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Linear
Programming
12
Continued
Linear programming was invented in the
1940s by George Dantzig as a result of a
military research project on planning how
to distribute men, weapons, and supplies
efficiently to the various fronts during
World War II. (Here, the word
programming means creating a plan that
solves a problem; it is not a reference to
computer programming.)
Many businesses, industries,
and government agencies use linear
programming successfully.
• Airlines use it to schedule flights in a
way that minimizes costs without
overloading a pilot or crew with too
many hours.
• Ecological organizations and
government agencies use it to determine
the best way to keep our water and air
clean.
continued
© British Retail Photography / Alamy
W
HAT WE WILL DO I n T h i s C h a p t e r
• You may well find yourself working in a small
business. Over half of the American workforce
does, according to the U.S. Small Business
Administration. Small businesses have limited
resources, and how they allocate their resources
can make or break the business. We will look at
how businesses can use linear programming to
find the best way to do this.
• Starbucks stores and smaller local coffee shops
seem to be everywhere. They blend coffee beans
from different parts of the world. Like them, many
businesses sell a product that involves blending
resources. These businesses have many factors to
consider: the desired quality of the blend, the
availability of the raw materials, and the cost.
We will investigate how linear programming is
used do this.
• Many businesses use the same resources to make
different products. A craftsman uses the same
wood to make different tables and chairs. A
fashion designer uses the same fabrics to make
different garments. We will learn how linear
programming is used to find the best way to
allocate resources to different products.
12-1
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Linear Programming
• Businesses use it to find the best way to assign personnel to jobs.
• Supermarket chains use it to determine which warehouses should ship
which products to which stores.
• Investment companies use it to create portfolios with the best mix of stocks
and bonds.
• Refineries use it to decide what crude oil to buy and to determine what
products to produce with the oil.
12.2
Introduction to the Simplex Method
Objective
•
Be able to set up a linear programming problem as a matrix problem rather
than a geometry problem
With the simplex method of linear programming, you start by modeling the problem in the same way that you do with the geometric method from Section 12.1.
Then, instead of graphing the region of solutions, you
• convert the constraints to equations
• convert the problem to a matrix problem
• find the possible solution that corresponds to that matrix
Finally, you solve the matrix problem, as we will show in Section 12.3.
We’ll use the craftsman problem from Section 12.1, summarized below, to introduce the procedure and explain why it works. Here’s the model, from Section 12.1:
Independent Variables
x1 number of coffee tables
x2 number of end tables
Constraints
C1: 6x1 5x2 40
C2: 200x1 100x2 1,000
C3: x1 0
C4: x2 0
the time constraint
the money constraint
Objective Function
z 240x1 160x2
z measures profit
We’ve labeled the constraints C1 through C4, for reference. Because there are frequently more than two variables, the simplex method uses x1, x2, and so on rather
than x and y to represent the variables.
Converting the Constraints to Equations
Recall that the constraint
C1: 6x1 5x2 40
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12.2
Introduction to the Simplex Method
12-3
was derived from
(coffee table hours) (end table hours) 40
If the total number of hours is less than 40, then some “slack” or unused hours are
left, and we could say
(coffee table hours) (end table hours) (unused hours) 40
C1 can be rewritten as
C1: 6x1 5x2 s1 40
where s1 measures unused hours and s1 0.
The variable s1 is called a slack variable because it “takes up the slack” between the hours used (which can be less than 40) and the hours available (which
are exactly 40). The introduction of slack variables converts the constraint inequalities to equations, allowing us to use matrices to solve the problem.
The constraint
C2: 200x1 100x2 1,000
was derived from
(coffee table money) (end table money) 1,000
This could be rephrased as
(coffee table money) (end table money) (unused money) 1,000
Thus, C2 can be rewritten as
C2: 200x1 100x2 s2 1,000
where s2 is a slack variable that measures unused money and s2 0. This new variable “takes up the slack” between the money used (which can be less than $1,000)
and the money available (which is exactly $1,000).
Constraints C3: x1 0 and C4: x2 0 remind us that variables x1 and x2 count
things that cannot be negative (coffee tables and end tables, respectively). The simplex method requires that all variables are nonnegative, and we do not convert
these constraints to equations.
We have rewritten our model:
Independent Variables
x1 number of coffee tables
x2 number of end tables
(x1, x2, s1, s2 0)
Slack Variables
s1 unused hours
s2 unused money
Constraints
C1: 6x1 5x2 s1 40
C2: 200x1 100x2 s2 1,000
the time constraint
the money constraint
Objective Function
z 240x1 160x2
z measures profit
Converting the Problem to a Matrix Problem
We rewrite the constraints so that each includes all of the variables (x1, x2, s1, s2,
and z).
The first constraint:
C1: 6x1 5x2 s1 40
using the variables x1, x2, and s1
C1: 6x1 5x2 1s1 0s2 0z 40
using all of the variables
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This equation becomes the first row of our matrix:
[6
5
1
0
0
40]
The second constraint:
C1: 200x1 100x2 0s1 1s2 0z 1,000
using all of the variables
[200
the matrix’s second row
100
0
1
0
1,000]
We rewrite the objective function with all of the variables on the left side of the
equation.
the objective function
z 240x1 160x2
moving all of the variables
240x1 160x2 0s1 0s2 1z 0 to the left side
the matrix’s third row
[240 160
0
0
1
0]
The resulting matrix is
x1
x2
6
5
£ 200
100
240 160
s1
1
0
0
s2 z
0 0
40
1 0 1,000 §
0 1
0
We call this matrix the first simplex matrix.
Finding the Possible Solution That Corresponds
to the Matrix
Some of the columns in our first simplex matrix contain a single 1 and several
0’s (the s1, s2, and z columns). We find the value of the variable heading each of
these columns by reading down the column, turning right at the 1, and stopping
at the end of the row. Other columns (the x1 and x2 columns) do not contain a
single 1 and several 0’s. The value of the variables heading these columns
is zero.
x1
x2
6
5
£ 200
100
240 160
x1 0 x2 0
s1
1
0
0
s2
0
1
0
z
S s1 40
0
40
0 1,000 § S s2 1,000
S z0
1
0
The possible solution that corresponds to the matrix is
(x1, x2, s1, s2) (0, 0, 40, 1,000) with z 0
Why do the numbers read from the matrix in this manner yield a possible
solution? Clearly, (x1, x2) (0, 0) satisfies all four constraints, so it is a possible
solution. If we substitute this solution into C1 and C2 and the objective function,
we get the following:
C1: 6x1 5x2 s1 40
C2: 200x1 100x2 s2 1,000
200 0 100 0 s2 1,000
6 0 5 0 s1 40
s2 1,000
s1 40
z 240x1 160x2 240 0 160 0 0
Thus, (x1, x2, s1, s2) (0, 0, 40, 1,000) with z 0 satisfies all the constraints and
the objective function and is a possible solution.
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12.2
Introduction to the Simplex Method
12-5
The possible solution means “make no coffee tables, make no end tables,
have 40 unused hours and 1,000 unused dollars, and make no profit.” When we
solved this same problem with the geometric method in Section 12.1, one of the
corner points was (0, 0) (see Figure 12.34). We have just found that corner point
with the simplex method rather than the geometric method (the geometric method
does not find values of the slack variables).
y
P2 (0, 8)
P3 (
5
2,
5)
P4 (5, 0)
x
P1 (0, 0)
FIGURE 12.34
The first simplex matrix yields a corner point.
As we will see in the next section, the simplex method requires us to explore
a series of matrices, just as the geometric method requires us to explore a series of
corner points. Each simplex matrix will provide us with a corner point of the
region of possible solutions, without our actually graphing that region. The last
simplex matrix will provide us with the optimal corner point, that is, the point that
solves the problem.
SIMPLEX METHOD STEPS
Step 1 Model the problem.
a. List the independent variables.
b. List the constraints, and translate them into linear inequalities.
c. Find the objective, and translate it into a linear equation.
Step 2 Convert each constraint from an inequality to an equation.
a. Use one slack variable for each constraint.
b. Determine what each slack variable measures.
Step 3 Rewrite the objective function with all variables on the left side.
Step 4 Make a matrix out of the rewritten constraints and the rewritten
objective function. This is the first simplex matrix. The constraints go
in the first rows, and the objective function goes in the last row.
