Explanation Given: Consider the linear programming problem as follows. Minimize w = 15 y 1 + 12 y 2 subject to 3 y 1 + 5 y 2 ≥ 20 3 y 1 + y 2 ≥ 18 with y 1 ≥ 0 , y 2 ≥ 0 Calculation: Obtain the augmented matrix for the given linear programming problem. [ 3 5 20 3 1 18 15 12 0 ] Observe that the last row is the coefficients of the objective function w = 15 y 1 + 12 y 2 and the upper two rows are the coefficients of the two linear constraints 3 y 1 + 5 y 2 ≥ 20 and 3 y 1 + y 2 ≥ 18 with their constants. Obtain the transpose of the augmented matrix obtained above as follows. [ 3 3 15 5 1 12 20 18 0 ] Observe that the last row becomes the coefficients of the objective function that is z = 20 x 1 + 18 x 2 and the upper two rows becomes the coefficients as well as constant of the two linear constraints that are 3 x 1 + 3 x 2 ≤ 15 and 5 x 1 + x 2 ≤ 12 . Hence, the dual of the linear programming problem is obtained as follows. Maximize z = 20 x 1 + 18 x 2 subject to 3 x 1 + 3 x 2 ≤ 15 5 x 1 + x 2 ≤ 12 with x 1 ≥ 0 , x 2 ≥ 0 Add slack variable to the above linear programming problem. Maximize z = 20 x 1 + 18 x 2 subject to 3 x 1 + 3 x 2 + s 1 = 15 5 x 1 + x 2 + s 2 = 12 with x 1 ≥ 0 , x 2 ≥ 0 , s 1 ≥ 0 , s 2 ≥ 0 Rewrite the objective function as, z − 20 x 1 − 18 x 2 = 0 . Now, obtain the simplex tableau, x 1 x 2 s 1 s 2 z [ 3 3 1 0 0 15 5 1 0 1 0 12 − 20 − 18 0 0 1 0 ] Here, the basic variables are s 1 and s 2 . Thus, the basic solution is x 1 = 0 , x 2 = 0 , s 1 = 15 , s 2 = 12 and z = 0 . This is not the optimal solution as only the slack variables are nonzero. Look in the bottom row, the coefficient of x 1 is most negative, so choose first column as a pivot column and bring x 1 into the solution. Divide the entries of the last column by the respective entries in the pivot column as follows. x 1 x 2 s 1 s 2 z [ 3 3 1 0 0 15 5 1 0 1 0 12 − 20 − 18 0 0 1 0 ] → 15 / 3 → 12 / 5 Observe that 12 5 is the least. Hence, the pivot element is chosen as the second element of the pivot column in the above tableau. x 1 x 2 s 1 s 2 z [ 3 3 1 0 0 15 5 1 0 1 0 12 − 20 − 18 0 0 1 0 ] Simplify the simplex tableau with the help of pivot method until the last row is completely filled with non-negative entries. Apply the operation 5 R 1 − 3 R 2 → R 1 in the above tableau as follows. x 1 x 2 s 1 s 2 z [ 0 12 5 − 3 0 39 5 1 0 1 0 12 − 20 − 18 0 0 1 0 ] Apply the operation R 3 + 4 R 2 → R 3 in the above tableau as follows

10th Edition

ISBN: 8220102020252

Author: RITCHEY

Publisher: PEARSON

See similar textbooks

Concept explainers

Statements and Logic

Logic is the analysis of general patterns of thoughts without regard to any specific sense of context.

Videos

Question

Chapter 4.3, Problem 2YT

To determine

To solve: The minimization problem.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 4.3, Problem 1YT

Chapter 4.3, Problem 1WE

Students have asked these similar questions

Find the area of this sector please show work

11 fr² Find√² du In(x+u)

Find area of sector for number 14

Chapter 4 Solutions

EBK FINITE MATHEMATICS AND CALCULUS WIT

Ch. 4.1 - Prob. 1YT Ch. 4.1 - Prob. 2YT Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Convert each inequality into an equation by adding...Ch. 4.1 - For Exercises 58, (a) determine the number of...Ch. 4.1 - Prob. 6E Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

The minimization problem.

Solution Summary: The author explains how to solve the minimization problem by obtaining the augmented matrix for the linear programming problem.

Concept explainers

Videos

To solve: The minimization problem.

Want to see the full answer?

Chapter 4 Solutions