Explanation Given: Minimum w = 15 y 1 + 12 y 2 Subject to y 1 + 2 y 2 ≤ 12 3 y 1 + y 2 ≥ 18 y 1 + y 2 = 10 and y 1 ≥ 0 , y 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. The minimization problem is solved by converting the objective function into maximization problem. Now, multiply on both sides of the objective function by − 1 . Maximize z = − w = − 15 y 1 − 12 y 2 The constraintcan be written as the following pair of inequalities, y 1 + y 2 ≤ 10 y 1 + y 2 ≥ 10 As y 1 + y 2 = 10 means when y 1 + y 2 ≤ 10 and simultaneously y 1 + y 2 ≥ 10 . Subject to y 1 + 2 y 2 + s 1 = 12 3 y 1 + y 2 − s 2 = 18 y 1 + y 2 + s 3 = 10 y 1 + y 2 − s 4 = 10 and y 1 ≥ 0 , y 2 ≥ 0 , s 1 ≥ 0 , s 2 ≥ 0 , s 3 ≥ 0 , 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 [ 1 2 1 0 0 0 12 3 1 0 − 1 0 0 18 1 1 0 0 1 0 10 1 1 0 0 0 − 1 10 15 12 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 = 12 , s 2 = − 18 , s 3 = 10 , and s 4 = − 10 . However, because two variables are negative, the solution is not feasible. Now find the basic feasible solution. In basic solution given by tableau, s 2 and s 4 have negative value. The only non-zero entry in its column is the − 1 in row 1. Choose any positive entry in row 2 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 are considered by the smallest quotient. 12 1 = 12 , 18 3 = 6 , 10 1 = 10 , and 10 1 = 10 . The smallest quotient is 6 so the pivot is 3 in row 2 of column 1. Multiply by 1 3 to row 2 in order to make pivot element to 1 for easier calculations. y 1 y 2 s 1 s 2 s 3 s 4 [ 1 2 1 0 0 0 12 1 1 3 0 − 1 3 0 0 6 1 1 0 0 1 0 10 1 1 0 0 0 − 1 10 15 12 0 0 0 0 0 ] Then pivoting in the usual way leads to the tableau, by using operations R 1 − R 2 , R 3 − R 2 , R 4 − R 2 and R 5 − 15 R 2 to convert all elements of that column to zero except pivot element. y 1 y 2 s 1 s 2 s 3 s 4 [ 0 5 3 1 1 3 0 0 6 1 1 3 0 − 1 3 0 0 6 0 2 3 0 1 3 1 0 4 0 2 3 0 1 3 0 − 1 4 0 7 0 5 0 0 − 90 ] Still, it is not the basic feasible solution as s 4 is negative

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ISBN: 8220101336323

Author: MULLINS

Publisher: PEARSON

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Chapter 7.7, Problem 20E

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

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Chapter 7.7, Problem 19E

Chapter 7.7, Problem 21E

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