Explanation Given: Maximum, z = 10 x 1 + 9 x 2 Subject to: x 1 + x 2 = 30 x 1 + x 2 ≥ 25 2 x 1 + x 2 ≤ 40 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 = 10 x 1 + 9 x 2 Constraint: x 1 + x 2 = 30 can be written as the following pair of inequalities, x 1 + x 2 ≥ 30 x 1 + x 2 ≤ 30 As x 1 + x 2 = 30 means when x 1 + x 2 ≤ 50 and simultaneously x 1 + x 2 ≥ 30 . Subjected to: x 1 + x 2 − s 1 = 30 x 1 + x 2 + s 2 = 30 x 1 + x 2 − s 3 = 25 2 x 1 + x 2 + s 4 = 40 and x 1 ≥ 0 , x 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 is the initial simplex table. x 1 x 2 s 1 s 2 s 3 s 4 [ 1 1 − 1 0 0 0 30 1 1 0 1 0 0 30 1 1 0 0 − 1 0 25 2 1 0 0 0 1 40 − 10 − 9 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 x 1 = 0 , x 2 = 0 , x 3 = 0 , s 1 = − 30 , s 2 = 30 , s 3 = − 25 and s 4 = 40 . 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 value. The only non-zero entry in its column is the − 1 in 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 choose the second column as the pivot column in order to ease our calculation, and the pivot row is considered by the smallest quotient. 30 1 = 30 , 30 1 = 30 , 25 1 = 25 and 40 1 = 40 . The smallest quotient is 25 so the pivot is 1 in row 3 of column 2. Then pivoting in the usual way leads to the tableau, by using operations R 1 − R 3 , R 2 − R 3 , R 4 − R 3 and R 5 + 9 R 3 to convert all elements of that column to zero except pivot element. x 1 x 2 s 1 s 2 s 3 s 4 [ 0 0 − 1 0 1 0 5 0 0 0 1 1 0 5 1 1 0 0 − 1 0 25 1 0 0 0 1 1 15 − 1 0 0 0 − 9 0 225 ] Still, our solution is not feasible as s 1 has negative value

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Author: Lial, HUNGERFO

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Chapter 7.7, Problem 18E

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To calculate: The minimum value of equation z=10x1+9x2 using two-stage simplex method subjected to,

x1+x2=30x1+x2≥252x1+x2≤40

and x1≥0,x2≥0.

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Chapter 7.7, Problem 17E

Chapter 7.7, Problem 19E

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Chapter 7 Solutions

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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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The minimum value of equation z = 10 x 1 + 9 x 2 using two-stage simplex method subjected to, x 1 + x 2 = 30 x 1 + x 2 ≥ 25 2 x 1 + x 2 ≤ 40 and x 1 ≥ 0 , x 2 ≥ 0 .

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To calculate: The minimum value of equation z=10x1+9x2 using two-stage simplex method subjected to, x1+x2=30x1+x2≥252x1+x2≤40 and x1≥0,x2≥0.

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Chapter 7 Solutions

To calculate: The minimum value of equation z=10x1+9x2 using two-stage simplex method subjected to,

x1+x2=30x1+x2≥252x1+x2≤40

and x1≥0,x2≥0.