To determine To calculate: Minimization of the problem with the help of dual method. Explanation Given: Minimize, w = 4 y 1 + 5 y 2 Subject to, 10 y 1 + 5 y 2 ≥ 100 , 20 y 1 + 10 y 2 ≥ 150 , y 1 ≥ 0 , y 2 ≥ 0 Formula used: Calculation: Consider the minimization problem which is: Minimize, w = 4 y 1 + 5 y 2 Subject to, 10 y 1 + 5 y 2 ≥ 100 , 20 y 1 + 10 y 2 ≥ 150 , y 1 ≥ 0 , y 2 ≥ 0 In the above, there are two variables which are y 1 and y 2 and the two constraints are: 10 y 1 + 5 y 2 ≥ 100 , 20 y 1 + 10 y 2 ≥ 150 The objective function is: w = 4 y 1 + 5 y 2 By the dual method, the minimization problem converted into maximization problem and inequalities are reversed. All the changes are listed in the tabular form: Given problem Dual problem m variables n variables n constraints m constraints Coefficients from objective function Constraint constants Constraint constants Coefficients from objective function Greater than inequalities Lesser than inequalities Minimization Maximization First, write into the augmented matrix of the system of inequalities and the coefficient of the objective function is written in the last row. Then, write into the augmented matrix. [ 10 5 100 20 10 150 4 5 0 ] Here, the last row shows the coefficients of the objective function and last columns shows the constants. Take transpose of the above-augmented matrix to convert dual minimization to dual maximization problem. Then, [ 10 20 4 5 10 5 100 150 0 ] After converting, the objective function changes to: z = 100 x 1 + 150 x 2 And, the obtained constants are written in the last columns which are 4 and 5 . Obtained constraints are: 10 x 1 + 20 x 2 ≤ 4 , 5 x 1 + 10 x 2 ≤ 5 , This gives maximization problem which is: Maximize, z = 100 x 1 + 150 x 2 Subject to, 10 x 1 + 20 x 2 ≤ 4 , 5 x 1 + 10 x 2 ≤ 5 , x 1 ≥ 0 , x 2 ≥ 0 Then, solve the above maximization problem by simplex method, as the requirements of simplex method are fulfilled which is: (1) All variables are non-negative, which is x i ≥ 0 . (2) Objective function has to be maximized. (3) All constraints have ≤ sign. (4) The constants on the right-hand side in the constraints are all non-negative, which is denoted by b . Further, to convert linear inequality into the linear equation, take a slack variable which is done by adding a non-negative variable to each constraint which is: 10 x 1 + 20 x 2 + s 1 = 4 , 5 x 1 + 10 x 2 + s 2 = 5 where s 1 and s 2 is greater than or equal to zero as the above constraints are less than or equal to zero. Then, rewrite constraints by adding the non-negative slack variables which are: Maximize, z = 100 x 1 + 150 x 2 Subject to, 10 x 1 + 20 x 2 + s 1 = 4 , 5 x 1 + 10 x 2 + s 2 = 5 , x 1 ≥ 0 , x 2 ≥ 0 , s 1 ≥ 0 , s 2 ≥ 0 Then, convert the linear programming into a system of linear equations in which all the coefficient of variables is written on the left of the equal sign and the constants are on the right. Then convert the objective function into linear equations which are: − 100 x 1 − 150 x 2 + z = 0 Then, write the constraint and objective function into the augmented matrix which is in the form of a simplex table. x 1 x 2 s 1 s 2 [ 10 20 1 0 4 5 10 0 1 5 − 100 − 150 0 0 0 ] Here, the first row denotes the first constraints and the second row denotes the second constraints and the last row denotes the objective function. In the last row, all the entries except the last and second last entries are called indicators. The simplex table represents a system of three linear equations in five variables as there are more variable than equations then the system is dependent and it has infinitely many solutions. In the simplex, final solution will get when all the variables are non-negative, it will be done by row operation. x 1 x 2 s 1 s 2 [ 10 20 1 0 4 5 10 0 1 5 − 100 − 150 0 0 0 ] In the simplex, choose the pivot entry as the pivot is the most negative indicator which is − 150 . Then, the column containing the most negative indicator is called pivot column, which is x 2 Now all the positive entry of pivot column is divided by the last column which is constant which is: First, 4 ÷ 20 = 1 5 Second, 5 ÷ 10 = 1 2 After the row with the smallest quotient is called the pivot row, which is 1 5 . The entry in the pivot row and pivot column is called pivot. x 1 x 2 s 1 s 2 [ 10 20 1 0 4 5 10 0 1 5 − 100 − 150 0 0 0 ] In this, the pivot is 20 at the pivot is selected then do the row operations in the initial simplex table to get the new simplex table by eliminating pivot column all value to zero except the pivot which is change by row transform into 1

Mathematics with Applications In the Management, Natural, and Social Sciences (12th Edition)

12th Edition

ISBN: 9780134767628

Author: Margaret L. Lial, Thomas W. Hungerford, John P. Holcomb, Bernadette Mullins

Publisher: PEARSON

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Chapter 7, Problem 46RE

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Chapter 7, Problem 45RE

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Chapter 7 Solutions

Mathematics with Applications In the Management, Natural, and Social Sciences (12th Edition)

Ch. 7.1 - Checkpoint 1 Graph the given...Ch. 7.1 - Prob. 2CP Ch. 7.1 - Prob. 3CP Ch. 7.1 - Prob. 4CP Ch. 7.1 - Prob. 5CP Ch. 7.1 - Checkpoint 6 Graph the feasible region of the...Ch. 7.1 - Prob. 7CP Ch. 7.1 - Match the inequality with its graph, which is one...Ch. 7.1 - Match the inequality with its graph, which is one...Ch. 7.1 - Match the inequality with its graph, which is one...

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