LP2As the EEM of Engineering Department of Taxila.ppt

6
6
 Graphing Systems of Linear
Graphing Systems of Linear
Inequalities in Two Variables
Inequalities in Two Variables
 Linear Programming Problems
Linear Programming Problems
 Graphical Solutions of Linear
Graphical Solutions of Linear
Programming Problems
Programming Problems
 The Simplex Method:
The Simplex Method:
Standard Maximization Problems
 The Simplex Method:
The Simplex Method:
Standard Minimization Problems
Linear Programming: A Geometric Approach
Linear Programming: A Geometric Approach

6.1
6.1
Graphing Systems of Linear Inequalities
in Two Variables
in Two Variables
x
x
y
y
4
4x
x + 3
+ 3y
y =
= 12
12
12 12
7 7
( , )
P 12 12
7 7
( , )
P
x
x –
– y
y =
= 0
0
4 3 12
0
x y
x y
 
 
4 3 12
0
x y
x y
 
 
4
4
3
3
2
2
1
1
–
–1
1 1
1 2
2 3
3

Graphing Linear Inequalities
Graphing Linear Inequalities
 We’ve seen that
We’ve seen that a linear
a linear equation
equation in two variables
in two variables x
x and
and y
y
has a
has a solution set
solution set that may be exhibited
that may be exhibited graphically
graphically as
as points
points
on a straight line
on a straight line in the
in the xy
xy-plane
-plane.
.
 There is also a simple
There is also a simple graphical representation
graphical representation for
for linear
linear
inequalities
inequalities of two variables
of two variables:
:
0
ax by c
  
0
ax by c
  
0
ax by c
  
0
ax by c
  
0
ax by c
  

Procedure for Graphing Linear Inequalities
Procedure for Graphing Linear Inequalities
1.
1. Draw the
Draw the graph
graph of the
of the equation
equation obtained for the given
obtained for the given
inequality by
inequality by replacing the inequality sign with an
replacing the inequality sign with an
equal sign
equal sign.
.
✦ Use a
Use a dashed or dotted line
dashed or dotted line if the problem involves a
if the problem involves a
strict inequality
strict inequality,
, <
< or
or >
>.
.
✦ Otherwise, use a
Otherwise, use a solid line
solid line to indicate that
to indicate that the line
the line
itself constitutes part of the solution
itself constitutes part of the solution.
.
2.
2. Pick a test point
Pick a test point lying in one of the half-planes
lying in one of the half-planes
determined by the line sketched in
determined by the line sketched in step 1
step 1 and
and substitute
substitute
the values of
the values of x
x and
and y
y into the given
into the given inequality
inequality.
.
✦ Use the
Use the origin
origin whenever possible.
whenever possible.
3.
3. If the
If the inequality is satisfied
inequality is satisfied, the graph of
, the graph of the inequality
the inequality
includes the half-plane
includes the half-plane containing the
containing the test point
test point.
.
✦ Otherwise, the solution includes the half-plane not
Otherwise, the solution includes the half-plane not
containing the test point.
containing the test point.

Examples
Examples
 Determine the
Determine the solution set
solution set for the
for the inequality
inequality 2
2x
x + 3
+ 3y
y 
 6
6.
.
Solution
Solution
 Replacing
Replacing the
the inequality
inequality 
 with an
with an equality
equality =
=, we obtain
, we obtain
the equation
the equation 2
2x
x + 3
+ 3y
y = 6
= 6, whose graph is:
, whose graph is:
x
x
y
y
7
7
5
5
3
3
1
1
–
–1
1
–
–5
5 –
–3
3 –
–1
1 1
1 3
3 5
5
2
2x
x + 3
+ 3y
y = 6
= 6

Examples
Examples
 Determine the
Determine the solution set
solution set for the
for the inequality
inequality 2
2x
x + 3
+ 3y
y 
 6
6.
.
Solution
Solution
 Picking the
Picking the origin
origin as a
as a test point
test point, we find
, we find 2(0) + 3(0)
2(0) + 3(0) 
 6
6,
,
or
or 0
0 
 6
6, which is
, which is false
false.
.
 Thus, the
Thus, the solution set
solution set is:
is:
x
x
y
y
7
7
5
5
3
3
1
1
–
–1
1
–
–5
5 –
–3
3 –
–1
1 1
1 3
3 5
5
2
2x
x + 3
+ 3y
y = 6
= 6
2
2x
x + 3
+ 3y
y 
 6
6
(0, 0)
(0, 0)

 The
The solution set
solution set of a
of a system of linear inequalities
system of linear inequalities in two
in two
variables
variables x
x and
and y
y is the
is the set of all points
set of all points (
(x
x,
, y
y)
) that
that satisfy
satisfy
each inequality
each inequality of the system.
of the system.
 The
The graphical solution
graphical solution of such a system may be obtained
of such a system may be obtained
by
by graphing the solution set for each inequality
graphing the solution set for each inequality
independently and then
independently and then determining the region in common
determining the region in common
with each solution set.
with each solution set.

–
–5
5 –
–3
3 1
1 3
3 5
5
Examples
Examples
 Graph
Graph x
x – 3
– 3y
y > 0
> 0.
.
Solution
Solution
 Replacing
Replacing the
the inequality
inequality >
> with an
with an equality
equality =
=, we obtain
, we obtain
the equation
the equation x
x – 3
– 3y
y = 0
= 0, whose graph is:
, whose graph is:
x
x
y
y
3
3
1
1
–
–1
1
–
–3
3
x
x – 3
– 3y
y = 0
= 0

Examples
Examples
 Graph
Graph x
x – 3
– 3y
y > 0
> 0.
.
Solution
Solution
 We use a
We use a dashed line
dashed line to indicate
to indicate the line itself will
the line itself will not
not be
be
part of the solution
part of the solution, since we are dealing with a
, since we are dealing with a strict
strict
inequality
inequality >
>.
.
x
x
y
y
x
x – 3
– 3y
y = 0
= 0
–
–5
5 –
–3
3 1
1 3
3 5
5
3
3
1
1
–
–1
1
–
–3
3

–
–5
5 –
–3
3 1
1 3
3 5
5
3
3
1
1
–
–1
1
–
–3
3
Examples
Examples
 Graph
Graph x
x – 3
– 3y
y > 0
> 0.
.
Solution
Solution
 Since the origin lies on the line, we
Since the origin lies on the line, we cannot use the origin
cannot use the origin
as a
as a testing point
testing point:
:
x
x
y
y
x
x – 3
– 3y
y = 0
= 0
(0, 0)
(0, 0)

Examples
Examples
 Graph
Graph x
x – 3
– 3y
y > 0
> 0.
.
Solution
Solution
 Picking instead
Picking instead (3, 0)
(3, 0) as a
as a test point
test point, we find
, we find (3) – 2(0) > 0
(3) – 2(0) > 0,
,
or
or 3 > 0
3 > 0, which is
, which is true
true.
.
 Thus, the
Thus, the solution set
solution set is:
is:
y
y
x
x – 3
– 3y
y = 0
= 0
x
x – 3
– 3y
y > 0
> 0
–
–5
5 –
–3
3 1
1 3
3 5
5
3
3
1
1
–
–1
1
–
–3
3
x
x
(3, 0)
(3, 0)

Example
Example
 Determine the solution set for the system
Determine the solution set for the system
Solution
Solution
 The
The intersection
intersection of the
of the solution regions
solution regions of the two
of the two
inequalities
inequalities represents the
represents the solution to the system
solution to the system:
:
4 3 12
0
x y
x y
 
 
x
x
y
y
4
4
3
3
2
2
1
1
4
4x
x + 3
+ 3y
y 
 12
12
4
4x
x + 3
+ 3y
y = 12
= 12
–
–1
1 1
1 2
2 3
3

Example
Example
Solution
Solution
 The
The intersection
intersection of the
of the two
inequalities
:
4 3 12
0
x y
x y
 
 
x
x
y
y
x
x –
– y
y 
 0
0 x
x –
– y
y = 0
= 0
4
4
3
3
2
2
1
1
–
–1
1 1
1 2
2 3
3

Example
Example
Solution
Solution
 The
The intersection
intersection of the
of the two
inequalities
:
4 3 12
0
x y
x y
 
 
x
x
y
y
4
4x
x + 3
+ 3y
y = 12
= 12
x
x –
– y
y = 0
= 0
4 3 12
0
x y
x y
 
 
4
4
3
3
2
2
1
1
–
–1
1 1
1 2
2 3
3
12 12
7 7
( , )
P

Bounded and Unbounded Sets
Bounded and Unbounded Sets
 The
The solution set
solution set of a system of linear inequalities
of a system of linear inequalities
is
is bounded
bounded if it
if it can be enclosed by a circle
can be enclosed by a circle.
.
 Otherwise, it is
Otherwise, it is unbounded
unbounded.
.

