Objective Function: Maximize Z

To solve this linear programming problem using the simplex method, we first convert the inequalities into equations by introducing slack variables. Then, we construct the initial simplex tableau and proceed with the simplex iterations until we reach the optimal solution.

Problem Formulation

Objective Function: $Z = x_1 + 2x_2 + x_3$

Subject to the constraints: $2x_1 + x_2 - x_3 + s_1 = 2$ $-2x_1 + x_2 - 5x_3 + s_2 = -6$ $4x_1 + x_2 + x_3 + s_3 = 6$ $x_1, x_2, x_3, s_1, s_2, s_3 \geq 0$

Here, $s_1$, $s_2$, and $s_3$ are the slack variables added to convert the inequalities into equations.

Initial Simplex Tableau

The initial simplex tableau is constructed as follows:

\text{Basis} & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & \text{RHS} \\ \hline s_1 & 2 & 1 & -1 & 1 & 0 & 0 & 2 \\ s_2 & -2 & 1 & -5 & 0 & 1 & 0 & -6 \\ s_3 & 4 & 1 & 1 & 0 & 0 & 1 & 6 \\ \hline Z & -1 & -2 & -1 & 0 & 0 & 0 & 0 \\ \end{array}$$ ### Steps of the Simplex Method 1. **Identify the entering variable**: The most negative coefficient in the objective function row indicates the entering variable. Here, $$-2$$ (corresponding to $$x_2$$) is the most negative. 2. **Identify the leaving variable**: Calculate the ratio of the RHS to the coefficients of the entering variable in the constraint rows. The smallest positive ratio indicates the leaving variable. - Row 1: $$\frac{2}{1} = 2$$ - Row 2: Not applicable since the ratio is negative. - Row 3: $$\frac{6}{1} = 6$$ Thus, the leaving variable is $$s_1$$. 3. **Pivot**: Perform row operations to make the coefficient of the entering variable 1 in the pivot row and 0 in all other rows. ### Pivot Operations After performing the necessary row operations (pivoting): $$\begin{array}{c|cccccc|c} \text{Basis} & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & \text{RHS} \\ \hline x_2 & 2 & 1 & -1 & 1 & 0 & 0 & 2 \\ s_2 & -4 & 0 & -3 & -1 & 1 & 0 & -8 \\ s_3 & 2 & 0 & 2 & -1 & 0 & 1 & 4 \\ \hline Z & -3 & 0 & -3 & 2 & 0 & 0 & 4 \\ \end{array}$$ Continue the simplex iterations until there are no negative coefficients in the objective function row. The final tableau will provide the optimal solution. ### Final Tableau Continuing the iterations, we get: $$\begin{array}{c|cccccc|c} \text{Basis} & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & \text{RHS} \\ \hline x_1 & 1 & 0.5 & -0.5 & 0.5 & 0 & 0 & 1 \\ x_2 & 0 & 0.5 & -0.5 & 0.5 & 0 & 0 & 1 \\ s_3 & 0 & -1 & 1 & -2 & 0 & 1 & 2 \\ \hline Z & 0 & 0.5 & -1.5 & 0.5 & 0 & 0 & 5 \\ \end{array}$$ ### Solution The optimal solution is: $$x_1 = 1,$$ $$x_2 = 1,$$ $$x_3 = 0.$$ The maximum value of $$Z$$ is $$5$$. Would you like more details or have any questions? Here are some related questions for further practice: 1. How do you determine the entering and leaving variables in the simplex method? 2. What are slack variables and why are they used in linear programming? 3. How do you interpret the final tableau in the simplex method? 4. Can the simplex method be used for minimization problems? If so, how? 5. What are the assumptions of linear programming? 6. What is the dual problem in linear programming? 7. How does sensitivity analysis relate to the simplex method? 8. What are some limitations of the simplex method? **Tip**: Always ensure that your initial simplex tableau is correctly set up, as errors in the initial setup can lead to incorrect solutions.