Explanation Given: Minimum w = 7 y 1 + 6 y 2 + y 3 Subject to y 1 + y 2 + y 3 ≥ 5 − y 1 + y 2 ≤ − 4 2 y 1 + y 2 + 3 y 3 ≥ 15 and y 1 ≥ 0 , y 2 ≥ 0 , y 3 ≥ 0 Calculation: To write the constraints as equations, add slack variable for less than equal to ( ≤ ) term and subtract a surplus variable from more than equal to ( ≥ ) term. The minimization problem is solved through converting the objective function into maximization problem. For that, multiply both sides of objective function by − 1 Maximize z = − w = − 10 y 1 − 8 y 2 − 15 y 3 And also multiply second constraint by − 1 So, constraint changes to y 1 − y 2 ≥ 4 So, Subject is y 1 + y 2 + y 3 − s 1 = 5 y 1 − y 2 − s 2 = 4 2 y 1 + y 2 + 3 y 3 + s 3 = 15 and y 1 ≥ 0 , y 2 ≥ 0 , y 3 ≥ 0 , s 1 ≥ 0 , and 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. y 1 y 2 y 3 s 1 s 2 s 3 [ 1 1 1 − 1 0 0 5 1 − 1 0 0 − 1 0 4 2 1 3 0 0 − 1 15 7 6 1 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 y 1 = 0 , y 2 = 0 , y 3 = 0 , s 1 = − 5 , s 2 = − 4 and s 3 = − 15 However, because two variables are negative, the solution is not feasible. Now find the basic feasible solution. In basic solution given by tableau, s 1 , s 2 and s 3 have negative values. The only non-zero entry in its column is the − 1 in the row 1. Choose any positive entry in row 2 except the entry in the far right. The column that is chosen will be the pivot column. Here, second column is the pivot column and the pivot row are considered by the smallest quotient. 5 1 = 5 , 4 1 = 4 and 15 2 = 7.5 The smallest quotient is 4 so the pivot is 1 in row 2 of column 1. Then pivoting in the usual way leads to the tableau, after the use of operations R 1 − R 2 , R 3 − R 2 and R 4 − 7 R 2 to convert all elements of that column to zero except pivot element. y 1 y 2 y 3 s 1 s 2 s 3 [ 0 2 1 − 1 1 0 1 1 − 1 0 0 − 1 0 4 0 3 3 0 2 − 1 7 0 13 1 0 7 0 − 28 ] Still this is not the basic feasible solution as s 1 and s 3 is still negative

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 26E

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

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Chapter 7.7, Problem 25E

Chapter 7.7, Problem 27E

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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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