Example
Example
 The solution to the problem we just discussed is
The solution to the problem we just discussed is
unbounded
unbounded, since
, since the solution set
the solution set cannot be
cannot be
enclosed in a circle
enclosed in a circle:
:
x
x
y
y
4
4x
x + 3
+ 3y
y = 12
= 12
12 12
7 7
( , )
P
x
x –
– y
y = 0
= 0
4 3 12
0
x y
x y
 
 
4
4
3
3
2
2
1
1
–
–1
1 1
1 2
2 3
3

7
7
5
5
3
3
1
1
–
–1
1 1
1 3
3 5
5 9
9
Example
Example
Solution
Solution
 The
The intersection
intersection of the
solution regions of the four
of the four
inequalities
:
6 0 2 8 0 0 0
x y x y x y
       
x
x
y
y
2 8 0
x y
  
6 0
x y
  
(2,4)
P

Example
Example
Solution
Solution
 Note that the solution to this problem is
Note that the solution to this problem is bounded
bounded, since
, since it
it
can be enclosed by a circle
can be enclosed by a circle:
:
6 0 2 8 0 0 0
x y x y x y
       
–
–1
1 1
1 3
3 5
5 9
9
x
x
y
y
7
7
5
5
3
3
1
1
6 0
x y
  
(2,4)
P
2 8 0
x y
  

6.2
6.2
0
0
x
y


1.2
P x y
 
3 300
x y
 
2 180
x y
 
Maximize
Subject to

Linear Programming Problem
Linear Programming Problem
 A linear programming problem consists of a
A linear programming problem consists of a
linear objective function
linear objective function to be
to be maximized or
maximized or
minimized
minimized subject to certain
subject to certain constraints
constraints in the
in the
form of
form of linear equations or inequalities
linear equations or inequalities.
.

Applied Example 1:
Applied Example 1: A Production Problem
A Production Problem
 Ace Novelty wishes to produce
Ace Novelty wishes to produce two types of souvenirs
two types of souvenirs:
:
type-A
type-A will result in a profit of
will result in a profit of $1.00
$1.00, and
, and type-B
type-B in a
in a
profit of
profit of $1.20
$1.20.
.
 To manufacture a
To manufacture a type-A
type-A souvenir requires
souvenir requires 2
2 minutes on
minutes on
machine I
machine I and
and 1
1 minute on
minute on machine II
machine II.
.
 A
A type-B
type-B souvenir requires
souvenir requires 1
1 minute on
minute on machine I
machine I and
and 3
3
minutes on
minutes on machine II
machine II.
.
 There are
There are 3
3 hours available on
hours available on machine I
machine I and
and 5
5 hours
hours
available on
available on machine II
machine II.
.
 How many souvenirs
How many souvenirs of each type should Ace make in
of each type should Ace make in
order to
order to maximize its profit
maximize its profit?
?

Applied Example 1:
Solution
Solution
 Let’s first
Let’s first tabulate
tabulate the given information:
the given information:
 Let
Let x
x be the number of
be the number of type-A
type-A souvenirs and
souvenirs and y
y the number
the number
of
of type-B
type-B souvenirs to be made.
souvenirs to be made.
Type-A
Type-A Type-B
Type-B Time Available
Time Available
Profit/Unit
Profit/Unit $1.00
$1.00 $1.20
$1.20
Machine I
Machine I 2 min
2 min 1 min
1 min 180 min
180 min
Machine II
Machine II 1 min
1 min 3 min
3 min 300 min
300 min

Applied Example 1:
Solution
Solution
 Let’s first
 Then, the
Then, the total profit
total profit (in dollars) is given by
(in dollars) is given by
which is the
which is the objective function
objective function to be
to be maximized
maximized.
.
1.2
P x y
 
Type-A
Type-A Type-B
Time Available
Profit/Unit
Profit/Unit $1.00
$1.00 $1.20
$1.20
Machine I
Machine I 2 min
2 min 1 min
1 min 180 min
180 min
Machine II
Machine II 1 min
1 min 3 min
3 min 300 min
300 min

Applied Example 1:
Solution
Solution
 Let’s first
 The total amount of
The total amount of time
time that
that machine I
machine I is used is
is used is
and must not exceed
and must not exceed 180
180 minutes.
minutes.
 Thus, we have the
Thus, we have the inequality
inequality
2x y

2 180
x y
 
Type-A
Type-A Type-B
Time Available
Profit/Unit
Profit/Unit $1.00
$1.00 $1.20
$1.20
Machine I
Machine I 2 min
2 min 1 min
1 min 180 min
180 min
Machine II
Machine II 1 min
1 min 3 min
3 min 300 min
300 min

Applied Example 1:
Solution
Solution
 Let’s first
 The total amount of
The total amount of time
time that
that machine II
machine II is used is
is used is
and must not exceed
and must not exceed 300
300 minutes.
minutes.
 Thus, we have the
Thus, we have the inequality
inequality
3
x y

3 300
x y
 
Type-A
Type-A Type-B
Time Available
Profit/Unit
Profit/Unit $1.00
$1.00 $1.20
$1.20
Machine I
Machine I 2 min
2 min 1 min
1 min 180 min
180 min
Machine II
Machine II 1 min
1 min 3 min
3 min 300 min
300 min

Applied Example 1:
Solution
Solution
 Let’s first
 Finally, neither
Finally, neither x
x nor
nor y
y can be
can be negative
negative, so
, so
0
0
x
y


Type-A
Type-A Type-B
Time Available
Profit/Unit
Profit/Unit $1.00
$1.00 $1.20
$1.20
Machine I
Machine I 2 min
2 min 1 min
1 min 180 min
180 min
Machine II
Machine II 1 min
1 min 3 min
3 min 300 min
300 min

Applied Example 1:
Solution
Solution
 In short, we want to
In short, we want to maximize
maximize the
the objective function
objective function
subject to
subject to the
the system of inequalities
system of inequalities
 We will discuss the
We will discuss the solution
solution to this problem in
to this problem in section 6.4
section 6.4.
.
0
0
x
y


1.2
P x y
 
3 300
x y
 
2 180
x y
 

Applied Example 2:
Applied Example 2: A Nutrition Problem
A Nutrition Problem
 A nutritionist advises an individual who is suffering from
A nutritionist advises an individual who is suffering from
iron
iron and
and vitamin B
vitamin B deficiency to take at least
deficiency to take at least 2400
2400
milligrams (mg) of
milligrams (mg) of iron
iron,
, 2100
2100 mg of
mg of vitamin B
vitamin B1
1, and
, and 1500
1500
mg of
mg of vitamin B
vitamin B2
2 over a period of time.
over a period of time.
 Two vitamin pills are suitable,
Two vitamin pills are suitable, brand-A
brand-A and
and brand-B
brand-B.
.
 Each
Each brand-A
brand-A pill costs
pill costs 6
6 cents and contains
cents and contains 40
40 mg of
mg of iron
iron,
,
10
10 mg of
mg of vitamin B
vitamin B1
1, and
, and 5
5 mg of
mg of vitamin B
vitamin B2
2.
.
 Each
Each brand-B
brand-B pill costs
pill costs 8
8 cents and contains
cents and contains 10
10 mg of
mg of iron
iron
and
and 15
15 mg each of
mg each of vitamins B
vitamins B1
1 and
and B
B2
2.
.
 What combination of pills
What combination of pills should the individual purchase
should the individual purchase
in order to
in order to meet
meet the minimum iron and vitamin
the minimum iron and vitamin
requirements
requirements at the
at the lowest cost
lowest cost?
?

