4.3 The Simplex Method and the Standard

Maximization Problem

Question 1: What is a standard maximization problem?

Question 2: What are slack variables?

Question 3: How do you find a basic feasible solution?

Question 4: How do you get the optimal solution to a standard maximization problem

with the Simplex Method?

In Section 4.2, we examined several linear programming problems. The common theme

to these problems is the number of decision variables. To be able to solve these linear

programming problems graphically, they must have exactly two decision variables.

In this section you will learn how to solve linear programming problems with two or more

decision variables. This strategy, called the Simplex Method, will allow us to solve the

problems from section 4.2 as well as other maximization problems with more than 2

variables. Since large numbers of decision variables are common in business and

industry, the Simplex Method and similar variations are standard tools used to analyze

linear optimization problems with hundreds of decision variables and hundreds of

constraints.

1

Question 1: What is a standard maximization problem?

The Simplex Method is easiest to apply to a type of linear programming problem called

the standard maximization problem.

A standard maximization problem is a type of linear

programming problem in which the objective function is to be

maximized and has the form

1 1 2 2 n nz a x a x a x

where 1, , na a are real numbers and 1, , nx x are decision

variables. The decision variables must represent non-

negative values. The other constraints for the standard

maximization problem have the form

1 1 2 2 n nb x b x b x c

where 1, , nb b and c are real numbers and 0c .

There are other types of linear programming problems (we’ll examine some of

these in the next section), but in this section all of the problems are standard

maximization problems. For instance, the craft brewery problem is a standard

maximization problem. Even though the letter describing the variable P is not the

same as z, it still fits the standard maximization form:

2

1 2

1 2

1 2

1 2

1 2

Maximize 100 80

subject to

50,000

69.75 85.25 4,000,000

23.8 10.85 1,000,000

0, 0

P x x

x x

x x

x x

x x

Since this example has only two decision variables 1x and 2x , this example does not

exploit the full power of the Simplex Method. However this problem and similar

problems are useful for demonstrating the Simplex Method and we’ll focus on them to

begin with. Once the strategy for solving the standard maximization problem has been

established, we’ll extend the strategy to more complex problems.

All constraints have the form

1 1 2 2b x b x c where c is

nonnegative.

Decision variables are nonnegative

The objective function is maximized

The objective function has the

form 1 1 2 2z a x a x

3

Question 2: What are slack variables?

The graphical strategy for solving linear programming problems relies on the idea that

the maximum value of the objective function will occur at a corner point of a bounded

feasible region. The Simplex Method is no different, but we need to work with equations

for the border of the feasible region in matrix form. The trouble is that augmented

matrices are designed to solve equations and we have inequalities. For the graphical

strategy, we simply converted each inequality to an equation. For the Simplex Method,

we’ll introduce slack variables to change the inequalities to equalities.

Let’s start from one of the linear programming problems from section 4.2:

Maximize 5 6 subject to

2 4

2 4

0, 0

z x y

x y

x y

x y

This problem is a standard maximization problem with the decision variables x and y.

The goal of utilizing slack variables is to change the two inequalities to equalities. We do

this by adding some unknown amount to the left hand side of each inequality.

For instance, let’s look at the first inequality 2 4x y . This inequality corresponds to

an infinite number of ordered pairs in the xy-plane. However, we can also think of this

inequality as an equation. To do this you need to realize that the left hand side is less

than or equal to 4. If we were to add a non-negative amount to the left side, we could

increase it enough to guarantee that the left side is equal to 4. We’ll call this positive

amount 1s and refer to it as the slack variable for the first inequality. The value for 1s is

such that

12 4x y s .

In effect, the variable 1s takes up the slack in the two sides of the inequality.

4

We can apply the same reasoning to the second inequality 2 4x y . If we add a

different nonnegative amount 2s to the left side, the inequality changes to the equality

22 4x y s .

Written in this format, each inequality is now an equality with the variables (including the

slack variables) on the left side and the constant on the right side.

The objective function 5 6z x y already includes an equal sign, but not in the same

format as the equations derived from the constraints. Subtract 5x and 6y from both sides

of the objective function to put all of the variables on the left side of the equal sign. This

leaves us with the equation

5 6 0x y z .

If we put these three equations together, we get a system of three equations in five

variables. Write the constraint equations on top of the equation corresponding to the

objective function to give

1

2

2 4

2 4

5 6 0

x y s

x y s

x y z

This system can be written as an augmented matrix with three rows and six columns:

1 2

2 1 1 0 0 4

1 2 0 1 0 4

5 6 0 0 1 0

x y s s z

The vertical line corresponds to the equal sign in the equations and the horizontal line

helps to distinguish the constraints from the objective function.

This matrix,

5

1 2

2 1 1 0 0 4

1 2 0 1 0 4

5 6 0 0 1 0

x y s s z

is called the initial simplex tableau and is the starting point for the Simplex Method. The

term tableau is used when dealing with matrices in the context of the Simplex Method.

The bottom row obtained from the objective function is called the indicator row. The

variables above each column are not always shown, but we do so here to clarify how

the variables match up with the coefficients in the augmented matrix.

Example 1 Find the Initial Simplex Tableau

Find the initial simplex tableau for the craft brewery linear programming

problem:

1 2

1 2

1 2

1 2

1 2

Maximize 100 80

subject to

50,000

69.75 85.25 4,000,000

23.8 10.85 1,000,000

0, 0

P x x

x x

x x

x x

x x

Solution This problem is a standard maximization problem. To convert it

to a system of equations, we need to add three different slack variables

to the three different inequalities.

Using the slack variables 1s , 2s , and 3s , we get three equations,

1 2 1

1 2 2

1 2 3

50,000

69.75 85.25 4,000,000

23.8 10.85 1,000,000

x x s

x x s

x x s

6

To these equations, we need to add the objective function. If we move

all of the variables in 1 2100 80P x x to the left side of the equal sign,

we are left with 1 2100 80 0x x P .

