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Solve each linear programming problem by the simplex method. $$ \begin{array}{ll} \text { Maximize } & P=4 x+6 y+5 z \\ \text { subject to } & x+y+z \leq 20 \\ & 2 x+4 y+3 z \leq 42 \\ & 2 x+3 z \leq 30 \\ x & \geq 0, y \geq 0, z \geq 0 \end{array} $$

Short Answer

Expert verified
The optimal solution for the given linear programming problem is \(P = 63\), with \(x = 0\), \(y = 10.5\), and \(z = 0\).

Step by step solution

01

Introduce Slack Variables

To convert inequalities into equalities, we introduce new non-negative variables called slack variables: \(x + y + z + s_1 = 20\) \(2x + 4y + 3z + s_2 = 42\) \(2x + 3z + s_3 = 30\) And add these variables to the objective function: \(P = 4x + 6y + 5z\)
02

Set up the Initial Simplex Tableau

Create a tableau with columns for the objective function and each of the constraints: \[ \begin{array}{c|cccccc|c} & P & x & y & z & s_1 & s_2 & s_3 & \\ \hline P & 1 & -4 & -6 & -5 & 0 & 0 & 0 & 0 \\ s_1 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 20 \\ s_2 & 0 & 2 & 4 & 3 & 0 & 1 & 0 & 42 \\ s_3 & 0 & 2 & 0 & 3 & 0 & 0 & 1 & 30 \\ \hline \end{array} \]
03

Check for Optimality

Determine if the current solution is optimal. If there are any negative values in the bottom row of the tableau (apart from the P column), the solution is not optimal. In this case, our current solution is not optimal as the bottom row contains negative values.
04

Choose the Pivot Column

Select the column that has the most significant negative value in the bottom row. In our tableau, it is the y-column.
05

Choose the Pivot Row

Find the smallest non-negative ratio of the entries in the column b (the column with 20, 42, and 30) to the corresponding entries in the pivot column: \(20/1 = 20\) \(42/4 = 10.5\) There is no coefficient for y in row 3, so its ratio cannot be computed. The minimum ratio is 10.5, which corresponds to the second row, so our pivot row is row 2.
06

Update the Tableau

Perform the pivot operation: Divide the pivot row by the pivot element (row 2 by 4): \(s_2 \rightarrow s_2/4\) \[ \begin{array}{c|cccccc|c} & P & x & y & z & s_1 & s_2 & s_3 & \\ \hline P & 1 & -4 & -6 & -5 & 0 & 0 & 0 & 0 \\ s_1 & 0 & 1/4 & 0 & 1/4 & 1 & -1/4 & 0 & 5 \\ y & 0 & 1/2 & 1 & 3/4 & 0 & 1/4 & 0 & 10.5 \\ s_3 & 0 & 2 & 0 & 3 & 0 & 0 & 1 & 30 \\ \hline \end{array} \] Update the remaining rows with row operations: \(P \rightarrow P + 6y\) \(s_1 \rightarrow s_1 - \frac{1}{4}y\) \(s_3 \rightarrow s_3 - 2y\) After performing these operations, the new tableau looks like this: \[ \begin{array}{c|cccccc|c} & P & x & y & z & s_1 & s_2 & s_3 & \\ \hline P & 1 & 0 & 0 & -1 & 0 & 1.5 & 0 & 63 \\ s_1 & 0 & 0 & 0 & -1/2 & 1 & -3/8 & 0 & -0.5 \\ y & 0 & 1/2 & 1 & 3/4 & 0 & 1/4 & 0 & 10.5 \\ s_3 & 0 & 1 & 0 & 0 & 0 & -1/2 & 1 & 9 \\ \hline \end{array} \]
07

Check for Optimality Again

Now that we have updated the tableau, we check for the optimality of the solution. Since there are no longer any negative values in the bottom row (excluding the P column), we have reached the optimal solution. The optimal solution is \(P = 63\), with: \(x = 0\) \(y = 10.5\) \(z = 0\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Programming
Linear programming (LP) is a mathematical method used for optimizing a linear objective function, subject to linear equality and inequality constraints. Its applications range from business and economics to engineering and the social sciences. In the context of your textbook exercise, you encountered a maximization problem formulated as follows:

Maximize the linear function \(P=4x+6y+5z\), subject to the constraints \(x+y+z \leq 20\), \(2x+4y+3z \leq 42\), and \(2x+3z \leq 30\) with \(x, y, z \geq 0\). The goal is to find the maximum value of \(P\) given these restrictions.