Step 5 Determine the possible solution that corresponds to the matrix.
a. Columns containing only a 1 and 0’s: turn at 1.
b. Other columns: the value of the variable is 0.
The remaining steps, which build on this work to find the maximal solution,
will be discussed in Section 12.3.
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EXAMPLE
1
SETTING UP A SIMPLEX METHOD PROBLEM AS A MATRIX
PROBLEM Find the first simplex matrix and the corresponding solution for the
objective function
1
z x1 x2
2
subject to the constraints
C1: 2x1 x2 15
C2: x1 x2 10
C3: x1 0
C4: x2 0
SOLUTION
Step 1
Model the problem. No modeling needs to be done, because the problem is stated
in the language of mathematics rather than as a real-world problem.
Step 2
Convert each constraint from an inequality to an equation.
C1: 2x1 x2 15 S 2x1 x2 s1 15
C2: x1 x2 10 S x1 x2 s2 10
Rewrite the objective function with all the variables on the left side.
1
1
z x1 x2 S x1 x2 z 0
2
2
Step 3
Step 4
Make a matrix out of the rewritten constraints and the rewritten objective function.
x2 s1 s2 z
x1
2
1 1 0 0
15 S s1 15
£ 1
1 0 1 0
10 § S s2 10
Sz0
1>2 1 0 0 1
0
x1 0
Step 5
x2 0
Determine the possible solution that corresponds to the matrix. The s1, s2, and
z columns contain zeros and a single 1, so we find the values of those variables by
reading down the column and turning at 1. The x1 and x2 columns are not that type
of column, so the value of those variables is zero. The possible solution is (x1, x2,
s1, s2) (0, 0, 15, 10), and z 0.
Remember that this is only a possible solution. In the next section, we will
see how to build on this work to find the maximal solution.
12.2 Exercises
In Exercises 1– 4, convert the linear inequality to a linear
equation by introducing a slack variable.
In Exercises 5–10, (a) translate the information into a linear
inequality, (b) convert the linear inequality into a linear
equation by introducing a slack variable, and (c) determine
what each variable (including the slack variable) measures.
1. 3x1 2x2 5
2.
512x1 339x2 254
5.
3. 3.41x1 9.20x2 6.16x3 45.22
4. 99.52x1 21.33x2 102.15x3 50.14
Selected exercises available online at www.webassign.net/brookscole
A shopper has $42.15 in cash. She wants to purchase
meat at $6.99 a pound, cheese at $3.15 a pound, and
bread at $1.98 a loaf. The shopper must pay for these
items in cash.
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12.2
6. A sharp dresser is at his favorite store, where
everything is on sale. His budget will allow him to
spend up to $425. Shirts are on sale for $19.99 each,
slacks for $74.98 a pair, and sweaters for $32.98
each.
7.
A plumber is starting an 8-hour day. Laying pipe takes
her 5 minutes per foot, and installing elbows takes her
4 minutes per elbow.
8. An electrician is starting a 7 1/2-hour day. Installing
conduit takes him 2.5 minutes per foot, connecting lines
takes him 1 minute per connection, and installing
circuit breakers takes him 4.5 minutes per circuit
breaker.
9. A mattress warehouse has 50,000 cubic feet of space.
Twin beds are 24 cubic feet, double beds are 36 cubic
feet, queen-size beds are 56 cubic feet, and king-size
beds are 72 cubic feet.
10.
A nursery has 1000 square feet of outdoor display
space. Trees need 16 square feet each, shrubs need
10 square feet each, and flowering plants need 1 square
foot each.
In Exercises 11–16, find the possible solution that corresponds to
the given matrix.
x1 x2 s1 s2
5 0 3 1
11. £ 3 0 19 0
21 1 48 0
x1
1
12. £ 0
0
x2 s1
31 0
0 0
1 1
x1
1.9
3.2
13. ≥
5.5
2.1
14.
x1
0
0
≥
1
0
s2
9
15
0
z
0
1
0
7.4
4.9 §
20
s1
0
0
1
0
s2
0
1
0
0
s3
1
0
0
0
s2 s3
0 62
0 5
0 0
1 1
z
0
1
0
0
x2
0.3
0.7
0.8
3.2
x2 s1
1 7
0 5
0 0
0 9
x1 x2 s1 s2 s3
15.
1
0
≥0
0
0
0
0
0
1
0
6
8
3
4
4
z
0 12
1 22 §
0 19
0
0
1
0
0
0
1
0
0
0
z
0
0
0
1
0.5
9.3
¥
7.8
9.6
0.5
9.3
¥
7.8
9.6
s4
z
76
13
50
9
12
0
0
0
0
1
25
46
32 ¥
73
63
x1
32
7
16. E10
65
24
x2 s1
0 1
0 0
0 0
1 0
0 0
s2 s3 s4
0 0
7
0 1
3
1 0
5
0 0 9
0 0
2
z
0
0
0
0
1
Exercises
12-7
25
46
32U
73
44
In Exercises 17–22, find the first simplex matrix and the possible
solution that corresponds to it.
17. Objective function: 18. Objective function:
z 2x1 4x2
z 2.4x1 1.3x2
Constraints:
Constraints:
3x1 4x2 40
6.4x1 x2 360
4x1 7x2 50
x1 9.5x2 350
x1 0
x1 0
x2 0
x2 0
19. Objective function:
20. Objective function:
z 12.10x1 43.86x2
z 3.52x 1 4.72x 2
Constraints:
Constraints:
112x1 3x2 370
2x1 4x2 7,170
x1 x2 70
32x1 19x2 1960
47x1 19x2 512
x1 5
x1 0
x1 0
x2 0
x2 0
21. Objective function:
z 4x1 7x2 9x3
Constraints:
5x1 3x2 9x3 10
12x1 34x2 100x3 10
52x1 7x2 12x3 10
x1 0
x2 0
x3 0
22. Objective function:
z 9.1x1 3.5x2 8.22x3
Constraints:
x1 16x2 9.5x3 1210
72x1 3.01x2 50x3 1120
57x1 87x2 742x3 309
x1 0
x2 0
x3 0
In Exercises 23–28, find the first simplex matrix and the possible
solution that corresponds to it. Do not try to solve the problem.
23. Five friends, each of whom is an experienced baker,
form a company that will make bread and cakes and
sell them to local restaurants and specialty stores.
Each loaf of bread requires 50 minutes of labor and
ingredients costing $0.90 and can be sold for a $1.20
profit. Each cake requires 30 minutes of labor and
ingredients costing $1.50 and can be sold for a $4.00
profit. The partners agree that no one will work more
than 8 hours a day. Their financial resources do not
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allow them to spend more than $190 per day on
ingredients. How many loaves of bread and how
many cakes should they make each day to maximize
their profit?
24.
A craftswoman produces two products: floor lamps
and table lamps. Production of one floor lamp requires
75 minutes of her labor and materials that cost $25.
Production of one table lamp requires 50 minutes of
labor and materials that cost $20. The craftswoman
wants to work no more than 40 hours each week, and
her financial resources allow her to pay no more than
$900 for materials each week. If she can sell as many
lamps as she can make and if her profit is $40 per floor
lamp and $32 per table lamp, how many floor lamps
and how many table lamps should she make each week
to maximize her weekly profit?
25. Pete’s Coffees sells two blends of coffee beans: Yusip
Blend and Exotic Blend. Yusip Blend is one-half Costa
Rican beans and one-half Ethiopian beans, and Exotic
Blend is one-quarter Costa Rican beans and threequarters Ethiopian beans. Profit on the Yusip Blend is
$3.50 per pound, while profit on the Exotic Blend is
$4 per pound. Each day, the shop can obtain 200 pounds
of Costa Rican beans and 330 pounds of Ethiopian
beans, and it uses those beans only in the two blends. If
it can sell all that it makes, how many pounds of Yusip
Blend and of Exotic Blend should Pete’s Coffees
prepare each day to maximize profit?
26.
Pete’s Coffees sells two blends of coffee beans:
Morning Blend and South American Blend. Morning
Blend is one-third Mexican beans and two-thirds
Colombian beans, and South American Blend is twothirds Mexican beans and one-third Colombian beans.
Profit on the Morning Blend is $3 per pound, while
profit on the South American Blend is $2.50 per pound.