Combine all of the equations together with the objective function on the

bottom and we get

1 2 1

1 2 2

1 2 3

1 2

50,000

69.75 85.25 4,000,000

23.8 10.85 1,000,000

100 80 0

x x s

x x s

x x s

x x P

A 4 x 7 augmented matrix for this system of equations is

1 2 1 2 3

1 1 1 0 0 0 50,000

69.75 85.25 0 1 0 0 4,000,000

23.8 10.85 0 0 1 0 1,000,000

100 80 0 0 0 1 0

x x s s s z

This is the initial simplex tableau for the craft brewery linear

programming problem.

7

Question 3: How do you find a basic feasible solution?

The initial simplex tableau allows us to calculate the locations of the corner points as

well as any other points where the lines corresponding to the equations cross or cross

the axes. Let’s start with the initial simplex tableau

1 2

2 1 1 0 0 4

1 2 0 1 0 4

5 6 0 0 1 0

x y s s z

.

Since there are more variables than equations, there is not a unique solution to the

system of equations represented by the tableau. There are five variables and three

equations so we need 5 3 or 2 parameters to write the solution. Any two of the

variables may be parameters, but in the case of the initial simplex tableau it is

convenient to choose the parameters as x and y.

To see why, rewrite the augmented matrix as a system of equations,

1

2

2 4

2 4

5 6 0

x y s

x y s

x y z

.

The variables 1s , 2s , and z each have a coefficient of 1 and appear nowhere else in the

system. In the matrix, this corresponds to the columns consisting of all zeros except for

a single entry of 1. These variables are the basic variables for the initial simplex tableau.

The other variables, x and y, are the nonbasic variables for the initial simplex tableau.

To find possible corner points, we’ll solve for the basic variables in term of the nonbasic

variables.

Solve for the basic variables variables in terms of x and y:

8

1

2

4 2

4 2

5 6

s x y

s x y

z x y

One possible solution to this system is found by setting the nonbasic variables x and y

equal to zero. In this situation,

1

2

4 2 0 0 4

4 0 2 0 4

5 0 6 0 0

s

s

z

.

This situation corresponds to the corner point of the feasible region at , 0,0x y with

an objective function value of 0z . At this corner point, the slack variables take on the

values 1 4s and 2 4s .

Figure 1 - The corner point corresponding to a solution of the initial simplex tableau with s1 = 4, s2 = 4 and z = 0.

Other corner points can be obtained by using row operations to make other variables

basic instead of 1s , 2s , and z. In general, z should always be a basic variable so that it

is easy to determine the value of the objective function.

9

Suppose we start from the initial simplex tableau

1 2

2 1 1 0 0 4

1 2 0 1 0 4

5 6 0 0 1 0

x y s s z

and multiply the second row by 12 and place the result in the second row

12 22 becomes R R . This gives us a 1 in the second row, second column:

1 12 2

1 2

2 1 1 0 0 4

0 0 2

5 6 0 0 1 0

1

x y s s z

.

If we make the rest of the second column zeros, the variable y will become a basic

variable. To do this, we’ll use two row operations.

First, multiply the second row by -1 and add it to the first row:

1 12 2 2

1

3 12 2

1 0 0 21 :

2 1 1 0 0 4:

0 1 0 2

R

R

This sum will be placed in the first row. Second, multiply the second row by 6 and add it

to the third row:

2

3

6 : 3 6 0 3 0 12

: 5 6 0 0 1 0

2 0 0 3 1 12

R

R

.

This sum is placed in the third row. Putting these rows into the tableau yields

10

3 12 2

1 12 2

1 2

0 1 0 2

1 0 0 2

2 0 0 3 1 12

x y s s z

.

In this tableau, the variables y, 1s , and z are basic, and x and 2s are nonbasic

variables. Notice that by placing the 1 in the second column in the second row, the

second slack variable 2s (which originally was in the second row) became nonbasic. In

general, when a 1 is placed in a column to make a variable basic, the slack variable

corresponding to the row it was placed in becomes nonbasic. As with the initial simplex

tableau, we’ll write this matrix as a system of equations and set the nonbasic variables

equal to zero.

The corresponding system of equations is

3 11 22 2

1 122 2

2

2

2

2 3 12

x s s

x y s

x s z

.

If we solve each equation for a different basic variable we get

3 11 22 2

1 122 2

2

2

2

12 2 3

s x s

y x s

z x s

.

If we set the nonbasic variables x and 2s equal to zero, we find that 1 2s , 2y , and

12z . This corresponds to the corner point at , 0,2x y with an objective function

value of 12z .

11

Figure 2 - The corner point at (0,2) corresponding to s1 = 2, y = 2, and z = 12.

It is easier to find this corner point using the tableau without transforming it to a system

of equations. The key is to realize that when nonbasic variables are set equal to zero,

their coefficients in the matrix disappear. For instance, if we take the tableau we

transformed with row operations and cross out the columns of the nonbasic variables

we get:

3 12 2

1 12 2

1 2

0 1 0 2

1 0 0 2

2 0 0 3 1 12

x y s s z

By ignoring these columns, we can see that 1 2s , 2y , and 12z . Combining this

observation with our knowledge of setting the nonbasic variables equal to zero gives us

the corner point at 0,2 and objective function value 12z .

We cannot arbitrarily make a variable into a basic variable. For instance, we might make

x a basic variable by putting a 1 in the second row and first column and zeros in the rest

of the column. Starting from the original simplex tableau, we must carry out two row

operations:

12

To find the corner point corresponding to this matrix, cover the nonbasic variables and

read off the values of the basic variables:

2

1

1

0 3 1 2 0 4 4

1 2 0 1 0 4 4

0 4 0 5 1 20 20

x y s

x

z

s

s

z

Figure 3 - A solution to the tableau that is not a corner point of the feasible region.

On the surface this might not seem all that different from earlier tableaus, but this one

violates one of the assumptions for slack variables. In a standard maximization problem,

the slack variables are assumed to be non-negative. In the solution we found from this

1 2

2 1 1 0 0 4

1 2 0 1 0 4

5 6 0 0 1 0

x y s s z

1 2

0 3 1 2 0 4

1 2 0 1 0 4

0 4 0 5 1 20

x y s s z

2 1

1

2becomes

R R

R

2 3

3

5becomes

R R

R

13

tableau, the value for 1s is negative. This tells us that this solution does not correspond

to a corner point of the feasible region.

