ch04_part1

Linear Programming: Simplex Method

4.1 The Simplex Tableau;Pivoting

4.2 The Simplex Method:Solving MaximumProblems in StandardForm

4.3 Solving MinimumProblems in StandardForm Using the DualityPrinciple

4.4 The Simplex Method with Mixed Constraints

• Chapter Review• Chapter Project• Mathematical Questions

from Professional Exams

OUTLINE

That hiking trip to Pinnacles isstill going to happen. But when youtold the others who are going aboutyour trail mix of unsalted peanutsand raisins, they all said “borrring.” Itwas decided to use the unsaltedpeanuts and raisins mixed withM&Ms and salted mini-pretzels. Nowit’s more complicated to determine in

what proportions these four ingredi-ents should be mixed. And thereremain the issues of carbohydrates,protein, calories, and fat. Fortunately,you also finished the simplex methodbefore midterm, so you can handlethe problem just fine. The ChapterProject at the end of the chapter willguide you.

T

4C H A P T E R

194

The Simplex Tableau; Pivoting 195

A L O O K B A C K , A L O O K F O R WA R D

Introduction

If we have a large number of either variables or constraints, it is still true that if anoptimal solution exists, it will be found at a corner point of the set of feasible points. Infact, we could find these corner points by writing all the equations corresponding tothe inequalities of the problem and then proceeding to solve all possible combinationsof these equations. We would, of course, have to discard any solutions that are not fea-sible (because they do not satisfy one or more of the constraints). Then we could eval-uate the objective function at the remaining corner points.

Just how difficult is this procedure? Well, if there were just 4 variables and 7 con-straints, we would have to solve all possible combinations of 4 equations chosen from aset of 7 equations — that would be 35 solutions in all. Each of these solutions wouldthen have to be tested for feasibility. So even for this relatively small number of vari-ables and constraints, the work would be quite tedious. In the real world of applica-tions, it is fairly common to encounter problems with hundreds, even thousands, ofvariables and constraints. Of course, such problems must be solved by computer. Evenso, choosing a more efficient problem-solving strategy might reduce the computer’srunning time from hours to seconds, or, for very large problems, from years to hours.

A more systematic approach would involve choosing a solution at one corner pointof the feasible set, then moving from there to another corner point at which the objec-tive function has a better value, and continuing in this way until the best possible valueis found. The simplex method is a very efficient and popular way of doing this.

In Chapter 3 we described a geometric method (using graphs)

for solving linear programming problems. Unfortunately, this

method is useful only when there are no more than two vari-

ables and the number of constraints is small.

Since most practical linear programming problems

involve systems of several hundred linear inequalities con-

taining several hundred variables, more sophisticated tech-

niques need to be used. One of these methods is the simplex

method, the subject of this chapter.

The simplex method is a way to solve linear program-

ming problems involving many inequalities and variables.

This method was developed by George Dantzig in 1946 and

is particularly well suited for computerization. In 1984,

Narendra Karmarkar of Bell Laboratories discovered a way

of solving large linear programming problems that

improves on the simplex method.

A discussion of LINDO, a software package that closely

mimics the simplex method, may be found in Appendix B.

PREPARING FOR THIS SECTION Before getting started, review the following:

4.1

> Row Operations (Section 2.3, pp. 67–68) > Linear programming (Section 3.2, pp. 171–178)

OBJECTIVES 1 Determine a maximum problem is in standard form

2 Set up the initial simplex tableau

3 Perform pivot operations

4 Analyze a tableau

The Simplex Tableau; Pivoting

196 Chapter 4 Linear Programming: Simplex Method

Standard Form of a Maximum Problem

A linear programming problem in which the objective function is to be maximized isreferred to as a maximum problem. Such problems are said to be in standard formprovided two conditions are met:

Standard Form of a Maximum Problem

Condition 1 All the variables are nonnegative.

Condition 2 Every other constraint is written as a linear expression that is less thanor equal to a positive constant.

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EXAMPLE 1 Determining Whether a Maximum Problem Is in Standard Form

Determine which of the following maximum problems are in standard form.*

(a) Maximize

z � 5x1 � 4x2

subject to the constraints

3x1 � 4x2 � 120

4x1 � 3x2 � 20

x1 � 0, x2 � 0

(b) Maximize

z � 8x1 � 2x2 � 3x3


4x1 + 8x2 � 120

3x2 � 4x3 � 120

x1 � 0, x2 � 0

(c) Maximize

z � 6x1 � 8x2 � x3


3x1 � x2 � 10

4x1 � x2 � 5

x1 � x2 � x3 � �3

x1 � 0, x2 � 0, x3 � 0

*Due to the nature of solving linear programming problems using the simplex method, we shall find it convenientto use subscripted variables throughout this chapter.

Determine a maximumproblem is in standardform

1


(d) Maximize

z � 8x1 � x2


3x1 � 4x2 � 2

x1 � x2 � 6

x1 � 0, x2 � 0

(a) This is a maximum problem containing two variables x1 and x2. Since both vari-ables are nonnegative and since the other constraints

are each written as linear expressions less than or equal to a positive constant, weconclude the maximum problem is in standard form.

(b) This is a maximum problem containing three variables x1, x2, and x3. Since the vari-able x3 is not given as nonnegative, the maximum problem is not in standard form.

(c) This is a maximum problem containing three variables x1, x2, and x3. Each variableis nonnegative. The set of constraints

3x1 � x2 � 10

4x1 � x2 � 5

x1 � x2 � x3 � �3

contains x1 � x2 � x3 � �3, which is not a linear expression that is less than orequal to a positive constant. The maximum problem is not in standard form.Notice, however, that by multiplying this constraint by �1, we get

�x1 � x2 � x3 � 3

which is in the desired form. Although the maximum problem as stated is notin standard form, it can be modified to conform to the requirements of the stan-dard form.

(d) The maximum problem contains two variables x1 and x2, each of which is nonneg-ative. Of the other constraints, the first one, 3x1 � 4x2 � 2 does not conform. Themaximum problem is not in standard form. Notice that we cannot modify thisproblem to place it in standard form. Even though multiplying by �1 will changethe � to �, in so doing the 2 will change to �2. ◗

NOW WORK PROBLEMS 1 AND 11.

Slack Variables and the Simplex Tableau

In order to apply the simplex method to a maximum problem, we need to first

1. Introduce slack variables.2. Construct the initial simplex tableau.

3x1 � 4x2 � 120

Lessthan orequal

PositiveLinearexpressions

4x1 � 3x2 � 20

SOLUTION

Set up the initial simplextableau

2


We will show how these steps are done by working with a specific maximum problemin standard form. (This problem is the same as Example 1, Section 3.3, page xxx.)

Maximize

P � 3x1 � 4x2


2x1 � 4x2 � 120

2x1 � 2x2 � 80

x1 � 0, x2 � 0

First, observe that this maximum problem is in standard form.Next, recall that when we say that 2x1 � 4x2 � 120, we mean that there is a number

greater than or equal to 0, which we designate by s1, such that

2x1 � 4x2 � s1 � 120, s1 � 0

This number s1 is a variable. It must be nonnegative since it is the difference between120 and a number that is less than or equal to 120. We call it a slack variable since it“takes up the slack” between the left and right sides of the inequality.

Similarly, for the constraint 2x1 � 2x2 � 80, we introduce the slack variable s2:

2x1 � 2x2 � s2 � 80, s2 � 0

Finally, we write the objective function P � 3x1 � 4x2 as

P � 3x1 � 4x2 � 0

In effect, we have now replaced our original system of constraints and the objectivefunction by a system of three equations containing five variables, P, x1, x2, s1, and s2:

2x1 � 4x2 � s1 � 120 �2x1 � 2x2 � s2 � 80Constraints

P � 3x1 � 4x2 � 0 Objective function

where x1 � 0 x2 � 0 s1 � 0 s2 � 0

To solve the maximum problem is to find the particular solution (P, x1, x2, s1, s2) thatgives the largest possible value for P. The augmented matrix for this system is givenbelow:

If we write the augmented matrix in the form given next, we have the initial sim-plex tableau for the maximum problem:

(1)

The bottom row of the initial simplex tableau represents the objective function andis called the objective row. The rows above it represent the constraints. We separate the

P

00

1

x1

22

�3

BV

s1

s2

P

x2

42

�4

s1

10

0

s2

01

0

12080

0

RHS

P

00

1

x1

22

�3

x2

42

�4

s1

10

0

s2

01

0

12080

0


objective row from these rows with a horizontal rule. Notice that we have written thesymbol for each variable above the column in which its coefficients appear. The nota-tion BV stands for basic variables. These are the variables that have a coefficient of 1and 0 elsewhere in their column. The notation RHS stands for right-hand side, that is,the numbers to the right of the equal sign in each equation.

So far, we have seen this much of the simplex method:

For a maximum problem in standard form:

1. The constraints are changed from inequalities to equations by the introductionof additional variables — one for each constraint and all nonnegative — calledslack variables.

2. These equations, together with one that describes the objective function, areplaced in the initial simplex tableau.

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EXAMPLE 2 Setting Up the Initial Simplex Tableau

The following maximum problems are in standard form. For each one introduce slackvariables and set up the initial simplex tableau.

(a) Maximize

P � 3x1 � 2x2 � x3


3x1 � x2 � x3 � 30

5x1 � 2x2 � x3 � 24

x1 � x2 � 4x3 � 20

x1 � 0 x2 � 0 x3 � 0

(b) Maximize

P � x1 � 4x2 � 3x3 � x4


2x1 � x2 � 10

3x1 � x2 � x3 � 2x4 � 18

x1 � x2 � x3 � x4 � 14

x1 � 0 x2 � 0 x3 � 0 x4 � 0

(a) We write the objective function in the form

P � 3x1 � 2x2 � x3 � 0

For each constraint we introduce a nonnegative slack variable to obtain the follow-ing system of equations:

SOLUTION


3x1 � x2 � x3 � s1 � 30

5x1 � 2x2 � x3 � s2 � 24

x1 � x2 � 4x3 � s3 � 20

where x1 � 0 x2 � 0 x3 � 0

s1 � 0 s2 � 0 s3 � 0

These equations, together with the objective function P, give the initial simplextableau:

(b) We write the objective function in the form

P � x1 � 4x2 � 3x3 � x4 � 0

For each constraint we introduce a nonnegative slack variable to obtain the systemof equations

2x1 � x2 � s1 � 10

3x1 � x2 � x3 � 2x4 � s2 � 18

where x1 � x2 � x3 � x4 � s3 � 14

x1 � 0 x2 � 0 x3 � 0 x4 � 0

s1 � 0 s2 � 0 s3 � 0

These equations, together with the objective function P, give the initial simplextableau:

Notice that in each initial simplex tableaux an identity matrix appears under thecolumns headed by P and the slack variables. Notice too that the right-hand column(RHS) always contains nonnegative constants.

