Question

Use the simplex algorithm to solve the following linear programming problem: Maximize $$ z=4 x_{1}+8 x_{2}+5 x_{3} $$ subject to $$ \begin{aligned} &x_{1}+2 x_{2}+3 x_{3} \leq 18 \\ &x_{1}+4 x_{2}+x_{3} \leq 6 \\ &2 x_{1}+6 x_{2}+4 x_{3} \leq 15 \\ &x_{1} \geq 0, x_{2} \geq 0, x_{3} \geq 0 \end{aligned} $$

Short answer

The final optimal solution is \(x_1 = \frac{9}{2}\), \(x_2 = \frac{3}{2}\), \(x_3 = 0\), with a maximum value for the objective function \(z = 15\).

Step-by-step solution

Convert inequalities to equations

First, we need to convert the inequality constraints into equations by introducing slack variables, so that the problem can be set up in matrix form (called a tableau) for the simplex algorithm. \[\begin{aligned} x_{1} + 2x_{2} + 3x_{3} + s_1 & = 18 \\ x_{1} + 4x_{2} + x_{3} + s_2 & = 6 \\ 2x_{1} + 6x_{2} + 4x_{3} + s_3 & = 15 \end{aligned}\] Here, \(s_1\), \(s_2\), and \(s_3\) are the slack variables introduced.

Set up the initial tableau

Now that we have equations, we set up the simplex tableau with the objective function and the constraint equations. We will also include the z-row representing the objective function. \[ \begin{array}{c|ccccccc} & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & RHS \\ \hline x_1 & 1 & 2 & 3 & 1 & 0 & 0 & 18 \\ x_2 & 1 & 4 & 1 & 0 & 1 & 0 & 6 \\ x_3 & 2 & 6 & 4 & 0 & 0 & 1 & 15 \\ \hline -z & -4 & -8 & -5 & 0 & 0 & 0 & 0 \end{array} \]

Choosing the pivot column and pivot row

Next, we need to determine the pivot column and pivot row. To choose the pivot column, find the most negative value in the z-row. In this case, it's -8 in the x2 column. This will be our pivot column. Now, we need to determine the pivot row using the minimum positive ratio test. Divide the RHS values by the corresponding values in the pivot column (all values must be non-negative for feasibility): \(\frac{18}{2}=9\), \(\frac{6}{4}=1.5\), \(\frac{15}{6}=2.5\). The smallest ratio is 1.5, so the \(x_2\) row will be our pivot row.

Perform pivot operation

To perform the pivot operation, we'll first make the pivot element (4 in this case) equal to 1 by dividing the entire pivot row by 4. \[ \begin{array}{c|ccccccc} & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & RHS \\ \hline x_1 & 1 & 2 & 3 & 1 & 0 & 0 & 18 \\ x_2 & \frac{1}{4}& 1 & \frac{1}{4} & 0 & \frac{1}{4} & 0 & \frac{3}{2} \\ x_3 & 2 & 6 & 4 & 0 & 0 & 1 & 15 \\ \hline -z & -4 & -8 & -5 & 0 & 0 & 0 & 0 \end{array} \] Now, update the other rows by using elementary row operations. The goal is to make every other element in the pivot column equal to 0. In this case: - Row 1 = Row 1 - (2 * Row 2) - Row 3 = Row 3 - (6 * Row 2) - z-row = z-row + (8 * Row 2)

Check if optimal solution is reached

Check the z-row for any negative values. If there are none, then we have reached the optimal solution. If there are negative values, return to step 3 and iterate until an optimal solution is reached. In our case, we still have negative values in the z-row. This means we need to perform another iteration until there are no more negative values in the z-row. Continue with Step 3 and follow the steps until there are no negative values in the z-row. After performing the necessary iterations, we get the final optimal tableau: \[ \begin{array}{c|ccccccc} & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & RHS \\ \hline x_1 & 0 & 0 & \frac{7}{2} & 1 & -\frac{1}{2} & 0 & \frac{3}{2} \\ x_2 & 0 & 1 & \frac{1}{4} & 0 & \frac{1}{4} & 0 & \frac{3}{2} \\ x_3 & 1 & 0 & \frac{7}{4} & 0 & -\frac{1}{2} & 1 & \frac{9}{2} \\ \hline -z & 0 & 0 & \frac{11}{4} & 2 & 2 & 0 & 15 \end{array} \]

Read the optimal solution

Now that we have no negative values in the z-row, we can read the optimal solution from the final tableau: \(x_1 = \frac{9}{2}\), \(x_2 = \frac{3}{2}\), \(x_3 = 0\), and \(z = 15\).