Applied Example 2:
A Nutrition Problem
Solution
Solution
 Let’s first
 Let
Let x
x be the number of
be the number of brand-A
brand-A pills and
pills and y
y the number of
the number of
brand-B
brand-B pills to be
pills to be purchased
purchased.
.
Brand-A
Brand-A Brand-B
Brand-B Minimum Requirement
Minimum Requirement
Cost/Pill
Cost/Pill 6
6¢
¢ 8
8¢
¢
Iron
Iron 40 mg
40 mg 10 mg
10 mg 2400 mg
2400 mg
Vitamin B
Vitamin B1
1 10 mg
10 mg 15 mg
15 mg 2100 mg
2100 mg
Vitamin B
Vitamin B2
2 5mg
5mg 15 mg
15 mg 1500 mg
1500 mg

Applied Example 2:
A Nutrition Problem
Solution
Solution
 Let’s first
 The
The cost
cost C
C (in cents) is given by
(in cents) is given by
and is the
and is the objective function
objective function to be
to be minimized
minimized.
.
Brand-A
Brand-A Brand-B
Minimum Requirement
Cost/Pill
Cost/Pill 6
6¢
¢ 8
8¢
¢
Iron
Iron 40 mg
40 mg 10 mg
10 mg 2400 mg
2400 mg
Vitamin B
Vitamin B1
1 10 mg
10 mg 15 mg
15 mg 2100 mg
2100 mg
Vitamin B
Vitamin B2
2 5mg
5mg 15 mg
15 mg 1500 mg
1500 mg
6 8
C x y
 

Applied Example 2:
A Nutrition Problem
Solution
Solution
 Let’s first
 The amount of
The amount of iron
iron contained in
contained in x
x brand-A
brand-A pills and
pills and y
y
brand-B
brand-B pills is given by
pills is given by 40
40x
x + 10
+ 10y
y mg, and this must be
mg, and this must be
greater than or equal to
greater than or equal to 2400
2400 mg.
mg.
 This translates into the
This translates into the inequality
inequality
Brand-A
Brand-A Brand-B
Minimum Requirement
Cost/Pill
Cost/Pill 6
6¢
¢ 8
8¢
¢
Iron
Iron 40 mg
40 mg 10 mg
10 mg 2400 mg
2400 mg
Vitamin B
Vitamin B1
1 10 mg
10 mg 15 mg
15 mg 2100 mg
2100 mg
Vitamin B
Vitamin B2
2 5mg
5mg 15 mg
15 mg 1500 mg
1500 mg
40 10 2400
x y
 

Applied Example 2:
A Nutrition Problem
Solution
Solution
 Let’s first
 The amount of
The amount of vitamin B
vitamin B1
1 contained in
contained in x
x brand-A
brand-A pills and
pills and
y
y brand-B
pills is given by 10
10x
x + 15
+ 15y
greater or equal to
greater or equal to 2100
2100 mg.
mg.
inequality
Brand-A
Brand-A Brand-B
Minimum Requirement
Cost/Pill
Cost/Pill 6
6¢
¢ 8
8¢
¢
Iron
Iron 40 mg
40 mg 10 mg
10 mg 2400 mg
2400 mg
Vitamin B
Vitamin B1
1 10 mg
10 mg 15 mg
15 mg 2100 mg
2100 mg
Vitamin B
Vitamin B2
2 5mg
5mg 15 mg
15 mg 1500 mg
1500 mg
10 15 2100
x y
 

Applied Example 2:
A Nutrition Problem
Solution
Solution
 Let’s first
 The amount of
The amount of vitamin B
vitamin B2
2 contained in
contained in x
x brand-A
brand-A pills and
pills and
y
y brand-B
pills is given by 5
5x
x + 15
+ 15y
greater or equal to
greater or equal to 1500
1500 mg.
mg.
inequality
Brand-A
Brand-A Brand-B
Minimum Requirement
Cost/Pill
Cost/Pill 6
6¢
¢ 8
8¢
¢
Iron
Iron 40 mg
40 mg 10 mg
10 mg 2400 mg
2400 mg
Vitamin B
Vitamin B1
1 10 mg
10 mg 15 mg
15 mg 2100 mg
2100 mg
Vitamin B
Vitamin B2
2 5mg
5mg 15 mg
15 mg 1500 mg
1500 mg
5 15 1500
x y
 

Applied Example 2:
A Nutrition Problem
Solution
Solution
 In short, we want to
In short, we want to minimize
minimize the
objective function
subject to
subject to the
 We will discuss the
We will discuss the solution
solution to this problem in
to this problem in section 6.4
section 6.4.
.
5 15 1500
x y
 
6 8
C x y
 
40 10 2400
x y
 
10 15 2100
x y
 
0
0
x
y



6.3
6.3
Graphical Solutions
Graphical Solutions
of Linear Programming Problems
of Linear Programming Problems
200
200
100
100
100
100 200
200 300
300
x
x
y
y
S
S
10 15 2100
x y
 
10 15 2100
x y
 
40 10 2400
x y
 
40 10 2400
x y
 
5 15 1500
x y
 
5 15 1500
x y
 
C
C(120, 60)
(120, 60)
D
D(300, 0)
(300, 0)
A
A(0, 240)
(0, 240)
B
B(30, 120)
(30, 120)

Feasible Solution Set and Optimal Solution
Feasible Solution Set and Optimal Solution
 The
The constraints
constraints in a
in a linear programming problem
linear programming problem form a
form a
system of linear inequalities
system of linear inequalities, which have a
, which have a solution set
solution set S
S.
.
 Each point in
Each point in S
S is a
is a candidate
candidate for the
for the solution
solution of the linear
of the linear
programming problem and is referred to as a
programming problem and is referred to as a feasible
feasible
solution
solution.
.
 The set
The set S
S itself is referred to as a
itself is referred to as a feasible set
feasible set.
.
 Among all the points in the set
Among all the points in the set S
S, the point(s) that
, the point(s) that
optimizes the objective function
optimizes the objective function of the linear programming
of the linear programming
problem is called an
problem is called an optimal solution
optimal solution.
.

Theorem 1
Theorem 1
Linear Programming
Linear Programming
 If a linear programming problem has a
If a linear programming problem has a solution
solution,
,
then it must occur at a
then it must occur at a vertex
vertex, or
, or corner point
corner point, of
, of
the
the feasible set
feasible set S
S associated with the problem.
associated with the problem.
 If the
If the objective function
objective function P
P is
is optimized
optimized at
at two
two
adjacent vertices
adjacent vertices of
of S
S, then it is optimized
, then it is optimized at
at every
every
point
point on the line segment
on the line segment joining these vertices, in
joining these vertices, in
which case there are
which case there are infinitely many solutions
infinitely many solutions to
to
the problem.
the problem.

Theorem 2
Theorem 2
Existence of a Solution
Existence of a Solution
 Suppose we are given a linear programming
Suppose we are given a linear programming
problem with a
problem with a feasible set
feasible set S
S and an
and an objective
objective
function
function P
P =
= ax
ax +
+ by
by.
.
a.
a. If
If S
S is
is bounded
bounded, then
, then P
P has both a
has both a maximum and
maximum and
a minimum value
a minimum value on
on S
S.
.
b.
b. If
If S
S is
is unbounded
unbounded and both
and both a
a and
and b
b are
are
nonnegative
nonnegative, then
, then P
P has a
has a minimum value
minimum value on
on S
S
provided that the constraints defining
provided that the constraints defining S
S include
include
the inequalities
the inequalities x
x 
 0
0 and
and y
y 
 0
0.
.
c.
c. If
If S
S is the
is the empty set
empty set, then the linear
, then the linear
programming problem has
programming problem has no solution
no solution: that is,
: that is, P
P
has
has neither a maximum nor a minimum
neither a maximum nor a minimum value.
value.

The Method of Corners
The Method of Corners
1.
1. Graph
Graph the
the feasible set
feasible set.
.
2.
2. Find the
Find the coordinates
coordinates of all
of all corner points
corner points
(vertices) of the feasible set.
(vertices) of the feasible set.
3.
3. Evaluate the
Evaluate the objective function
objective function at
at each corner
each corner
point
point.
.
4.
4. Find the
Find the vertex
vertex that renders the
that renders the objective
objective
function
function a
a maximum
maximum or a
or a minimum
minimum.
.
✦ If there is only
If there is only one such vertex
one such vertex, it constitutes a
, it constitutes a
unique solution
unique solution to the problem.
to the problem.
✦ If there are two
If there are two such adjacent vertices
such adjacent vertices, there
, there
are
are infinitely many optimal solutions
infinitely many optimal solutions given by
given by
the points on the line segment determined by
the points on the line segment determined by
these vertices.
these vertices.