NOW WORK PROBLEM 17.

The Pivot Operation

Before going any further in our discussion of the simplex method, we need to discussthe matrix operation known as pivoting. The first thing one does in a pivot operation isto choose a pivot element. However, for now the pivot element will be specified inadvance. The method of selecting pivot elements in the simplex tableau will be shownin the next section.

P

000

1

x1

231

�1

BV

s1

s2

s3

P

x2

111

�4

x3

011

�3

x4

021

�1

s1

100

0

s2

010

0

s3

001

0

101814

0

RHS

P

000

1

x1

351

�3

BV

s1

s2

s3

P

x2

121

�2

x3

114

�1

s1

100

0

s2

010

0

s3

001

0

302420

0

RHS

◗

Perform pivot operations 3


The steps for pivoting utilize two variations of the three row operations for matrices,namely:

Pivoting

To pivot a matrix about a given element, called the pivot element, is to applycertain row operations so that the pivot element is replaced by a 1 and all otherentries in the same column, called the pivot column, become 0s.

�

Steps for Pivoting

STEP 1 In the pivot row where the pivot element appears, divide each entry by thepivot element, which we assume is not 0. This causes the pivot element tobecome 1.

STEP 2 Obtain 0s elsewhere in the pivot column by performing row operationsusing the revised pivot row.

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Row Operations Used in Pivoting

STEP 1 Replace the pivot row by a positive multiple of that same row.STEP 2 Replace a row by the sum of that row and a multiple of the pivot row.

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Warning! Step 2 requires row operations that must involve the pivot row.We continue with the initial simplex tableau given in Display (1), page 198, to illus-

trate the pivot operation.

EXAMPLE 3 Performing a Pivot Operation

Perform a pivot operation on the initial simplex tableau given in Display (1) andrepeated below in Display (2), where the pivot element is circled, and the pivot row andpivot column are marked by arrows:

(2)

In this tableau the pivot column is column x2 and the pivot row is row s1. Step 1 of thepivoting procedure tells us to divide the pivot row by 4, so we use the row operation

R1 �14

r1

P

00

1

x1

22

�3

BV

s1

s2

P

x2

42

�4

s1

10

0

s2

01

0

12080

0

RHS

SOLUTION


For Step 2 the pivot row is row 1. To obtain 0s elsewhere in the pivot column, wemultiply row 1 by �2 and add it to row 2; then we multiply row 1 by 4 and add it torow 3. The row operations specified are

R2 � �2r1 � r2 R3 � 4r1 � r3

The new tableau looks like this:

(3)

This completes the pivot operation since the pivot column (column x2) has a 1 inthe pivot row x2 and has 0s everywhere else. ◗

This process should look familiar. It is similar to the one used in Chapter 2 to obtainthe reduced row-echelon form of a matrix.

Analyzing a Tableau

Just what has the pivot operation done? To see, look again at the initial simplex tableau, inDisplay (2). Observe that the entries in columns P, s1, and s2 form an identity matrix (I3, tobe exact). This makes it easy to solve for P, s1, and s2 using the other variables as parameters:

2x1 � 4x2 � s1 � 120 or s1 � �2x1 � 4x2 � 120

2x1 � 2x2 � s2 � 80 or s2 � �2x1 � 2x2 � 80

P � 3x1 � 4x2 � 0 or P � 3x1 � 4x2

The variables s1, s2, and P are the original basic variables (BV) listed in the tableau.After pivoting, we obtain the tableau given in Display (3). Notice that in this form, it

is easy to solve for P, x2, and s2 in terms of x1 and s1.

x2 � � x1 � s1 � 30

s2 � �x1 � s1 � 20 (4)

P � x1 � s1 � 120

The variables x2, s2, and P are the new basic variables of the tableau. The variables x1

and s1 are the nonbasic variables. The result of pivoting is that x2 becomes a basic vari-able, while s1 becomes a nonbasic variable.

Notice in Equations (4) that if we let the value of the nonbasic variables x1 and s1 equal0, then the basic variables P, x2, and s2 equal the entries across from them in the right-handside (RHS) of the tableau in Display (3). For the tableau in Display (3) the current value ofthe objective function is P � 120, obtained for x1 � 0, s1 � 0. The values of x2 and s2 arex2 � 30, s2 � 20. Because P � x1 � s1 � 120, x1 � 0, and s1 � 0, the value of P can beincreased beyond 120 when x1 � 0 and s1 � 0. So we have not maximized P yet.

12

14

12

BV

x2

s2

P

�

0

0

1

1

�1

1

0

0

0

1

0

P x1 x2 s1 s2

30

20

120

RHS12

14

1

12

0

0

1

2

�3

1

2

�4

0

0

0

1

0

P x1 x2 s1 s2

30

80

0

RHS12

14

Analyze a tableau 4


We summarize this discussion below.

Analyzing a Tableau

To obtain the current values of the objective function and the basic variables in atableau, follow these steps:

STEP 1 From the tableau, write the equation corresponding to each row.STEP 2 Solve the bottom equation for P and the remaining equations for

the basic variables.STEP 3 Set each nonbasic variable equal to zero to obtain the current values

of P and the basic variables.

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EXAMPLE 4 Performing a Pivot Operation; Analyzing a Tableau

Perform another pivot operation on the tableau given in Display (3). Use the circledpivot element in the tableau below. Then analyze the new tableau.

Since the pivot element happens to be a 1 in this case, we skip Step 1. For Step 2 weperform the row operations

R1 � � r2 � r1 R3 � r2 � r3

The result is

In the tableau given in Display (5), the new basic variables are x2, x1, and P. Thevariables s1 and s2 are the nonbasic variables. The result of pivoting caused x1 tobecome a basic variable and s2 to become a nonbasic variable. Finally, the equationsrepresented by Display (5) can be written as

P � �12

s1 � s2 � 140

x1 � 12

s1 � s2 � 20

x2 � �12

s1 �12

s2 � 20

BV

x2

x1

P

�

�0

0

1

0

1

�1

1

0

0

1

1

P x1 x2 s1 s2

20

20

140

RHS12

1212

12

12

BV

x2

s2

P

�

0

0

1

1

�1

1

0

0

0

1

0

P x1 x2 s1 s2

30

20

120

RHS12

14

1

12

SOLUTION

(5)


If we let the nonbasic variables s1 and s2 equal 0, then the current values of the basicvariables are P � 140, x2 � 20, and x1 � 20. The second pivot has improved the valueof P from 120 to 140.

Because P � � s1 � s2 � 140 and s1 � 0 and s2 � 0, the value of P cannot increasebeyond 140 (any values of s1 and s2, other than 0, reduce the value of P). So we havemaximized P. ◗

COMMENT: The pivot operation can be done using a graphing utility such as the TI83Plus. The augmented matrix contained in the initial tableau is entered into the calcu-lator. Once the pivot element has been identified, the elementary row operations builtinto the graphing utility are used to obtain a new tableau with a pivot element of oneand all other elements in the pivot column replaced with zeros. This is best seen usingan example. ◗

12

EXAMPLE 5 Performing a Pivot Operation Using a Graphing Utility

The following maximum problem is in standard form,

MaximizeP � 2x1 � 3x2


3x1 � 2x2 � 200

x1 � 3x2 � 150

x1 � 0 x2 � 0

By introducing nonnegative slack variables s1 and s2, and by rewriting the objectivefunction in the form

P � 2x1 � 3x2 � 0

we obtain the system of equations

3x1 � 2x2 � s1 � 200

x1 � 3x2 � s2 � 150

P � 2x1 � 3x2 � 0

where x1 � 0 x2 � 0 s1 � 0 s2 � 0

The initial simplex tableau is

(6)

The pivot element 3 is circled and the pivot row and pivot column are marked with anarrow.

At this point we enter the initial tableau (6) into the graphing utility. See Figure 1.The matrix has 3 rows and 6 columns and has been named matrix [A]. Using the pivotelement 3 as indicated above, we follow the steps below using the elementary row oper-ations in the calculator to obtain the new tableau.

P

00

1

x1

31

�2

BV

s1

s2

s2

x2

23

�3

s1

10

0

s2

01

0

200150

0

RHS

SOLUTION


STEP 1 Make the pivot element one by dividing the pivot row by 3. On a TI 83Plus, theoperation is:

matrix math E: *row( (hit enter)

*row(1/3, [A], 2)

STEP 2 Obtain 0s elsewhere in the pivot column by performing row operations usingthe revised pivot row from Step 1. To obtain 0s elsewhere in the pivot column,we multiply row 2 by �2 and add it to row 1, then we multiply row 2 by 3 andadd it to row 3. On a TI 83Plus, the operations are:

matrix math F: *row�( (hit enter)

*row�(�2, ANS, 2, 1)

matrix math F: *row�( (hit enter)

*row�(3, ANS, 2, 3)

Figure 2 shows the result. In Figure 2 the pivot column has a 1 in the pivot row and 0severywhere else, as the pivot operation is complete.

FIGURE 1 The initial tableau.

FIGURE 2 The simplex tableauafter one complete iteration.

◗

EXERCISE 4.1 Answers to Odd-Numbered Problems Begin on Page AN-00.

In Problems 1–10, determine which maximum problems are in standard form. Do not attempt tosolve them!

1. Maximize

P � 2x1 � x2


x 1 � x 2 � 52x 1 � 3x 2 � 2

x 1 � 0 x 2 � 0

2. Maximize

P � 3x1 � 4x2


3x 1 � x 2 � 6x 1 � 4x 2 � 74

x 1 � 0 x 2 � 0

3. Maximize

P � 3x1 � x2 � x3


x 1 � x 2 � x 3 � 62x 1 � 3x 2 � 4x 3 � 10

x 1 � 0

4. Maximize

P � 2x1 � x2 � 4x3


2x 1 � x 2 � x 3 � 10x 2 � 0

5. Maximize

P � 3x1 � x2 � x3


x 1 � x 2 � x 3 � 82x 1 � x 2 � 4x 3 � 6

x 1 � 0 x 2 � 0

6. Maximize

P � 2x1 � x2 � 4x3


2x 1 � x 2 � x 3 � �1x 1 � 0 x 2 � 0

7. Maximize

P � 2x1 � x2


x 1 � x 2 � �62x 1 � x 2 � 4

x 1 � 0 x 2 � 0

8. Maximize

P � 3x1 � x2


x 1 � 3x 2 � 42x 1 � x 2 � 1

x 1 � 0 x 2 � 0


9. Maximize

P � 2x1 � x2 � 3x3


10. Maximize

P � 2x1 � 2x2 � 3x3


x 1 � x 2 � x 3 � 6x 1 � 4

x 1 � 0 x 2 � 0 x 3 � 0

x 1 � x 2 � x 3 � 10x 2 � x 3 � 4

x 1 � 0 x 2 � 0 x 3 � 0

In Problems 11 – 16, each maximum problem is not in standard form. Determine if the problem canbe modified so as to be in standard form. If it can, write the modified version.

