5.6 Maximization and 5.6 Maximization and Minimization with Mixed Minimization with Mixed

Problem ConstraintsProblem Constraints

Introduction to the Big M Introduction to the Big M MethodMethod In this section, a generalized version of In this section, a generalized version of

the simplex method that will solve both the simplex method that will solve both maximization and minimization maximization and minimization problems with any combination of problems with any combination of

constraints will be constraints will be presented.presented.

Definition: Initial Simplex Definition: Initial Simplex TableauTableau For a system tableau to be considered an For a system tableau to be considered an initial initial

simplex tableausimplex tableau, it must satisfy the following two , it must satisfy the following two requirements: requirements: 1. A variable can be selected as a basic variable only 1. A variable can be selected as a basic variable only

if it corresponds to a column in the tableau that has if it corresponds to a column in the tableau that has exactly one nonzero element and the nonzero exactly one nonzero element and the nonzero element in the column is not in the same row as the element in the column is not in the same row as the nonzero element in the column of another basic nonzero element in the column of another basic variable. variable.

2. The remaining variables are then selected as non-2. The remaining variables are then selected as non-basic variables to be set equal to zero in determining basic variables to be set equal to zero in determining a basic solution. a basic solution.

3. The basic solution found by setting the non-basic 3. The basic solution found by setting the non-basic variables equal to zero is feasible. variables equal to zero is feasible.

Key Steps of the big M Key Steps of the big M methodmethod Big M Method: Introducing slack, surplus, and artificial Big M Method: Introducing slack, surplus, and artificial

variables to form the modified problem variables to form the modified problem 1. If any problem constraints have negative constants on the 1. If any problem constraints have negative constants on the

right side, multiply both sides by -1 to obtain a constraint with right side, multiply both sides by -1 to obtain a constraint with a nonnegative constant. (remember to reverse the direction of a nonnegative constant. (remember to reverse the direction of the inequality if the constraint is an inequality).the inequality if the constraint is an inequality).

2. Introduce a 2. Introduce a slack variableslack variable for each constraint of the form for each constraint of the form

3. Introduce a 3. Introduce a surplus variable surplus variable and an and an artificial variable artificial variable in in each constraint. each constraint.

4. Introduce an 4. Introduce an artificial variableartificial variable in each = constraint. in each = constraint. 5. For each artificial variable 5. For each artificial variable a, a, add add –Ma–Ma to the objective to the objective

function. Use the same constant M for all artificial variables. function. Use the same constant M for all artificial variables.

An example: An example:

MaximizeMaximize subject to :subject to :

1 2 3

1 2 3

1 2 3

1 2 3

1, 2 3

1 2 3

2 5

3 4 10

2 4 5 20

3 15

, 0

3 2x x x

x x x

x x x

x x x

x x x

P x x x

Solution: Solution:

1 2 33 4 10x x x

2. 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:

1 2 33 15x x x

1)1) Notice that the second constraint has a Notice that the second constraint has a negative number negative number on the right hand on the right hand

side. To make that number positive, side. To make that number positive, multiply both sides by -1 and reverse multiply both sides by -1 and reverse the the direction of the inequality: direction of the inequality:

Solution continued: Solution continued:

1 2 3

1 2 3 1 1

2 5

2 5

x x x

x x x s a

3)3) Introduce a surplus variable and Introduce a surplus variable and an artificial variable for the an artificial variable for the constraint:constraint:

Solution continued: Solution continued:

4) 4) Do the same procedure for the other Do the same procedure for the other constraint: constraint:

1 2 3 2 23 4 10x x x s a

5) Introduce surplus variable for less than or equal to constraint:

1 2 3 32 4 5 20x x x s

Solution continued: Solution continued:

6)6) Introduce the third artificial variable for the Introduce the third artificial variable for the equation constraint: equation constraint:

1 2 3 33 15x x x a

7) For each of the three artificial variables, we will add –Ma to the objective function:

1 2 3 1 2 33 2P x x x Ma Ma Ma

Final resultFinal result

The modified problem The modified problem is:is:

1 2 3 1 2 33 2P x x x Ma Ma Ma

Maximize

subject to the constraints:

1 2 3 1 1

1 2 3 2 2

1 2 3 3

1 2 3 3

2 5

3 4 10

2 4 5 20

3 15

x x x s a

x x x s a

x x x s

x x x a

Key steps for solving a problem Key steps for solving a problem using the big M methodusing the big M method

Now that we have learned the procedure Now that we have learned the procedure for finding the modified problem for a linear for finding the modified problem for a linear programming problem, we will turn our programming problem, we will turn our attention to the procedure for actually attention to the procedure for actually solving such problems. The procedure is solving such problems. The procedure is called the called the Big M Method. Big M Method.

Big M Method: solving the Big M Method: solving the problemproblem 1. Form the preliminary simplex tableau for 1. Form the preliminary simplex tableau for

the modified problem. the modified problem. 2. Use row operations to eliminate the M’s 2. Use row operations to eliminate the M’s

in the bottom row of the preliminary in the bottom row of the preliminary simplex tableau in the columns simplex tableau in the columns corresponding to the artificial variables. corresponding to the artificial variables. The resulting tableau is the The resulting tableau is the initial simplex initial simplex tableau. tableau.

3. Solve the modified problem by applying 3. Solve the modified problem by applying the simplex method to the initial simplex the simplex method to the initial simplex tableau found in the second step. tableau found in the second step.

Big M method: continued: Big M method: continued:

4. Relate the optimal solution of the 4. Relate the optimal solution of the modified problem to the original problem.modified problem to the original problem. A) if the modified problem has no optimal A) if the modified problem has no optimal

solution, the original problem has no optimal solution, the original problem has no optimal solution. solution.

B) if all artificial variables are 0 in the optimal B) if all artificial variables are 0 in the optimal solution to the modified problem, delete the solution to the modified problem, delete the artificial variables to find an optimal solution to artificial variables to find an optimal solution to the original problem the original problem

C) if any artificial variables are nonzero in the C) if any artificial variables are nonzero in the optimal solution, the original problem has no optimal solution, the original problem has no optimal solution. optimal solution.

An example to illustrate the An example to illustrate the Big M method: Big M method: MaximizeMaximize

321 24 xxxP

1

6

4

321

31

32

xxx

xx

xxSubject toSubject to

Solution:Solution:

Form the preliminary simplex tableau for the Form the preliminary simplex tableau for the modified problem: Introduce slack variables, modified problem: Introduce slack variables,

artificial variables and variable Martificial variables and variable M. .

024

1

6

4

21321

22321

131

132

PMaMaxxx

asxxx

axx

sxx

Solution:Solution:

Use row operations to eliminate M’s in Use row operations to eliminate M’s in the bottom row of the preliminary simplex the bottom row of the preliminary simplex

tableautableau. . (-M) R2 + R4 = R4 (-M) R2 + R4 = R4 (-M)R3 + R4 = R4(-M)R3 + R4 = R4

Solution:Solution:

Solve the modified problem by Solve the modified problem by applying the simplex method: applying the simplex method:

The basic variables are P,a,a,s 211

The basic solution is feasible:

Solution:Solution:

Use the following operations to solve Use the following operations to solve the problem:the problem:

441

443

441

442

332

1

3

12

1

RRR)(

RRMR

RRR

RRR)M(

RRR

Solution:Solution:

..

221014100

501110010

600

400

010

001

101

1102211321

M)M(

Pasasxxx

Solution: Solution:

0

22

6

4

3

1

2

x

P

x

x