The solution we have just found, 4,0 , matches a point on a line, but is not along the

border of the feasible region. If we are not careful about picking the basic and nonbasic

variables, we may obtain a point that is not related to the solution.

If we want to make the variable x a basic variable, it is better to use row operations to

transform the original matrix

1 2

2 1 1 0 0 4

1 2 0 1 0 4

5 6 0 0 1 0

x y s s z

to 1 12 2

1

3 12 2

2

2

7 52

1 0 0 2

0 1 0 1

0 0 1 10

x y s s z

By covering the columns under y and 1s , we observe that this new tableau corresponds

to the solution 2x and 0y and the objective function value 10z . This does match

the corner point (2, 0) on the feasible region.

Example 2 Change Basic Variables

The basic variables for the initial simplex tableau

1 2

2 1 1 0 0 4

1 2 0 1 0 4

5 6 0 0 1 0

x y s s z

are 1s , 2s , and z. Use row operations to change the tableau so that the

basic variables are y, 1s , and z. What corner point of the basic feasible

solution does this new tableau correspond to?

14

Solution We want to change the basic variables from 1s , 2s , and z to y,

1s , and z. This means we need to make 2s nonbasic and y basic. The

other two variables, 1s and z, should remain basic variables.

To make y a basic variable, we need to place two 0’s and a 1 in the

column corresponding to y. The only question is where should we put

the one? The first row of the second column already has a one so all we

need to do is to use row operations to place zeros in the rest of the

column.

If you examine the new matrix, you’ll notice that y is a basic variable,

but 1s is not. Instead, 2s is a basic variable.

If we place the one in the first row, the slack variable corresponding to

that row, 1s , becomes a nonbasic variable. Since we want 1s to remain

basic and 2s to become nonbasic, the one should be placed in the

second row of the first column. Start from the original tableau

1

2

1

3

2 : 4 2 2 0 0 8

: 1 2 0 1 0 4

3 0 2 1 0 4

6 : 12 6 6 0 0 24

: 5 6 0 0 1 0

7 0 6 0 1 24

R

R

R

R

1 2

2 1 1 0 0 4

3 0 2 1 0 4

7 0 6 0 1 24

x y s s z

1 3

3

6becomes

R R

R

1 2

2

2becomes

R R

R

15

1 2

2 1 1 0 0 4

1 2 0 1 0 4

5 6 0 0 1 0

x y s s z

In general, we get the one by multiplying the row by a constant:

Now we can add multiples of rows to put zeros in the rest of the second

column.

Since the second, third, and fifth columns contains two zeros and a one,

the basic variables are now y, 1s , and z. The point in the feasible region

corresponding to this tableau is found by covering the nonbasic variable

columns x and 2s . This shows the corner point to be at 0,2 with an

objective function value of 12z .

1 12 2 2

1

3 12 2

2

3

1 0 0 21 :

2 1 1 0 0 4:

0 1 0 2

6 : 3 6 0 3 0 12

: 5 6 0 0 1 0

2 0 0 3 1 12

R

R

R

R

3 12 2

1 12 2

1 2

0 1 0 2

1 0 0 2

2 0 0 3 1 12

x y s s z

2 3

3

6becomes

R R

R

2 1

1

1becomes

R R

R

12 2

1

1

2

2 1 1 0 0 4

1 0 0 2

5 6 0 0 1 0

x y s s z

12

1 12 2

1 2 0 1 0 4

1 0 0 2

1

22

2

becomesR

R

16

Question 4: How do you get the optimal solution to a standard maximization problem

with the Simplex Method?

In the previous question, we learned how to make different variables into basic and

nonbasic variables. This allowed us to calculate the locations of corner points on the

feasible region and other points of intersection. For simple linear programming

problems, there can be many such points. If we naively make different variables into

basic variables, it may take a very long time to find the corner point that optimizes the

objective function.

The Simplex Method is an algorithm for solving standard maximization problems. By

following the steps below, you ensure that the points you find by changing basic

variables are corner points of the feasible region. In addition, the algorithm is iterative.

In other words, it is carried out several times with the result of each iteration providing

the starting point to the next iteration. The beauty of the Simplex Method is that each

iteration yields a larger value for the objective function and eventually leads to the

largest possible value for the objective function.

To get a feel for how the Simplex Method works, let’s solve the standard maximization

problem

Maximize 5 6

subject to

2 4

2 4

0, 0

z x y

x y

x y

x y

We’ll start by converting this problem to an appropriate system of linear equations. Add

a different slack variable to each constraint to yield

1

2

2 4

2 4

x y s

x y s

17

To convert the objective function to an appropriate form, move all of the terms with

variables to the left side of the equation. This is done by subtracting 5x and 6y from

both sides. This yields the equation 5 6 0x y z .

These equations form the system of linear equations

1

2

2 4

2 4

5 6 0

x y s

x y s

x y z

.

The augmented matrix for this system,

1 2

2 1 1 0 0 4

1 2 0 1 0 4

5 6 0 0 1 0

x y s s z

,

is the initial simplex tableau. It is the starting point for carrying out the Simplex Method

algorithm. The bottom row of the tableau is called the indicator row. The algorithm starts

by determining which variable will become a basic variable in the first iteration of the

algorithm. The variable that corresponds to the most negative entry in the indicator row

will become a basic variable.

For this matrix, the most negative entry is -6 in the second column. The variable y that

corresponds to the column will become the basic variable. This column is the pivot

column.

If we pick the column with the most negative entry in the indicator row to be the pivot

column, the objective function will increase by the largest amount. Remember, those

negative entries correspond to positive coefficients in the original objective function. For

the objective function 5 6z x y , the most negative entry in the indicator row, -6,

corresponds to the largest positive coefficient, 6, in the objective function. Increasing the

value of y in the potential solution from 0 to some positive value leads to a larger

increase in the objective function value z than increasing the value of x from 0 to the

18

same positive value. This is due to the fact that the coefficient on y is larger than the

coefficient of x. In effect, it is more efficient to change y values than x values in the

solution.

