that $x_1 = \frac{6}{7}$ and $x_2 = \frac{12}{7}$}\\ \text{We can determine the feasible solutions by setting one variable to 0 in each constraint }\\ \text{Hence, we have }\\ (0,2), (6,0),(0,3),(2,0)\\ \text{(0,2) and (2,0) are feasible solutions, since they satisfy all the constraints and (0,3)}\\ \text{and (6,0) are infeasible solutions because they do not satisfy all the constraints}\\ b. \text{ Next, we find the optimal solution using simplex method}\\ \text{The first tableau is given by}\\ \begin{matrix} & & x_1 & x_2 & x_3 & x_4 \\ & & 2 & 3 & 0 & 0\\\hline x_3 & 0 & 1 & 3 & 1 & 0 & 6\\ x_4 & 0 & 3 & 2 & 0 & 1 & 6\\ & & -2 & -3 & 0 & 0 & 0 \end{matrix}\\ \text{Applying row reduction techniques to each row and the simplex algorithm, we get}\\ \text{tableau 2}\\ \begin{matrix} & & x_1 & x_2 & x_3 & x_4 \\ & & 2 & 3 & 0 & 0\\\hline x_2 & 3 & \frac{1}{3} & 1 & \frac{1}{3} & 0 & 2\\ x_4 & 0 & \frac{7}{3} & 0 & -\frac{2}{3} & 1 & 2\\ & & -1 & 0 & 1 & 0 & 6 \end{matrix}\\ \text{The final tableau is given by}\\ \begin{matrix} & & x_1 & x_2 & x_3 & x_4 \\ & & 2 & 3 & 0 & 0\\\hline x_2 & 3 & 0 & 1 & \frac{3}{7} & -\frac{1}{7} & \frac{12}{7}\\ x_1 & 2 & 1 & 0 & -\frac{2}{7} & \frac{3}{7} & \frac{6}{7}\\ & & 0 & 0 & \frac{5}{7} & \frac{3}{7} & \frac{48}{7} \end{matrix}\\ \text{Hence the feasible solution which is optimal is $x_1=\frac{12}{7}$, $x_2=\frac{6}{7}$ and $z=\frac{48}{7}$}$