Section 3.4 Simplex Method

In this section, we will only solve linear programming problems that are so called standard maxi-

mization problems.

A linear programming problem is a standard maximization problem if:

1. The variables are constrained to be nonnegative

2. All of the problem constraint ineqaulities are a nonnegative constant

3. The objective function is to be maximized

The first step in the simplex method is to convert the system of linear inequalities to a system of linear

equalities.

Example 1: Convert the inequality 3x� 5y 12 into an equality.

For the simplex method, we convert all the ineqaulities into equalities and treat the objective function

as a variable.

Simplex Method:

1. Set up the initial simplex tableau in the following manner:

Introduce a slack variable for each of the constraints and write it as an equality.

Rewrite the objective function in the form · · ·+ P = 0.

Put this system of linear equations in an augmented matrix with the objective function as

the bottom row.

Example 2: Consider the linear programming problem that is Example 1 of Section 3.3 in the

book:

Maximize P = 60x+ 80y

Subject to 2x+ 4y 80

4x+ 2y 84

2x+ 2y 50

x � 0 , y � 0

(a) Convert the linear programming problem into a system of equalities.

(b) Now write the system of equations as a simplex tableau.

2. Examine the entries in the bottom row of the augmented matrix. If no negative entries remain,

then a solution has been found. (Reading the solution will be discussed in Step 4)

3. Select and perform the pivot:

Find the most negative entry in the bottom row. This is the pivot column. If the is a tie for

the most negative entry, use any of the columns with the most negative entry.

Form the ratio q = a/b where a is the entry in the column of the matrix and b is the entry

in the pivot column above the horizontal line. Skip this ratio for negative values of b.

Find the smallest nonnegative ratio. This is the pivot row.

Pivot about the entry in the pivot column and pivot row.

Go to Step 2.

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Example 3: For the simplex tableau formed in Example 2, use pivots until there are no more

negative entries in the bottom row of the matrix.

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4. Read the solution:

Identify the nonunit columns. These nonbasic variables are set to zero.

Read each row to find the values of the basic variables.

Example 4: Read the solution from the final tableau in Example 3.

Example 5: Consider the following simplex tableau. Use the simplex method to find the next tableau.

x1 x2 s1 s2 s3 z

1 0 0 1 1 0 5

3 0 1 -2 0 0 3

7 1 0 1 0 0 6

-3 0 0 -4 2 1 10

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Example 6: Use the simplex method to solve the following linear programming problem:

Maximize R = 5x+ 6y + 4z

Subject to 0.5x+ y + 0.25z 5000

�x+ 3y + 3z 0

�2y + z 0

x � 0 , y � 0 , z � 0

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System :

.g× + y+ .

252-+5=5000

- X +3g +32-+52=0

- Zy tz 753=0

- 5x - by- 4ztR=o

Inotral Simplex Tableau :

EYESIGHT- 1 3 3 0 1 0 0 0

O- 2 I 0 O 1 0 0

÷6 - 4 0 0 0 1 / 0

- 6 is most negative so pivot on column 2 .

q , = 5000/1=5000 ; 92--0/3=0 ; skip 93 .

lfz is the smallest, so pivot on row 2

.

Pivot on row 2 column 2 :

nEe÷¥FEE!3EfEoo- 2/3 0 3 0 2/3 1 0 0

÷7 0 2 0 2 0 1 0

Example 6: Continued......

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-7 is the most negative , so pivot on column l.

q ,= 5000/15167=6000 ; skip qzwd q ,

.

9 ,is the only one , so Pivot on row I

.

Pivot on row I column I :

ITRI0 -9110615 -45 O O 6005

:.IE#siYi:/::::÷4%0445-415 0 1 42000

- 4340 is the most negative . so pivot on column 3.

Skip % ; qz= 2000/(21/0)=2000017 ; 93=500013 . 93 is the

smallest, so pivot on row 3 .

Pivot on row 3 column 3 :

!tsEEE¥f÷oo's

0 0 l' 13 46 5112 0 500°13

00059=12431241147500/3- 1112 Is the most negative , so pivot on row 5 .

skip q , ;qz= (2500/3)/111127=10000 ;qz= (5000/3)/(1/6)=10000 . qzarndlfzAre the same , so pivot on either row 2 or 3

.

Pivot on row 2 column 5 :

Example 6: Continued......

Example 7: Given the following simplex tableau, determine the maximum of the objective function

P .

x1 x2 x3 s1 s2 s3 z

1 0 0 0.4 -0.7 0 0 4

0 0 1 -0.2 0.3 -0.1 0 2

0 1 0 1.2 -0.2 0.3 0 5

0 0 0 3.4 2 9.6 1 145

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thereforeO 120 2 I -7/2 0 10000

0 -2 I 0 0 1 O 0

O I O 10 0 312 I 50000

No more negative values at the bottom

Solution ;

X=loooo;uy=O ;Z=0

Si=O ; 52=10000 ;S3=O

R= soooo

mail.IE#

Example 8: You manage an ice cream factory that makes two flavors: Vanilla and Mocha. Each quart

of Vanilla requires 2 eggs and 3 cups of cream. Each quart of Mocha requires 1 egg and 4 cups of cream.

You have in stock 500 eggs and 1200 cups of cream. You make a profit of $3 on each quart of Vanilla

and $5 on each quart of Mocha.

(a) How many quarts of each flavor should you make in order to earn the largest profit?

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(b) Are there any leftover resources? Be specific.

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