11. Maximize

P � x1 � x2


12. Maximize

P � 2x1 � 3x2


13. Maximize

P � x1 � x2 � x3


x 1 � x 2 � x 3 � 64x 1 � 3x 2 � 12

x 1 � 0 x 2 � 0 x 3 � 0

�4x 1 � 2x 2 � �8x 1 � x 2 � 6

x 1 � 0 x 2 � 0

3x 1 � 4x 2 � �6x 1 � x 2 � 4

x 1 � 0 x 2 � 0

14. Maximize

P � 2x1 � x2 � 3x3


15. Maximize

P � 2x1 � x2 � 3x3


16. Maximize

P � x1 � x2 � x3


2x 1 � x 2 � 3x 3 � 8x 1 � x 2 � 6

x 3 � 4x 1 � 0 x 2 � 0 x 3 � 0

�x 1 � x 2 � x 3 � �62x 1 � 3x 2 � �12

x 3 � 2x 1 � 0 x 2 � 0 x 3 � 0

x 1 � x 2 � x 3 � �8x 1 � x 2 � �6

x 1 � 0 x 2 � 0 x 3 � 0

In Problems 17 – 24, each maximum problem is in standard form. For each one introduce slack vari-ables and set up the initial simplex tableau.

17. Maximize

P � 2x1 � x2 � 3x3


18. Maximize

P � 3x1 � 2x2 � x3


19. Maximize

P � 3x1 � 5x2


2.2x 1 � 1.8x 2 � 50.8x 1 � 1.2x 2 � 2.5

x 1 � x 2 � 0.1x 1 � 0 x 2 � 0

3x 1 � 2x 2 � x 3 � 10x 1 � x 2 � 3x 3 � 12

2x 1 � x 2 � x 3 � 6x 1 � 0 x 2 � 0 x 3 � 0

5x 1 � 2x 2 � x 3 � 206x 1 � x 2 � 4x 3 � 24

x 1 � x 2 � 4x 3 � 16x 1 � 0 x 2 � 0 x 3 � 0

20. Maximize

P � 2x1 � 3x2


21. Maximize

P � 2x1 � 3x2 � x3


22. Maximize

P � x1 � 4x2 � 2x3


3x 1 � x 2 � x 3 � 10x 1 � x 2 � 3x 3 � 5

x 1 � 0 x 2 � 0 x 3 � 0

x 1 � x 2 � x 3 � 503x 1 � 2x 2 � x 3 � 10

x 1 � 0 x 2 � 0 x 3 � 0

1.2x 1 � 2.1x 2 � 0.50.3x 1 � 0.4x 2 � 1.5

x 1 � x 2 � 0.7x 1 � 0 x 2 � 0


23. Maximize

P � 3x1 � 4x2 � 2x3


24. Maximize

P � 2x1 � x2 � 3x3


2x 1 � x 2 � x 3 � 2x 1 � x 2 � 4

2x 1 � x 2 � x 3 � 5x 1 � 0 x 2 � 0 x 3 � 0

3x 1 � x 2 � 4x 3 � 5x 1 � x 2 � 5

2x 1 � x 2 � x 3 � 6x 1 � 0 x 2 � 0 x 3 � 0

In Problems 25 – 28, each maximum problem can be modified so as to be in standard form. Write themodified version and, for each one, introduce slack variables and set up the initial simplex tableau.

25. Maximize

P � x1 � 2x2 � 5x3


26. Maximize

P � 2x1 � 4x2 � x3


�2x 1 � 3x 2 � x 3 � �8�3x 1 � x 2 � 2x 3 � �12

2x 1 � x 2 � x 3 � 10x 1 � 0 x 2 � 0 x 3 � 0

�x 1 � 2x 2 � 3x 3 � �103x 1 � x 2 � x 3 � �12

x 1 � 0 x 2 � 0 x 3 � 0

27. Maximize

P � 2x1 � 3x2 � x3 � 6x4


28. Maximize

P � x1 � 5x2 � 3x3 � 6x4


x 1 � x 2 � 2x 3 � 2x 4 � �8�x 1 � x 2 � x 3 � x 4 � �10

x 1 � x 2 � x 3 � x 4 � 12x 1 � 0 x 2 � 0 x 3 � 0 x 4 � 0

�x 1 � x 2 � 2x 3 � x 4 � 10x 1 � x 2 � x 3 � x 4 � �8x 1 � x 2 � x 3 � x 4 � 9

x 1 � 0 x 2 � 0 x 3 � 0 x 4 � 0

In Problems 29 – 33, perform a pivot operation on each tableau. The pivot element is circled. Usingthe new tableau obtained, write the corresponding system of equations. Indicate the current values ofthe objective function and the basic variables.

29. 30. P

00

1

x1

12

�2

BV

s1

s2

P

x2

45

�1

s1

10

0

s2

01

0

10050

0

RHSP

00

1

x1

13

�1

BV

s1

s2

P

x2

22

�2

s1

10

0

s2

01

0

300480

0

RHS

31. 32. P

000

1

x1

121

�1

BV

s1

s2

s3

P

x2

23

�2

�2

x3

113

�3

s1

100

0

s2

010

0

s3

001

0

612

0

0

RHSP

000

1

x1

123

�1

BV

s1

s2

s3

P

x2

2�1

2

�2

x3

414

�3

s1

100

0

s2

010

0

s3

001

0

243218

0

RHS

33. P

0000

1

x1

�3200

�1

BV

s1

s2

s3

s4

P

x2

00

�3�3

�2

x3

1010

�3

x4

0101

�4

s1

1000

0

s2

0100

0

s3

0010

0

s4

0001

0

20242824

0

RHS


4.2

OBJECTIVES 1 Solve maximum problems in standard form using the simplex method

2 Determine if a tableau is final, requires additional pivoting, or indicates no solution

The Simplex Method: Solving Maximum Problems in Standard Form

We are now ready to state the details of the simplex method for solving a maximumproblem. This method requires that the problem be in standard form and that theproblem be placed in an initial simplex tableau with slack variables.

Solve maximum problems in standard form using the simplex method

1

Let’s go through the process.

The Simplex Method for Solving a Maximum Problem in Standard Form

STEP 1

STEP 2

STEP 3

STEP 4

STEP 5

STEP 6

Begin with the initial simplex tableau of a maximum problem in standard form.

The pivot element is the entry in the pivot row and pivot column. (Note that the pivot element can never be in the objective row.) Pivot and repeat the process from Step 2 until a STOP is obtained.

If it is nonnegative, STOP.A solution has been found.

If all the entries in thepivot column are zero or

negative, STOP. The problem isunbounded and has no solution.

The pivot row is the onehaving the smallest

nonnegative quotient.(In case of ties, choose any

one of the tied entries.)

For each positive entry above the objective row in the

pivot column, compute thevalue of the entry in the far

right column (RHS) divided by the corresponding

positive entry of the pivot column. (Note that the objective row is ignored)

If it is negative, the pivotcolumn is found. (In case ofties, choose any one of the

tied entries.)

Identify thesmallest entry in

the objective row.

�

The Simplex Method: Solving Maximum Problems in Standard Form 209

EXAMPLE 1 Using the Simplex Method

Maximize

P � 3x1 � 4x2


STEP 1 This is a maximum problem in standard form. To obtain the initial simplextableau, we proceed as follows: The objective function is written in the form

P � 3x1 � 4x2 � 0

After introducing slack variables s1 and s2, the constraints take the form

2x1 � 4x2 � s1 � 120

2x1 � 2x2 � s2 � 80

where x1 � 0 x2 � 0

s1 � 0 s2 � 0


STEP 2 The smallest entry in the objective row is �4.

STEP 3 Since �4 is negative, the pivot column is column x2.

STEP 4 For each positive entry above the objective row in the pivot column, form thequotient of the corresponding RHS entry divided by the positive entry.

BV Positive entry, x2 RHS Quotient

s1 4 120 120 � 4 � 30s2 2 80 80 � 2 � 40

STEP 5 The smallest nonnegative value is 30, so the pivot row is row s1. The tableaubelow shows the pivot element (circled) and the current values.

(1)

P

00

1

x1

22

�3

BV

s1

s2

P

x2

42

�4

s1

10

0

s2

01

0

12080

0

RHS

s1

s2

P

�

�

�

120

80

0

Current values

s1

10

0

P

00

1

x1

22

�3

BV

s1

s2

P

x2

42

�4

s2

01

0

12080

0

RHS

2x 1 � 4x 2 � 120

2x 1 � 2x 2 � 80

x 1 � 0 x 2 � 0

SOLUTION


STEP 6 The pivot element is 4. Next, we pivot by using the row operations:

1. R1 � r1

2. R2 � �2r1 � r2, R3 � 4r1 � r3

After pivoting, we have this tableau:

(2)

Notice that x2 is now a basic variable, replacing s1. The objective row is

P � x1 � s1 � 120

P has a current value of 120. Because P � x1 � s1 � 120 and x1 � 0 and s1 � 0,the value of P can be increased beyond 120, when x1 � 0 and s1 � 0. So wehave not maximized P yet.

We continue with the simplex method at Step 2.STEP 2 The smallest entry in the objective row is �1.STEP 3 Since �1 is negative, the pivot column is column x1.STEP 4 For each positive entry above the objective row in the pivot column, form the

quotient of the corresponding RHS entry by the positive entry.

BV Positive entry, x2 RHS Quotient

x2 30 30 � � 60s2 1 20 20 � 1 � 20

STEP 5 The smallest nonnegative value is 20, so the pivot row is row s2. The tableaubelow shows the pivot element (circled).