Applied Example 1:
 Recall
Recall Applied Example 1
Applied Example 1 from the
from the last section (3.2)
last section (3.2), which
, which
required us to find the
required us to find the optimal quantities
optimal quantities to produce of
to produce of
type-A
type-A and
and type-B
type-B souvenirs in order to
souvenirs in order to maximize profits
maximize profits.
.
 We
We restated
restated the problem as a
the problem as a linear programming problem
linear programming problem
in which we wanted to
in which we wanted to maximize
maximize the
objective function
subject to
subject to the
 We can now
We can now solve the problem
solve the problem graphically.
graphically.
0
0
x
y


1.2
P x y
 
3 300
x y
 
2 180
x y
 

200
200
100
100
100
100 200
200 300
300
Applied Example 1:
 We first
We first graph the feasible set
graph the feasible set S
S for the problem.
for the problem.
✦ Graph the
Graph the solution
solution for the inequality
for the inequality
considering only
considering only positive values
positive values for
for x
x and
and y
y:
:
2 180
x y
 
x
x
y
y
2 180
x y
 
(90, 0)
(90, 0)
(0, 180)
(0, 180)
2 180
x y
 

200
200
100
100
Applied Example 1:
 We first
S for the problem.
for the problem.
✦ Graph the
Graph the solution
for the inequality
considering only
positive values for
for x
x and
and y
y:
:
3 300
x y
 
100
100 200
200 300
300
x
x
y
y
3 300
x y
 
(0, 100)
(0, 100)
(300, 0)
(300, 0)
3 300
x y
 

200
200
100
100
Applied Example 1:
 We first
S for the problem.
for the problem.
✦ Graph the
Graph the intersection
intersection of the solutions to the inequalities,
of the solutions to the inequalities,
yielding the
yielding the feasible set
feasible set S
S.
.
(Note that the
(Note that the feasible set
feasible set S
S is
is bounded
bounded)
)
100
100 200
200 300
300
x
x
y
y
S
S 3 300
x y
 
2 180
x y
 

200
200
100
100
Applied Example 1:
 Next, find the
Next, find the vertices
vertices of the
of the feasible set
feasible set S
S.
.
✦ The
The vertices
vertices are
are A
A(0, 0)
(0, 0),
, B
B(90, 0)
(90, 0),
, C
C(48, 84)
(48, 84), and
, and D
D(0, 100)
(0, 100).
.
100
100 200
200 300
300
x
x
y
y
S
S
C
C(48, 84)
(48, 84)
3 300
x y
 
2 180
x y
 
D
D(0, 100)
(0, 100)
B
B(90, 0)
(90, 0)
A
A(0, 0)
(0, 0)

200
200
100
100
Applied Example 1:
 Now, find the
Now, find the values
values of
of P
P at the
at the vertices
vertices and
and tabulate
tabulate them:
them:
100
100 200
200 300
300
x
x
y
y
S
S
C
C(48, 84)
(48, 84)
3 300
x y
 
2 180
x y
 
D
D(0, 100)
(0, 100)
B
B(90, 0)
(90, 0)
A
A(0, 0)
(0, 0)
Vertex
Vertex P
P =
= x
x + 1.2
+ 1.2 y
y
A
A(0, 0)
(0, 0) 0
0
B
B(90, 0)
(90, 0) 90
90
C
C(48, 84)
(48, 84) 148.8
148.8
D
D(0, 100)
(0, 100) 120
120

200
200
100
100
Applied Example 1:
 Finally,
Finally, identify
identify the
the vertex
vertex with the
with the highest value
highest value for
for P
P:
:
✦ We can see that
We can see that P
P is
is maximized
maximized at the vertex
at the vertex C
C(48, 84)
(48, 84)
and has a value of
and has a value of 148.8
148.8.
.
100
100 200
200 300
300
x
x
y
y
S
S 3 300
x y
 
2 180
x y
 
D
D(0, 100)
(0, 100)
B
B(90, 0)
(90, 0)
A
A(0, 0)
(0, 0)
Vertex
Vertex P
P =
= x
x + 1.2
+ 1.2 y
y
A
A(0, 0)
(0, 0) 0
0
B
B(90, 0)
(90, 0) 90
90
C
C(48, 84)
(48, 84) 148.8
148.8
D
D(0, 100)
(0, 100) 120
120
C
C(48, 84)
(48, 84)

Applied Example 1:
 Finally,
Finally, identify
identify the
the vertex
vertex with the
with the highest value
highest value for
for P
P:
:
✦ We can see that
We can see that P
P is
is maximized
maximized at the vertex
at the vertex C
C(48, 84)
(48, 84)
and has a value of
and has a value of 148.8
148.8.
.
✦ Recalling what the symbols
Recalling what the symbols x
x,
, y
y, and
, and P
P represent, we
represent, we
conclude that
conclude that ACE Novelty
ACE Novelty would
would maximize its profit
maximize its profit at
at
$148.80
$148.80 by producing
by producing 48
48 type-A
souvenirs and 84
84 type-B
type-B
souvenirs.
souvenirs.

Applied Example 2:
A Nutrition Problem
 Recall
Recall Applied Example 2
Applied Example 2 from the
from the last section (3.2)
last section (3.2), which
, which
asked us to determine the
asked us to determine the optimal combination
optimal combination of
of pills
pills to
to
be purchased in order to
be purchased in order to meet
meet the minimum
the minimum iron
iron and
and
vitamin
vitamin requirements
requirements at the
at the lowest cost
lowest cost.
.
 We
We restated
restated the problem as a
the problem as a linear programming problem
linear programming problem
in which we wanted to
in which we wanted to minimize
minimize the
objective function
subject to
subject to the
 We can now
We can now solve the problem
solve the problem graphically.
graphically.
5 15 1500
x y
 
6 8
C x y
 
40 10 2400
x y
 
10 15 2100
x y
 
, 0
x y 

200
200
100
100
Applied Example 2:
A Nutrition Problem
 We first
S for the problem.
for the problem.
✦ Graph the
Graph the solution
for the inequality
considering only
positive values for
for x
x and
and y
y:
:
100
100 200
200 300
300
x
x
y
y
40 10 2400
x y
 
40 10 2400
x y
 
(60, 0)
(60, 0)
(0, 240)
(0, 240)

200
200
100
100
Applied Example 2:
A Nutrition Problem
 We first
S for the problem.
for the problem.
✦ Graph the
Graph the solution
for the inequality
considering only
positive values for
for x
x and
and y
y:
:
100
100 200
200 300
300
x
x
y
y
10 15 2100
x y
 
10 15 2100
x y
 
(210, 0)
(210, 0)
(0, 140)
(0, 140)

200
200
100
100
Applied Example 2:
A Nutrition Problem
 We first
S for the problem.
for the problem.
✦ Graph the
Graph the solution
for the inequality
considering only
positive values for
for x
x and
and y
y:
:
100
100 200
200 300
300
x
x
y
y
5 15 1500
x y
 
5 15 1500
x y
 
(300, 0)
(300, 0)
(0, 100)
(0, 100)

200
200
100
100
Applied Example 2:
A Nutrition Problem
 We first
S for the problem.
for the problem.
✦ Graph the
Graph the intersection
intersection of the solutions to the inequalities,
of the solutions to the inequalities,
yielding the
yielding the feasible set
feasible set S
S.
.
(Note that the
(Note that the feasible set
feasible set S
S is
is unbounded
unbounded)
)
100
100 200
200 300
300
x
x
y
y
S
S
10 15 2100
x y
 
40 10 2400
x y
 
5 15 1500
x y
 

200
200
100
100
Applied Example 2:
A Nutrition Problem
 Next, find the
Next, find the vertices
vertices of the
of the feasible set
feasible set S
S.
.
✦ The
The vertices
vertices are
are A
A(0, 240)
(0, 240),
, B
B(30, 120)
(30, 120),
, C
C(120, 60)
(120, 60), and
, and
D
D(300, 0)
(300, 0).
.
100
100 200
200 300
300
x
x
y
y
S
S
10 15 2100
x y
 
40 10 2400
x y
 
5 15 1500
x y
 
C
C(120, 60)
(120, 60)
D
D(300, 0)
(300, 0)
A
A(0, 240)
(0, 240)
B
B(30, 120)
(30, 120)

Applied Example 2:
A Nutrition Problem
 Now, find the
Now, find the values
values of
of C
C at the
at the vertices
vertices and
and tabulate
tabulate them:
them:
200
200
100
100
100
100 200
200 300
300
x
x
y
y
S
S
10 15 2100
x y
 
40 10 2400
x y
 
5 15 1500
x y
 
C
C(120, 60)
(120, 60)
D
D(300, 0)
(300, 0)
A
A(0, 240)
(0, 240)
B
B(30, 120)
(30, 120)
Vertex
Vertex C
C = 6
= 6x
x + 8
+ 8y
y
A
A(0, 240)
(0, 240) 1920
1920
B
B(30, 120)
(30, 120) 1140
1140
C
C(120, 60)
(120, 60) 1200
1200
D
D(300, 0)
(300, 0) 1800
1800

Applied Example 2:
A Nutrition Problem
 Finally,
Finally, identify
identify the
the vertex
vertex with the
with the lowest value
lowest value for
for C
C:
:
✦ We can see that
We can see that C
C is
is minimized
minimized at the vertex
at the vertex B
B(30, 120)
(30, 120)
and has a value of
and has a value of 1140
1140.
.
200
200
100
100
100
100 200
200 300
300
x
x
y
y
S
S
10 15 2100
x y
 
40 10 2400
x y
 
5 15 1500
x y
 
C
C(120, 60)
(120, 60)
D
D(300, 0)
(300, 0)
A
A(0, 240)
(0, 240)
Vertex
Vertex C
C = 6
= 6x
x + 8
+ 8y
y
A
A(0, 240)
(0, 240) 1920
1920
B
B(30, 120)
(30, 120) 1140
1140
C
C(120, 60)
(120, 60) 1200
1200
D
D(300, 0)
(300, 0) 1800
1800
B
B(30, 120)
(30, 120)

Applied Example 2:
A Nutrition Problem
 Finally,
Finally, identify
identify the
the vertex
vertex with the
with the lowest value
lowest value for
for C
C:
:
✦ We can see that
We can see that C
C is
is minimized
minimized at the vertex
at the vertex B
B(30, 120)
(30, 120)
and has a value of
and has a value of 1140
1140.
.
✦ Recalling what the symbols
Recalling what the symbols x
x,
, y
y, and
, and C
C represent, we
represent, we
conclude that the individual should
conclude that the individual should purchase
purchase 30
30 brand-A
brand-A
pills and
pills and 120
120 brand-B
brand-B pills at a
pills at a minimum cost
minimum cost of
of $11.40
$11.40.
.

