Explanation Given: Maximum, z = 6 x 1 + 8 x 2 Subject to: 3 x 1 + 12 x 2 ≥ 48 2 x 1 + 4 x 2 ≤ 60 and x 1 ≥ 0 , x 2 ≥ 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 = 6 x 1 + 8 x 2 Subject to: 3 x 1 + 12 x 2 − s 1 = 48 2 x 1 + 4 x 2 + s 2 = 60 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 [ 3 12 − 1 0 48 2 4 0 1 60 − 6 − 8 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 = − 48 and s 2 = 60 . 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 row 2. 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, the first column is the pivot column and the pivot row is considered by the smallest quotient. 48 3 = 12 and 60 2 = 30 The smallest quotient is 12 so the pivot is 3 in row 1 of column 1. Multiply by 1 3 to row 1 in order to make pivot element to 1 for easier calculations. x 1 x 2 s 1 s 2 [ 1 4 − 1 3 0 12 2 4 0 1 60 − 6 − 8 0 0 0 ] Then pivoting in the usual way leads to the tableau, by using operations R 2 − 2 R 1 and R 3 + 6 R 1 to convert all elements of that column to zero except pivot element

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ISBN: 9780135240687

Author: Lial

Publisher: PEARSON

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Chapter 7.7, Problem 12E

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