STEP 6 The pivot element is 1. Next, we pivot by using the row operations:

R1 � � r2 � r1 R3 � r2 � r3

The result is this tableau:

We continue with the simplex method at Step 2.

x2

x1

P

�

�

�

20

20

140

Current valuesBV

x2

x1

P

�

�0

0

1

0

1

�1

1

0

0

1

1

P x1 x2 s1 s2

20

20

140

RHS12

1212

12

12

BV

x2

s2

P

�

0

0

1

1

�1

1

0

0

0

1

0

P x1 x2 s1 s2

30

20

120

RHS12

14

1

12

x2

s2

P

�

�

�

30

20

120

Current values

12

12

x2

s2

P

�

�

�

30

20

120

Current valuesBV

x2

s2

P

�

0

0

1

1

�1

1

0

0

0

1

0

P x1 x2 s1 s2

30

20

120

RHS12

14

1

12

14

(3)


STEP 2 The smallest entry in the objective row is 0.STEP 3 Since it is nonnegative, we STOP. To see why we STOP, write the equation

from the objective row, namely,

P � � s1 � s2 � 140

Since s1 � 0 and s2 � 0, any positive value of s1 or s2 would make the value ofP smaller than 140. By choosing s1 � 0 and s2 � 0, we obtain the largest possi-ble value for P, namely, 140. If we write the equations from the second andthird rows, substituting 0 for s1 and s2, we have

so x2 � 20 if s1 � 0, s2 � 0

so x1 � 20 if s1 � 0, s2 � 0

The maximum value of P is

P � 140and it occurs at

x1 � 20 x2 � 20 s1 � 0 s2 � 0 ◗

The tableau in Display (3) is called the final tableau because with this tableauthe maximum value is found.

To summarize, each tableau obtained by applying the simplex method provides infor-mation about the current status of the solution:

1. The RHS entry in the objective row gives the current value of the objective function.2. The remaining entries in the RHS column give the current values of the corre-

sponding basic variables.

We can trace what happened each time we pivoted. The maximum problem justsolved is the same as one we solved using geometrical methods in Section 3.3. Refer toExample 1, page 180, and look at Figure 21, the graph of the set of feasible solutions.Now look at Display (1), where the current values are s1 � 120, s2 � 80, and P � 0. Thevalues of x1 and x2 are each 0. So, we are at the origin of the set of feasible solutions.After the first pivot, the current values are x2 � 30, s2 � 20, and P � 120 [see Display(2)]. The values of the remaining variables are x1 � 0 and s1 � 0. Now we are at thepoint (x1, x2) � (0, 30) of the set of feasible solutions. After the next pivot, the currentvalues are x2 � 20, x1 � 20, and P � 140. See Display (3). The remaining variables havethe values s1 � 0 and s2 � 0. Now we are at the point (20, 20) of the set of feasible solu-tions. So the pivot process moves us from one corner point to another corner point,until we reach the corner point that maximizes P.

Next, we discuss the analysis of a simplex tableau in more detail.

x1 �12

s1 � s2 � 20

x2 � �12

s1 �12

s2 � 20

12

SUMMARY

EXAMPLE 2 Analyzing Tableaus

Determine whether each tableau

(1) is a final tableau. [If it is, give the solution.](2) requires additional pivoting. [If so, identify the pivot element.](3) indicates no solution.

2


(a)

(b)

(c)

(a) The smallest entry in the objective row is �2, which is negative. The pivot columnis column x2. Since all the entries in the pivot column are zero or negative, theproblem is unbounded and has no solution.

(b) The objective row contains no negative entries, so this is a final tableau. Thesolution is

P � 220 when x1 � 40 x2 � 20 s1 � 0 s2 � 0

(c) The smallest entry in the objective row is �4, which is negative. The pivot columnis column x1. Since both entries in the pivot column are positive, compute the quo-tients of the RHS divided by the corresponding entries in the pivot column

60 � 1 � 60 125 � 3 � 41

Since 41 60, x2 is the pivot row and 3 is the pivot element. ◗

NOW WORK PROBLEM 1.

23

23

P

00

1

x1

10

�4

BV

x1

x2

P

x2

01

0

s1

10

0

s2

0.5 0.25

2

60125

10

RHS

P

00

1

x1

10

0

BV

x1

x2

P

x2

01

0

s1

2�1

1

s2

�11

2

4020

220

RHS

P

00

1

x1

01

0

BV

s1

x1

P

x2

0�1

�2

s1

10

0

s2

11

1

4020

20

RHS

SOLUTION

SOLUTION

EXAMPLE 3 Solving a Maximum Problem Using the Simplex Method

Maximize

P � 6x1 � 8x2 � x3


3x1 � 5x2 � 3x3 � 20

x1 � 3x2 � 2x3 � 9

6x1 � 2x2 � 5x3 � 30

x1 � 0 x2 � 0 x3 � 0

Note that the problem is in standard form. By introducing slack variables s1, s2, and s3,the constraints take the form


3x1 � 5x2 � 3x3 � s1 � 20

x1 � 3x2 � 2x3 � s2 � 9

6x1 � 2x2 � 5x3 � s3 � 30

where x1 � 0 x2 � 0 x3 � 0

s1 � 0 s2 � 0 s3 � 0

SinceP � 6x1 � 8x2 � x3 � 0

the initial simplex tableau is

The pivot column is found by locating the column containing the smallest entry inthe objective row (�8 in column x2). The pivot row is obtained by dividing each entryin the RHS column by the corresponding entry in the pivot column and selecting thesmallest nonnegative quotient. The pivot row is row s2. The pivot element is 3, which iscircled. After pivoting, the new tableau is

s1

s2

s3

P

�

�

�

�

20

9

30

0

�

�

�

4

3

15

20

9

30

�

�

�

5

3

2

Current values RHS � x2

P

000

1

x1

316

�6

BV

s1

s2

s3

P

x2

532

�8

x3

325

�1

s1

100

0

s2

010

0

s3

001

0

209

30

0

RHS

�

�

�

3.75

9

4.5

s1

x2

s3

P

�

�

�

�

5

3

24

24

5

3

24

�

�

�

Current values RHS � x1 BV

s1

x2

s3

P

��0

0

0

1

0

1

0

0

1

0

0

0

P x1 x2

0

0

1

0

s3x3 s1

5

3

24

24

RHS1213

4313

�

�

s2

531323

163

4

3

1

3

16

3

103

113

133

83

The value of P has improved to 24, but the negative entry, , in the objective rowindicates further improvement is possible. We determine the next pivot element to be .After pivoting, we obtain the tableau

43

�103

�

�

s1

x2

s3

P

�

�

�

�

4 4 6

�

�

Current values RHS � s2 BV

x1

x2

s3

P

��

�

0

0

0

1

0

1

0

0

1

0

0

0

P x1 x2

0

0

1

0

5 �4 6

s3x3 s1

4

RHS1434

3414

�

s2

5434

3

4

7

4

7

3

2

3

7

4

15

4

73

2

74

732

154

32

52

72

The value of P has improved to , but, since we still observe a negative entry in theobjective row, we pivot again. (Remember, in finding the pivot row, we ignore theobjective row and any rows in which the pivot column contains a negative number orzero— in this case, row x1, containing � , is ignored.) The new tableau is 5

4

732


This is a final tableau since all the entries in the objective row are nonnegative. Theobjective (bottom) row yields the equation

The maximum value of P is P = , obtained when x3 � 0, s1 = 0, and s3 = 0. From therows x1, x2, and s2 of the final tableau we have the equations

Using x3 � 0, s1 � 0, and s3 � 0 in these equations, we find

(4)

The solution of the maximum problem is P � , obtained when x1 � , x2 � , and x3 � 0.

Note: this solution may also be found by looking at the current values of the finaltableau. ◗

NOW WORK PROBLEM 9.

54

5512

752

x 1 �5512

x 2 �54 s2 �

23

s2 � �56

x 3 �23

s1 �16

s3 �23

x 2 � �18

x 3 �14

s1 �18

s3 �54

x 1 � �1924

x 3 �1

12 s1 �

524

s3 �5512

752

P � �194

x 3 �32

s1 �14

s3 �752

s1

x2

s3

P

�

�

�

�

Current valuesBV

x1

x2

s2

P

�

�

�

0

0

0

1

0

1

0

0

1

0

0

0

P x1 x2

0

0

1

0

s3x3 s1 RHS

18

1816

23

56

23

14

s2

5

4

2

3

55

12

75

2

54

752

194

5512

524

112

1924

14

32

EXAMPLE 4 Solving a Maximum Problem Using the Simplex Method

MaximizeP � 4x1 � 4x2 � 3x3


x1 � 0 x2 � 0 x3 � 0

The problem is in standard form. We introduce slack variables s1, s2, s3, and write theconstraints as

where

2x1 � 3x2 � x3 � s1 � 6

x1 � 2x2 � 3x3 � s2 � 6

x1 � x2 � x3 � s3 � 5

x1 � 0 x2 � 0 x3 � 0

s1 � 0 s2 � 0 s3 � 0

2x 1 � 3x 2 � x 3 � 6

x 1 � 2x 2 � 3x 3 � 6

x 1 � x 2 � x 3 � 5

SOLUTION


Since P � 4x1 � 4x2 � 3x3 � 0, the initial simplex tableau is

Since �4 is the smallest negative entry in the objective row, we have a tie for the pivotcolumn between the columns x1 and x2. We choose (arbitrarily) as pivot column the col-umn x1. The pivot row is row s1. (Do you see why?) The pivot element is 2, which iscircled. After pivoting, we obtain the following tableau:

The pivot element is , which is circled. After pivoting, we obtain

This is a final tableau. The solution is P � , obtained when x1 � , x2 � 0, x3 � . ◗

NOW WORK EXAMPLE 4 AGAIN, THIS TIME CHOOSING COLUMN X2 AS THE PIVOT

COLUMN FOR THE FIRST PIVOT.

65

125

665

x1

x3

s3

P

�

�

�

�

Current valuesBV

x1

s2

s3

P

�

0

0

0

1

1

0

0

0

P x1 x2

0

0

1

0

0

1

0

0

s3x3 s1 RHS1525

3515

75

65

1255

6655

115

s2

� ��

�

15

25

95

25

751535

665

1256575

52

�

�

�

3

6

5

6

6

5

�

�

�

2

1

1

P

000

1

x1

211

�4

BV

s1

s2

s3

P

x2

321

�4

x3

131

�3

s2

010

0

s3

001

0

665

0

RHSs1

100

0

RHS � x1

EXAMPLE 5 Maximizing Profit

Mike’s Famous Toy Trucks specializes in making four kinds of toy trucks: a deliverytruck, a dump truck, a garbage truck, and a gasoline truck. Three machines — a metalcasting machine, a paint spray machine, and a packaging machine — are used in theproduction of these trucks. The time, in hours, each machine works to make each typeof truck and the profit for each truck are given in Table 1. The maximum time availableper week for each machine is given as: metal casting, 4000 hours; paint spray, 1800hours; and packaging, 1000 hours. How many of each type truck should be producedto maximize profit? Assume that every truck made is sold.