6.4
6.4
The Simplex Method:
The Simplex Method:
x
x y
y u
u v
v P
P Constant
Constant
1
1 0
0 3/5
3/5 –
–1/5
1/5 0
0 48
48
0
0 1
1 –
–1/5
1/5 2/5
2/5 0
0 84
84
0
0 0
0 9/25
9/25 7/25
7/25 1
1 148
148 4/5
4/5

The Simplex Method
The Simplex Method
 The
The simplex method
simplex method is an
is an iterative procedure
iterative procedure.
.
 Beginning at a
Beginning at a vertex
vertex of the
of the feasible region
feasible region S
S, each
, each
iteration
iteration brings us to another
brings us to another vertex
vertex of
of S
S with an
with an improved
improved
value of the
value of the objective function
objective function.
.
 The
The iteration
iteration ends when the
ends when the optimal solution
optimal solution is reached.
is reached.

A Standard Linear Programming Problem
A Standard Linear Programming Problem
 A
A standard maximization problem
standard maximization problem is one in which
is one in which
1.
1. The
The objective function
objective function is to be
is to be maximized
maximized.
.
2.
2. All the
All the variables
variables involved in the problem are
involved in the problem are
nonnegative
nonnegative.
.
3.
3. All other
All other linear constraints
linear constraints may be written so
may be written so
that the expression involving the variables is
that the expression involving the variables is less
less
than or equal to
than or equal to a nonnegative constant
a nonnegative constant.
.

Setting Up the Initial Simplex Tableau
Setting Up the Initial Simplex Tableau
1.
1. Transform the
Transform the system of linear
system of linear inequalities
inequalities
into a
into a system of linear
system of linear equations
equations by
by
introducing
introducing slack variables
slack variables.
.
2.
2. Rewrite the
Rewrite the objective function
objective function
in the form
in the form
where all the
where all the variables
variables are on the
are on the left
left and the
and the
coefficient
coefficient of
of P
P is
is +1
+1. Write this equation
. Write this equation
below the equations in
below the equations in step 1
step 1.
.
3.
3. Write the
Write the augmented matrix
augmented matrix associated with
associated with
this system of linear equations.
this system of linear equations.
1 1 2 2 n n
P c x c x c x
  


1 1 2 2 0
n n
c x c x c x P
   

  

Applied Example 1:
 Recall the production problem discussed in
Recall the production problem discussed in section 6.3
section 6.3,
,
which required us to
which required us to maximize
maximize the
objective function
subject to
subject to the
 This is a
This is a standard maximization problem
standard maximization problem and may be
and may be
solved by the
solved by the simplex method
simplex method.
.
 Set up
Set up the initial
the initial simplex tableau
simplex tableau for this linear
for this linear
programming problem.
programming problem.
, 0
x y 
6
1.2
5
P x y P x y
   
or equivalently,
3 300
x y
 
2 180
x y
 

Applied Example 1:
Solution
Solution
 First, introduce the
First, introduce the slack variables
slack variables u
u and
and v
v into the
into the
inequalities
inequalities
and turn these into
and turn these into equations
equations, getting
, getting
 Next, rewrite the
Next, rewrite the objective function
objective function in the form
in the form
2 180
3 300
x y u
x y v
  
  
3 300
x y
 
2 180
x y
 
6
0
5
x y P
   

Applied Example 1:
Solution
Solution
 Placing the restated
Placing the restated objective function
objective function below the system of
below the system of
equations of the
equations of the constraints
constraints we get
we get
 Thus, the
Thus, the initial tableau
initial tableau associated with this system is
associated with this system is
2 180
3 300
6
0
5
x y u
x y v
x y P
  
  
   
x
x y
y u
u v
v P
P Constant
Constant
2
2 1
1 1
1 0
0 0
0 180
180
1
1 3
3 0
0 1
1 0
0 300
300
–
–1
1 –
– 6/5
6/5 0
0 0
0 1
1 0
0

The Simplex Method
The Simplex Method
1.
1. Set up the
Set up the initial simplex tableau
initial simplex tableau.
.
2.
2. Determine whether the
Determine whether the optimal solution
optimal solution has
has
been reached by
been reached by examining all entries
examining all entries in the
in the last
last
row
row to the
to the left
left of the
of the vertical line
vertical line.
.
a.
a. If all the entries are
If all the entries are nonnegative
nonnegative, the
, the optimal
optimal
solution
solution has
has been reached
been reached. Proceed to
. Proceed to step 4
step 4.
.
b.
b. If there are one or more
If there are one or more negative entries
negative entries, the
, the
optimal solution
optimal solution has not
has not been reached
been reached.
.
Proceed to
Proceed to step 3
step 3.
.
3.
3. Perform the
Perform the pivot operation
pivot operation. Return to
. Return to step 2
step 2.
.
4.
4. Determine the
Determine the optimal solution(s)
optimal solution(s).
.

Applied Example 1:
 Recall again the
Recall again the production problem
production problem discussed previously.
discussed previously.
 We have already performed
We have already performed step 1
step 1 obtaining the
obtaining the initial
initial
simplex tableau
simplex tableau:
:
 Now, complete the
Now, complete the solution
solution to the problem.
to the problem.
x
x y
y u
u v
v P
P Constant
Constant
2
2 1
1 1
1 0
0 0
0 180
180
1
1 3
3 0
0 1
1 0
0 300
300
–
–1
1 –
– 6/5
6/5 0
0 0
0 1
1 0
0

Applied Example 1:
Solution
Solution
Step 2.
Step 2. Determine whether the
optimal solution has been
has been
reached.
reached.
✦ Since
Since there
there are
are negative entries
negative entries in the last row of the
in the last row of the
tableau, the
tableau, the initial solution
initial solution is
is not
not optimal
optimal.
.
x
x y
y u
u v
v P
P Constant
Constant
2
2 1
1 1
1 0
0 0
0 180
180
1
1 3
3 0
0 1
1 0
0 300
300
–
–1
1 –
– 6/5
6/5 0
0 0
0 1
1 0
0

Applied Example 1:
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ Since the entry
Since the entry –
– 6/5
6/5 is
is the
the most negative entry
most negative entry to the left
to the left
of the vertical line in the last row of the tableau, the
second column
second column in the tableau is the
in the tableau is the pivot column
pivot column.
.
x
x y
y u
u v
v P
P Constant
Constant
2
2 1
1 1
1 0
0 0
0 180
180
1
1 3
3 0
0 1
1 0
0 300
300
–
–1
1 –
– 6/5
6/5 0
0 0
0 1
1 0
0

Applied Example 1:
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ Divide each
Divide each positive number
positive number of the
of the pivot column
pivot column into the
into the
corresponding entry
corresponding entry in the
in the column of constants
column of constants and
and
compare
compare the
the ratios
ratios thus obtained.
thus obtained.
✦ We see that the
We see that the ratio
ratio 300/3 = 100
300/3 = 100 is
is less than
less than the
the ratio
ratio
180/1 = 180
180/1 = 180, so
, so row 2
row 2 is the
is the pivot row
pivot row.
.
x
x y
y u
u v
v P
P Constant
Constant
2
2 1
1 1
1 0
0 0
0 180
180
1
1 3
3 0
0 1
1 0
0 300
300
–
–1
1 –
– 6/5
6/5 0
0 0
0 1
1 0
0
180
1
300
3
180
100



Applied Example 1:
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ The
The entry
entry 3
3 lying in the
lying in the pivot column
pivot column and the
and the pivot row
pivot row
is the
is the pivot element
pivot element.
.
x
x y
y u
u v
v P
P Constant
Constant
2
2 1
1 1
1 0
0 0
0 180
180
1
1 3
3 0
0 1
1 0
0 300
300
–
–1
1 –
– 6/5
6/5 0
0 0
0 1
1 0
0