Each day, the shop can obtain 100 pounds of Mexican
beans and 80 pounds of Colombian beans, and it uses
those beans only in the two blends. If it can sell all that it
makes, how many pounds of Morning Blend and of
12.3
South American Blend should Pete’s Coffees prepare
each day to maximize profit?
27. A furniture-manufacturing firm makes sofas and chairs,
each of which is available in several styles. Each sofa,
regardless of style, requires 8 hours in the upholstery
shop and 4 hours in the carpentry shop and can be sold
for a profit of $450. Each chair requires 6 hours in the
upholstery shop and 3.5 hours in the carpentry shop
and can be sold for a profit of $375. There are nine
people working in the upholstery shop and five in the
carpentry shop, each of whom can work no more than
40 hours per week. How many sofas and how many
chairs should the firm make each week to maximize its
profit? What is that maximum profit? Would this leave
any extra time in the upholstery shop or the carpentry
shop? If so, how much?
28. City Electronics Distributors handles two lines of televisions: the Packard and the Bell. The company purchases up to $57,000 worth of television sets from
the manufacturers each month, which it stores in its
9,000-cubic-foot warehouse. The Packards come in
36-cubic-foot packing crates, and the Bells come in
30-cubic-foot crates. The Packards cost City Electronics
$200 each and can be sold to a retailer for a $200 profit,
while the Bells cost $250 each and can be sold for a
$260 profit. How many sets should City Electronics
order each month to maximize profit? What is that
maximum profit? Would this leave any unused storage
space or money? If so, how much?
Answer the following questions using complete
sentences and your own words.
•
Concept Questions
29.
30.
Why does the simplex method exist? That is, why do
we not use the geometric method for all linear programming problems?
What do the simplex method and the geometric method
have in common?
The Simplex Method: Complete Problems
Objectives
•
•
Understand the row operations
Be able to use the row operations to solve a linear programming problem
In Section 12.2, we saw how to convert a linear programming problem to a matrix
problem. In this section, we will see how to use the simplex method to solve a linear programming problem. We’ll start by continuing where we left off with the
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12.3
The Simplex Method: Complete Problems
12-9
craftsman problem. We have modeled the problem, introduced slack variables, and
rewritten the objective function so all variables are on the left side.
Independent Variables
Slack Variables
x1 number of coffee tables
x2 number of end tables
(x1, x2, s1, s2 0)
s1 unused hours
s2 unused money
Constraints
C1: 6x1 5x2 s1 40
C2: 200x1 100x2 s2 1,000
the time constraint
the money constraint
Objective Function
240x1 160x2 0s1 0s2 1z 0
z measures profit
We have also made a matrix out of the rewritten constraints and the rewritten
objective function (the “first simplex matrix”) and determined the possible solution that corresponds to that matrix. The first simplex matrix was as follows:
x1
x2
s1 s2 z
6
5
1 0 0
40
d C1
£ 200
100 0 1 0 1,000 § d C2
d objective function
240 160 0 0 1
0
The corresponding possible solution was
(x1, x2, s1, s2) (0, 0, 40, 1,000)
with z 0
This solution is a corner point of the region of possible solutions (see
Figure 12.35); it is point P1 (0, 0) (the geometric method does not find values of the
slack variables). However, it is not the optimal corner point; it is a possible solution but not the maximal solution. Our goal in the remaining steps of the simplex
method is to find the maximal solution. This process is called pivoting.
y
P2 (0, 8)
P3 (
5
2,
5)
P4 (5, 0)
x
P1 (0, 0)
FIGURE 12.35
The possible solution (x1, x2, s1, s2) (0, 0, 40, 1,000) is the corner point P1 (0, 0).
Deciding Where to Pivot
Step 1
Look at the last row, the objective function row. Select the most negative entry in
that row. The column containing that entry is the pivot column. If the last row
contains no negative entries, then no pivoting is necessary; the possible solution
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Linear Programming
that corresponds to the matrix is the maximal solution.
x1
6
£ 200
240
x2
s1
5
1
100 0
160 0
s2
0
1
0
z
0
40
0 1,000 §
d objective function
1
0
The only negative entries in the last row are 240 and 160; the most negative is
240. This entry is in the first column, so our pivot column is the first column. We
will pivot on one of the entries of the pivot column.
x1
x2
s1
6
5
1
£ 200
100 0
240 160 0
c
Pivot column
Step 2
s2
0
1
0
z
0
40
0 1,000 §
1
0
Divide the last entry in each constraint row by the corresponding entry in the pivot
column, if that entry is positive. The row that yields the smallest nonnegative such
quotient is the pivot row.
6
5
£ 200
100
240 160
c
Pivot column
1
0
0
0
1
0
0
0
1
40
1,000 §
0
d 40>6 6.67
d 1,000>200 5
d not a constraint row
Our quotients are 6.67 and 5; the smallest nonnegative quotient is 5, which is in
row two. Our pivot row is the second row.
Step 3
Pivot on the entry in the pivot row and pivot column.
6
5
1 0 0
40
£ 200
100 0 1 0 1,000 §
240 160 0 0 1
0
c
Pivot column
d pivot row
We will pivot on the entry “200.”
The Row Operations
Remember, each of the matrix’s rows represents an equation. And the entire matrix
is just a different way of writing a system of equations. When you solve a system
of equations with the elimination method, you are allowed to do things like adding
multiples of equations together. When you pivot on a matrix with the simplex
method, you are allowed to do things like adding multiples of rows together. These
actions are called the row operations.
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12.3
The Simplex Method: Complete Problems
12-11
THE ROW OPERATIONS
1. Multiply or divide a row by any number (except 0). This is equivalent to
multiplying or dividing an equation by a number, when using the elimination
method.
2. Add one row to another row, or add a multiple of one row to another row. This
is equivalent to adding one equation to another equation or adding a multiple of
an equation to another equation.
3. Interchange any two rows. This is equivalent to interchanging two
equations.
Pivoting
1.
Use the row operations to make the pivot entry a 1. The pivot entry is 200. We’ll
turn it into a 1 by dividing row 2 by 200. (This is row operation 1 in the above list.)
This gives us
6
5
1
0
0
40
£ 200>200 100>200 0>200 1>200 0>200 1000>200 §
240
160
0
0
1
0
d dividing row
2 by 200
6
5
1
0
0 40
£ 1
0.5
0 0.005 0 5 §
240 160 0
0
1 0
Use the row operations to make each of the other entries in the pivot column 0. The
other entries in the pivot column are 6 and 240. First, we will turn the 6 into a 0
by adding –6 times row 2 to row 1. (This is row operation 2 in the above list.) This
gives us
2.
6 # 1 6
1
£
240
6 # 0.5 5
0.5
160
6 # 0 1
0
0
0
2
1 .03
£ 1
0.5
0 0.005
240 160 0
0
0
0
1
6 # 0.005 0 6 # 0 0
0.005
0
0
1
d 6 row 2 row 1
10
6 # 5 40
5
§
0
5 §
0
Now we’ll turn the –240 into a 0 by adding 240 times row 2 to row 3. (This is row
operation 2 in the above list.) This gives us
0
£
1
#
240 1 12402
0
£1
0
2
0.5
#
240 0.5 1160 2
2
1 .03
0.5 0 0.005
40 0
1.2
0
0
1
1
0
#
240 0 0
10
5 §
1200
.03
0.005
#
240 0.005 0
0
0
#
240 0 1
10
5
§
#
240 5 0
d 240 row 2 row 3
The above matrix is called the second simplex matrix. It marks the end of our first
pivot.
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CHAPTER 12
Linear Programming
The simplex method requires us to explore a series of matrices, just as the
geometric method requires us to explore a series of corner points. Each simplex matrix provides us with a corner point of the region of possible solutions. The final
simplex matrix will provide us with the optimal corner point, the point that solves
the problem.
The second simplex matrix
x1 x2 s1
s2
z
0
2
1 0.03 0
10
£ 1 0.5 0 0.005 0
5 §
0 40 0
1.2
1 1,200
provides us with the corner point:
(x1, x2, s1, s2) (5, 0, 10, 0) with z 1,200
This is corner point P4 (see Figure 12.36).
y
P2 (0, 8)
P3 (
5
2
, 5)
P4 (5, 0)
x
P1 (0, 0)
FIGURE 12.36
The simplex method produced corner point
P4 (5, 0) as a possible solution.