Let x1, x2, x3, and x4 denote the number of delivery trucks, dump trucks, garbage trucks,and gasoline trucks, respectively, to be made. If P denotes the profit to be maximized,we have this problem:

SOLUTION

�

�

�

x1

s2

s3

P

�

�

�

�

3

3

2

12

3

3

2

6

4

�

�

�

Current values RHS � x3 BV

x1

s2

s3

P

�

0

0

0

1 2 2 2

1

0

0

0

P x1 x2

0

0

1

0

3

3

2

12

0

1

0

0

s3x3 s1 RHS1252

1212

12

52

65

12

s2

��12

12

321212


Maximize

P � 0.5x1 � x2 � 1.5x3 � 2x4


x1 � 0 x2 � 0 x3 � 0 x4 � 0

Since this problem is in standard form, we introduce slack variables s1, s2, and s3, writethe initial simplex tableau, and solve:

This is a final tableau. The maximum profit is P � $1900, and it is attained for

x1 � 0 x2 � 0 x3 � 200 x4 � 800 ◗

The practical considerations of the situation described in Example 5 are that deliv-ery trucks and dump trucks are too costly to produce or too little profit is being gainedfrom their sale. Since the slack variable s1 has a value of 2000 for maximum P and, sinces1 represents the number of hours the metal casting machine is idle, it may be possibleto release this machine for other duties. Also note that both the paint spray and pack-aging machines are operating at full capacity. This means that to increase productivity,more paint spray and packaging capacity is required.


s1

x4

x3

P

�

�

�

�

2000

800

200

1900

Current valuesP

000

1

x1

10.50

0.5

BV

s1

x4

x3

P

x4

010

0

x3

001

0

s2

0 1�1

0.5

s3

�2�1 2

1

s1

100

0

2000800200

1900

RHS x2

0.25

1.5 1�0.5

s1

x4

s3

P

�

�

�

�

2200

900

100

1800

Current valuesP

000

1

x1

10.50

0.5

x3

�0.5

BV

s1

x4

s3

P

x4

010

0

s2

�1 0.5�0.5

1

s1

100

0

s3

001

0

2200900100

1800

RHS x2

0.5

10.50.5

1 0.75�0.25

s1

s2

s3

P

�

�

�

�

4000

1800

1000

0

Current valuesP

000

1

210.5

x1

�0.5

x3

�1.5

BV

s1

s2

s3

P

x4

221

�2

s2

010

0

s1

100

0

s3

001

0

400018001000

0

RHS

2.51.50.5

x2

�1

211

2x 1 � 2.5x 2 � 2x 3 � 2x 4 � 4000

x 1 � 1.5x 2 � x 3 � 2x 4 � 1800

0.5x 1 � 0.5x 2 � x 3 � x 4 � 1000

Delivery Truck Dump Truck Garbage Truck Gasoline Truck Maximum Time

Metal Casting 2 hours 2.5 hours 2 hours 2 hours 4000 hours

Paint Spray 1 hour 1.5 hours 1 hour 2 hours 1800 hours

Packaging 0.5 hour 0.5 hour 1 hour 1 hour 1000 hours

Profit $0.50 $1.00 $1.50 $2.00

TABLE 1


Use Excel to solve Example 5.Maximize

P � 0.5x1 � x2 � 1.5x3 � 2x4


x1 � 0 x2 � 0 x3 � 0 x4 � 0

STEP 1 Set up an Excel spreadsheet containing the variables, the objective function,and the constraints as follows:

A B C

1. Variables2. Delivery Truck, x1 03. Dump Truck, x2 04. Garbage Truck, x3 05. Gasoline Truck, x4 0

7. Objective9. Maximize Profit �0.5*B2�1*B3�1.5*B4�2*B5

11. Constraints12. Amount Used Maximum13. Metal Casting �2*B2�2.5*B3�2*B4�2*B5 400014. Paint Spray �B2�1.5*B3�1*B4�2*B5 180015. Packaging �0.5*B2�0.5*B3�B4�B5 1000

You should obtain:

2x 1 � 2.5x 2 � 2x 3 � 2x 4 � 4000

x 1 � 1.5x 2 � x 3 � 2x 4 � 1800

0.5x 1 � 0.5x 2 � x 3 � x 4 � 1000

SOLUTION


STEP 2 Click on Tools and then Solver to get the screen below.

• Set Target Cell: is the cell of the optimization function.• Equal To: Allows you to maximize or minimize the objective function.• By Changing Cells: The variable cells.• Subject to the Constraints: The system of inequalities.

STEP 3 Input Set Target Cell, B9, Equal To, Max, By Changing Cells, B2 to B9. Justclick on the appropriate cells.

STEP 4 Add the constraints.

• Click cursor into Subject to the Constraints entry box.• Press Add button. You should see:

• Cell Reference is the cell containing the formula.• Constraint is the cell containing the maximum.• You must also enter the constraint that the variables must be greater than or equal to

zero.• To enter the Metal Casting constraint, put the cursor in the Cell Reference box and

click on cell B13; then put the cursor in the Constraint box and click on C13.


• Click on Add button and repeat this for Paint Spray and Packaging constraints.• To add the constraint that the variables are greater than or equal to zero:

• Click on the variable cells.• Select �� in the middle box.• Type in zero in the constraint box.

• When you are finished, click on OK.

STEP 5 Your Solver box should look like this:

STEP 6 Click on Options box and make sure Assume Linear Model is checked.


The solution is in the Excel spread sheet below.

VariablesDelivery Truck, x1 0Dump Truck, x2 0Garbage Truck, x3 200Gasoline Truck, x4 800

ObjectiveMaximize Profit 1900

Constraints

Amount Used Maximum

Metal Casting 2000 4000Paint Spray 1800 1800Packaging 1000 1000

To get a maximum profit of $1900, you should produce 0 delivery trucks, 0 dumptrucks, 200 garbage trucks, and 800 gasoline trucks. Only 2000 hours of metal castingare used. The other constraints are used at full capacity. ◗

NOW WORK PROBLEM 25 USING EXCEL.

Analyzing the Simplex Method

To justify some of the steps in the simplex method, we analyze more carefully what wedid in Example 5.

The reason we choose the most negative entry in the objective row is that it is thenegative of the largest coefficient in the objective function:

P � 0.5x1 � x2 � 1.5x3 � 2x4

In the initial tableau we set x1 � x2 � x3 � x4 � 0, which results in a profit P � 0.Of course, this profit can easily be improved by manufacturing some trucks.The profit per gasoline truck, x4, is $2.00, while the profit from the other trucks is$0.50, $1.00, and $1.50, respectively. So it is more effective to increase x4 than toincrease x1, x2, or x3. But by how much can x4 be increased? That is determined bythe constraints.

Referring to Table 1, we see that each gasoline truck takes 2 hours to cast, 2 hours topaint, and 1 hour to package. The limited time available for each task restricts thenumber of trucks that can be made. So we look at the limits for producing x4:

STEP 7 Find the solution: Click on Solve and highlight Answer.


metal casting constraint: 4000 hours available, 2 hours per truck; 2000 gasolinetrucks can be manufactured.

paint spraying constraint: 1800 hours available, 2 hours per truck; 900 gasolinetrucks can be manufactured.

packaging constraint: 1000 hours available, 1 hour per truck; 1000 gasoline truckscan be manufactured.

The paint spraying constraint indicates that we can make at most 900 gasoline trucks.So we choose row s2 to be the pivot row. This will increase x4 to as large an amount aspossible. After pivoting, the tableau indicates that if we make 900 gasoline trucks, butno other trucks (x1 � x2 � x3 � 0), we will make a profit P of $1800. However, thenegative value in the objective row indicates that the profit can be improved if weincrease x3.

Increasing production of x3, while increasing profit, will also impact the amount of x4

that can be produced, since some painting time will need to be freed up to produce x3.This is illustrated in the final tableau, which shows the maximum profit is obtained byproducing x3 � 200 garbage trucks and x4 � 800 gasoline trucks.

Each iteration of the simplex method consists in choosing an “entering” variablefrom the nonbasic variables and a “leaving” variable from the basic variables, using aselective criterion so that the value of the objective function is not decreased (some-times the value may remain unchanged). The iterative procedure stops when no vari-able can enter from the nonbasic variables.

The reasoning behind the simplex method for standard maximum problems is nottoo complicated, and the process of “moving to a better solution” is made quite easysimply by following the rules. Briefly, the pivoting strategy works like this:

The choice of the pivot column forces us to pivot the variable that apparentlyimproves the value of the objective function most effectively.

The choice of the pivot row prevents us from making this variable too large to befeasible.

�

Geometry of the Simplex Method

The maximum value (provided it exists) of the objective function will occur at one ofthe corner points of the set of feasible points. The simplex method is designed to movefrom corner point to corner point of the set of feasible points, at each stage improvingthe value of the objective function until a solution is found. More precisely, the geome-try behind the simplex method is outlined below.

1. A given tableau corresponds to a corner point of the set of feasible points.2. The operation of pivoting moves us to an adjacent corner point, where the objective

function has a value at least as large as it did at the previous corner point.3. The process continues until the final tableau is reached — which produces a corner

point that maximizes the objective function.

Though drawings depicting this process can be rendered in only two or threedimensions (that is, when the objective function has only two or three variables in it),the same interpretation can be shown to hold regardless of the number of variablesinvolved. Let’s look at an example.


EXAMPLE 6 The Geometry of the Simplex Method

Maximize

P � 3x1 � 5x2


The feasible region is shown in Figure 3.Below, we apply the simplex method, indicating the corner point corresponding to

each tableau and the value of the objective function there. You should supply thedetails.

Corner Value ofPoint P � 3x1 � 5x2

Tableau (x1, x2) at the Corner Point

(0, 0) 0

(0, 40) 200

(40, 20) 220

(Maximum value) ◗

The Unbounded Case

So far in our discussion, it has always been possible to continue to choose pivotelements until the problem has been solved. But it may turn out that all the entries in acolumn of a tableau are 0 or negative at some stage. If this happens, it means that theproblem is unbounded and a maximum solution does not exist.

