Let's solve the linear programming problem step-by-step using the simplex method. --- ## **Step 1: Introduction to the Simplex Method** **The simplex method** is a popular algorithm used to solve linear programming problems (LPPs) where the objective is to maximize or minimize a linear function subject to linear constraints. The method moves from one feasible solution (vertex of the feasible region) to another, improving the objective function at each step, until the optimal solution is reached. --- ## **Step 2: Convert the Problem to Standard Form** Given: - **Maximize** \( z = 2x_1 - x_2 + 2x_3 \) - Subject to: 1. \( 2x_1 + x_2 \leq 10 \) 2. \( x_1 + 2x_2 - 2x_3 \leq 20 \) 3. \( x_2 + 2x_3 \leq 5 \) 4. \( x_1, x_2, x_3 \geq 0 \) **Step 2.1: Add slack variables** \( s_1, s_2, s_3 \geq 0 \) to convert inequalities into equations: \[ \begin{align*} 2x_1 + x_2 + s_1 &= 10 \\ x_1 + 2x_2 - 2x_3 + s_2 &= 20 \\ x_2 + 2x_3 + s_3 &= 5 \\ \end{align*} \] --- ## **Step 3: Set Up the Initial Simplex Tableau** Variables: \( x_1, x_2, x_3, s_1, s_2, s_3 \) \[ \begin{array}{c|cccccc|c} \text{Basic} & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & \text{RHS} \\ \hline s_1 & 2 & 1 & 0 & 1 & 0 & 0 & 10 \\ s_2 & 1 & 2 & -2 & 0 & 1 & 0 & 20 \\ s_3 & 0 & 1 & 2 & 0 & 0 & 1 & 5 \\ \hline z & -2 & 1 & -2 & 0 & 0 & 0 & 0 \\ \end{array} \] --- ## **Step 4: Apply the Simplex Method Iterations** **Step 4.1: Identify the Entering Variable** Look for the most negative coefficient in the objective function row (bottom row): -2 (\(x_1\)), **1 (\(x_2\)), -2 (\(x_3\))** Most negative: \(x_1\) or \(x_3\) (both -2). Let's choose \(x_1\). **Step 4.2: Identify the Leaving Variable** Calculate ratios (RHS / coefficient of entering variable): - Row 1: \(10/2 = 5\) - Row 2: \(20/1 = 20\) - Row 3: \(5/0 = \infty\) (can't divide by 0) Smallest non-negative ratio: **5** (Row 1, \(s_1\)). So, \(s_1\) leaves, \(x_1\) enters. --- **Step 4.3: Pivot Operation (Row Operations)** Make the coefficient of \(x_1\) in Row 1 equal to 1: - Row 1: Divide by 2 \[ \text{New Row 1}: \quad x_1 + 0.5x_2 + 0s_3 + 0.5s_1 = 5 \] Update the other rows to make their \(x_1\) coefficient zero: - Row 2: Row2 - (1)*Row1 - \( [1,2,-2,0,1,0|20] - [1,0.5,0,0.5,0,0|5] = [0,1.5,-2,-0.5,1,0|15] \) - Row 3: already 0 in \(x_1\) column, no change - Row z: Rowz + 2*Row1 - \( [-2,1,-2,0,0,0|0] + 2*[1,0.5,0,0.5,0,0|5] = [0,2,-2,1,0,0|10] \) \[ \begin{array}{c|cccccc|c} \text{Basic} & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & \text{RHS} \\ \hline x_1 & 1 & 0.5 & 0 & 0.5 & 0 & 0 & 5 \\ s_2 & 0 & 1.5 & -2 & -0.5 & 1 & 0 & 15 \\ s_3 & 0 & 1 & 2 & 0 & 0 & 1 & 5 \\ \hline z & 0 & 2 & -2 & 1 & 0 & 0 & 10 \\ \end{array} \] --- **Step 4.4: Repeat: Next Entering Variable** Most negative in z row: -2 (\(x_3\)) **Leaving variable:** - Row 2: \(15 / 2 = 7.5\) (since -2, must be positive, but negative so not eligible) - Row 3: \(5 / 2 = 2.5\) (since 2 is positive, eligible) - Row 1: \(5 / 0 = \infty\) (not eligible) So, smallest is 2.5 (Row 3, \(s_3\)), so \(s_3\) leaves, \(x_3\) enters. --- **Step 4.5: Pivot on Row 3, \(x_3\)** Divide Row 3 by 2: \[ 0x_1 + 0.5x_2 + 1x_3 + 0s_1 + 0s_2 + 0.5s_3 = 2.5 \] Update other rows: - Row 1: Row1 - (0)*Row3 (no change) - Row 2: Row2 - (-2)*Row3 = [0,1.5,-2,-0.5,1,0|15] + [0,1,-2,0,0,1|5]*2 = [0,3.5,0,-0.5,1,1|20] - Row z: Rowz + 2*Row3: [0,2,-2,1,0,0|10] + [0,1,2,0,0,1|5]*2 = [0,4,0,1,0,2|20] So, updated tableau: \[ \begin{array}{c|cccccc|c} \text{Basic} & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & \text{RHS} \\ \hline x_1 & 1 & 0.5 & 0 & 0.5 & 0 & 0 & 5 \\ s_2 & 0 & 1.5 & -2 & -0.5 & 1 & 0 & 15 \\ x_3 & 0 & 0.5 & 1 & 0 & 0 & 0.5 & 2.5 \\ \hline z & 0 & 4 & 0 & 1 & 0 & 2 & 20 \\ \end{array} \] --- **Step 4.6: Next Entering Variable** Check the z-row: All coefficients are non-negative. **Optimality Reached!** --- ## **Step 5: State the Final Solution** From the final tableau: - \(x_1 = 5\) - \(x_2 = 0\) - \(x_3 = 2.5\) - \(z = 20\) --- ## **Final Step: Short Summary** **The optimal solution is \(x_1 = 5\), \(x_2 = 0\), \(x_3 = 2.5\), which gives the maximum value \(z = 20\).**