Because x1 is the number of coffee tables, x2 is the number of end tables, s1 is
the number of unused hours, and s2 is the amount of unused money, this possible
solution means “make five coffee tables and no end tables, have ten unused hours
and no unused money, and make a profit of $1,200.”
When to Stop Pivoting
Is (5, 0, 10, 0) the optimal corner point, the point that solves the craftsman’s problem? We answer this question by determining whether it is possible to pivot again.
1.
Select the most negative entry in the last row, the objective function row. The column containing that entry is the pivot column. If the last row contains no negative
entries, then no pivoting is necessary; the possible solution that corresponds to the
matrix is the maximal solution.
s2
z
x1 x2 s1
0
2
1 0.03 0
10
£ 1 0.5 0 0.005 0
5 §
0 40 0
1.2
1 1,200
d objective function
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12.3
The Simplex Method: Complete Problems
12-13
Because there is a negative entry in the objective function row, we must pivot
again. This negative entry is in the “x2” column, so our pivot column is the “x2”
column.
x1 x2 s1
s2
z
0
2
1 0.03 0
10
£ 1 0.5 0 0.005 0
5 §
0 40 0
1.2
1 1,200
c
pivot column
2.
Divide the last entry in each constraint row by the corresponding entry in the
pivot column if that entry is positive. The row that yields the smallest nonnegative
such quotient is the pivot row.
0
2
£ 1 0.5
0 40
1 0.03
0 0.005
0
1.2
0
10
0
5 §
1 1,200
d 10>2 5
d 5>0.5 10
d not a constraint row
c
pivot column
Our quotients are 5 and 10; the smallest nonnegative quotient is 5, which is in row
one. Our pivot row is the first row.
3.
Pivot on the entry in the pivot row and pivot column.
0
2
1 0.03 0
10
£ 1 0.5 0 0.005 0
5 §
0 40 0
1.2
1 1,200
c
pivot column
d pivot row
We will pivot on the entry “2.”
Pivoting Again
1.
Use the row operations to make the pivot entry a 1.
0
£1
0
2.
1 0.5
0.5
0
40 0
0.015 0
5
0.005 0
5 §
1.2
1 1,200
d row 1 2
Use the row operations to make each of the other entries in the pivot column 0.
x1
0
£1
0
x2
s1
s2
1
0.5
0.015
0 0.25 0.0125
0
20
0.6
z
0
5
0
2.5 § d 0.5 row 1 row 2
1 1,400 d 40 row 1 row 3
This matrix is our third and last simplex matrix, because it is not possible to
pivot further—the bottom row contains no negative entries. The solution that corresponds to this matrix is the optimal corner point:
(x1, x2, s1, s2) (2.5, 5, 0, 0)
with z 1,400
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CHAPTER 12
Linear Programming
meaning “each week make 2.5 coffee tables and 5 end tables, have no unused
hours and no unused money, and make a profit of $1,400.” See Figure 12.37.
y
P2 (0, 8)
P3 ( 25 , 5)
P4 (5, 0)
x
P1 (0, 0)
FIGURE 12.37
The simplex method produced corner
point P3(52, 5) as the optimal solution.
Why the Simplex Method Works
In the craftsman problem, our first simplex matrix was
x1
6
£ 200
240
x2
s1
5
1
100 0
160 0
s2
0
1
0
z
0
40
0 1,000 §
1
0
and its corresponding possible solution was
(x1, x2, s1, s2) (0, 0, 40, 1,000)
with z 0
This means “make no tables and generate no profit.” We selected the first column
as our pivot column because it was the column with the most negative number in
the bottom. Why do we select the column with the most negative number in the
bottom?
The possible solution corresponding to the first simplex matrix involves no
profit (z 0). Certainly, the profit can be increased by making some tables. If
the craftsman were to make only one type of table, which table would be best?
Coffee tables generate more profit per table ($240 versus $160), so the craftsman
would be better off making coffee tables. This choice of coffee tables over end
tables or of $240 over $160 was made in the pivoting process when we chose the
first column as the pivot column. That choice was made because 240 is larger
than 160.
After selecting the pivot column, we selected the second row as our pivot row
because 5 is the smallest nonnegative quotient. Why do we select the row with the
smallest nonnegative quotient?
6
£ 200
240
5 1 0 0
100 0 1 0
160 0 0 1
40
1,000 §
0
d 40>6 6 23
d 1000>200 5
d not a constraint row
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12.3
The Simplex Method: Complete Problems
12-15
If the craftsman were to make only coffee tables, he should make the largest
amount allowed by the constraints when 0 is substituted for x2 (that is, when no end
tables are made).
Time Constraint
Money Constraint
C1: 6x1 5x2 40
6x1 5 0 40
6x1 40
x
C2: 200x1 100x2 1,000
200x1 100 0 1,000
200x1 1,000
2
40 20
6
3
3
6
x1 1,000
5
200
The time constraint will not be violated as long as x1 6 32 ; the money constraint
will not be violated as long as x1 5. The craftsman can make at most x1 5 coffee tables and not violate either constraint. This choice of 5 over 6 23 was made in
the pivoting process when we chose the second row as the pivot row. That choice
was made because 5 is less than 6 23 .
We have just used logic to determine that if the craftsman makes only one
type of table, he should make coffee tables, and that he could make at most five coffee tables without violating his constraints. The simplex method leads to the same
conclusion. The second simplex matrix was
s2
z
x1 x2 s1
0
2
1 0.03 0
10
£ 1 0.5 0 0.005 0
5 §
0 40 0
1.2
1 1,200
and its corresponding possible solution was
(x1, x2, s1, s2) (5, 0, 10, 0)
with z 1,200
This does suggest that the craftsman make five coffee tables and no end tables.
This matrix was not the final simplex matrix, so this possible solution was not the
optimal solution. That is, the craftsman will increase his profit if he makes more
than one type of table.
The last row of this matrix, the objective function row, is
0
40
0
1.2
1
1,200
We pivoted again because of the presence of a negative number in this row. This
pivoting resulted in an increased profit. Why does the presence of a negative
number imply that if we pivot again, the profit will be increased? The last row
represents the equation
0x1 (40x2) 0s1 1.2s2 1z 1,200
which can be rewritten as
z 1,200 40x2 1.2s2
by solving for z. According to this last equation, the profit is z 1,200 if x2 and s2
are both zero. If x2 is positive, then the profit could be larger than 1,200. Thus, it is
possible to achieve a larger profit by changing x2 from zero to a positive number.
Because x2 measures the number of end tables, this means that the craftsman
should make some end tables. Our last pivot did in fact result in his making some
end tables.
The following list of steps includes those steps developed in Section 12.2,
as well as those developed in this section. Notice that you can check your work at
step 7; if you have made an arithmetic error during your pivot, catch it here before
you go any further.
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CHAPTER 12
Linear Programming
SIMPLEX METHOD STEPS
1. Model the problem.
a. List the independent variables.
b. List the constraints and translate them into linear inequalities.
c. Find the objective and translate it into a linear equation.
2. Convert each constraint from an inequality to an equation.
a. Use one slack variable for each constraint.
b. Determine what each slack variable measures.
3. Rewrite the objective function with all variables on the left side.
4. Make a matrix out of the rewritten constraints and the rewritten objective function. This is the first simplex matrix.
5. Determine the possible solution that corresponds to the matrix.
a. Columns with zeros and a single one—turn at 1.
b. Other columns—the value of the variable is zero.
6. Pivot to find a better possible solution.
a. The pivot column is the column with the most negative entry in the last row.
In case of a tie, choose either.
b. Divide the last entry of each constraint row by the entry in the pivot column, if that entry is positive. The pivot row is the row with the smallest
nonnegative such quotient. In case of a tie, choose either.
c. Pivot on the selected row and column; that is, use the row operations to
transform the selected column into a column with 1 in the pivot row and 0’s
in all other rows. This gives a new simplex matrix.
7. Determine the possible solution that corresponds to the matrix (as in step 5).
✔ Check your work by seeing whether the solution substitutes into the
objective function and constraints.
