Let's solve the given linear programming problem step by step. --- ### **Problem Statement:** Maximize: \[ Z = -15x_1 + 30x_2 + 25x_3 \] Subject to: \[ \begin{cases} 4x_1 + 2x_2 + 3x_3 \leq 75 \quad \quad \quad (1) \\ 4x_1 + 6x_2 + 2x_3 \leq 60 \quad \quad \quad (2) \\ x_1 + x_2 - x_3 \leq 80 \quad \quad \quad (3) \\ x_1, x_2, x_3 \geq 0 \end{cases} \] --- ### **Step 1: Convert inequalities into equations with slack variables** Introduce slack variables \( s_1, s_2, s_3 \) to convert inequalities into equalities: \[ \begin{cases} 4x_1 + 2x_2 + 3x_3 + s_1 = 75 \\ 4x_1 + 6x_2 + 2x_3 + s_2 = 60 \\ x_1 + x_2 - x_3 + s_3 = 80 \\ x_1, x_2, x_3, s_1, s_2, s_3 \geq 0 \end{cases} \] Objective function remains the same: \[ Z = -15x_1 + 30x_2 + 25x_3 \] --- ### **Step 2: Set up the initial simplex tableau** Since the problem is maximization, and the initial basic feasible solution is at \( x_1 = x_2 = x_3 = 0 \), with slack variables: \[ \begin{array}{c|cccccc|c} & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & \text{RHS} \\ \hline s_1 & 4 & 2 & 3 & 1 & 0 & 0 & 75 \\ s_2 & 4 & 6 & 2 & 0 & 1 & 0 & 60 \\ s_3 & 1 & 1 & -1 & 0 & 0 & 1 & 80 \\ \hline Z & -15 & 30 & 25 & 0 & 0 & 0 & 0 \\ \end{array} \] --- ### **Step 3: Identify entering and leaving variables** **Most positive coefficient in Z-row**: \( 30 \) for \( x_2 \). *Hence, \( x_2 \) will enter the basis.* **Determine leaving variable:** Calculate the minimum ratio of RHS to positive coefficient in \( x_2 \): - Row 1: \( 75 / 2 = 37.5 \) - Row 2: \( 60 / 6 = 10 \) - Row 3: \( 80 / 1 = 80 \) Smallest ratio is **10** in row 2, so **\( s_2 \)** leaves the basis. --- ### **Step 4: Pivot on the element in row 2, column \( x_2 \): 6** **Pivot element**: 6 in row 2, column \( x_2 \). **Normalize row 2:** Divide row 2 by 6: \[ R_2: \quad (4/6, 6/6, 2/6, 0, 1/6, 0, 60/6) = \left(\frac{2}{3}, 1, \frac{1}{3}, 0, \frac{1}{6}, 0, 10\right) \] **Update the tableau:** - Use row operations to make other entries in column \( x_2 \) zero. --- ### **Step 5: Update rows** **Row 1:** \[ R_1 := R_1 - (2/3) \times R_2 \] Calculate: \[ \begin{cases} 4 - (2/3) \times 4 = 4 - (2/3) \times 4 = 4 - \frac{8}{3} = \frac{12}{3} - \frac{8}{3} = \frac{4}{3} \\ 2 - (2/3) \times 6 = 2 - 4 = -2 \\ 3 - (2/3) \times 2 = 3 - \frac{4}{3} = \frac{9}{3} - \frac{4}{3} = \frac{5}{3} \\ 1 - (2/3) \times 0 = 1 \\ 0 - (2/3) \times (1/6) = 0 - \frac{2}{3} \times \frac{1}{6} = 0 - \frac{2}{18} = -\frac{1}{9} \\ 0 - (2/3) \times 0 = 0 \\ 75 - (2/3) \times 10 = 75 - \frac{20}{3} = \frac{225}{3} - \frac{20}{3} = \frac{205}{3} \end{cases} \] Updated row 1: \[ \left(\frac{4}{3}, 0, \frac{5}{3}, 1, -\frac{1}{9}, 0, \frac{205}{3}\right) \] --- **Row 3:** \[ R_3 := R_3 - 1 \times R_2 \] Calculate: \[ \begin{cases} 1 - \frac{2}{3} = \frac{1}{3} \\ 1 - 1 = 0 \\ -1 - \frac{1}{3} = -\frac{4}{3} \\ 0 - 0 = 0 \\ 1 - \frac{1}{6} = \frac{6}{6} - \frac{1}{6} = \frac{5}{6} \\ 1 - 0 = 1 \\ 80 - 10 = 70 \end{cases} \] Updated row 3: \[ \left(\frac{1}{3}, 0, -\frac{4}{3}, 0, \frac{5}{6}, 1, 70 \right) \] **Update Z-row:** \[ R_Z := R_Z + 30 \times R_2 \] Calculate: \[ \begin{cases} -15 + 30 \times \frac{2}{3} = -15 + 20 = 5 \\ 30 + 30 \times 1 = 60 \\ 25 + 30 \times \frac{1}{3} = 25 + 10 = 35 \\ 0 + 30 \times 0 = 0 \\ 0 + 30 \times \frac{1}{6} = 0 + 5 = 5 \\ 0 + 30 \times 0 = 0 \\ 0 + 30 \times 10 = 300 \end{cases} \] Updated Z-row: \[ (5, 60, 35, 0, 5, 0, 300) \] --- ### **Step 6: Current tableau** \[ \begin{array}{c|cccccc|c} & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & \text{RHS} \\ \hline s_1 & \frac{4}{3} & 0 & \frac{5}{3} & 1 & -\frac{1}{9} & 0 & \frac{205}{3} \\ x_2 & \frac{2}{3} & 1 & \frac{1}{3} & 0 & \frac{1}{6} & 0 & 10 \\ s_3 & \frac{1}{3} & 0 & -\frac{4}{3} & 0 & \frac{5}{6} & 1 & 70 \\ \hline Z & 5 & 60 & 35 & 0 & 5 & 0 & 300 \\ \end{array} \] --- ### **Step 7: Check for optimality** - The only positive coefficient in Z-row is 60 for \( x_2 \), but \( x_2 \) is already in the basis. - No other positive coefficients; thus, **current solution is optimal**. --- ### **Step 8: Determine the solution** From the tableau: - \( x_2 = 10 \) (since row 2 corresponds to \( x_2 \)) - \( s_1, s_3 \) are slack variables. Compute \( x_1 \) and \( x_3 \): From row 2: \[ x_2 =

