Explanation Given: Minimum, z = 4 y 1 + 5 y 2 Subject to: 10 y 1 + 5 y 2 ≥ 100 20 y 1 + 10 y 2 ≥ 160 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. For that multiply, both sides of the objective function by − 1 Maximize, z = − w = − 6 y 1 − 8 y 2 Subject to: 10 y 1 + 5 y 2 − s 1 = 100 20 y 1 + 10 y 2 − s 2 = 160 and y 1 ≥ 0 , y 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. y 1 y 2 s 1 s 2 [ 10 5 − 1 0 100 20 10 0 − 1 160 4 5 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 = − 100 and s 2 = − 160 However, because bothvariablesare 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 row1. Choose any positive entry in a row 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. 100 10 = 10 and 160 20 = 8 The smallest quotient is 8 so the pivot is 20 in row 2 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 [ 10 5 − 1 0 100 1 1 2 0 − 1 20 8 4 5 0 0 0 ] Then, pivoting in the usual way leads to the tableau, by using operations R 1 − 10 R 2 and R 3 − 4 R 1 to convert all elements of that column to zero except pivot element

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Chapter 7.7, Problem 14E

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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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