8. Determine whether the current possible solution maximizes the objective
function.
a. If the last row of the new simplex matrix has no negative entries, then the
problem is solved, and the current solution is the maximal solution.
b. If the last row of the new simplex matrix has one or more negative entries,
then pivot again. Return to step 6.
9. Interpret the final solution. Express the solution in words, as a solution to a
real-world problem rather than a mathematical problem.
EXAMPLE
1
THE LEATHER FACTORY AND RESOURCE ALLOCATION The
Leather Factory has one sewing machine with which it makes coats and vests.
Each coat requires 50 minutes on the sewing machine and uses 12 square feet of
leather. Each vest requires 30 minutes on the sewing machine and uses 8 square
feet of leather. The sewing machine is available 8 hours a day, and the Leather
Factory can obtain 118 square feet of leather a day. The coats sell for $175 each,
and the vests sell for $100 each. Find the daily production level that would yield
the maximum revenue.
SOLUTION
1.
Model the problem.
Independent Variables
x1 number of coats made each day
x2 number of vests made each day
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12.3
The Simplex Method: Complete Problems
12-17
Constraints
C1: sewing machine hours 8 hours
(coat sewing time)
(vest sewing time)
time
number
time
number
° per ¢ · ° of
¢ ° per ¢ · ° of ¢
coat
coats
vest
vests
30
x2
50
x1
C2: leather used 118 square feet
(coat leather)
(vest leather)
480 minutes
480 minutes
480 minutes
480 minutes
118 square feet
leather
number
leather
number
° per ¢ · ° of
¢ ° per ¢ · ° of ¢ 118 square feet
coat
coats
vest
vests
8
x2
118
12
x1
Objective Function
maximize z revenue coat revenue vest revenue
price
number
price
number
z ° per ¢ · ° of
¢ ° per ¢ · ° of ¢
coat
coats
vest
vests
175
x1
100
x2
2.
Convert each constraint from an inequality to an equation.
C1: 50x1 30x2 480
50x1 30x2 s1 480
where s1 unused sewing machine time
C2: 12x1 8x2 118
12x1 8x2 s2 118
where s2 unused leather
3.
Rewrite the objective function with all variables on the left side.
z 175x1 100x2
175x1 100x2 0s1 0s2 1z 0
4.
5.
6.
Make a matrix out of the rewritten constraints and the rewritten objective function.
The first simplex matrix is as follows:
x1
x2
s1 s2 z
50
30
1 0 0 480
£ 12
8
0 1 0 118 §
175 100 0 0 1
0
Determine the possible solution that corresponds to the first simplex matrix. The
solution is (x1, x2, s1, s2) (0, 0, 480, 118) with z 0. This means “make no product and make no profit.”
Pivot to find a better possible solution.
x1
x2
s1 s2 z
50
30
1 0 0 480
£ 12
8
0 1 0 118 §
175 100 0 0 1
0
c
pivot column
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CHAPTER 12
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Our pivot column is the first column, because 175 is the most negative entry in
the bottom row. To select the pivot row, divide the last entry in each constraint row
by the entry in the pivot column, if that entry is positive, and select the row with
the smallest nonnegative such quotient.
50
£ 12
175
d 480>50 9.6
30
1 0 0 480
8
0 1 0 118 §
100 0 0 1
0
d 118>12 9.8333
d not a constraint row
Because 9.6 is the smallest nonnegative quotient, our pivot row is the first row.
50
30
1 0 0 480
£ 12
8
0 1 0 118 §
175 100 0 0 1
0
c
pivot column
d pivot row
We will pivot on the 50.
1
0.6
£ 12
8
175 100
s1
x1 x2
1 0.6 0.02
£ 0 0.8 0.24
0 5
3.5
7.
0.02
0
0
s2
0
1
0
0 0 9.6
1 0 118 §
0 1
0
z
0
9.6
0
2.8 §
1 1,680
d row 1 50
d 12 row 1 row 2
d 175 row 1 row 3
Determine the possible solution that corresponds to the matrix. The possible solution is (x1, x2, s1, s2) (9.6, 0, 0, 2.8) with z 1,680.
Check your work by seeing if the solution substitutes into the objective
function and constraints:
C 1: 50x1 30x2 s1 50 · 9.6 30 · 0 0 480
C 2: 12x1 8x2 s2 12 · 9.6 8 · 0 2.8 118
Objective function: z 175x1 100x2 175 · 9.6 100 · 0
1,680
If any of these substitutions failed, we would know to stop and find our error.
8.
Determine whether the current possible solution maximizes the objective function.
The last row contains no negative entries, so we don’t need to pivot again, and the
problem is finished. Some linear programming problems involve only one pivot,
some involve two, and some involve more. This problem involved only one pivot.
9.
Interpret the final solution. Recall that x1 measures the number of coats made per
day, x2 measures the number of vests, s1 measures the number of slack machine
hours, and s2 measures the amount of unused leather. Thus, the Leather Factory
should make 9.6 coats per day and no vests. (If workers spend all day making
coats, they will make 9.6 coats. They’ll finish the tenth coat the next day.) This
will result in no unused machine hours and 2.8 square feet of unused leather and
will generate a maximum revenue of $1,680 per day. These profit considerations
indicate that the Leather Factory should increase the cost of its leather vests if it
wants to sell vests.
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12.3 Exercises
In Exercises 1–4, determine where to pivot.
1.
2.
3.
4.
x1 x2 s1 s2 z
5
0
3
1 0
£ 3
0 19 0 10
21 1 48 0 0
x1
1
£0
0
x2 s1 s2
31 0 9
20 0 15
1 1 0
x1
19
32
≥
55
21
x1
0
0
≥
1
0
z
0 7.4
1 4.9 §
0 20
x2 s1 s2 s3
3 0 0 1
7 0 1 0
8 1 0 0
32 0 0 0
x2 s1
1 7
0 5
0 0
0 9
12
22 §
19
z
0
0
0
1
5
93
¥
8
96
s2 s3 z
0 62 0 5
1 5 0 93
¥
0 0 0 78
1 1 1 96
In Exercises 5–8, (a) determine where to pivot, (b) pivot,
and (c) determine the solution that corresponds to the
resulting matrix.
5.
6.
7.
8.
x1 x2 s1
1
2 1
£ 4
1 0
6 4 0
s2 z
0 0 3
1 0 2§
0 1 0
x1 x2 s1 s2
8
2 1 0
£ 5
1 0 1
2 4 0 0
z
0
0
1
x1 x2 s1 s2
1 1 5 0
£ 3 0 1 1
6 0 2 0
x1 x2 s1 s2
0 2 1 5
£1 3 0 9
0 4 0 2
4
3§
0
z
0 3
0 12 §
1 6
z
0
0
2
8
6 §
12
Use the simplex method to solve Exercises 9–16. (Some of these
exercises were started in Section 12.2.)
9. Five friends, all of whom are experienced bakers, form a
company that will make bread and cakes and sell them
to local restaurants and specialty stores. Each loaf of
bread requires 50 minutes of labor and ingredients
costing $0.90 and can be sold for $1.20 profit. Each cake
requires 30 minutes of labor and ingredients costing
$1.50 and can be sold for $4.00 profit. The partners
agree that no one will work more than eight hours a day.
Their financial resources do not allow them to spend
more than $190 per day on ingredients. How many
loaves of bread and how many cakes should they make
each day to maximize their profit? What is the
maximum profit? Will this leave any extra time or
money? If so, how much?
10. A craftswoman produces two products: floor lamps and
table lamps. Production of one floor lamp requires
75 minutes of her labor and materials that cost $25.
Production of one table lamp requires 50 minutes of
labor and materials that cost $20. The craftswoman
wants to work no more than 40 hours each week, and
her financial resources allow her to pay no more than
$900 for materials each week. If she can sell as many
lamps as she can make and if her profit is $40 per floor
lamp and $32 per table lamp, how many floor lamps
and how many table lamps should she make each week
to maximize her weekly profit? What is that maximum
profit? Will this leave any unused time or money? If so,
how much?
11. A furniture manufacturing firm makes sofas and
chairs, each of which is available in several styles.