Certainly! Here's a step-by-step guide on how to use Excel Solver to solve the given linear programming problem, along with explanations for each step. --- ### **Problem Recap** **Objective:** Maximize \( Z = -15x_1 + 30x_2 + 25x_3 \) **Constraints:** \[ \begin{cases} 4x_1 + 2x_2 + 3x_3 \leq 75 \quad (1) \\ 4x_1 + 6x_2 + 2x_3 \leq 60 \quad (2) \\ x_1 + x_2 - x_3 \leq 80 \quad (3) \\ x_1, x_2, x_3 \geq 0 \end{cases} \] --- ### **Step-by-step Guide to Use Excel Solver** #### **Step 1: Set up the spreadsheet** 1. **Organize your variables:** | Cell | Description | Example Values (initial) | |---------|----------------------------------------------|---------------------------| | B1 | \( x_1 \) | 0 | | C1 | \( x_2 \) | 0 | | D1 | \( x_3 \) | 0 | 2. **Input the coefficients for the constraints:** | Cell | Constraint 1 (Left Side) | Cell | Constraint 2 (Left Side) | Cell | Constraint 3 (Left Side) | |---------|------------------------------------------------|---------|------------------------------------------------|---------|------------------------------------------------| | B2 | `=4*B1 + 2*C1 + 3*D1` | C2 | `=4*B1 + 6*C1 + 2*D1` | D2 | `=B1 + C1 - D1` | 3. **Input the RHS of constraints:** | Cell | RHS for Constraint 1 | Cell | RHS for Constraint 2 | Cell | RHS for Constraint 3 | |---------|------------------------|---------|------------------------|---------|------------------------| | B3 | 75 | C3 | 60 | D3 | 80 | 4. **Set up the formulas in the cells:** - For constraints, in **B2, C2, D2**: formulas as above. - For the objective function \( Z \): | Cell | Objective Function \( Z \) | Formula | |---------|----------------------------|--------------------------------------------------| | B4 | `= -15*B1 + 30*C1 + 25*D1` | (this cell will calculate the current Z) | --- #### **Step 2: Open Solver** - Go to **Data** tab → click **Solver**. - If Solver is not visible, you may need to add it via **Excel Options** → **Add-ins** → **Solver Add-in**. --- #### **Step 3: Set the Solver parameters** - **Set Objective:** select cell with \( Z \) (e.g., B4). - **To:** select **Max**. - **By Changing Variable Cells:** select cells for \( x_1, x_2, x_3 \) (B1, C1, D1). --- #### **Step 4: Add constraints** - Click **Add** to input constraints: | Constraint | Cell Reference | Relation | Constraint Value | |--------------|------------------|----------|------------------| | Constraint 1 | B2 | <= | 75 | | Constraint 2 | C2 | <= | 60 | | Constraint 3 | D2 | <= | 80 | | Non-negativity | B1, C1, D1 | >= | 0 | - Repeat for each constraint, ensuring all are added. --- #### **Step 5: Choose solving method** - In the Solver window, at the bottom, select **Simplex LP** (for linear programming). --- #### **Step 6: Solve the problem** - Click **Solve**. - When Solver finds a solution, it will display the optimal values for \( x_1, x_2, x_3 \). --- ### **Step 7: Interpret the results** - The cells B1, C1, D1 will contain the optimal values of \( x_1, x_2, x_3 \). - Cell B4 will show the maximum value of \( Z \). --- ### **Summary** Using Excel Solver simplifies solving LP problems by automating the optimization process: - Set up decision variables and constraints in the spreadsheet. - Define the objective function. - Use