To solve such problems, the simplex method, developed by George Dantzig in 1947, is a systematic, tabular way to calculate the optimal solution to an LP problem. It involves several steps beginning with setting up an initial tableau, introducing slack variables, and performing a series of pivoting operations. During each pivot, the method seeks to improve the objective value until optimality is achieved. The ability to tackle this kind of problem effectively is vital in fields like logistics, finance, and operations management.
Slack Variables
In linear programming, constraints are typically inequalities, and one of the key steps to solving an LP problem using the simplex method is to convert these inequalities into equalities. This transformation is done by introducing slack variables.

Slack variables serve a valuable purpose. They measure the 'leftover' capacity of the resources after some portion has been allocated to existing activities. For each inequality constraint in the LP, a slack variable is added to transform it into an equation. In the exercise, the original constraints:
  • \(x + y + z \leq 20\)
  • \(2x + 4y + 3z \leq 42\)
  • \(2x + 3z \leq 30\)
become:
  • \(x + y + z + s_1 = 20\) (where \(s_1\) is the slack for the first constraint)
  • \(2x + 4y + 3z + s_2 = 42\) (where \(s_2\) is the slack for the second constraint)
  • \(2x + 3z + s_3 = 30\) (where \(s_3\) is the slack for the third constraint)
In the simplex tableau, slack variables start with an initial value equal to the right-hand side of the equation, and this value changes as the simplex algorithm iterates to find the optimal solution, which should maximize the objective function without violating any of the constraints.
Optimality in Simplex Tableau
Optimality in a simplex tableau is a crucial concept to understand as it signifies when the algorithm has reached the best solution possible within the constraints of a linear programming problem. As the simplex method progresses, each iteration updates the tableau and moves the solution closer to optimality.

The check for optimality is relatively straightforward. After each iteration, you look at the bottom row of the simplex tableau, sometimes known as the 'objective function row' or 'cost row'. If you find negative values in this row, excluding the cell for the value of the objective function itself (denoted by P in your exercise), the current solution is not optimal. Why negative values, you ask? It's because they indicate that increasing the corresponding variable could still improve the total value of the objective function.

Through a series of pivoting operations—choosing a pivot column with the most negative coefficient and a pivot row where entry into the basic feasible solution is determined by the smallest non-negative ratio test—the tableau systematically pushes these negative values out. When all negative values are eradicated from the objective function row (except for P), the solution has reached optimality. The final values then present you with the best feasible solution to the LP problem. In the exercise provided, that optimal solution is characterised by \(P=63\), \(x=0\), \(y=10.5\), and \(z=0\).

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Most popular questions from this chapter

Solve each linear programming problem by the simplex method. $$ \begin{array}{ll} \text { Maximize } & P=3 x+3 y+4 z \\ \text { subject to } & x+y+3 z \leq 15 \\ & 4 x+4 y+3 z \leq 65 \\ x & \geq 0, y \geq 0, z \geq 0 \end{array} $$

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Solve each linear programming problem by the simplex method. $$ \begin{aligned} \text { Maximize } & P=2 x+6 y+6 z \\ \text { subject to } & 2 x+y+3 z \leq 10 \\ & 4 x+y+2 z \leq 56 \\ & 6 x+4 y+3 z \leq 126 \\ & 2 x+y+z \leq 32 \\ & x \geq 0, y \geq 0, z \geq 0 \end{aligned} $$

Consider the linear programming problem $$ \begin{aligned} \text { Minimize } & C=-2 x+5 y \\ \text { subject to } & x+y \leq 3 \\ & 2 x+y \leq 4 \\ & 5 x+8 y \geq 40 \\ & x \geq 0, y \geq 0 \end{aligned} $$ a. Sketch the feasible set. b. Find the solution(s) of the linear programming problem, if it exists.

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