Set up the initial simplex tableau that could be used to solve the following problem by the simplex method.
Newspaper ads cost $500 each, and no more than 40 can run per month. Internet banner ads cost $20 each, and no more than 600 can run per month. TV ads cost $1000 each, with a maximum of 15 available each month. Approximately 5000 women will see each newspaper ad, 3000 will see each Internet banner, and 12,000 will see each TV ad. How much of each type of advertising should be used if the store wants to maximize its ad exposure?
Let [tex]$x_1$[/tex] = the number of newspaper ads. Let [tex]$x_2$[/tex] = the number of Internet banner ads. Let [tex]$x_3$[/tex] = the number of TV ads. Complete the initial simplex tableau below.
[tex]$\left[\begin{array}{rrrrrrrr} x _{ 1 } & x _{ 2 } & x _{ 3 } & s _{ 1 } & s _{ 2 } & s _{ 3 } & s _{ 4 } & z \\ 500 & \square & \square & 1 & \square & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 & 0 \\ \hline & \square & \square & 0 & 0 & 0 & 1\end{array}\right]$[/tex]
Answer (1)
Define variables and objective function.
Introduce slack variables to convert inequalities to equalities.
Set up the initial simplex tableau with coefficients from the constraints and objective function.
The completed initial simplex tableau is: x 1 1 0 − 5000 x 2 0 0 − 3000 x 3 0 1 − 12000 s 1 1 0 0 s 3 0 1 0 z 0 0 1 R H S 40 15 0 .
Explanation
Problem Analysis and Setup We are given a linear programming problem where the objective is to maximize ad exposure to women, subject to constraints on the number of newspaper and TV ads. We need to set up the initial simplex tableau to solve this problem. The problem provides the following information:
Let x 1 = the number of newspaper ads.
Let x 2 = the number of Internet banner ads.
Let x 3 = the number of TV ads.
The objective is to maximize z = 5000 x 1 + 3000 x 2 + 12000 x 3 .
The constraints are:
x 1 ≤ 40 (Newspaper ad limit)
x 3 ≤ 15 (TV ad limit)
x 1 , x 2 , x 3 ≥ 0 (Non-negativity)
We need to convert these inequalities into equalities by introducing slack variables.
Converting Inequalities to Equalities First, we introduce slack variables s 1 and s 3 to convert the inequalities into equalities:
x 1 + s 1 = 40 x 3 + s 3 = 15
Also, we rewrite the objective function as:
z − 5000 x 1 − 3000 x 2 − 12000 x 3 = 0
Setting up the Simplex Tableau Now, we set up the initial simplex tableau. The tableau will have the following structure:
[ x 1 x 2 x 3 s 1 s 3 z R H S ]
From the constraint x 1 ≤ 40 , we have x 1 + s 1 = 40 . So, the corresponding row in the tableau is:
[ 1 0 0 1 0 0 40 ]
From the constraint x 3 ≤ 15 , we have x 3 + s 3 = 15 . So, the corresponding row in the tableau is:
[ 0 0 1 0 1 0 15 ]
The objective function is z − 5000 x 1 − 3000 x 2 − 12000 x 3 = 0 . So, the corresponding row in the tableau is:
[ − 5000 − 3000 − 12000 0 0 1 0 ]
Completing the Initial Simplex Tableau Now, we fill in the missing values in the provided tableau:
x 1 5000 1 0 0 − 5000 x 2 3000 0 1 0 − 3000 x 3 12000 0 0 1 − 12000 s 1 1 1 0 0 0 s 3 0 0 0 1 0 z 0 0 0 0 1 R H S 0 40 0 15 0
However, there's an error in the provided tableau. The first row should represent the objective function coefficients after moving them to the left side of the equation. Also, there's an extra row for x 2 which is not constrained. The correct tableau should be:
x 1 1 0 − 5000 x 2 0 0 − 3000 x 3 0 1 − 12000 s 1 1 0 0 s 3 0 1 0 z 0 0 1 R H S 40 15 0
So, the completed initial simplex tableau is:
x 1 1 0 − 5000 x 2 0 0 − 3000 x 3 0 1 − 12000 s 1 1 0 0 s 3 0 1 0 z 0 0 1 R H S 40 15 0
Final Answer Based on the analysis, the completed initial simplex tableau is:
x 1 1 0 − 5000 x 2 0 0 − 3000 x 3 0 1 − 12000 s 1 1 0 0 s 3 0 1 0 z 0 0 1 R H S 40 15 0
Therefore, the missing values are:
The entry in the first row, second column is 0.
The entry in the first row, third column is 0.
The entry in the first row, fifth column is 0.
The entry in the last row, first column is -5000.
The entry in the last row, second column is -3000.
The entry in the last row, third column is -12000.
Examples
Linear programming is a powerful tool used in various real-world scenarios. For instance, a delivery company can use linear programming to optimize delivery routes, minimizing fuel costs and delivery time. Similarly, a manufacturing company can use it to determine the optimal production levels of different products, maximizing profit while considering resource constraints. In finance, it can be used to create an investment portfolio that maximizes returns while minimizing risk. These examples demonstrate the versatility and practical applications of linear programming in optimizing complex decision-making processes.
