Explanation Given: Maximize: z = − 5 x 1 + 4 x 2 − 2 x 3 Subject to: − 2 x 2 + 5 x 3 ≥ 8 4 x 1 − x 2 + 3 x 3 ≤ 12 and x 1 ≥ 0 , x 2 ≥ 0 , x 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. Maximize: z = − 5 x 1 + 4 x 2 − 2 x 3 Subject to: − 2 x 2 + 5 x 3 − s 1 = 8 4 x 1 − x 2 + 3 x 3 + s 2 = 12 x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 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 x 3 s 1 s 2 [ 0 − 2 5 − 1 0 8 4 − 1 3 0 1 12 5 − 4 2 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 = − 8 and s 2 = 12 . However, because one variable is negative, the solution is not feasible. Now find the basic feasible solution. In basic solution given by tableau s 1 has negative value. The only non-zero entry in its column is the − 1 in the row 1. Choose any positive entry in row 1 except the entry in the far right. The column that the chosen entry is in will be the pivot column. Here third column is the pivot column and the pivot row is considered by the smallest quotient. 8 5 = 1.6 and 12 3 = 4 So, pivot element is 5 in row 1 of column 3. Multiply by 1 5 to ease our calculation. x 1 x 2 x 3 s 1 s 2 [ 0 − 2 5 1 − 1 5 0 8 5 4 − 1 3 0 1 12 5 − 4 2 0 0 0 ] Now use operation R 2 − 3 R 1 and R 3 − 2 R 1 to convert all column element to zero except the pivot element

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 21E

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To calculate: The maximum value of equation z=−5x1+4x2−2x3 using two stage simplex method subjected to,

−2x2+5x3≥84x1−x2+3x3≤12

and x1≥0,x2≥0,x3≥0

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Chapter 7.7, Problem 20E

Chapter 7.7, Problem 22E

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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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The maximum value of equation z = − 5 x 1 + 4 x 2 − 2 x 3 using two stage simplex method subjected to, − 2 x 2 + 5 x 3 ≥ 8 4 x 1 − x 2 + 3 x 3 ≤ 12 and x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0

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To calculate: The maximum value of equation z=−5x1+4x2−2x3 using two stage simplex method subjected to, −2x2+5x3≥84x1−x2+3x3≤12 and x1≥0,x2≥0,x3≥0

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Chapter 7 Solutions

To calculate: The maximum value of equation z=−5x1+4x2−2x3 using two stage simplex method subjected to,

−2x2+5x3≥84x1−x2+3x3≤12

and x1≥0,x2≥0,x3≥0