We know that we’ll need to make all of the entries but one in that column equal to zero

via row operations. The other entry will be a one. The row where we will put the one is

called the pivot row, and the entry that becomes a one is called the pivot. To find which

row will become the pivot row, we form quotients from the entries in the pivot column

and the last column:

14

2

1

4

2

2 1 0 0 4

1 0 1 0 2

5 6 0 0 1 0

pivot

co

4

l

1

umn

2 4

x y s s z

The row whose quotient is positive and smallest is the pivot row. In this case, the

second row is the pivot row so the pivot is the number 2 in this row.

As we saw earlier, picking the pivot row carefully guarantees that the potential solution

stays in the feasible region and all variables are non-negative. The variables must

remain non-negative for a standard linear programming problem.

Let’s look at this idea more closely. Look at the initial tableau from the standard linear

programming problem:

14

2

1

4

2

2 1 0 0 4

1 0 1 0 2

5 6 0

4

4

1

0

2

0 1

x y s s z

We already know that y will become a basic variable based on the numbers in the

indicator row. By making y a basic variable, either 1s or 2s will join x as a nonbasic

variable. The question is, which will it be?

19

Suppose that 1s and x will be the nonbasic variables in the next tableau, and will be set

equal to zero. In this case, we can use the first row of the tableau to determine the

corresponding value of y:

1 4

4

4

4

2 1

2 0

1

1 1

1

1

0

4

x y s

y

y

y

The value of 2s is found from the second row of the tableau by setting the nonbasic

variables equal to zero and 4y :

2

2

2

1 2 4

1 0 2 4 4

4

x y s

s

s

This is not a reasonable solution since the slack variable 2s is negative.

Let’s look at making 2s and x the nonbasic variables. In this case, we can use the

second row of the tableau to determine the corresponding value of y:

21 1

2 0 1

2

2

02

4

4

42

x y s

y

y

The value of 1s is found from the first row of the tableau,

1

1

1

2 1 4

2 0 1 2 4

2

x y s

s

s

Write the first row of the tableau as an equation

Set 0x and 1 0s

Solve for y

Write the second row of the tableau as an equation

Set 0x and 4y

Solve for 2s

Write the second row of the tableau as an equation

Set 0x and 2 0s

Solve for y

Write the first row of the tableau as an equation

Set 0x and 2y

Solve for 1s

20

All variables in this case are non-negative. Note that the values for y in each case are

precisely the quotients we use to choose which row becomes the pivot row. By

choosing a value for y that is smaller (the smaller quotient), we get a situation in which

all variables are non-negative.

Now that we know that the pivot is the 2 in the second row, second column of

1 2

2 1 1 0 0 4

1 0 1 0 4

5 6 0 0 1

2

0

x y s s z

we’ll use row operations to change it to a one. The rest of the pivot column will be

changed to zeros using row operations. This change in basic variables will lead to the

largest increase in the value of the objective function and keep all variables non-

negative.

As shown in Question 3, change the pivot to a one by multiplying the second row by 12 :

Once the pivot is in place, use row operations to place zeros in the rest of the column:

122

2

becomesR

R

12

1 12 2

1 2 0 1 0 4

1 0 0 2

12 2

1

1

2

2 1 1 0 0 4

1 0 0 2

5 6 0 0 1 0

x y s s z

21

In the Simplex Method, the optimal solution is obtained when there are no negative

numbers in the indicator row.

Since the transformed matrix has a negative number in the indicator row, the solution

, 0,2x y is not optimal. The value of the objective function, 12z , can be made

greater by picking the first column to be the pivot column. To find the pivot row, examine

the quotients formed from dividing entries in the last column by corresponding entries in

the pivot column:

32

12

41

2

2

32

1

1 2

32

122

0 1 0

1 0 0 4

2 0 0 3 1 12

pivo

2

2

tcolumn

x y s s z

The smallest quotient is 43 in the first row. The pivot, 3

2 , is in the first row, first column.

As in the first iteration, we must use row operations to change the pivot to a one and

change the rest of the pivot column to zeros.

To change the pivot to a one, multiply the first row by 23 :

2 3

3

6becomes

R R

R

1 12 2 2

1

3 12 2

2

3

1 0 0 21 :

2 1 1 0 0 4:

0 1 0 2

6 : 3 6 0 3 0 12

: 5 6 0 0 1 0

2 0 0 3 1 12

R

R

R

R

3 12 2

1 12 2

1 2

0 1 0 2

1 0 0 2

2 0 0 3 1 12

x y s s z

2 1

1

1becomes

R R

R

22

To complete the transformation of the pivot column, we must use row operations to put

zeros in the rest of the pivot column:

For this tableau, the basic variables are x, y, and z. The nonbasic variables are 1s and

2s . The indicator row has no negative entries so this tableau is the final tableau. The

optimal solution is read from this tableau by setting the nonbasic variables equal to

zero.

If we cover the nonbasic variables,

2 1 43 3 3

1 2 43 3 3

74 443 3

1

3

2

1 0 0

0 1 0

0 0 1

x y s s z

,

we see that this tableau corresponds to 4 43 3, ,x y and an optimal value of 44

3z .

This is the same value we found graphically in Section 4.2

The process outlined here is summarized in the steps shown here.

11 22

2

becomes

R R

R

1 3

3

2becomes

R R

R

1 1 1 211 2 3 6 32

1 12 2 2

1 2 43 3 3

84 21 3 3 3

3

74 443 3 3

0 0:

1 0 0 2:

0 1 0

2 : 2 0 0

: 2 0 0 3 1 12

0 0 1

R

R

R

R

2 1 43 3 3

1 2 43 3 3

74 443 3

1

3

2

1 0 0

0 1 0

0 0 1

x y s s z

213

1

becomes

R

R

3 12 2

23

2 1 43 3 3

0 1 0 2

1 0 0

2 1 43 3 3

1 12 2

1 2

1 0 0

1 0 0 2

2 0 0 3 1 12

x y s s z

23

The Simplex Method

1. Make sure the linear programming problem is a standard

maximization problem.