For example, consider the tableau

P

00

1

x1

�11

�2

BV

s1

s2

P

x2

1�1

�1

s2

01

0

22

0

RHSs1

10

0

00

1

10

0

x1

x2

P

01

0

( )

(Final tableau)

�11

2

4020

220

2�1

1

P x1BV x2 s2 RHSs1

( )

00

1

s1

x2

P

01

0

� 2040

200

10

02121

25

2121

21�

P x1BV x2 s2 RHSs1

P

00

1

x1

11

�3

BV

s1

s2

P

x2

12

�5

s2

01

0

6080

0

RHSs1

10

0

x 1 � 0 x 2 � 0

x 1 � 2x 2 � 80

x 1 � x 2 � 60

x1

80

60

20

x2

x1 + x2 = 60

x1 + 2x2 = 80

(40, 20)

(60, 0)(0, 0)

(0, 40)

804020

FIGURE 3

SOLUTION


The pivot element is in column x1 row s2. After pivoting, the tableau becomes

Now the only negative entry in the objective (bottom) row is in column x2, and it isimpossible to choose a pivot element in that column. To see why, notice that the con-straints are �x1 � x2 � 2, x1 � x2 � 2, x1 � 0, and x2 � 0. The set of feasible pointsshown in Figure 4, is unbounded, so the maximum problem has no solution.

Summary of the Simplex Method

The general procedure for solving a maximum problem in standard form using thesimplex method can be outlined as follows:

1. The maximum problem is stated in standard form as

MaximizeP � c1x1 � c2x2 � . . . � cn xn


where b1 � 0, b2 � 0, . . . , bm � 0.

2. Introduce slack variables s1, s2, . . . , sm so that the constraints take the form of equalities:

3. Write the objective function in the form

P � c1x1 � c2x2 � . . . � cn xn � 0

4. Set up the initial simplex tableau

P

00

0

1

x1

a11

a21

am1

�c1

x2

a12

a22

am2

�c2

xn

a1n

a2n

amn

�cn

BV

s1

s2

sm

P

. . .

. . .

. . .

. . .

. . .

. . .

. . .

. . .

. . .

. .

.

. .

.

. .

.

. .

.

. .

.

. .

.

. .

.

. .

.

. .

.

s1

10

0

0

s2

01

0

0

sm

00

1

0

b1

b2

bm

0

RHS

x 1 � 0, x 2 � 0, . . . , x n � 0

s1 � 0, s2 � 0, . . . , sm � 0

a11x 1 � a12x 2 � � a1nx n � s1 � b1

a21x 1 � a22x 2 � � a2nx n � s2 � b2

am1x 1 � am2x 2 � � amnx n � sm � bm

a11x 1 � a12x 2 � � a1nx n � b1

a21x 1 � a22x 2 � � a2nx n � b2

am1x 1 � am2x 2 � � amnx n � bm

x 1 � 0, x 2 � 0, . . . , x n � 0

P

00

1

x1

01

0

BV

s1

x1

P

x2

0�1

�3

s2

11

2

42

4

RHSs1

10

0

1 2 3 4 5 6 7 8 9 10

1234567

x1

x2

(2, 0)

(0, 2)

–x1 + x2 = 2

x1 – x2 = 2

FIGURE 4


5. Pivot until(a) All the entries in the objective row are nonnegative. This is a final tableau from

which a solution can be read.Or until(b) The pivot column is a column whose entries are negative or zero. In this case

the problem is unbounded and there is no solution.

The flowchart in Figure 5 illustrates the steps to be used in solving standard maxi-mum linear programming problems.

Is themaximum linear

programming problem in standard

form?

Can the problem be

modified so as to bein standard

form?

STOPSee

Section 4.4

NO NO

Introduceslack variables;

set up simplex tableau

YES

Are there negativeentries in the

objective row ?

YES

Are there positive

entries in the pivot column ?

STOPThis isa final

tableau.

NO

STOPUnbounded;no solution.

NO

YES

Select pivotelement and

pivot.

Modify it

YES

FIGURE 5


In Problems 1– 8, determine which of the following statements is true about each tableau:(a) It is the final tableau.(b) It requires additional pivoting.(c) It indicates no solution to the problem.If the answer is (a), write down the solution; if the answer is (b), indicate the pivot element.

1. 2. BV

x1

x2

P

�

�0

0

1

1

1

1

0

0

0

1

0

P x1 x2 s1 s2

20

20

140

RHS12

12

12

12

BV

s1

x2

P

�0

0

1

1

0

1 1

1

�1

0

1

0

P x1 x2 s1 s2

20

30

120

RHS12

12

14


In Problems 9 – 24, use the simplex method to solve each maximum problem.

3. 4.BV

s1

x1

P

�0

0

1

1

0

0

0

1

0

P x1 x2 s1 s2 RHS1747

114127

327

127

327

2567

18621

BV

s1

s2

P

0

0

1

1

0

0

0

1

0�8

3

�12

P x1 x2 s1 s2

10

8

0

RHS

74

14

12

5. 6.

7. 8.

P

00

1

x1

12

�2

BV

s1

s2

P

x2

31

�5

s2

01

0

3012

0

RHSs1

10

0

P

00

1

x1

10

5

BV

x1

s1

P

x2

�2�2

�10

s2

44

4

2436

20

RHSs1

01

12

P

000

1

x1

001

0

BV

x2

s2

x1

P

x2

100

0

s2

010

0

661

110

RHSs3

14

�1

�5

s1

�211

�10

P

000

1

x1

2�1

1

10

BV

x2

s2

s1

P

x2

100

0

s2

010

0

s3

�15

�1

�15

851

120

RHSs1

001

0

9. Maximize

P � 5x1 � 7x2

subject to

10. Maximize

P � x1 � 5x2

subject to

11. Maximize

P � 5x1 � 7x2

subject to

x 1 � 0 x 2 � 0

x 1 � 2x 2 � 2

2x 1 � x 2 � 2x 1 � 0 x 2 � 0

2x 1 � x 2 � 10

x 1 � 2x 2 � 10x 1 � 0 x 2 � 0

2x 1 � 3x 2 � 123x 1 � x 2 � 12

12. Maximize

P � 5x1 � 4x2

subject to

13. Maximize

P � 3x1 � x2

subject to

14. Maximize

P � 3x1 � 5x2

subject to

x 1 � 0 x 2 � 0

2x 1 � x 2 � 4

x 1 � 2x 2 � 6

x 1 � 0 x 2 � 0

x 1 � x 2 � 2

2x 1 � 3x 2 � 123x 1 � x 2 � 12x 1 � 0 x 2 � 0

x 1 � x 2 � 2

2x 1 � 3x 2 � 6

15. Maximize

P � 2x1 � x2 � x3

subject to

16. Maximize

P � 4x1 � 2x2 � 5x3

subject to

17. Maximize

P � 2x1 � x2 � 3x3

subject to

x 1 � 0 x 2 � 0 x 3 � 0

x 1 � 2x 2 � x 3 � 25

3x 1 � 2x 2 � 3x 3 � 30x 1 � 0 x 2 � 0 x 3 � 0

x 1 � 3x 2 � 2x 3 � 30

2x 1 � x 2 � 3x 3 � 12x 1 � 0 x 2 � 0 x 3 � 0

�2x 1 � x 2 � 2x 3 � 4

x 1 � 2x 2 � x 3 � 2

18. Maximize

P � 6x1 � 3x2 � 2x3

subject to

19. Maximize

P � 2x1 � 4x2 � x3 � x4

subject to

20. Maximize

P � 2x1 � 4x2 � x3

subject to

�x 1 � 2x 2 � 3x 3 � 6�x 1 � 4x 2 � 5x 3 � 5

x 1 � 5x 2 � 7x 3 � 7x 1 � 0 x 2 � 0 x 3 � 0

x 1 � 0 x � 0 x 3 � 0

2x 1 � 2x 2 � 3x 3 � 302x 1 � 2x 2 � x 3 � 12

x 1 � 0 x 2 � 0 x 3 � 0 x 4 � 0

2x 1 � x 2 � 2x 3 � 3x 4 � 12

2x 2 � x 3 � 2x 4 � 202x 1 � x 2 � 4x 3 � 16


27. Scheduling Products A, B, and C are sold door-to-door.Product A costs $3 per unit, takes 10 minutes to sell (on theaverage), and costs $0.50 to deliver to the customer. ProductB costs $5, takes 15 minutes to sell, and is left with thecustomer at the time of sale. Product C costs $4, takes 12 minutes to sell, and costs $1.00 to deliver. During any weeka salesperson is allowed to draw up to $500 worth of A, B,and C (at cost) and is allowed delivery expenses not to exceed$75. If a salesperson’s selling time is not expected to exceed 30 hours (1800 minutes) in a week, and if the salesperson’sprofit (net after all expenses) is $1 each on a unit of A or B and$2 on a unit of C, what combination of sales of A, B, and C willlead to maximum profit and what is this maximum profit?

28. Resource Allocations Suppose that a large hospital classi-fies its surgical operations into three categories according totheir length and charges a fee of $600, $900, and $1200,respectively, for each of the categories. The average time ofthe operations in the three categories is 30 minutes, 1 hour,and 2 hours, respectively; the hospital has four operatingrooms, each of which can be used for 10 hours per day. If thetotal number of operations cannot exceed 60, how many ofeach type should the hospital schedule to maximize itsrevenues?

29. Mixture The Lee refinery blends high and low octane gaso-line into three intermediate grades: regular, premium, andsuper premium. The regular grade consists of 60% highoctane and 40% low octane, the premium consists of 70%high octane and 30% low octane, and the super premiumconsists of 80% high octane and 20% low octane. The compa-ny has available 140,000 gallons of high octane and 120,000gallons of low octane, but can mix only 225,000 gallons.Regular gas sells for $1.20 per gallon, premium sells for$1.30 per gallon, and super premium sells for $1.40 per gallon.How many gallons of each grade should the company mix inorder to maximize revenues?

Jean Cutting Sewing Finishing

I 8 12 4

II 12 18 8

III 18 24 12

There are 5200 minutes of cutting time, 6000 minutes ofsewing time, and 2200 minutes of finishing time each day.The company can sell all the jeans it makes and make a profitof $3 on each Jean I, $4.50 on each Jean II, and $6 on eachJean III. Determine the number of jeans in each categorythat should be made each day to maximize profits.