Applied Example 1:
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ Convert the
Convert the pivot element
pivot element into a
into a 1
1.
.
1
2
3 R
x
x y
y u
u v
v P
P Constant
Constant
2
2 1
1 1
1 0
0 0
0 180
180
1
1 3
3 0
0 1
1 0
0 300
300
–
–1
1 –
– 6/5
6/5 0
0 0
0 1
1 0
0

Applied Example 1:
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ Convert the
into a 1
1.
.
1
2
3 R
x
x y
y u
u v
v P
P Constant
Constant
2
2 1
1 1
1 0
0 0
0 180
180
1/3
1/3 1
1 0
0 1/3
1/3 0
0 100
100
–
–1
1 –
– 6/5
6/5 0
0 0
0 1
1 0
0

Applied Example 1:
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ Use elementary
Use elementary row operations
row operations to convert the
to convert the pivot
pivot
column
column into a
into a unit column
unit column.
.
1 2
6
3 2
5
R R
R R


x
x y
y u
u v
v P
P Constant
Constant
2
2 1
1 1
1 0
0 0
0 180
180
1/3
1/3 1
1 0
0 1/3
1/3 0
0 100
100
–
–1
1 –
– 6/5
6/5 0
0 0
0 1
1 0
0

Applied Example 1:
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ Use elementary
pivot
column
column into a
into a unit column
unit column.
.
1 2
6
3 2
5
R R
R R


x
x y
y u
u v
v P
P Constant
Constant
5/3
5/3 0
0 1
1 –
–1/3
1/3 0
0 80
80
1/3
1/3 1
1 0
0 1/3
1/3 0
0 100
100
–
–3/5
3/5 0
0 0
0 2/5
2/5 1
1 120
120

Applied Example 1:
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ This
This completes an iteration
completes an iteration.
.
✦ The
The last row
last row of the tableau contains a
of the tableau contains a negative number
negative number,
,
so an
so an optimal solution
has not
not been reached
been reached.
.
✦ Therefore, we
Therefore, we repeat
repeat the
the iteration step
iteration step.
.
x
x y
y u
u v
v P
P Constant
Constant
5/3
5/3 0
0 1
1 –
–1/3
1/3 0
0 80
80
1/3
1/3 1
1 0
0 1/3
1/3 0
0 100
100
–
–3/5
3/5 0
0 0
0 2/5
2/5 1
1 120
120

Applied Example 1:
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation again.
again.
✦ Since the entry
Since the entry –
– 3/5
3/5 is
is the
to the left
of the vertical line in the last row of the tableau, the first
first
column
column in the tableau is now the
in the tableau is now the pivot column
pivot column.
.
x
x y
y u
u v
v P
P Constant
Constant
5/3
5/3 0
0 1
1 –
–1/3
1/3 0
0 80
80
1/3
1/3 1
1 0
0 1/3
1/3 0
0 100
100
–
–3/5
3/5 0
0 0
0 2/5
2/5 1
1 120
120

Applied Example 1:
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ Divide each
of the pivot column
into the
corresponding entry
and
compare the ratios
compare the ratios thus obtained.
thus obtained.
✦ We see that the
ratio 80/(5/3) = 48
80/(5/3) = 48 is
is less than
less than the
the ratio
ratio
100/(1/3) = 300
100/(1/3) = 300, so
, so row 1
row 1 is the
is the pivot row
pivot row now.
now.
x
x y
y u
u v
v P
P Constant
Constant
5/3
5/3 0
0 1
1 –
–1/3
1/3 0
0 80
80
1/3
1/3 1
1 0
0 1/3
1/3 0
0 100
100
–
–3/5
3/5 0
0 0
0 2/5
2/5 1
1 120
120
80
5/3
100
1/3
48
300


Ratio

Applied Example 1:
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ The
The entry
entry 5
5/
/3
3 lying in the
and the pivot
pivot
row
row is the
pivot element.
.
x
x y
y u
u v
v P
P Constant
Constant
5/3
5/3 0
0 1
1 –
–1/3
1/3 0
0 80
80
1/3
1/3 1
1 0
0 1/3
1/3 0
0 100
100
–
–3/5
3/5 0
0 0
0 2/5
2/5 1
1 120
120

Applied Example 1:
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ Convert the
into a 1
1.
.
x
x y
y u
u v
v P
P Constant
Constant
5/3
5/3 0
0 1
1 –
–1/3
1/3 0
0 80
80
1/3
1/3 1
1 0
0 1/3
1/3 0
0 100
100
–
–3/5
3/5 0
0 0
0 2/5
2/5 1
1 120
120
3
1
5 R

Applied Example 1:
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ Convert the
into a 1
1.
.
x
x y
y u
u v
v P
P Constant
Constant
1
1 0
0 3/5
3/5 –
–1/5
1/5 0
0 48
48
1/3
1/3 1
1 0
0 1/3
1/3 0
0 100
100
–
–3/5
3/5 0
0 0
0 2/5
2/5 1
1 120
120
3
1
5 R

Applied Example 1:
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ Use elementary
pivot
column
column into a
into a unit column
unit column.
.
1
2 1
3
3
3 1
5
R R
R R


x
x y
y u
u v
v P
P Constant
Constant
1
1 0
0 3/5
3/5 –
–1/5
1/5 0
0 48
48
1/3
1/3 1
1 0
0 1/3
1/3 0
0 100
100
–
–3/5
3/5 0
0 0
0 2/5
2/5 1
1 120
120

Applied Example 1:
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ Use elementary
pivot
column
column into a
into a unit column
unit column.
.
1
2 1
3
3
3 1
5
R R
R R


x
x y
y u
u v
v P
P Constant
Constant
1
1 0
0 3/5
3/5 –
–1/5
1/5 0
0 48
48
0
0 1
1 –
–1/5
1/5 2/5
2/5 0
0 84
84
0
0 0
0 9/25
9/25 7/25
7/25 1
1 148
148 4/5
4/5

Applied Example 1:
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ The
The last row
last row of the tableau contains
of the tableau contains no
no negative
negative
numbers
numbers, so an
, so an optimal solution
has been reached
been reached.
.
x
x y
y u
u v
v P
P Constant
Constant
1
1 0
0 3/5
3/5 –
–1/5
1/5 0
0 48
48
0
0 1
1 –
–1/5
1/5 2/5
2/5 0
0 84
84
0
0 0
0 9/25
9/25 7/25
7/25 1
1 148
148 4/5
4/5

Applied Example 1:
Solution
Solution
Step 4.
Step 4. Determine the
Determine the optimal solution
optimal solution.
.
✦ Locate the
Locate the basic variables
basic variables in the final tableau.
in the final tableau.
In this case, the
In this case, the basic variables
basic variables are
are x
x,
, y
y, and
, and P
P.
.
 The
The optimal value
optimal value for
for x
x is
is 48
48.
.
 The
The optimal value
optimal value for
for y
y is
is 84
84.
.
 The
The optimal value
optimal value for
for P
P is
is 148.8
148.8.
.
✦ Thus, the firm will
Thus, the firm will maximize profits
maximize profits at
at $148.80
$148.80 by
by
producing
producing 48
48 type-A
souvenirs and 84
84 type-B
type-B souvenirs.
souvenirs.
This
This agrees
agrees with the results obtained in
with the results obtained in section 6.3
section 6.3.
.
x
x y
y u
u v
v P
P Constant
Constant
1
1 0
0 3/5
3/5 –
–1/5
1/5 0
0 48
48
0
0 1
1 –
–1/5
1/5 2/5
2/5 0
0 84
84
0
0 0
0 9/25
9/25 7/25
7/25 1
1 148
148 4/5
4/5

6.5
6.5
The Simplex Method:
The Simplex Method:
30
30
–
–1/50
1/50
3/100
3/100
x
x
0
0
1
1
0
0
v
v
450
450
11/10
11/10
–
–3/20
3/20
w
w
0
0
0
0
1
1
u
u
1140
1140
1
1
120
120
13/25
13/25
0
0
2/25
2/25
1/50
1/50
0
0
–
–1/50
1/50
Constant
Constant
P
P
y
y
30
30
–
–1/50
1/50
3/100
3/100
x
x
0
0
1
1
0
0
v
v
450
450
11/10
11/10
–
–3/20
3/20
w
w
0
0
0
0
1
1
u
u
1140
1140
1
1
120
120
13/25
13/25
0
0
2/25
2/25
1/50
1/50
0
0
–
–1/50
1/50
Constant
Constant
P
P
y
y
Solution
Solution for the
for the
primal problem
primal problem

Minimization with
Minimization with 
 Constraints
Constraints
 In the last section we developed the
In the last section we developed the simplex method
simplex method to
to
solve linear programming problems that satisfy
solve linear programming problems that satisfy three
three
conditions
conditions:
:
1.
1. The
is to be maximized
maximized.
.
2.
2. All the
All the variables involved
variables involved are
are nonnegative
nonnegative.
.
3.
3. Each
Each linear constraint
linear constraint may be written so that the
may be written so that the
expression involving the variables is
expression involving the variables is less than or equal to
less than or equal to
a nonnegative constant
a nonnegative constant.
.
 We will now see how the simplex method can be used to
We will now see how the simplex method can be used to
solve
solve minimization problems
minimization problems that
that meet the second and
meet the second and
third conditions
third conditions listed above.
listed above.