Each sofa, regardless of style, requires eight hours in
the upholstery shop and four hours in the carpentry
shop and can be sold for a profit of $450. Each chair
requires six hours in the upholstery shop and 3.5 hours
in the carpentry shop and can be sold for a profit of
$375. There are nine people working in the upholstery
shop and five in the carpentry shop, each of whom can
work no more than 40 hours per week. How many
sofas and how many chairs should the firm make each
week to maximize its profit? What is that maximum
profit? Would this leave any extra time in the
upholstery shop or the carpentry shop? If so, how
much?
12. City Electronics Distributors handles two lines of
televisions: the Packard and the Bell. The company
purchases up to $57,000 worth of television sets from
the manufacturers each month, which it stores in its
9,000-cubic-foot warehouse. The Packards come in
36-cubic-foot packing crates, and the Bells come in 30cubic-foot crates. The Packards cost City Electronics
$200 each and can be sold to a retailer for a $200 profit,
while the Bells cost $250 each and can be sold for a
$260 profit. How many sets should City Electronics
order each month to maximize profit? What is that
maximum profit? Would this leave any unused storage
space or money? If so, how much?
Selected exercises available online at www.webassign.net/brookscole
12-19
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12-20
CHAPTER 12
Linear Programming
13. J & M Winery makes two jug wines: House White and
Premium White, which it sells to restaurants. House
White is a blend of 75% French colombard grapes and
25% sauvignon blanc grapes, and Premium White is
75% sauvignon blanc grapes and 25% French
colombard grapes. J & M also makes a Sauvignon
Blanc, which is 100% sauvignon blanc grapes. Profit on
the House White is $1.00 per liter, profit on the
Premium White is $1.50 per liter, and profit on the
Sauvignon Blanc is $2.00 per liter. This season, J & M
can obtain 30,000 pounds of French colombard grapes
and 20,000 pounds of sauvignon blanc grapes. It takes
2 pounds of grapes to make 1 liter of wine. If J & M can
sell all that it makes, how many liters of House White,
of Premium White, and of Sauvignon Blanc should J &
M prepare to maximize profit? What is that maximum
profit? Would this leave any extra grapes? If so, what
amount?
14. J & M Winery makes two jug wines: House Red and
Premium Red, which it sells to restaurants. House Red
is a blend of 20% cabernet sauvignon grapes and
80% gamay grapes. Premium Red is 60% cabernet
sauvignon grapes and 40% gamay grapes. J & M
also makes a Cabernet Sauvignon, which is 100%
cabernet sauvignon grapes. Profit on the House Red is
$0.90 per liter, profit on the Premium Red is $1.60 per
liter, and profit on the Cabernet Sauvignon is
$2.50 per liter. This season, J & M can obtain
30,000 pounds of gamay grapes and 22,000 pounds of
cabernet sauvignon grapes. It takes 2 pounds of grapes
to make 1 liter of wine. If J & M can sell all that it
makes, how many liters of House Red, of Premium
Red, and of Cabernet Sauvignon should J & M prepare
to maximize profit? What is that maximum profit?
Would this leave any extra grapes? If so, what
amount?
15. Pete’s Coffees sells two blends of coffee beans: Smooth
Sipper and Kona Blend. Smooth Sipper is composed of
equal amounts of Kona, Colombian, and Arabian beans,
and Kona Blend is one-half Kona beans and one-half
Colombian beans. Profit on the Smooth Sipper is $3 per
pound, while profit on the Kona Blend is $4 per pound.
Each day, the shop can obtain 100 pounds of Kona beans,
200 pounds of Colombian beans, and 200 pounds of
Arabian beans. It uses those beans only in the two blends.
If it can sell all that it makes, how many pounds of
Smooth Sipper and of Kona Blend should Pete’s Coffees
prepare each day to maximize profit? What is that
maximum profit? Would this leave any extra beans? If so,
what amount?
16. Pete’s Coffees sells two blends of coffee beans:
African Blend and Major Thompson’s Blend. African
Blend is one-half Tanzanian beans and one-half
Ethiopian beans, and Major Thompson’s Blend is one-
quarter Tanzanian beans and three-quarters Colombian
beans. Profit on the African Blend is $4.25 per pound,
while profit on Major Thompson’s Blend is $3.50 per
pound. Each day, the shop can obtain 300 pounds of
Tanzanian beans, 200 pounds of Ethiopian beans, and
450 pounds of Colombian beans. It uses those beans
only in the two blends. If it can sell all that it makes,
how many pounds of African Blend and of Major
Thompson’s Blend should Pete’s Coffees prepare
each day to maximize profit? What is that maximum
profit? Would this leave any extra beans? If so, what
amount?
•
Concept Questions
17. Do Exercise 11 with the geometric method. Which of
the geometric method’s corner points were found with
the simplex method?
18. Do Exercise 12 with the geometric method. Which of
the geometric method’s corner points were found with
the simplex method?
Answer the following questions with complete
sentences and your own words.
19. Why does the simplex method exist? That is, why do
we not use the geometric method for all linear programming problems?
20. Compare and contrast the geometric method of linear
programming with the simplex method of linear
programming. Be sure to discuss both their similarities
and their differences. Also give advantages and disadvantages of each.
21. Explain how the maximal value of the objective
function can be found without looking at the number
in the lower right corner of the final simplex matrix.
22. Explain why the entries in the last row have had their
signs changed but the entries in all of the other rows
have not had their signs changed.
23. Choose one of Exercises 9–16, and explain why the
simplex method works in that exercise. Base your
explanation on the discussion in this section titled
“Why the Simplex Method Works.”
•
History Question
24.
Who invented the simplex method of linear programming?
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TECHNOLOGY AND THE SIMPLEX METHOD
One of the simplex method’s advantages over the geometric method is that the
simplex method is amenable to the use of a computer or graphing calculator.
In the real world, linear programming problems have too many constraints
and variables to do the work by hand. Instead, they are always done with a computer. In fact, George Dantzig, the inventor of linear programming, has stated that
he had to delay his invention until computers became sufficiently advanced.
The Row Operations on a Graphing Calculator
Many graphing calculators will perform the row operations. We will illustrate this
process with the craftsman problem from earlier in this section. The first simplex
matrix is
6
£ 200
240
5
100
160
1
0
0
0
1
0
0
0
1
40
1,000 §
0
We will discuss how to enter this matrix into the calculator and how to perform the
first two row operations on it.
ENTERING A MATRIX
To enter the above matrix on a TI, select “EDIT” from the “MATRIX” menu and
then matrix A from the “EDIT” menu, as described below.
TI-83/84
•
•
•
•
Press MATRX ( 2nd MATRIX on a TI-84).
Use the ➤ button to highlight “EDIT”.
Highlight option 1 “[A]”.
Press ENTER .
Casio
•
•
•
•
Press MENU .
Use the arrow buttons to highlight “MAT”.
Press EXE .
Use the arrow buttons to highlight “Mat A”.
Our matrix has three rows and six columns. Enter these dimensions by
typing
3
ENTER 6 ENTER
as shown in Figure 12.38. Enter the matrix’s elements by typing
FIGURE 12.38
Entering the matrix on a
TI-83/84.
6
ENTER 5 ENTER 1 ENTER 0 ENTER
Continue in this manner until you have entered all of the matrix’s elements. See
Figure 12.38.
When you’re done entering, type 2nd QUIT .
12-21
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Linear Programming
VIEWING A MATRIX
To view a previously entered matrix, do the following.
TI-8384
•
•
•
•
•
Press MATRX ( 2nd MATRIX on a TI-84).
Highlight “NAMES”.
Highlight option 1 “[A]”.
Press ENTER .
Once “[A]” is on the screen, press ENTER .
Casio
•
•
•
•
Press MENU .
Use the arrow buttons to highlight “MAT” and press EXE .
Use the arrow buttons to highlight “Mat A”.
Press EXE , and the matrix will appear on the screen.
Use the ➤ button to view the right side of the matrix. See Figure 12.39.
MULTIPLYING OR DIVIDING A ROW BY A NUMBER
Our first pivot is to divide row 2 by 200. On a graphing calculator, we can’t divide
a row; instead, we multiply it by 1200. To do this, make the screen read
“*row(1200,[A],2)”
FIGURE 12.39
Viewing the matrix on a
TI-83/84.
TI-83/84
Type 1 200 , .
• Press MATRX
( 2nd MATRIX on
a TI-84).
• Highlight “NAMES”.
• Highlight option
1 “[A]”.
