Explanation Given: Maximize, z = 2 x 1 + 4 x 2 Subject to, 3 x 1 + 2 x 2 ≤ 12 , 5 x 1 + x 2 ≥ 5 , x 1 ≥ 0 , x 2 ≥ 0 Formula used: Calculation: Consider the maximization problem which is: Maximize, z = 2 x 1 + 4 x 2 Subject to, 3 x 1 + 2 x 2 ≤ 12 , 5 x 1 + x 2 ≥ 5 , x 1 ≥ 0 , x 2 ≥ 0 Use two stages method as there are various steps which have to follow: Step (1): Modify each equation constraint by an equivalent pair of inequality constraint Step (2): Write each constraint with a positive constant. Step (3): If minimization problem, then convert into maximization by replace z = − w . Step (4): Add or subtract surplus variables according to the need of the constraints into equations. Step (5): Write initial simplex table. Step (6): Find a basic feasible solution if a solution exists, which is stage I. Step (7): If basic feasible solution is found, then do the simplex method further which is a stage and get the optimal solution. Further, to convert linear inequality into the linear equation then take a slack variable and surplus variable, which is done by adding and subtracting a non-negative variable to each constraint. If linear equality has less than or equal to sign, then use slack variable by add as the equation is less than or equal to the value if it is equal then the slack variable is zero. If linear equality has greater than or equal to sign then use surplus variable by subtracting as the equation is greater than or equal to the value, if it is equal then the surplus variable is zero which is: 3 x 1 + 2 x 2 + s 1 = 12 , 5 x 1 + x 2 − s 2 = 5 where s 1 and s 2 is greater than or equal to zero. Then, rewrite constraints by adding or subtracting the non-negative slack and surplus variables which are: Maximize, z = 2 x 1 + 4 x 2 Subject to, 3 x 1 + 2 x 2 + s 1 = 12 , 5 x 1 + x 2 − s 2 = 5 , x 1 ≥ 0 , x 2 ≥ 0 , s 1 ≥ 0 , s 2 ≥ 0 Then, convert the linear programming into a system of linear equations in which all the coefficient of variables are written on the left of the equal sign and the constants are on the right then convert the objective function into linear equations which are: − 2 x 1 − 4 x 2 + z = 0 Then, write the constraint and objective function into the augmented matrix which is in the form of a simplex table. x 1 x 2 s 1 s 2 [ 3 2 1 0 12 5 1 0 − 1 5 − 2 − 4 0 0 0 ] Here, the last row shows the coefficients of the objective function and last columns shows the constants. The basic variables correspond to columns which have an entry of 1 and rest is zero its value corresponds to the last column which is constant, and the non-basic variables correspond to the other columns and its value is zero. In the above, nonbasic variables are x 1 , x 2 and the basic variables are s 1 , s 2 . Here, x 1 = 0 , x 2 = 0 , s 1 = 12 , s 2 = − 5 In the above simplex, the table has not a basic feasible solution as the basic variable has the negative value so first, convert basic variables into nonnegative, then it is said to be basic feasible solution and stage I is completed. In the last row, all the entries except the last and second last entries are called indicators. The simplex table represents a system of three linear equations in five variables which is x 1 , x 2 , s 1 , s 2 , z as there are more variables than equations then the system is dependent, and it has infinitely many solutions. In the feasible solution as s 2 has negative value, the only non-zero entry in its column is − 1 . In the second row, select any positive entry except the entry which is far from the right that is 5 . So, take x 2 column as pivot column. Now all the positive entry of pivot column is divided by the last column which is constant. First, 12 ÷ 2 = 6 Second, 5 ÷ 1 = 5 After the row with the smallest quotient is called the pivot row, which is 5 and it occurs in the second row. The entry in the pivot row and pivot column is called Pivot. x 1 x 2 s 1 s 2 [ 3 2 1 0 12 5 1 0 − 1 5 − 2 − 4 0 0 0 ] In this, the pivot is 1

Mathematics with Applications In the Management, Natural, and Social Sciences (12th Edition)

12th Edition

ISBN: 9780134767628

Author: Margaret L. Lial, Thomas W. Hungerford, John P. Holcomb, Bernadette Mullins

Publisher: PEARSON

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Chapter 7, Problem 56RE

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Mathematics with Applications In the Management, Natural, and Social Sciences (12th Edition)

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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