2. Convert each inequality to an equality by adding a slack

variable. Each inequality must have a different slack

variable. Each constraint will now be an equality of the

form

1 1 2 2 n nb x b x b x s c

where s is the slack variable for the constraint. If more

than one slack variable is needed, use subscripts like

1 2, ,s s

3. Rewrite the objective function 1 1 2 2 n nz a x a x a x by

moving all of the variables to the left side. After rewriting

the equation, the function will have the form

1 1 2 2 0n na x a x a x z .

4. Convert the equations from steps 2 and 3 to an initial

simplex tableau. Put the equation from step 3 in the

bottom row of the tableau and all other equations above it.

The bottom row is called the indicator row.

5. Find the entry in the indicator row that is most negative. If

two of the entries are most negative and equal, pick the

entry that is farthest to the left. The column with this entry

is called the pivot column.

24

6. For each row except the last row, divide the entry in the

last column by the entry in the pivot column. The row with

the smallest positive quotient is the pivot row. If more than

one row has the same smallest quotient, the higher of the

rows is the pivot row.

7. The pivot is the entry where the pivot row and pivot

column intersect. Multiply the pivot row by the reciprocal

of the pivot to change it to a one.

8. To change the rest of the pivot column to zeros, multiply

the pivot row by constants and add them to the other rows

in the tableau. Replace those rows with the appropriate

sums. When complete, the pivot should be a one, and the

rest of the pivot column should be zeros.

9. If the indicator does not contain any negative entries, this

tableau corresponds to the optimum solution. In this case,

cover the nonbasic variables (set the nonbasic variables

equal to zero), and read off the solution for the basic

variables. Otherwise, repeat steps 5 through 9 until the

indicator row contains no negative numbers.

The Simplex Method will reach an optimal solution for a standard maximization problem

with any number of variables and any number of inequalities. However, problems with

large numbers of variables and inequalities may require many iterations of steps 5

through 9 to reach the optimal solution. Take special care to avoid arithmetic errors

when carrying out the row operations. A graphing calculator can make the task of

carrying out the row operations fairly painless.

25

Example 3 Find the Optimal Solution

Find the optimal solution to the linear programming problem

1 2

1 2

1 2

1 2

1 2

Maximize 2 5

subject to

2 3 15

7 6 63

5 3

0, 0

z x x

x x

x x

x x

x x

Solution Although this example has three constraints (in addition to the

non-negativity constraints), it still has two decision variables. Each

constraint needs a slack variable to convert it to an equation. Using the

slack variables 1s , 2s , and 3s , the constraints become

1 2 1

1 2 2

1 2 3

2 3 15

7 6 63

5 3

x x s

x x s

x x s

Move all of the variable terms in the objective function to the left side of

the equation to yield

  1 22 5 0x x z  

These equations become the initial simplex tableau

1 2 1 2 3

2 3 1 0 0 0 15

7 6 0 1 0 0 63

1 5 0 0 1 0 3

2 5 0 0 0 1 0

x x s s s z

26

The bottom row of the initial simplex tableau is the indicator row. The

most negative entry of the indicator row is -5. This means the pivot

column is the second column in the tableau.

To determine the pivot row, compute the quotients formed by dividing

entries in the last column by entries in the pivot column,

1 2 1 2 3

15

63

3

6

2 1 0 0 0 57 0 1 0 0 10.51 5 0 0 1 0 3

2 5 0 0 0 1 0

3

6

15

63

x x s s s z

Any negative quotients, such as 35 in the third row, are discarded. The

quotient formed in the first row, 153 , is the smallest quotient so the pivot

is the 3 in the first row, second column.

In the first iteration of the simplex method, we must use row operations

to change the pivot to a one and all other entries in the pivot column to

zero.

The pivot is changed to a one by multiplying the first row by 13 . Carrying

out the row operation yields

Three row operations are necessary to put the zeros in the rest of the

pivot column:

113

1

becomes

R

R

13

2 13 3

2 3 1 0 0 0 15

1 0 0 0 5

1 2 1 2 3

2 13 31 0 0 0 5

7 6 0 1 0 0 63

1 5 0 0 1 0 3

2 5 0 0 0 1 0

x x s s s z

27

The new tableau corresponds to the solution 1 2, 0,5x x and is not

optimal since there is a negative entry in the indicator row.

The new pivot column is the first column due to the negative number in

the indicator row. To find the new pivot row, compute the quotient

formed from dividing the entries in the last column by the entries in the

new pivot column:

1 2 1 2 3

2 13 3

7 53 3

543 3

3311

1 0 0 0 5

0 2 1 0 0 3

0

33

0 1 0 28

0 0 0 1 25

11

x x s s s z

The second row is the only row with a valid quotient so it is the pivot

row. The pivot is the 11 in the second row, first column of the tableau.

1 2

2

6becomes

R R

R

1 3

3

5becomes

R R

R

1 4

4

5becomes

R R

R

1

2

10 51 3 3

3

7 53 3

10 51 3 3

4

543 3

6 : 4 6 2 0 0 0 30

: 7 6 0 1 0 0 63

11 0 2 1 0 0 33

5 : 5 0 0 0 25

: 1 5 0 0 1 0 3

0 0 1 0 28

5 0 0 0 255 :

2 5 0 0 0 1 0:

0 0 0 1 25

R

R

R

R

R

R

1 2 1 2 3

2 13 3

7 53 3

543 3

1 0 0 0 5

11 0 2 1 0 0 33

0 0 1 0 28

0 0 0 1 25

x x s s s z

28

Multiply the pivot row by 111 to change the pivot to a one:

To complete this Simplex Method iteration, we must place zero above

and below the pivot using row operations.

Three row operations are required to place the zeros above and below

the pivot.

The indicator row does not contain any negative numbers so the

optimal solution has been reached.