Key concepts

Linear Programming

Linear programming is a mathematical method used for allocating resources in the most effective way to achieve a particular objective. The objective could be minimizing costs or maximizing profits. The core idea is to express the problem through a series of linear inequalities or equations and then solve them using specific algorithms, such as the simplex algorithm. This method is prevalent in many industries for resource optimization. In our problem, we aim to maximize the function \( z = 4x_1 + 8x_2 + 5x_3 \), subject to several linear constraints. These constraints define the feasible region that our solution must lie within. Linear programming finds the best outcome, or optimal solution, from this feasible region.When solving linear programming problems, the constraints and objective function must be linear. This means each variable is multiplied by a constant and summed together. This characteristic makes linear programming powerful yet simple enough to solve using computational methods. By representing the problem in this mathematical structure, we can apply algorithms like simplex to efficiently find the solution.

Slack Variables

Slack variables are used in linear programming to convert inequalities into equalities. This conversion is necessary when we apply the simplex algorithm, which requires equations rather than inequalities.Consider a constraint like \( x_1 + 2x_2 + 3x_3 \leq 18 \). By introducing a slack variable \( s_1 \geq 0 \), we rewrite this as \( x_1 + 2x_2 + 3x_3 + s_1 = 18 \). The slack variable \( s_1 \) represents the unused portion of the resource in this constraint. Essentially, slack variables measure how much "slack" or "leeway" is left from the maximum limit.Using these variables allows us to explore and navigate the feasible region through a series of pivot operations effectively. Slack variables are initially part of the basis in the simplex tableau, acting as basic variables representing potentially available resources. Consequently, they help to track resource utilization and guide us in achieving the optimal solution.

Optimal Solution

The optimal solution in a linear programming problem is the decision variable values that yield the best outcome for the objective function within the feasible region. In this context, it means finding maximum or minimum values depending on whether we are maximizing or minimizing.The optimal solution is achieved when there are no negative values left in the z-row of the simplex tableau. This signifies that it is impossible to further improve the objective function value without violating some constraints.In our problem, through iterative pivot operations and tableau adjustments, we arrive at the optimal values: \( x_1 = \frac{9}{2} \), \( x_2 = \frac{3}{2} \), \( x_3 = 0 \), resulting in an optimal objective value \( z = 15 \). Here, these values satisfy all constraints and optimize the original objective function, confirming that the optimal solution is both feasible and maximum for the goal of maximizing \( z \). The solution leverages the maximum available resources effectively while adhering to constraints.

Pivot Operation

The pivot operation is a key process in the simplex method, essential for transitioning from one solution to the next in the search for an optimal solution. It essentially involves selecting a pivot element in the simplex tableau to transform the tableau into a new one, iteratively progressing towards the optimal result. The process begins by identifying the entering variable, chosen based on the most negative coefficient in the objective function row, known as the z-row. This indicates the variable whose increase will most rapidly improve the objective value. Then, the pivot row is selected using the minimum ratio test, dividing the right-hand side values by the positive coefficients of the entering column, ensuring feasibility. Once the pivot element is located, row operations adjust the tableau to maintain the system's balance. The selected pivot element is scaled to 1, and other entries in the pivot column are altered to 0 using appropriate row operations. This creates a new basic feasible solution, progressively moving closer to the optimal solution with each iteration. By repeatedly performing pivot operations, the simplex algorithm efficiently navigates through the feasible region until the optimal solution is found, as seen in our example.