Example
Example
 Solve the following
Solve the following linear programming problem
linear programming problem:
:
 This problem involves the
This problem involves the minimization
minimization of the objective
of the objective
function and so is
function and so is not
not a
a standard maximization problem
standard maximization problem.
.
 Note, however, that
Note, however, that all the other conditions
all the other conditions for a
for a standard
standard
maximization
maximization hold true
hold true.
.
2 3
Minimize C x y
 
5 4 32
2 10
, 0
subject to x y
x y
x y
 
 


Example
Example
 We can use the
We can use the simplex method
simplex method to solve this problem by
to solve this problem by
converting
converting the
objective function from
from minimizing
minimizing C
C to its
to its
equivalent
equivalent of
of maximizing
maximizing P
P = –
= – C
C.
.
 Thus, the
Thus, the restated
restated linear programming problem
linear programming problem is
is
 This problem can now be solved using the
This problem can now be solved using the simplex method
simplex method
as discussed in
as discussed in section 6.4
section 6.4.
.
2 3
Maximize P x y
 
5 4 32
2 10
, 0
subject to x y
x y
x y
 
 


Example
Example
Solution
Solution
Step 1.
Step 1. Set up the
Set up the initial simplex tableau
initial simplex tableau.
.
✦ Turn the
Turn the constraints
constraints into
into equations
equations adding to them the
adding to them the
slack variables
slack variables u
u and
and v
v. Also
. Also rearrange
rearrange the
the objective
objective
function
function and place it below the constraints:
and place it below the constraints:
✦ Write the
Write the coefficients
coefficients of the system in a
of the system in a tableau
tableau:
:
5 4 32
2 10
2 3 0
x y u
x y v
x y P
  
  
   
2 3
Maximize P x y
 
x
x y
y u
u v
v P
P Constant
Constant
5
5 4
4 1
1 0
0 0
0 32
32
1
1 2
2 0
0 1
1 0
0 10
10
–
–2
2 –
–3
3 0
0 0
0 1
1 0
0

Example
Example
Solution
Solution
Step 2.
Step 2. Determine whether the
has been
reached.
reached.
✦ Since
Since there
there are
are negative entries
negative entries in the last row of the
in the last row of the
tableau, the
tableau, the initial solution
initial solution is
is not
not optimal
optimal.
.
x
x y
y u
u v
v P
P Constant
Constant
5
5 4
4 1
1 0
0 0
0 32
32
1
1 2
2 0
0 1
1 0
0 10
10
–
–2
2 –
–3
3 0
0 0
0 1
1 0
0

Example
Example
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ Since the entry
Since the entry –
– 3
3 is
is the
to the left
second column
second column in the tableau is the
in the tableau is the pivot column
pivot column.
.
2 3
Maximize P x y
 
x
x y
y u
u v
v P
P Constant
Constant
5
5 4
4 1
1 0
0 0
0 32
32
1
1 2
2 0
0 1
1 0
0 10
10
–
–2
2 –
–3
3 0
0 0
0 1
1 0
0

Example
Example
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ Divide each
of the pivot column
into the
corresponding entry
and
compare the ratios
thus obtained.
✦ We see that the
ratio 10/2 = 5
10/2 = 5 is
is less than
less than the
the ratio
ratio
32/4 = 8
32/4 = 8, so
, so row 2
row 2 is the
is the pivot row
pivot row.
.
2 3
Maximize P x y
 
32
4
10
2
8
5


x
x y
y u
u v
v P
P Constant
Constant
5
5 4
4 1
1 0
0 0
0 32
32
1
1 2
2 0
0 1
1 0
0 10
10
–
–2
2 –
–3
3 0
0 0
0 1
1 0
0
Ratio

Example
Example
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ The
The entry
entry 2
2 lying in the
and the pivot row
pivot row
is the
pivot element.
.
2 3
Maximize P x y
 
x
x y
y u
u v
v P
P Constant
Constant
5
5 4
4 1
1 0
0 0
0 32
32
1
1 2
2 0
0 1
1 0
0 10
10
–
–2
2 –
–3
3 0
0 0
0 1
1 0
0

Example
Example
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ Convert the
into a 1
1.
.
2 3
Maximize P x y
 
x
x y
y u
u v
v P
P Constant
Constant
5
5 4
4 1
1 0
0 0
0 32
32
1
1 2
2 0
0 1
1 0
0 10
10
–
–2
2 –
–3
3 0
0 0
0 1
1 0
0
1
2
2 R

Example
Example
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ Convert the
into a 1
1.
.
2 3
Maximize P x y
 
x
x y
y u
u v
v P
P Constant
Constant
5
5 4
4 1
1 0
0 0
0 32
32
1/2
1/2 1
1 0
0 1/2
1/2 0
0 5
5
–
–2
2 –
–3
3 0
0 0
0 1
1 0
0
1
2
2 R

Example
Example
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ Use elementary
pivot
column
column into a
into a unit column
unit column.
.
2 3
Maximize P x y
 
1 2
3 2
4
3
R R
R R


x
x y
y u
u v
v P
P Constant
Constant
5
5 4
4 1
1 0
0 0
0 32
32
1/2
1/2 1
1 0
0 1/2
1/2 0
0 5
5
–
–2
2 –
–3
3 0
0 0
0 1
1 0
0

Example
Example
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ Use elementary
pivot
column
column into a
into a unit column
unit column.
.
2 3
Maximize P x y
 
1 2
3 2
4
3
R R
R R


x
x y
y u
u v
v P
P Constant
Constant
3
3 0
0 1
1 –
–2
2 0
0 12
12
1/2
1/2 1
1 0
0 1/2
1/2 0
0 5
5
–
–1/2
1/2 0
0 0
0 3/2
3/2 1
1 15
15

Example
Example
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ This
This completes an iteration
completes an iteration.
.
✦ The
The last row
last row of the tableau contains a
of the tableau contains a negative number
negative number,
,
so an
so an optimal solution
has not
not been reached
been reached.
.
✦ Therefore, we
Therefore, we repeat
repeat the
the iteration step
iteration step.
.
2 3
Maximize P x y
 
x
x y
y u
u v
v P
P Constant
Constant
3
3 0
0 1
1 –
–2
2 0
0 12
12
1/2
1/2 1
1 0
0 1/2
1/2 0
0 5
5
–
–1/2
1/2 0
0 0
0 3/2
3/2 1
1 15
15

Example
Example
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ Since the entry
Since the entry –1/2
–1/2 is
is the
to the left
of the vertical line in the last row of the tableau, the first
first
column
column in the tableau is now the
in the tableau is now the pivot column
pivot column.
.
2 3
Maximize P x y
 
x
x y
y u
u v
v P
P Constant
Constant
3
3 0
0 1
1 –
–2
2 0
0 12
12
1/2
1/2 1
1 0
0 1/2
1/2 0
0 5
5
–
–1/2
1/2 0
0 0
0 3/2
3/2 1
1 15
15

Example
Example
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ Divide each
of the pivot column
into the
corresponding entry
and
compare the ratios
thus obtained.
✦ We see that the
ratio 12/3 = 4
12/3 = 4 is
is less than
less than the
the ratio
ratio
5/(1/2) = 10
5/(1/2) = 10, so
, so row 1
row 1 is now the
is now the pivot row
pivot row.
.
2 3
Maximize P x y
 
x
x y
y u
u v
v P
P Constant
Constant
3
3 0
0 1
1 –
–2
2 0
0 12
12
1/2
1/2 1
1 0
0 1/2
1/2 0
0 5
5
–
–1/2
1/2 0
0 0
0 3/2
3/2 1
1 15
15
12
3
5
1/2
4
10