The solution to the linear programming problem is easily observed by

covering the columns corresponding to the nonbasic variables 1s and 2s

:

22 13

1

becomes

R R

R

72 33

3

becomes

R R

R

42 43

4

becomes

R R

R

2 4 222 3 33 333

2 11 3 3

7 233 33

7 71472 3 33 333

7 53 3 3

74133 33

84 442 3 33 333

544 3 3

47 433 33

0 0 0 2:

1 0 0 0 5:

0 1 0 0 7

0 0 0 7:

0 0 1 0 28:

0 0 1 0 35

0 0 0 4:

0 0 0 1 25:

0 0 0 1 29

R

R

R

R

R

R

1 2 1 2 3

7 233 33

2 111 11

74133 33

47 433 33

0 1 0 0 7

1 0 0 0 3

0 0 1 0 35

0 0 0 1 29

x x s s s z

1211

2

becomes

R

R

111

2 111 11

11 0 2 1 0 0 33

1 0 0 0 3

1 2 1 2 3

2 13 3

2 111 11

7 53 3

543 3

1 0 0 0 5

1 0 0 0 3

0 0 1 0 28

0 0 0 1 25

x x s s s z

29

1 2 1 2 3

7 233 33

2 111 11

74133 33

47 433 33

0 1 0 0 7

1 0 0 0 3

0 0 1 0 35

0 0 0 1 29

x x s s s z

The optimal solution to the linear programming problem is 29z and

occurs at 1 2, 3,7x x .

Example 4 Find the Optimal Production Level

The linear programming problem for the craft brewery is

1 2

1 2

1 2

1 2

1 2

Maximize 100 80

subject to

50,000

69.75 85.25 4,000,000

23.8 10.85 1,000,000

0, 0

P x x

x x

x x

x x

x x

where 1x is the number of barrels of pale ale, and 2x is the number of

barrels of porter. Use the Simplex Method to find the production level

that maximizes profit.

Solution In Example 1, we added slack variables and found the initial

simplex tableau for this standard maximization problem.

30

The initial simplex tableau is a 4 x 7 matrix,

1 2 1 2 3

1 1 1 0 0 0 50,000

69.75 85.25 0 1 0 0 4,000,000

23.8 10.85 0 0 1 0 1,000,000

100 80 0 0 0 1 0

x x s s s P

The most negative entry in the indictor row is -100. This entry means the

pivot column is the first column in the tableau. The quotients formed

from the last column and the pivot column,

1 2 1 2 3

50,000

4,000,000

1,00

1

69.75

20,00

80

3.

1 1 0 0 0 50,00085.25 0 1 0 0 57,34810.85 0 0 1 0 42,017

100 80

1

69.75

23.8

50,000

4,000,000

1,000,000

0 0 0 1 0

x x s s s P

indicate that the third row is the pivot row.

The third row must be multiplied by 123.8 to change the pivot to a one.

The entries in the first six columns are shown rounded to three decimal

places when necessary, and the entries in the last column are rounded

to the nearest tenth.

1323.8

3

becomes

R

R

123.8

23.8 10.85 0 0 1 0 1,000,000

1 0.456 0 0 0.042 0 42,016.8

1 2 1 2 3

1 1 1 0 0 0 50,000

69.75 85.25 0 1 0 0 4,000,000

1 0.456 0 0 0.042 0 42,016.8

100 80 0 0 0 1 0

x x s s s P

31

Now we need to use three row operations to place zeros in the rest of

the pivot column.

The negative number in the indicator row tells us that the second

column will be the new pivot column.

To locate the new pivot, form the quotients with the last column and

second column. The quotients for the individual rows are

1 2 1 2 3

0.544

53.45

7983.2

1,069,322

0.4

7.7

42,01 .8566

0 1 0 0.042 0 14,6750 0 1 2.931 0 20,0051 0 0 0.042 0 92,1420 34.412 0 0 4.

0.544

53.452

0.45

7983.2

1,069,32

2

7.

02 1 4, 201,680.7

7

42,016.86

x x s s s P

The numbers shown in the quotient are rounded numbers, and the

amount calculated is rounded to the nearest integer. This does not

affect our choice of the pivot row. If the quotients are close to each

other, carry more decimals so you do not choose the wrong pivot row

due to rounding. The new pivot is 0.544 in the first row, second column.

3 1

1

1becomes

R R

R

3 2

2

69.75becomes

R R

R

3 4

4

100becomes

R R

R

3

1

3

2

3

4

1 : 1 0.456 0 0 0.042 0 42,016.8

: 1 1 1 0 0 0 50,000

0 0.544 1 0 0.042 0 7983.2

69.75 : 69.75 31.798 0 0 2.931 0 2,930,672.3

: 69.75 85.25 0 1 0 0 4,000,000.0

0 53.452 0 1 2.931 0 1,069,327.7

100 : 100 45.588 0 0 4.202 0 4,201,

:

R

R

R

R

R

R

680.7

100 80 0 0 0 1 0

0 34.412 0 0 4.202 1 4,201,680.7

1 2 1 2 3

0 0.544 1 0 0.042 0 7983.2

0 53.452 0 1 2.931 0 1,069,327.7

1 0.456 0 0 0.042 0 42,016.8

0 34.412 0 0 4.202 1 4, 201,680.7

x x s s s P

32

The new pivot is changed to a one by multiplying the first row of the

tableau by 10.544 :

 

We use three row operations to put zeros in the rest of the pivot

column:

Since there are no negative numbers in the indicator row, we can read

the solution from this tableau. If we cover the nonbasic variables 1s and

3s ,we see that the solution corresponding to this tableau is

1 2, 35,328.2, 14,671.8x x and 4,706,563.7P or approximately

35,328 barrels of pale ale and 14,672 barrels of porter for a maximum

profit of about $4,706,564.

10.544

0 0.544 1 0 0.042 0 7983.2

0 1 1.838 0 0.077 0 14,671.8

1 2 1 2 3

0 1 1.838 0 0.077 0 14,671.8

0 53.452 0 1 2.931 0 1,069,327.7

1 0.456 0 0 0.042 0 42,016.8

0 34.412 0 0 4.202 1 4, 201,680.7

x x s s s P

110.544

1

becomes

R

R

1 2

2

53.425becomes

R R

R

1 3

3

0.456becomes

R R

R

1 4

4

34.412becomes

R R

R

1

2

1

3

1

4

53.452 : 0 53.452 98.236 0 4.128 0 784, 240.8

: 0 53.452 0 1 2.931 0 1,069,327.7

0 0 98.236 1 1.197 0 285,086.9

0.456 : 0 0.456 0.838 0 0.035 0 6688.6

: 1 0.456 0 0 0.042 0 42,016.8

1 0 0.838 1 0.077 0 35,328.2

34.412 : 0 34.