• Press ENTER .
Type , 2 ) ENTER .
This makes “*row(”
or “multR(” appear
on the screen.
This makes “1200,”
appear on the screen.
This makes “[A]”
or “A” appear on
the screen
This makes “,2)”
appear on the
screen.
➤
• Press MATRX
( 2nd MATRIX on
a TI-84).
• Use the ➤ button
to highlight “MATH”.
• Use the
button
to highlight option E
“*row(”.
• Press ENTER .
Your calculator will respond by performing the row operation. See Figure 12.40.
FIGURE 12.40
Dividing row 2 by 200 on a
TI-83/84.
On a Casio:
• Press F1 , which is now labeled “R-OP”.
• Press F2 , which is now labeled “xRw”.
• Press 1 200 EXE . This is the multiplier.
• Press 2 EXE . This selects the row that will be multiplied.
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Technology and the Simplex Method
12-23
STORING A MATRIX
Currently, we have the first simplex matrix stored as matrix [A]. This allows us to
avoid reentering the matrix if we make a mistake later and have to start over. We
will store the matrix in Figure 12.40 as matrix [B].
TI-8384
•
•
•
•
•
Press STO➤ .
Press MATRX ( 2ND MATRX on a TI-84).
Highlight “NAMES”.
Highlight option 1“[B]”.
Press ENTER .
Casio
The Casio automatically stores the result of any matrix calculation
as matrix A.
Adding a Multiple of One Row to Another Row
Our second pivot is to multiply row 2 by –6 and add the result to row 1. To do this,
make the screen read “*row+(–6,[B],2,1)”:
TI-8384
Type () 6 , .
• Press MATRX
( 2nd MATRX on
a TI-84).
• Highlight “NAMES”.
• Highlight “[B]”.
• Press ENTER .
Type , 2 , 1 )
ENTER .
This makes “*row(”
or “multR(“ appear
on the screen.
This makes “6,”
appear on the screen.
This makes “[B]”
or “B” appear on
the screen.
This makes “,2, 1)”
appear on the
screen.
➤
• Press MATRX
( 2nd MATRX on
a TI-84).
• Use the ➤ button
to highlight “MATH”.
• Use the
button
to highlight option F
“*row(”.
• Press ENTER .
Your calculator will respond by performing the row operation. See
Figure 12.41.
Store the matrix in Figure 12.41 as [C]. Store the result of the next row operation as [B], and then alternate between [C] and [B] in storing later results. This
keeps the previous matrix available, in case you made any errors.
FIGURE 12.41
Adding 6 # row 2 to row 1
on a TI-83/84.
On a Casio:
• Press F3 , which is now labeled “xRw”.
• Press () 6 EXE . This is the multiplier.
• Press 2 EXE . This selects the row that will be multiplied.
• Press 1. This selects the row that will be added.
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ADDING TWO ROWS
In the rest of the pivoting on the above matrix, we don’t need to add two rows. In
general, though, to add rows 1 and 2 of matrix [A] and place the result in row 2:
TI-8384
Make the screen read row+([A],1, 2).
Casio
Press F4 1 EXE 2 EXE .
Maintaining Accuracy
Sometimes, after a few pivots, the resulting simplex matrix contains nothing but
twelve decimal place numbers. This can make the work more cumbersome.
The work can be easier if you use fractions rather than decimals. After a
calculation, type “NFrac,” and your matrix will be displayed with fractions rather
than decimals. To find the “NFrac” command, do the following:
TI-8384
• Press MATH.
• Highlight option 1.
The decimal difficulty does not arise with the matrices discussed above. With
a different matrix, the decimal difficulty does come up. See Figure 12.42.
Using decimals, without the
“NFrac” command.
FIGURE 12.42
Using fractions, with the
“NFrac” command.
Fractions make the simplex method easier.
Using decimals, you would multiply row 1 by 1>1.984824623. Using fractions, you multiply row 1 by 199>395. A few pivots later, you get one of the results
shown in Figure 12.43.
Notice the “–8.1518E–10” in the decimal version in Figure 12.43. This is the
way a TI calculator displays scientific notation. It means
8.1518 # 10 10 0.0000000008158 0
In the fraction version, you get 0.
Using decimals, without the
“NFrac” command.
FIGURE 12.43
Using fractions, with the
“NFrac” command.
Fractions make the simplex method more accurate.
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Technology and the Simplex Method
12-25
Summary of TI-83/84 Row Operations
*row(1/200, [A], 2)
Multiply a row by
a number
The
multiplier
The row that is
being multiplied
The matrix
whose row is
being multiplied
*row(6, [B], 2, 1)
Add a multiple of a
row to another row
The matrix
whose row is
being multiplied
The
multiplier
The row that is
being added
The row that is
being multiplied
row([A], 1, 2)
The matrix
whose rows are
being added
Add two
rows
The row that is being
added and replaced
The row that is
being added
The Row Operations and Amortrix
Amortrix is one of the features of the text web site www.cengage.com/math/
johnson. This software will enable you to quickly and easily perform the row operations on a matrix.
We will illustrate this process with the craftsman problem from earlier in this
section. The first simplex matrix is
6
£ 200
240
5
100
160
1
0
0
0
1
0
0
0
1
40
1,000 §
0
We will discuss how to enter this matrix into the calculator and how to perform the
first three row operations on it.
ENTERING THE MATRIX
When you start Amortrix, a main menu appears. Click on the “Matrix Row Operations” option, and you will be asked for the dimensions of the matrix you want to
work on. The above matrix has three rows and six columns, so after “Number of
Rows” enter “3,” and after “Number of columns” enter “6.” Click on “Create New
Matrix,” and the program will display a 3 6 matrix containing nothing but zeros.
Use your mouse and the computer’s tab button to enter the above numbers.
PIVOTING
To pivot on 200, first divide row two by 200. To do this, click on the “Two” label to
the left of the matrix. This selects row two and causes the number 2 to appear in a
box below the matrix labeled “1st Row.” The cursor automatically moves to a box
labeled “Enter a constant.” Type 200 in that box, and press the “Divide” button.
Next, we need to multiply row 2 by –6 and add the result to row 1. To multiply
row 2 by –6, click on the “Two” label to the left of the matrix and type –6 in the
box labeled “Enter a constant.” When adding a multiple of one row to another row,
always click first on the row to be multiplied. To add the result to row 1, click on
the “One” label. To place the result in row 3, type 3 in the box labeled “Enter a row
number.” Press the “Multiply & Add” button.
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In a similar manner, multiply row 2 by 240 and add the result to row 3. This
completes the first pivot. The results are shown in Figure 12.44.
FIGURE 12.44
The craftsman problem after the first three row operations, using Amortrix.
EXERCISES
Use a graphing calculator or Amortrix to solve the following linear programming problems.
25. Objective function:
Maximize z 25x1 53x2 18x3 7x4
Constraints:
3x1 2x2 5x3 12x4 28
4x1 5x2 x3 7x4 32
x1 7x2 9x3 10x4 25
26. Objective function:
Maximize z 275x1 856x2 268x3 85x4
Constraints:
5.2x1 9.8x2 7.2x3 3.7x4 33.6
3.9x1 5.3x2 1.4x3 2.5x4 88.3
5.2x1 7.7x2 4.6x3 4.6x4 24.7
27. Objective function:
Maximize z 37x1 19x2 53x3 49x4
Constraints:
6.32x1 7.44x2 8.32x3 1.46x4 9.35x5 63
8.36x1 5.03x2 x3 5.25x5 32
1.14x1 9.42x2 9.39x3 10.42x4 9.32x5 14.7
28. Objective function:
Maximize z 17x1 26x2 85x3 63x4 43x5
Constraints:
72x1 46x2 73x3 26x4 54x5 185
37x1 84x2 45x3 83x4 85x5 237
29. Our friend the craftsman now owns his own shop. He
still makes coffee tables and end tables, but he makes
each in three different styles: antique, art deco, and
modern. The amount of labor and the cost of the
materials required by each product, along with
the profit they generate, are shown in Figure 12.45. The
craftsman now has two employees, each of whom can
work up to 40 hours a week. The craftsman himself
frequently has to work more than 40 hours a week, but
he will not allow himself to exceed 50 hours a week.
