Explanation Given: Minimum w = 32 y 1 + 40 y 2 Subject to 20 y 1 + 10 y 2 = 200 25 y 1 + 40 y 2 ≤ 500 18 y 1 + 24 y 2 ≥ 300 and y 1 ≥ 0 , y 2 ≥ 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 = − 32 y 1 − 40 y 2 Constraint 20 y 1 + 10 y 2 = 200 can be written as following pair of inequalities, 20 y 1 + 10 y 2 ≥ 200 20 y 1 + 10 y 2 ≤ 200 As 20 y 1 + 10 y 2 = 200 means when 20 y 1 + 10 y 2 ≥ 200 and simultaneously 20 y 1 + 10 y 2 ≤ 200 Subject to 20 y 1 + 10 y 2 + s 1 = 200 20 y 1 + 10 y 2 − s 2 = 200 25 y 1 + 40 y 2 + s 3 = 500 18 y 1 + 24 y 2 − s 4 = 300 and y 1 ≥ 0 , y 2 ≥ 0 , s 1 ≥ 0 , s 2 ≥ 0 , s 3 ≥ 0 and s 4 ≥ 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 s 1 s 2 s 3 s 4 [ 20 10 − 1 0 0 0 200 20 10 0 1 0 0 200 25 40 0 0 1 0 500 18 24 0 0 0 − 1 300 32 40 0 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 , s 1 = − 200 , s 2 = 200 , s 3 = 500 and s 4 = − 300 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 4 have 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, first column is the pivot column and the pivot row is considered by the smallest quotient. 200 20 = 10 , 200 20 = 10 , 500 25 = 20 and 300 18 = 16.66 The smallest quotient is 10 so the pivot is 20 in row 1 of column 1. Multiply by 1 20 to row 1 in order to make pivot element to 1 for easier calculations. y 1 y 2 s 1 s 2 s 3 s 4 [ 1 1 2 − 1 20 0 0 0 200 20 10 0 1 0 0 200 25 40 0 0 1 0 500 18 24 0 0 0 − 1 300 32 40 0 0 0 0 0 ] Then pivoting in the usual way leads to the tableau, after that use operations R 2 − 20 R 1 , R 3 − 25 R 1 , R 4 − 18 R 1 and R 5 − 32 R 1 to convert all elements of that column to zero except pivot element

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ISBN: 9780135904282

Author: Lial, HUNGERFO

Publisher: PEARSON

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Chapter 7.7, Problem 19E

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To calculate: The minimum value of the provided equation using two stage simplex method.

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Chapter 7.7, Problem 18E

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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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