Explanation Given: Minimum w = 2 y 1 + 5 y 2 − 3 y 3 Subject to y 1 + 2 y 2 + 3 y 3 ≥ 115 2 y 1 + y 2 + y 3 ≤ 200 y 1 + y 3 ≥ 50 and y 1 ≥ 0 , y 2 ≥ 0 , y 3 ≥ 0 Calculation: To write the constraints as equations, add a slack variable for less than equal to ( ≤ ) term and subtract a surplus variable for 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 = − 2 y 1 − 5 y 2 + 3 y 3 subject to y 1 + 2 y 2 + 3 y 3 − s 1 = 115 2 y 1 + y 2 + y 3 + s 2 = 200 y 1 + y 3 − s 3 = 50 and y 1 ≥ 0 , y 2 ≥ 0 , y 3 ≥ 0 , s 1 ≥ 0 , and s 2 ≥ 0 . Use the coefficients of above equations to write the initial simplex tableau while omitting the z column. Now, the equations and the objective functions can be written as the following augmented matrix, and this is the initial simplex table. y 1 y 2 y 3 s 1 s 2 s 3 [ 1 2 3 − 1 0 0 115 2 1 1 0 1 0 200 1 0 1 0 0 − 1 50 2 5 − 3 0 0 0 0 ] The variables whose columns have one entry, that is, ± 1 , and the rest are 0 are termed as basic variables and others as non basic. So, from the above table, the basic solution is y 1 = 0 , y 2 = 0 , y 3 = 0 , s 1 = − 12 , and s 2 = − 48 . 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 and s 3 have negative values. The only nonzero entry in its column is − 1 in row 3. Choose any positive entry in row 3, except the entry in the far right. The column that is chosen will be the pivot column. Here, the first column is the pivot column and the pivot row is considered by the smallest quotient. 115 1 = 115 , 200 2 = 100 , and 50 1 = 50 . We are free to choose any element as a pivot element as there is tie between smallest quotient. Then pivoting in the usual way leads to the following tableau, after the use of operations R 1 − R 3 , R 2 − 2 R 3 , and R 4 − 2 R 3 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 2 − 1 0 1 65 0 1 − 1 0 1 2 100 1 0 1 0 0 − 1 50 0 5 − 5 0 0 2 − 100 ] Still this is not the basic feasible solution as s 1 is negative. The only nonzero entry in its column is − 1 in row 1. Choose any positive entry in row 1, except the entry in the far right. The column with the chosen entry will be the pivot column. Here, the first column is the pivot column and the pivot row is considered by the smallest quotient. 65 2 = 32.5 and 100 1 = 100 This time 2 is the pivot element. Then, multiply with 1 2 in row 1 to ease the calculation. y 1 y 2 y 3 s 1 s 2 s 3 [ 0 1 1 − 1 2 0 1 2 65 2 0 1 − 1 0 1 2 100 1 0 1 0 0 − 1 50 0 5 − 5 0 0 2 − 100 ] Then, use operations R 2 − R 1 and R 4 − 5 R 1 . y 1 y 2 y 3 s 1 s 2 s 3 [ 0 1 1 − 1 2 0 1 2 65 2 0 0 − 2 1 2 1 3 2 135 2 1 0 1 0 0 − 1 50 0 0 − 10 5 2 0 − 1 2 − 525 2 ] This is the basic 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 25E

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To calculate: The equation using two-stage simplex method by using the information given below:

Minimum

w=2y1+5y2−3y3

Subject to

y1+2y2+3y3≥1152y1+y2+y3≤200y1+y3≥50

and y1≥0, y2≥0, y3≥0

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Chapter 7.7, Problem 24E

Chapter 7.7, Problem 26E

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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 equation using two-stage simplex method by using the information given below: Minimum w = 2 y 1 + 5 y 2 − 3 y 3 Subject to y 1 + 2 y 2 + 3 y 3 ≥ 115 2 y 1 + y 2 + y 3 ≤ 200 y 1 + y 3 ≥ 50 and y 1 ≥ 0 , y 2 ≥ 0 , y 3 ≥ 0

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To calculate: The equation using two-stage simplex method by using the information given below: Minimum w=2y1+5y2−3y3 Subject to y1+2y2+3y3≥1152y1+y2+y3≤200y1+y3≥50 and y1≥0, y2≥0, y3≥0

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Chapter 7 Solutions

To calculate: The equation using two-stage simplex method by using the information given below:

Minimum

w=2y1+5y2−3y3

Subject to

y1+2y2+3y3≥1152y1+y2+y3≤200y1+y3≥50

and y1≥0, y2≥0, y3≥0