His new bank loan allows him to spend up to $4000 a
week on materials. How many coffee tables and end
tables of each style should he make each week to
maximize his profit?
Style
Hours of
labor
Cost of
materials
Profit
Coffee table
antique
6.00
$230
$495
Coffee table
art deco
6.25
$220
$500
Coffee table
modern
5.00
$190
$430
End table
antique
5.25
$125
$245
End table
art deco
5.75
$120
$250
End table
modern
4.25
$105
$220
Item
FIGURE 12.45
The craftsman’s production data.
30. The five friends have been so successful with their
baking business that they have opened their own shop,
the Five Friends Bakery. The bakery sells two types of
breads: nine-grain and sourdough; two types of cakes:
chocolate and poppyseed; and two types of muffins:
blueberry and apple-cinnamon. Production data are
given in Figure 12.46. The five partners have hired four
workers. No one works more than 40 hours a week.
Their business success (together with their new bank
loan) allows them to spend up to $2000 per week on
ingredients. What weekly production schedule should
they follow to maximize their profit?
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Exercises
Labor
Cost of ingredients
Profit
Nine grain
bread
50 min. per loaf
$1.05 per loaf
$0.60 per loaf
Sour dough
bread
50 min. per loaf
$0.95 per loaf
$0.70 per loaf
Chocolate cake
35 min. per cake
$2.00 per cake
$2.50 per cake
Poppy seed cake
30 min. per cake
$1.55 per cake
$2.00 per cake
Blueberry muffins
15 min. per dozen
$1.60 per dozen
$16.10 per dozen
Apple cinnamon
muffins
15 min. per dozen
$1.30 per dozen
$14.40 per dozen
FIGURE 12.46
The Five Friends Bakery production data.
31. Fiat Lux, Inc., manufactures floor lamps, table lamps,
and desk lamps. Production data are given in Figure 12.47. There are seven employees in the wood
shop, five in the metal shop, three in the electrical shop,
and one in testing; each works no more than 40 hours
per week. If Fiat Lux, Inc., can sell all the lamps it
produces, how many should it produce each week?
Wood
shop
Metal
shop
Electrical
shop
Testing
Profit
Floor
lamps
20 mins.
30 mins.
15 mins.
5 mins.
$55
Table
lamps
25 mins.
15 mins.
12 mins.
5 mins.
$45
Desk
lamps
30 mins.
10 mins.
11 mins.
4 mins.
$40
FIGURE 12.47
Fiat Lux’s production data.
32. A furniture-manufacturing firm makes sofas, loveseats,
easy chairs, and recliners. Each piece of furniture is
constructed in the carpentry shop, then upholstered, and
finally coated with a protective coating. Production data
are given in Figure 12.48. Fifteen people work in the
upholstery shop, nine work in the carpentry shop, and
one person applies the protective coating, and each
person can work no more than 40 hours per week. How
many pieces of furniture should the firm manufacture
each week to maximize its profit?
Carpentry
Upholstery
Coating
Profit
Sofas
4 hours
8 hours
20 mins.
$450
Love seats
4 hours
7.5 hours
20 mins.
$350
Easy chairs
3 hours
6 hours
15 mins.
$300
Recliners
6 hours
6.5 hours
15 mins.
$475
FIGURE 12.48
12-27
Furniture production data.
33. City Electronics Distributors handles two lines of
televisions: the Packard and the Bell. The Packard
comes in three sizes: 18-inch, 24-inch, and rear
projection; the Bell comes in two sizes: 18-inch and
30-inch. City Electronics purchases up to $57,000
worth of television sets from manufacturers each
month and stores them in its 9000-cubic-foot
warehouse. Storage and financial data are given in
Figure 12.49. How many sets should City Electronics
order each month to maximize profit?
Set
Size
Cost
Profit
Packard 18''
30 cubic ft.
$150
$175
Packard 24''
36 cubic ft.
$200
$200
Packard rear
projection
65 cubic ft.
$450
$720
Bell 18''
28 cubic ft.
$150
$170
Bell 30''
38 cubic ft.
$225
$230
FIGURE 12.49
City Electronics inventory data.
34. J & M Winery makes two jug wines, House White and
Premium White, and two higher-quality wines,
Sauvignon Blanc and Chardonnay, which it sells to
restaurants, supermarkets, and liquor stores. House
White is a blend of 40% French colombard grapes,
40% chenin blanc grapes, and 20% sauvignon blanc
grapes, while Premium White is 75% sauvignon blanc
grapes and 25% French colombard grapes. J & M’s
Sauvignon Blanc is 100% sauvignon blanc grapes, and
its Chardonnay is 90% chardonnay grapes and 10%
chenin blanc grapes. Profit on the House White is
$1.00 per liter, profit on the Premium White is
$1.50 per liter, profit on the Sauvignon Blanc is $2.25 per
liter, and profit on the Chardonnay is $3.00 per liter.
This season, J & M can obtain 30,000 pounds of
French colombard grapes, 25,000 pounds of chenin
blanc grapes, and 20,000 pounds each of sauvignon
blanc grapes and chardonnay grapes. It takes 2 pounds
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CHAPTER 12
Linear Programming
of grapes to make 1 liter of wine. If the company can
sell all it makes, how many liters of its various
products should J & M prepare to maximize profit?
35. J & M Winery makes two jug wines, House Red and
Premium Red, and two higher-quality wines, Cabernet
Sauvignon and Zinfandel, which it sells to restaurants,
supermarkets, and liquor stores. House Red is a blend
of 20% pinot noir grapes, 30% zinfandel grapes, and
50% gamay grapes. Premium Red is 60% cabernet
sauvignon grapes and 20% each pinot noir and gamay
grapes. J & M’s Cabernet Sauvignon is 100% cabernet
sauvignon grapes, and its Zinfandel is 85% zinfandel
grapes and 15% gamay grapes. Profit on the House
Red is $0.90 per liter, profit on the Premium Red is
$1.60 per liter, profit on the Zinfandel is $2.25 per liter,
and profit on the Cabernet Sauvignon is $3.00 per liter.
This season, J & M can obtain 30,000 pounds each of
pinot noir, zinfandel, and gamay grapes and 22,000
pounds of cabernet sauvignon grapes. It takes 2 pounds
of grapes to make 1 liter of wine. If the company can
sell all that it makes, how many liters of its various
products should J & M prepare in order to maximize
profit?
•
Project
36. How many sheets of half-inch-thick plate glass and
three-eighths-inch thick plate glass should Glassco
make at each of its two plants to maximize its profit?
What is the maximum profit? Use the production data
and the results of Exercise 37 of Section 12.1.
REVIEW EXERCISES FOR SECTIONS 12.2 AND 12.3
Solve Exercises 1–4 with the simplex method of linear
programming.
1. Maximize z 4x1 7x2 8x3
Constraints:
x1 2x2 4x3 30
4x1 x2 3x3 20
x1, x2, x3 0
2. Maximize z 4.6x1 3.2x2
Constraints:
7x1 3x2 420
x1 x2 80
2x1 5x2 100
x1, x2 0
3. Maximize z 2x1 3x2 5x3
Constraints:
15x1 x2 2x3 45
x1 6x2 3x3 40
8x1 7x2 70
x1, x2, x3 0
4. Maximize z 8x1 6x2
Constraints:
x1 2x2 100
52x1 44x2 2288
25x1 20x2 1100
x1, x2 0
5. The Mowson Audio Co. makes stereo speaker
assemblies. They purchase speakers from a speaker
manufacturing firm and install them in their own
cabinets. Mowson’s model 110 speaker assembly,
which sells for $200, has a tweeter and a midrange
speaker. Their model 330 assembly, which sells for
$350, has two tweeters, a midrange speaker, and a
woofer. Mowson currently has in stock 90 tweeters,
60 midrange speakers, and 44 woofers. Use the
simplex method to determine how many speaker
assemblies Mowson should make in order to maximize
their income. What is that maximum income?
6. The Mowson Audio Co. (see Exercise 5) has
introduced a new speaker assembly. The new model
220 has a tweeter, a midrange speaker, and a woofer. It
sells for $280. If Mowson Audio has 140 tweeters,
90 midrange speakers, and 66 woofers in stock, how
many speaker assemblies should they make to
maximize their income? Interpret the solution in
addition to answering the question.