26. Process Utilization A company manufactures three types oftoys A, B, and C. Each requires rubber, plastic, and alu-minum as listed below:

Toy Rubber Plastic Aluminum

A 2 2 4

B 1 2 2

C 1 2 4

21. Maximize

P � 2x1 � x2 � x3

subject to

x 1 � 0 x 2 � 0 x 3 � 0

x 1 � 2x 2 � 4x 3 � 20

2x 1 � 4x 2 � 4x 3 � 603x 1 � 4x 2 � x 3 � 90

22. Maximize

P � x1 � 2x2 � 4x3

subject to

x 1 � 0 x 2 � 0 x 3 � 0

8x 1 � 5x 2 � 4x 3 � 30

�2x 1 � 6x 2 � x 3 � 5�2x 1 � 2x 2 � x 3 � 15

23. Maximize

P � x1 � 2x2 � 4x3 � x4

subject to

x 1 � 0 x 2 � 0 x 3 � 0 x 4 � 0

5x 1 � 4x 3 � 6x 4 � 204x 1 � 2x 2 � 2x 3 � 8x 4 � 40

24. Maximize

P � x1 � 2x2 � x3 � 3x4

subject to

x 1 � 0 x 2 � 0 x 3 � 0 x 4 � 0

2x 1 � 4x 2 � 5x 3 � 6x 4 � 244x 1 � 4x 2 � 2x 3 � 2x 4 � 4

25. Process Utilization A jean manufacturer makes three typesof jeans, each of which goes through three manufacturingphases — cutting, sewing, and finishing. The number ofminutes each type of product requires in each of the threephases is given below:

The company has available 600 units of rubber, 800 units ofplastic, and 1400 units of aluminum. The company makes aprofit of $4, $3, and $2 on toys A, B, and C, respectively.Assuming all toys manufactured can be sold, determine aproduction order so that profit is maximum.


30. Mixture Repeat Problem 29 under the additional assump-tion that the combined total number of gallons produced bythe refinery cannot exceed 200,000 gallons.

31. Investment A financial consultant has at most $90,000 toinvest in stocks, corporate bonds, and municipal bonds. Theaverage yields for stocks, corporate bonds, and municipalbonds is 10%, 8%, and 6%, respectively. Determine howmuch she should invest in each security to maximize thereturn on her investments, if she has decided that her invest-ment in stocks should not exceed half her funds, and thattwice her investment in corporate bonds should not exceedher investment in municipal bonds.

32. Investment Repeat Problem 31 under the additional assump-tion that no more than $25,000 can be invested in stocks.

33. Crop Planning A farmer has at most 200 acres of farmlandsuitable for cultivating crops A, B, and C. The costs for culti-vating crops A, B, and C are $40, $50, and $30 per acre,respectively. The farmer has a maximum of $18,000 availablefor land cultivation. Crops A, B, and C require 20, 30, and15 hours per acre of labor, respectively, and there is a maxi-mum of 4200 hours of labor available. If the farmer expectsto make a profit of $70, $90, and $50 per acre on crops A, B,and C, respectively, how many acres of each crop should heplant in order to maximize his profit?

34. Crop Planning Repeat Problem 33 if the farmer modified hisallocations as follows:

of copper; 2, 3, and 1 hour, respectively, to produce 1 unit of lead; and 1, 1, and 3 hours, respectively, to produce 1 unit of zinc. The oiling and separation phases of the processcan be in operation for a maximum of 10 hours a day, whilethe mixing phase can be in operation for a maximum of11 hours a day. The Alexander Company makes a profit of$45 per unit of copper, $30 per unit of lead, and $35 per unitof zinc. The demand for these metals is unlimited. How manyunits of each metal should be produced daily by use of theflotation process to achieve the highest profit?

37. Maximizing Profit A wood cabinet manufacturer producescabinets for television consoles, stereo systems, and radios,each of which must be assembled, decorated, and crated.Each television console requires 3 hours to assemble, 5 hoursto decorate, and 0.1 hour to crate, and returns a profit of$10. Each stereo system requires 10 hours to assemble, 8hours to decorate, and 0.6 hour to crate, and returns a profitof $25. Each radio requires 1 hour to assemble, 1 hour todecorate, and 0.1 hour to crate, and returns a profit of $3.The manufacturer has 30,000, 40,000, and 120 hoursavailable weekly for assembling, decorating, and crating,respectively. How many units of each product should bemanufactured to maximize profit?

38. Maximizing Profit The finishing process in the manufactureof cocktail tables and end tables requires sanding, staining,and varnishing. The time in minutes required for each fin-ishing process is given below:

35. Mixture Problem Nutt’s Nut Company has 500 pounds ofpeanuts, 100 pounds of pecans, and 50 pounds of cashewson hand. They package three types of 5-pound cans of nuts:can I contains 3 pounds of peanuts, 1 pound of pecans, and1 pound of cashews; can II contains 4 pounds of peanuts,

pound of pecans, and pound of cashews; and can III con-tains 5 pounds of peanuts. The selling price is $28 for can I,$24 for can II, and $20 for can III. How many cans of eachkind should be made to maximize revenue?

36. Maximizing Profit One of the methods used by theAlexander Company to separate copper, lead, and zinc fromores is the flotation separation process. This process consistsof three steps: oiling, mixing, and separation. These steps mustbe applied for 2, 2, and 1 hour, respectively, to produce 1 unit

12

12

Sanding Staining Varnishing

End table 8 10 4

Cocktail table 4 4 8

The equipment required for each process is used on onetable at a time and is available for 6 hours each day. If theprofit on each cocktail table is $20 and on each end table is$15, how many of each should be manufactured each day inorder to maximize profit?

39. Maximizing Profit A large TV manufacturer has warehousefacilities for storing its 52-inch color TVs in Chicago, NewYork, and Denver. Each month the city of Atlanta is shippedat most four hundred 52-inch TVs. The cost of transportingeach TV to Atlanta from Chicago, New York, and Denveraverages $20, $20, and $40, respectively, while the cost oflabor required for packing averages $6, $8, and $4, respec-tively. Suppose $10,000 is allocated each month for trans-portation costs and $3000 is allocated for labor costs. If theprofit on each TV made in Chicago is $50, in New York is$80, and in Denver is $40, how should monthly shippingarrangements be scheduled to maximize profit?

Cost of Cultivating MaximumCrop A Crop B Crop C Available

Cost $30 $40 $20 $12,000

Hours 10 20 18 3,600

Profit $50 $60 $40


Standard Form of a Minimum Problem

A linear programming problem in which the objective function is to be minimized isreferred to as a minimum problem. Such problems are said to be in standard formprovided the following three conditions are met:

PREPARING FOR THIS SECTION Before getting started, review the following:

4.3

OBJECTIVES 1 Determine a minimum problem is in standard form

2 Obtain the dual problem of a minimum problem in standard form

3 Solve a minimum problem in standard form using the duality principle

Solving Minimum Problems in Standard Form Using the Duality Principle*

Standard Form of a Minimum Problem

CONDITION 1 All the variables are nonnegative.CONDITION 2 All other constraints are written as linear expressions that are

greater than or equal to a constant.CONDITION 3 The objective function is expressed as a linear expression with

nonnegative coefficients.

�

*The solution of general minimum problems is discussed in Section 4.4. If you plan to cover Section 4.4, this section may be omitted without loss of continuity.

EXAMPLE 1 Determining a Minimum Problem Is in Standard Form

Determine which of the following minimum problems are in standard form.

(a) Minimize

C � 2x1 � 3x2


SOLUTION

(a) Since all three conditions are met, thisminimum problem is in standardform.

x 1 � 3x 2 � 24

2x 1 � x 2 � 18

x 1 � 0 x 2 � 0

(b) Minimize

C � 3x1 � x2 � 4x3


(b) Conditions 1 and 2 are met, butCondition 3 is not, since the coeffi-cient of x2 in the objective functionis negative. This minimum problemis not in standard form.3x 1 � x 2 � x 3 � 12

x 1 � x 2 � x 3 � 8

x 1 � 0 x 2 � 0 x 3 � 0

> Matrix Algebra (Section 2.4, pp. xx–xx)

Determine a minimumproblem is in standardform

1

Solving Minimum Problems in Standard Form Using the Duality Principle 229

NOW WORK PROBLEM 1.

The Duality Principle

One technique for solving a minimum problem in standard form was developed byJohn von Neumann and others. The solution (if it exists) is found by solving a relatedmaximum problem, called the dual problem. The next example illustrates how toobtain the dual problem.

(d) Minimize

C � 2x1 � x2 � 3x3


(d) Conditions 1, 2, and 3 are each met,so this minimum problem is instandard form.

�x 1 � 2x 2 � x 3 � �2

x 1 � x 2 � x 3 � 6

x 1 � 0 x 2 � 0 x 3 � 0

(c) Minimize

C � 2x1 � x2 � x3


SOLUTION

(c) Conditions 1 and 3 are met, butCondition 2 is not, since the firstconstraint

x1 � 3x2 � x3 � 12

is not written with a � sign. Theminimum problem as stated is not instandard form. Notice, however, thatby multiplying by �1, we can writethis constraint as

�x1 � 3x2 � x3 � �12

Written in this way, the minimumproblem is in standard form.

x 1 � 3x 2 � x 3 � 12

x 1 � x 2 � x 3 � 1

x 1 � 0 x 2 � 0 x 3 � 0

◗

EXAMPLE 2Obtaining the Dual Problem of a Minimum Problem in

Standard Form

Obtain the dual problem of the following minimum problem:

Minimize

C � 300x1 � 480x2


First notice that the minimum problem is in standard form. We begin by writing amatrix that represents the constraints and the objective function:

x1 x2

Constraint: x 1 � 3x 2 � 0.25

Constraint: 2x 1 � 2x 2 � 0.45

Objective function: C � 300x 1 � 480x 2

� 0.250.450 �� 1

2300

32

480

x 1 � 3x 2 � 0.25

2x 1 � 2x 2 � 0.45

x 1 � 0 x 2 � 0

SOLUTION

Obtain the dual problem of a minimumproblem in standard form

2


Now form the matrix that has as columns the rows of the above matrix by takingcolumn 1 above and writing it as row 1 below, taking column 2 above and writing it asrow 2 below, and taking column 3 above and writing it as row 3 below.

