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Unformatted text preview: 5.6 Maximization and Minimization with Mixed Problem Constraints Introduction to the Big M Method In this section, a generalized version of the simplex method that will solve both maximization and minimization problems with any combination of constraints will be . = presented. Definition: Initial Simplex Tableau For a system tableau to be considered an initial simplex tableau, it must satisfy the following two requirements: 1. A variable can be selected as a basic variable only if it corresponds to a column in the tableau that has exactly one nonzero element and the nonzero element in the column is not in the same row as the nonzero element in the column of another basic variable. 2. The remaining variables are then selected as nonbasic variables to be set equal to zero in determining a basic solution. 3. The basic solution found by setting the non-basic variables equal to zero is feasible. Key Steps of the big M method Big M Method: Introducing slack, surplus, and artificial variables to form the modified problem 1. If any problem constraints have negative constants on the right side, multiply both sides by -1 to obtain a constraint with a nonnegative constant. (remember to reverse the direction of the inequality if the constraint is an inequality). 2. Introduce a slack variable for each constraint of the form 3. Introduce a surplus variable and an artificial variable in each constraint. . . 4. Introduce an artificial variable in each = constraint. 5. For each artificial variable a, add Ma to the objective function. Use the same constant M for all artificial variables. An example: Maximize
subject to : P = 3 x1 - 2 x2 + x3
x1 - 2 x2 + x3 - x1 - 3 x2 + 4 x3 5 -10 2 x1 + 4 x2 + 5 x3 20 3 x1 - x2 - x3 = -15 x1, x2 , x3 0 Solution:
1) Notice that the second constraint has a negative number on the right hand side. To make that number positive, multiply both sides by -1 and reverse the direction of the inequality: x1 + 3 x2 - 4 x3
2. 10 The fourth constraint has a negative number on the right hand side so multiply both sides of this equation by -1 to change the sign of 5 to + 15: -3 x1 + x2 + x3 = 15 Solution continued:
Introduce a surplus variable and an artificial variable for the constraint:
3) x1 - 2 x2 + x3 5 x1 - 2 x2 + x3 - s1 + a1 = 5 Solution continued: 4) Do the same procedure for the other constraint: . x1 + 3 x2 - 4 x3 - s2 + a2 = 10
5) Introduce surplus variable for less than or equal to constraint: 2 x1 + 4 x2 + 5 x3 + s3 = 20 Solution continued: 6) Introduce the third artificial variable for the equation constraint: -3 x1 + x2 + x3 + a3 = 15
7) For each of the three artificial variables, we will add Ma to the objective function: P = 3 x1 - 2 x2 + x3 - Ma1 - Ma2 - Ma3 Final result The modified problem is: Maximize P = 3 x1 - 2 x2 + x3 - Ma1 - Ma2 - Ma3
subject to the constraints: x1 - 2 x2 + x3 - s1 + a1 = 5 x1 + 3 x2 - 4 x3 - s2 + a2 = 10 2 x1 + 4 x2 + 5 x3 + s3 = 20 -3 x1 + x2 + x3 + a3 = 15 Key steps for solving a problem using the big M method Now that we have learned the procedure for finding the modified problem for a linear programming problem, we will turn our attention to the procedure for actually solving such problems. The procedure is called the Big M Method. Big M Method: solving the problem 1. Form the preliminary simplex tableau for the modified problem. 2. Use row operations to eliminate the M's in the bottom row of the preliminary simplex tableau in the columns corresponding to the artificial variables. The resulting tableau is the initial simplex tableau. 3. Solve the modified problem by applying the simplex method to the initial simplex tableau found in the second step. Big M method: continued: 4. Relate the optimal solution of the modified problem to the original problem. A) if the modified problem has no optimal solution, the original problem has no optimal solution. B) if all artificial variables are 0 in the optimal solution to the modified problem, delete the artificial variables to find an optimal solution to the original problem C) if any artificial variables are nonzero in the optimal solution, the original problem has no optimal solution. An example to illustrate the Big M method: Maximize P = x1 + 4 x 2 + 2 x 3
Subject to x 2 + x3 4 x1 - x 3 = 6 x1 - x 2 - x 3 1 Solution: Form the preliminary simplex tableau for the modified problem: Introduce slack variables, artificial variables and variable M . x 2 + x3 + s1 = 4 x1 - x3 + a1 = 6 x1 - x2 - x3 - s 2 + a 2 = 1 - x1 - 4 x2 - 2 x3 + Ma1 + Ma2 + P = 0 Solution: Use row operations to eliminate M's in the bottom row of the preliminary simplex tableau. (-M) R2 + R4 = R4 (-M)R3 + R4 = R4 Solution: Solve the modified problem by applying the simplex method: The basic variables are 1 1 2 s ,a ,a , P The basic solution is feasible: Solution: Use the following operations to solve the problem: - 1R2 + R3 R3 ( 2 M + 1 )R2 + R4 R4 3R1 + R4 R4 MR3 + R4 R4 ( 1 )R1 + R4 R4 Solution: x1 0 . x2 1 x3 s1 a1 1 1 0 -1 0 0 0 s 2 a2 0 0 P 0 4 1 0 0 -1 0 0 1 0 0 0 6 -1 -1 1 0 - 5 M 1 22 1 4 ( M +1) 0 Solution: x2 = 4 x1 = 6 P = 22 x3 = 0 ...
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This note was uploaded on 03/27/2008 for the course MATH 150 taught by Professor Unk during the Winter '07 term at San Jacinto.
- Winter '07