:

R

R

R

R

R

R

412 63.243 0 2.658 0 504,883.0

0 34.412 0 0 4.202 1 4,201,680.7

0 0 63.243 0 1.544 1 4,706,563.7

1 2 1 2 3

0 1 1.838 0 0.077 0 14,671.8

0 0 98.236 1 1.197 0 285,086.9

1 0 0.838 1 0.077 0 35,328.2

0 0 63.243 0 1.544 1 4,706,563.7

x x s s s P

33

Example 3 and Example 4 both have two decision variables and were also done

graphically in Section 4.2. In the next example, we apply the Simplex Method to a

problem with three decision variables. Problems with more than two decision variables

are impossible to do with the graphical methods introduced in Section 4.2.

Example 5 Find the Optimal Investment Portfolio

Investing in stocks and mutual funds are a tradeoff. We want to

maximize the money we make from investments, but manage the risk of

losing money in poor investments. Typically, investments with higher

returns are riskier, and investments with lower returns are less risky.

Investors use several statistics to quantify the returns on their

investments. For instance, websites such as Yahoo Financial calculate

measures like average rate of return. This number indicates the

investor’s total returns on an investment annualized over some time

period. The higher the average rate of return, the more gains you

achieve from dividends or the sale of the stock or mutual fund.

There are also several measures of risk available to investors. One

commonly used indicator is called beta. The beta of a stock or mutual

fund is calculated using regression analysis. The details of the analysis

are not important for our purposes. However, understanding what the

beta tells you is an important investing tool.

A stock or mutual fund’s beta measures the volatility of the investment’s

price with respect to some benchmark like the S & P 500 or the Dow

Jones Industrial Average. A beta of 1 indicates that the investment’s

price will change with the benchmark’s changes. A beta less than 1

indicates that the investment’s price is less volatile than the benchmark.

In other words, changes in the benchmark will be larger than changes in

the investment’s price. A beta greater than 1 indicates that the

investment’s price is more volatile than the benchmark. In this case,

34

changes in the benchmark will be smaller than changes in the

investment’s price.

Suppose an investor is interested in investing in three different mutual

funds. The funds, their five year average annual return, and beta values

are shown (as of July 31, 2010) in the table below:

Fund 5 Year Average Annual Return

Beta

TIAA-CREF Mid Cap Value Fund (TCMVX) 1.72% 1.16

TIAA-CREF Growth & Income Fund (TIIRX) 2.88% 0.93

TIAA-CREF International Equity Fund (TIERX) 2.67% 1.04

The fund TIIRX has the highest rate of return and the lowest beta. The

fund TCMVX has the lowest rate of return and the highest beta. On the

surface, it would appear that the investor should put all of their money in

TIIRX since that would yield the highest return and the lowest risk.

Each investor has their own requirements for their investments and in

this case, the investor wants to diversify their holdings. This means the

investor wants to lower their overall risk by investing in several broad

categories of stocks. TIIRX seeks to invest 80% of its funds in securities

which offer long term returns. It invests approximately 20% in smaller

companies that offer an opportunity at growth. While this may seem like

a good mix, the investor wishes the sum of what is invested in TCMVX

and TIERX to be at least 25% of what is invested in TIINX.

If the investor has $100,000 to invest and wishes the average weighted

beta to be no larger than 1, how much should the investor put in each of

the three funds so that their returns are maximized?

35

Solution A good starting point for linear programming problems is the

variables. In this problem, we want to find the amount of money to

invest in each fund. It makes sense to define the variables as follows:

M: amount of money invested in Mid Cap Value Fund.

G: amount of money invested in Growth & Income Fund.

E: amount of money invested in International Equity Fund

We wish to maximize the return on the investment. The return on an

investment is found by multiplying the amount invested by the average

annual return. Since M, G, and E represent the amounts of money

invested, 0.0172M , 0.0288G , and 0.0267E represent the return on these

amounts. The objective function is the sum of these amounts or

0.0172 0.0288 0.0267R M G E ,

where R is the total return.

Three pieces of information can help us to write out constraints. First,

the investor has $100,000 to invest. This means the sum of the amounts

invested in each fund must be less than or equal to 100,000 or

sum of the

amounts invested

100,000M G E .

Second, the investor wishes the sum of what is invested in the Mid Cap

Value Fund and International Equity Fund to be at least 25% of what is

invested in Growth & Income Fund. Focusing on the word “sum” and

the phrase “25% of”, we write

sum of Mid Cap and 25% of Growth International Equity & Value FundFunds

0.25M E G

36

The word “sum” indicates that we must add the two quantities and the

phrase “25% of” indicates that we must multiply the quantity by 0.25.

The final piece of information dictates that the investor “wishes the

average weighted beta to be no larger than 1”. A weighted average tells

us that we need to add up some terms and then divide by a total

amount. The weighted sum of the risk is written as

1.16 0.93 1.04M G E . The term “weighted” indicates the presence of

the variable in each term. If we had simply added up the beta values,

they would have each contributed “equally” to the sum since each one

appears in their own term. By multiplying the beta value by the amount

invested in the corresponding fund, we are weighting the beta by the

amount invested. If we do this, then a fund with a large amount in it will

contribute more to the sum than a fund with very little invested in it. To

average the weighted sum, we divide by the total amount invested or

M G E . The constraint

1.16 0.93 1.041

M G E

M G E

reflects the investor’s wish that the average weighted beta be no more

than 1.

If we add non-negativity constraints, the linear programming problem

becomes

Maximize 0.0172 0.0288 0.0267

subject to

100,000

0.25

1.16 0.93 1.041

0, 0, 0

R M G E

M G E

M E G

M G E

M G E

M G E

37

This linear programming problem is not in standard maximum form

since the second and third constraints do not have the form

1 1 2 2 n nb x b x b x c

where 1, , nb b and c are real numbers and 0c . A little algebra

changes these constraints to the proper form.

If we start with the second constraint and subtract M and E from both

sides, we get the inequality

0 0.25G M E .