This matrix is called the transpose of the first matrix.From this matrix, create the following maximum problem:

Maximize

P � 0.25y1 � 0.45y2

subject to the conditions

The maximum problem is the dual of the given minimum problem. ◗

Notice that the dual of a minimum problem in standard form is a maximum prob-lem in standard form so it can be solved by using techniques discussed in the previoussection. The significance of this is expressed in the following principle:

y1 � 2y2 � 300

3y1 � 2y2 � 480

y1 � 0 y2 � 0

� 300480

0��130.25

220.45

Von Neumann Duality Principle

Suppose a minimum problem in standard form has a solution. The minimum valueof the objective function of the minimum problem in standard form equals themaximum value of the objective function of the dual problem, a maximum prob-lem in standard form.

�

So, one way to solve a minimum problem in standard form is to form the dualproblem and solve it. Another way to solve minimum problems (even those not instandard form) is given in Section 4.4.

The steps to use for obtaining the dual problem are listed below.

Steps for Obtaining the Dual Problem

STEP 1 Write the minimum problem in standard form.STEP 2 Construct a matrix that represents the constraints and the objective function.STEP 3 Interchange the rows and columns to form the matrix of the dual problem.STEP 4 Translate this matrix into a maximum problem in standard form.

�


EXAMPLE 3Obtaining the Dual Problem of a Minimum Problem

in Standard Form

Find the dual of the following minimum problem:

Minimize

C � 2x1 � 3x2

subject to

STEP 1 The minimum problem is in standard form.

STEP 2 The matrix that represents the constraints and the objective function is

STEP 3 Interchanging rows and columns, we obtain the matrix

STEP 4 This matrix represents the following maximum problem:

Maximize

P � 6y1 � 4y2 � 5y3

subject to

This maximum problem is in standard form and is the dual problem of the mini-mum problem. ◗

Some observations about Example 3:

1. The variables (x1, x2) of the minimum problem are different from the variables of itsdual problem (y1, y2, y3).

2. The minimum problem has three constraints and two variables, while the dualproblem has two constraints and three variables. (In general, if a minimum problemhas m constraints and n variables, its dual problem will have n constraints and mvariables.)

2y1 � y2 � y3 � 2

y1 � 2y2 � y3 � 3

y1 � 0 y2 � 0 y3 � 0

Constraint: 2y1 � y2 � y3 � 2

Constraint: y1 � 2y2 � y3 � 3

Objective function: P � 6y1 � 4y2 � 5y3

230��2

16

124

115

Constraint: 2x 1 � x 2 � 6

Constraint: x 1 � 2x 2 � 4

Constraint: x 1 � x 2 � 5

Objective function: C � 2x 1 � 3x 2

6450��

2112

1213

2x 1 � x 2 � 6

x 1 � 2x 2 � 4

x 1 � x 2 � 5

x 1 � 0 x 2 � 0

SOLUTION


3. The inequalities defining the constraints are � for the minimum problem and �for the maximum problem.

4. Since the coefficients in the objective function to be minimized are positive, the dualproblem has nonnegative numbers to the right of the � signs.

5. We follow the custom of denoting an objective function by C (for Cost), if it is to beminimized; and P (for Profit), if it is to be maximized.

NOW WORK PROBLEM 7.

EXAMPLE 4Solving a Minimum Problem in Standard Form by Using the

Duality Principle

Solve the maximum problem of Example 3 by the simplex method, and thereby obtainthe solution for the minimum problem.

We introduce slack variables s1 and s2 to obtain


The pivot element, 2, is circled. After pivoting, we obtain this tableau:

The pivot element, , is circled. After pivoting, we obtain this tableau:

This is a final tableau, so an optimal solution has been found. We read from it thatthe solution to the maximum problem is

P � 10 y1 � 0 y2 � 0 y3 � 2

The duality principle states that the minimum value of the objective function in theoriginal problem is the same as the maximum value in the dual; that is,

C � 10

P

00

1

y1

2�1

4

BV

y3

s2

P

y2

11

1

s2

01

0

21

10

RHSy3

10

0

s1

1�1

5

y3

s2

P

�

�

�

2

1

10

Current value

12

y1

s2

P

�

�

�

1

2

6

Current valueBV

y1

s2

P

�

0

0

1 �1

1

0

0

0

1

0

P y1 y2 s1 s2

1

2

6

RHS12

12

3

12

y3

12

�2

12

32

P

00

1

y1

21

�6

BV

s1

s2

P

y2

12

�4

s2

01

0

23

0

RHSy3

11

�5

s1

10

0

2y1 � y2 � y3 � s1 � 2, s1 � 0

y1 � 2y2 � y3 � s2 � 3, s2 � 0

SOLUTION

3


But which values of x1 and x2 will yield this minimum value? As it turns out, the valueof x1 is found in the objective row in column s1(x1 � 5) and x2 is found in the objectiverow in column s2(x2 � 0). As a consequence, the solution to the minimum problem canbe read from the right end of the objective row of the final tableau of the maximumproblem:

x1 � 5 x2 � 0 C � 10 ◗

We summarize how to solve a minimum linear programming problem below.

Solving a Minimum Problem in Standard Form Using the Duality Principle

STEP 1 Write the dual (maximum) problem.STEP 2 Solve this maximum problem by the simplex method.STEP 3 The minimum value of the objective function (C) will appear in the lower

right corner of the final tableau; it is equal to the maximum value of the dualobjective function (P). The values of the variables that give rise to the mini-mum value are located in the objective row in the slack variable columns.

�


EXAMPLE 5Solving a Minimum Problem in Standard Form Using

the Duality Principle

Minimize

C � 6x1 � 8x2 � x3

subject to

This minimum problem is in standard form. The matrix representing this problem is

We interchange rows and columns to get

353

20

1329

625

30

6810

Constraint: 3y1 � y2 � 6y3 � 6

Constraint: 5y1 � 3y2 �2y3 � 8

Constraint: 3y1 � 2y2 � 5y3 � 1

Objective function: P � 20y1 � 9y2 � 30y3

3166

5328

3251

209

300

Constraint: 3x1 � 5x2 � 3x3 � 20

Constraint: x1 � 3x2 �2x3 � 9

Constraint: 6x1 � 2x2 � 5x3 � 30

Objective function: C � 6x1 � 8x2 � x3

3x 1 � 5x 2 � 3x 3 � 20

x 1 � 3x 2 � 2x 3 � 9

6x 1 � 2x 2 � 5x 3 � 30

x 1 � 0 x 2 � 0 x 3 � 0

SOLUTION


The dual problem is:

Maximize

P � 20y1 � 9y2 � 30y3

subject to

We introduce nonnegative slack variables s1, s2, and s3. The initial tableau for this prob-lem is

The final tableau (as you should verify is)

The solution to the maximum problem is

For the minimum problem, the values of x1, x2, and x3 are read as the entries in theobjective row in the columns under s1, s2, and s3, respectively. The solution to the mini-mum problem is

x1 � 0 x2 � 0 x3 �

and the minimum value is C � . ◗

Use Excel to solve Example 5.

STEP 1 Set up the Excel spreadsheet. The spreadsheet below is presented with the for-mulas revealed.

203

203

P �203 y1 �

13 y2 � 0 y3 � 0

s1

x2

s3

P

�

�

�

�

Current valueBV

s1

s2

y1

P

0

0

0

1

�10

0

1

0

P x1 x2

�11 1

0

0

0

0

1

0

0

s3x3 s1

5

RHS

13

53

23

��

s2

5313

5

13

13

203

203

203

193

193

193

103

72

P

000

1

y1

353

�20

BV

s1

s2

s3

P

y2

132

�9

y3

625

�30

s2

010

0

s3

001

0

681

0

RHSs1

100

0

3y1 � y2 � 6y3 � 6

5y1 � 3y2 � 2y3 � 8

3y1 � 2y2 � 5y3 � 1

y1 � 0 y2 � 0 y3 � 0

SOLUTION


STEP 2 Set up Solver. Solver can be found under the toolbar command Tools.

The only differences between maximize and minimize problems are

(a) Equal To: must be min.(b) Subject to the Constraints: must be ��.

STEP 3 Click on Solve. The solution is given below.

Variablesx1 0x2 0x3 6.666666667


ObjectiveMinimize 6.666666667

ConstraintsAmountUsed Minimum

1 20 202 13.33333333 93 33.33333333 30

So x1 � 0, x2 � 0, x3 � 6.6667 with a minimum value of C � 6.6667. ◗


In Problems 1– 6, determine which of the following minimum problems are in standard form.

1. Minimize

C � 2x1 � 3x2


2. Minimize

C � 3x1 � 5x2


3. Minimize

C � 2x1 � x2


2x 1 � x 2 � 1�2x 1 � �3

x 1 � 0 x 2 � 0

3x 1 � x 2 � 4x 1 � 2x 2 � 3

x 1 � 0 x 2 � 0

4x 1 � x 2 � 2x 1 � x 2 � 1

x 1 � 0 x 2 � 0

4. Minimize

C � 2x1 � 3x2


5. Minimize

C � 3x1 � 7x2 � x3


6. Minimize

C � x1 � x2 � x3


x 1 � x 2 � 62x 1 � x 3 � 4

x 1 � 0 x 2 � 0 x 3 � 0

x 1 � x 3 � 62x 1 � x 2 � 4

x 1 � 0 x 2 � 0 x 3 � 0

x 1 � x 2 � 32x 1 � 3x 2 � 4

x 1 � 0 x 2 � 0

7. Minimize

C � 2x1 � 3x2

subject to

8. Minimize

C � 3x1 � 4x2

subject to

9. Minimize

C � 3x1 � x2 � x3

subject to

x 1 � x 2 � x 3 � 52x 1 � x 2 � 4

x 1 � 0 x 2 � 0 x 3 � 0

2x 1 � x 2 � 22x 1 � x 2 � 6

x 1 � 0 x 2 � 0

x 1 � x 2 � 22x 1 � 3x 2 � 6

x 1 � 0 x 2 � 0

10. Minimize

C � 2x1 � x2 � x3

subject to

11. Minimize

C � 3x1 � 4x2 � x3 � 2x4

subject to

12. Minimize

C � 2x1 � x2 � 4x3 � x4

subject to

2x 1 � x 2 � x 3 � x 4 � 802x 1 � 3x 2 � x 3 � 2x 4 � 100

x 1 � 0 x 2 � 0 x 3 � 0 x 4 � 0

x 1 � x 2 � x 3 � 2x 4 � 603x 1 � 2x 2 � x 3 � 2x 4 � 90

x 1 � 0 x 2 � 0 x 3 � 0 x 4 � 0

2x 1 � x 2 � x 3 � 4x 1 � 2x 2 � x 3 � 6

x 1 � 0 x 2 � 0 x 3 � 0

In Problems 7 – 12, write the dual problem of each minimum problem.

ch04_part1

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