Certainly! Here’s your solution, clearly explained step-by-step, as if I’m your Chegg tutor, **with no changes required to your process**: --- ### **1. Convert to Standard Form** Let's start by converting the given linear program into **standard form**. #### **Original Problem:** Minimize: \[ z = -3x_1 + x_2 + x_3 \] #### **Step 1: Convert to Maximization** Since the simplex method is typically used for maximization, we rewrite the objective: \[ \text{Maximize } Z = 3x_1 - x_2 - x_3 \] #### **Step 2: Standardize the Constraints** Introduce slack, surplus, and artificial variables as needed: \[ \begin{aligned} \text{Constraint 1 (≤):} & \quad x_1 - 2x_2 + x_3 + s_1 = 11 \\ \text{Constraint 2 (≥):} & \quad -4x_1 + x_2 + 2x_3 - s_2 + a_1 = 3 \\ \text{Constraint 3 (=):} & \quad -2x_1 + x_3 + a_2 = 1 \end{aligned} \] All variables must be non-negative: \[ x_1, x_2, x_3, s_1, s_2, a_1, a_2 \geq 0 \] #### **Step 3: Penalize Artificial Variables** When forming the objective, penalize artificial variables: \[ Z = 3x_1 - x_2 - x_3 - M a_1 - M a_2 \] --- ### **2. Initial Simplex Tableau** Using the constraints, set up the initial simplex tableau for the Big M method: \[ \begin{array}{c|rrrrrrrr|r} \text{Basis} & Z & x_1 & x_2 & x_3 & s_1 & s_2 & a_1 & a_2 & \text{RHS} \\ \hline Z & 1 & 6M - 3 & 1 - M & 1 - 3M & 0 & M & 0 & 0 & -4M \\ s_1 & 0 & 1 & -2 & 1 & 1 & 0 & 0 & 0 & 11 \\ a_1 & 0 & -4 & 1 & 2 & 0 & -1 & 1 & 0 & 3 \\ a_2 & 0 & -2 & 0 & 1 & 0 & 0 & 0 & 1 & 1 \\ \end{array} \] --- ### **3. Simplex Iterations** #### **Iteration 1** - **Entering variable:** \( x_3 \) (most negative coefficient in Z-row: \( 1 - 3M \)) - **Leaving variable:** \( a_2 \) (smallest positive ratio: \( 1/1 = 1 \)) Pivot on row \( a_2 \), column \( x_3 \): \[ \begin{array}{c|rrrrrrrr|r} \text{Basis} & Z & x_1 & x_2 & x_3 & s_1 & s_2 & a_1 & a_2 & \text{RHS} \\ \hline Z & 1 & -1 & 1 - M & 0 & 0 & M & 0 & 3M - 1 & -M - 1 \\ s_1 & 0 & 3 & -2 & 0 & 1 & 0 & 0 & -1 & 10 \\ a_1 & 0 & 0 & 1 & 0 & 0 & -1 & 1 & -2 & 1 \\ x_3 & 0 & -2 & 0 & 1 & 0 & 0 & 0 & 1 & 1 \\ \end{array} \] --- #### **Iteration 2** - **Entering variable:** \( x_2 \) (coefficient \( 1 - M \) is most negative) - **Leaving variable:** \( a_1 \) (minimum ratio: \( 1/1 = 1 \)) Pivot on row \( a_1 \), column \( x_2 \): \[ \begin{array}{c|rrrrrrr|r} \text{Basis} & Z & x_1 & x_2 & x_3 & s_1 & s_2 & \text{RHS} \\ \hline Z & 1 & -1 & 0 & 0 & 0 & 1 & -2 \\ s_1 & 0 & 3 & 0 & 0 & 1 & -2 & 12 \\ x_2 & 0 & 0 & 1 & 0 & 0 & -1 & 1 \\ x_3 & 0 & -2 & 0 & 1 & 0 & 0 & 1 \\ \end{array} \] *(Both artificial variables are now out of the basis—Phase I complete.)* --- #### **Iteration 3** - **Entering variable:** \( x_1 \) (only negative coefficient in Z-row: \( -1 \)) - **Leaving variable:** \( s_1 \) (minimum ratio: \( 12/3 = 4 \)) Pivot on row \( s_1 \), column \( x_1 \): \[ \begin{array}{c|rrrrrrr|r} \text{Basis} & Z & x_1 & x_2 & x_3 & s_1 & s_2 & \text{RHS} \\ \hline Z & 1 & 0 & 0 & 0 & \frac{1}{3} & \frac{1}{3} & 2 \\ x_1 & 0 & 1 & 0 & 0 & \frac{1}{3} & -\frac{2}{3} & 4 \\ x_2 & 0 & 0 & 1 & 0 & 0 & -1 & 1 \\ x_3 & 0 & 0 & 0 & 1 & \frac{2}{3} & -\frac{4}{3} & 9 \\ \end{array} \] --- ### **4. Conclusion** - All coefficients in the Z-row are **non-negative** → the solution is **optimal**. - Basic variables from the final tableau are: \[ \begin{aligned} x_1 &= 4 \\ x_2 &= 1 \\ x_3 &= 9 \\ Z &= 2 \end{aligned} \] #### **Final Answer:** - **Optimal Solution**: \[ x_1 = 4, \quad x_2 = 1, \quad x_3 = 9 \] - **Minimum Value of Original Objective**: \[ \min z = -Z = -2 \] --- **If you have any follow-up questions or want more explanation on any step, feel free to ask!**