Flip the inequality and rearrange the order of the terms to yield

0.25 0M G E .

This inequality has the proper form.

If we start with the third inequality and multiply both sides by M G E

we are left with

1.16 0.93 1.04M G E M G E .

Collect all of the terms on the left side to yield

0.16 0.07 0.04 0M G E

With these modifications, we can write out a standard maximization

problem:

38

Maximize 0.0172 0.0288 0.0267

subject to

100,000

0.25 0

0.16 0.07 0.04 0

0, 0, 0

R M G E

M G E

M G E

M G E

M G E

Now that the problem is in standard form, we can add slack variables to

the constraints and write out the initial simplex tableau.

The objective function must be rewritten with all terms on the left side of

the equal sign or

0.0172 0.0288 0.0267 0M G E R

Notice that the variables are written in a particular order. Since the

terms will match with columns in the tableau, we need to use this order

for each equation we write.

Add the slack variables 1s , 2s , and 3s to the inequalities

100,000

0.25 0

0.16 0.07 0.04 0

M G E

M G E

M G E

to give the equations

1

2

3

100,000

0.25 0

0.16 0.07 0.04 0

M G E s

M G E s

M G E s

Combine these equations with the equation from the objective function

to give

39

1 2 3

1 1 1 1 0 0 0 100,000

1 0.25 1 0 1 0 0 0

0.16 0.07 0.04 0 0 1 0 0

0.0172 0.0288 0.0267 0 0 0 1 0

M G E s s s R

.

From this initial simplex tableau we need to pick a pivot column and

row. To find the pivot column, locate the most negative entry in the

indicator row at the bottom. The entry -0.0288 is the most negative entry

so the second column is the pivot column.

The quotients for picking the pivot row are found by dividing the entries

in the last column by the corresponding entries in the pivot column,

1 2 3

100,1

0.25

000

0

1 1 1 0 0 0 100,0001 1 0 1 0 0 0

0.16 0.07 0.04 0 0 1 0 0

0.0172 0.0288 0.0267 0

100,000

0

1

0.25

0 0 1 0

M G E s s s R

Remember that quotients with a negative numerator or denominator are

ignored so the quotient for the third row isn’t even computed. The

second row contains the smallest quotient, so the pivot is 0.25 in the

second row, second column.

To change the pivot to a one, multiply the second row by 10.25 or 4:

2

2

4becomes

R

R

1 0.25 1 0 1 0 0 0

4

4 1 4 0 4 0 0 0

1 2 3

1 1 1 1 0 0 0 100,000

4 1 4 0 4 0 0 0

0.16 0.07 0.04 0 0 1 0 0

0.0172 0.0288 0.0267 0 0 0 1 0

M G E s s s R

40

With the pivot changed, row operations are needed to change the other

numbers in the pivot column to zeros.

After the first Simplex Method iteration, there is still a negative number

in the indicator row. This tableau does not correspond to an optimal

solution so we must select the first column as the pivot column and

determine the new pivot row. Only the first row has an admissible

quotient. The 5 in the first row, third column will be the new pivot.

To change the pivot to a one, multiply the first row by 15 :

   

   

2 1

1

1becomes

R R

R

2 3

3

0.07becomes

R R

R

2 4

4

0.0288becomes

R R

R

2

1

2

3

2

4

1 : 4 1 4 0 4 0 0 0

: 1 1 1 1 0 0 0 100,000

5 0 5 1 4 0 0 100,000

0.07 : 0.28 0.07 0.28 0 0.28 0 0 0

: 0.16 0.07 0.04 0 0 1 0 0

0.12 0 0.24 0 0.28 1 0 0

0.0288 : 0.1152 0.0288 0.1152 0 0.1152 0 0 0

: 0.0172 0.0288 0.0267 0 0 0 1 0

0.1324 0 0.14

R

R

R

R

R

R

19 0 0.1152 0 1 0

1 2 3

5 0 5 1 4 0 0 100,000

4 1 4 0 4 0 0 0

0.12 0 0.24 0 0.28 1 0 0

0.1324 0 0.1419 0 0.1152 0 1 0

M G E s s s R

15

1 45 5

5 0 5 1 4 0 0 100,000

1 0 1 0 0 20,000

1 2 3

1 45 51 0 1 0 0 20,000

4 1 4 0 4 0 0 0

0.12 0 0.24 0 0.28 1 0 0

0.1324 0 0.1419 0 0.1152 0 1 0

M G E s s s R

115

1

becomes

R

R

41

Row operations are required to change the rest of the column into

zeros.

After the row operations, there are no negative numbers in the indicator

row. Covering the first, fourth and fifth columns, we can read off the

solution , , 0, 80,000, 20,000M G E and 2838R . Remember, by

covering the first column we are assuming that 0M .

The optimal solution is to invest no money in the Mid Cap Value Fund,

$80,000 in the Growth & Income Fund, and $20,000 in the International

Equity Fund. This combination of investments results in the largest

return of $2838.

Even with more variables, the Simplex Method allows the standard maximization

problem to be solved. Adding more decision variables or constraints simply adds more

columns or rows to the initial simplex tableau. Although this makes for more writing, it

does not change how the Simplex Method is applied.

1641 5 5

2

4 45 5

1

3

1

4

4 0 4 0 0 80,0004 :

4 1 4 0 4 0 0 0:

0 1 0 0 0 80,000

0.24 : 0.24 0 0.24 0.048 0.192 0 0 4800

: 0.12 0 0.24 0 0.28 1 0 0

0.12 0 0 0.048 0.088 1 0 4800

0.1419 : 0.1419 0 0.1419 0.02838 0.11352 0 0 2838

: 0.1324 0 0.1419 0 0.1152 0 1 0

R

R

R

R

R

R

0.0095 0 0 0.02838 0.00168 0 1 2838

1 3

3

0.24becomes

R R

R

1 4

4

0.1419becomes

R R

R

1 2 3

1 45 5

4 45 5

1 0 1 0 0 20,000

0 1 0 0 0 80,000

0.12 0 0 0.048 0.088 1 0 4800

0.0095 0 0 0.02838 0.00168 0 1 2838

M G E s s s R

1 2

2

4becomes

R R

R

42