Explanation Given: Maximum, z = 5 x 1 + 10 x 2 Subject to: x 1 + 2 x 2 ≤ 18 x 1 + 3 x 2 ≥ 12 2 x 1 + 2 x 2 ≤ 30 and x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0 . Calculation: To write the constraints as equations, add a slack variable for less than equal to ( ≤ ) term and subtract a surplus variable from more than equal to ( ≥ ) term. Maximize z = 5 x 1 + 10 x 2 Subject to: x 1 + 2 x 2 + s 1 = 18 x 1 + 3 x 2 − s 2 = 12 2 x 1 + 2 x 2 + s 3 = 30 and x 1 ≥ 0 , x 2 ≥ 0 , s 1 ≥ 0 , s 2 ≥ 0 . And use the coefficient of above equations to write the initial simplex tableau while omitting the z column. Now, the equations and the objective function can be written as following augmented matrix, and this the initial simplex table. x 1 x 2 s 1 s 2 s 3 [ 1 2 1 0 0 18 1 3 0 − 1 0 12 2 2 0 0 1 30 − 5 − 10 0 0 0 0 ] The variable whose columns have one entry that is ± 1 and the rest are 0 termed as basic variables and others as non-basic. So, from above table, the basic solution is x 1 = 0 , x 2 = 0 , x 3 = 0 , s 1 = 18 , s 2 = − 12 and s 3 = 30 However, because one variable is negative, the solution is not feasible. Now, find the basic feasible solution. In basic solution given by tableau, s 2 has negative value. The only non-zero entry in its column is the − 1 in row 2. Choose any positive entry in row 2 except the entry in the far right. The column that the chosen entry is in will be the pivot column. Here, the first column is the pivot column and the pivot row is considered by the smallest quotient. 18 1 = 18 , 12 1 = 12 and 30 2 = 15 The smallest quotient is 12 so the pivot is 1 in row 2 of column 1. Use operation R 1 − R 2 , R 3 − 2 R 1 and R 4 + 5 R 1 to convert all elements of that column to zero except pivot element. x 1 x 2 s 1 s 2 s 3 [ 0 − 1 1 1 0 6 1 3 0 − 1 0 12 0 − 4 0 2 1 6 0 5 0 − 5 0 60 ] Now, this is thebasic feasible solution

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 7.7, Problem 13E

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To calculate: The minimum value of a set of equation using the two-stage simplex method.

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Chapter 7.7, Problem 12E

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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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