Example
Example
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ The
The entry
entry 3
3 lying in the
and the pivot row
pivot row
is the
pivot element.
.
x
x y
y u
u v
v P
P Constant
Constant
3
3 0
0 1
1 –
–2
2 0
0 12
12
1/2
1/2 1
1 0
0 1/2
1/2 0
0 5
5
–
–1/2
1/2 0
0 0
0 3/2
3/2 1
1 15
15

Example
Example
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ Convert the
into a 1
1.
.
x
x y
y u
u v
v P
P Constant
Constant
3
3 0
0 1
1 –
–2
2 0
0 12
12
1/2
1/2 1
1 0
0 1/2
1/2 0
0 5
5
–
–1/2
1/2 0
0 0
0 3/2
3/2 1
1 15
15
1
1
3 R

Example
Example
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ Convert the
into a 1
1.
.
x
x y
y u
u v
v P
P Constant
Constant
1
1 0
0 1/3
1/3 –
–2/3
2/3 0
0 4
4
1/2
1/2 1
1 0
0 1/2
1/2 0
0 5
5
–
–1/2
1/2 0
0 0
0 3/2
3/2 1
1 15
15
1
1
3 R

Example
Example
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ Use elementary
pivot
column
column into a
into a unit column
unit column.
.
1
2 1
2
1
3 1
2
R R
R R


x
x y
y u
u v
v P
P Constant
Constant
1
1 0
0 1/3
1/3 –
–2/3
2/3 0
0 4
4
1/2
1/2 1
1 0
0 1/2
1/2 0
0 5
5
–
–1/2
1/2 0
0 0
0 3/2
3/2 1
1 15
15

Example
Example
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ Use elementary
pivot
column
column into a
into a unit column
unit column.
.
1
2 1
2
1
3 1
2
R R
R R


x
x y
y u
u v
v P
P Constant
Constant
1
1 0
0 1/3
1/3 –
–2/3
2/3 0
0 4
4
0
0 1
1 –
–1/6
1/6 5/6
5/6 0
0 3
3
0
0 0
0 1/6
1/6 7/6
7/6 1
1 17
17

Example
Example
Solution
Solution
Step 3.
Step 3. Perform the
pivot operation.
.
✦ The
The last row
last row of the tableau contains
of the tableau contains no
no negative
negative
numbers
numbers, so an
, so an optimal solution
has been reached
reached.
.
x
x y
y u
u v
v P
P Constant
Constant
1
1 0
0 1/3
1/3 –
–2/3
2/3 0
0 4
4
0
0 1
1 –
–1/6
1/6 5/6
5/6 0
0 3
3
0
0 0
0 1/6
1/6 7/6
7/6 1
1 17
17

Example
Example
Solution
Solution
Step 4.
Step 4. Determine the
Determine the optimal solution
optimal solution.
.
✦ Locate the
Locate the basic variables
basic variables in the final tableau.
in the final tableau.
In this case, the
In this case, the basic variables
basic variables are
are x
x,
, y
y, and
, and P
P.
.
 The
The optimal value
optimal value for
for x
x is
is 4
4.
.
 The
The optimal value
optimal value for
for y
y is
is 3
3.
.
 The
The optimal value
optimal value for
for P
P is
is 17
17, which means that
, which means that
the
the minimized value
minimized value for
for C
C is
is –17
–17.
.
x
x y
y u
u v
v P
P Constant
Constant
1
1 0
0 1/3
1/3 –
–2/3
2/3 0
0 4
4
0
0 1
1 –
–1/6
1/6 5/6
5/6 0
0 3
3
0
0 0
0 1/6
1/6 7/6
7/6 1
1 17
17

The Dual Problem
The Dual Problem
 Another
Another special class
special class of
of linear programming problems
linear programming problems we
we
encounter in practical applications is characterized by the
encounter in practical applications is characterized by the
following
following conditions
conditions:
:
1.
1. The
is to be minimized
minimized.
.
2.
2. All the
All the variables involved
variables involved are
are nonnegative
nonnegative.
.
3.
3. All other
All other linear constraints
linear constraints may be written so that the
may be written so that the
expression involving the variables is
expression involving the variables is greater
greater than or
than or
equal to a nonnegative constant
equal to a nonnegative constant.
.
 Such problems are called
Such problems are called standard minimization
standard minimization
problems
problems.
.

The Dual Problem
The Dual Problem
 In solving this kind of
In solving this kind of linear programming problem
linear programming problem, it
, it
helps to note that each
helps to note that each maximization
maximization problem
problem is associated
is associated
with a
with a minimization
minimization problem
problem, and vice versa.
, and vice versa.
 The
The given problem
given problem is called the
is called the primal problem
primal problem, and the
, and the
related problem
related problem is called the
is called the dual problem
dual problem.
.

Example
Example
 Write the
Write the dual problem
dual problem associated with this problem:
associated with this problem:
 We first write down a
We first write down a tableau
tableau for the
for the primal problem
primal problem:
:
6 8
Minimize C x y
 
40 10 2400
10 15 2100
5 15 1500
, 0
subject to x y
x y
x y
x y
 
 
 

x
x y
y Constant
Constant
40
40 10
10 2400
2400
10
10 15
15 2100
2100
5
5 15
15 1500
1500
6
6 8
8
Primal
Primal
Problem
Problem

Example
Example
 Next, we
Next, we interchange
interchange the
the columns
columns and
and rows
rows of the tableau
of the tableau
and
and head
head the three columns of the resulting array with the
the three columns of the resulting array with the
three variables
three variables u
u,
, v
v, and
, and w
w, obtaining
, obtaining
x
x y
y Constant
Constant
40
40 10
10 2400
2400
10
10 15
15 2100
2100
5
5 15
15 1500
1500
6
6 8
8
u
u v
v w
w Constant
Constant
40
40 10
10 5
5 6
6
10
10 15
15 15
15 8
8
2400
2400 2100
2100 1500
1500

Example
Example
 Consider the resulting tableau as if it were the
Consider the resulting tableau as if it were the initial
initial
simplex tableau
simplex tableau for a
for a standard maximization problem
.
 From it we can reconstruct the required
From it we can reconstruct the required dual problem
dual problem:
:
u
u v
v w
w Constant
Constant
40
40 10
10 5
5 6
6
10
10 15
15 15
15 8
8
2400
2400 2100
2100 1500
1500
2400 2100 1500
Maximize P u v w
  
40 10 5 6
10 15 15 8
, , 0
subject to u v w
u v w
u v w
  
  

Dual
Dual
Problem
Problem

Theorem 1
Theorem 1
The Fundamental Theorem of Duality
The Fundamental Theorem of Duality
 A
A primal problem
primal problem has a
has a solution
solution if and only if the
if and only if the
corresponding
corresponding dual problem
dual problem has a
has a solution
solution.
.
 Furthermore, if a solution exists, then:
Furthermore, if a solution exists, then:
a.
a. The
The objective functions
objective functions of both the
of both the primal
primal and
and
the
the dual problem
dual problem attain the
attain the same optimal value
same optimal value.
.
b.
b. The
The optimal solution
optimal solution to the
to the primal problem
primal problem
appears under the
appears under the slack variables
slack variables in the last row
in the last row
of the final simplex tableau associated with the
of the final simplex tableau associated with the
dual problem
dual problem.
.

Example
Example
 Complete the solution
Complete the solution of the problem from our
of the problem from our last example
last example:
:
2400 2100 1500
Maximize P u v w
  
40 10 5 6
10 15 15 8
, , 0
subject to u v w
u v w
u v w
  
  

Dual
Dual
Problem
Problem

Example
Example
Solution
Solution
 The
The dual problem
dual problem associated with the given
associated with the given primal
primal
problem
problem is a
is a standard maximization problem
.
 Thus, we can proceed with the
Thus, we can proceed with the simplex method
simplex method.
.
 First, we
First, we introduce
introduce to the system of equations the
to the system of equations the slack
slack
variables
variables x
x and
and y
y, and
, and restate
restate the
the inequalities
inequalities as
as equations
equations,
,
obtaining
obtaining
40 10 5 6
10 15 15 8
2400 2100 1500 0
u v w x
u v w y
u v w P
   
   
    

Example
Example
Solution
Solution
 Next, we transcribe the
Next, we transcribe the coefficients
coefficients of the system of
of the system of
equations
equations
into an
into an initial simplex tableau
initial simplex tableau:
:
40 10 5 6
10 15 15 8
2400 2100 1500 0
u v w x
u v w y
u v w P
   
   
    
u
u v
v w
w x
x y
y P
P Constant
Constant
40
40 10
10 5
5 1
1 0
0 0
0 6
6
10
10 15
15 15
15 0
0 1
1 0
0 8
8
–
–2400
2400 –
–2100
2100 –
–1500
1500 0
0 0
0 1
1 0
0

Example
Example
Solution
Solution
 Continue with the
Continue with the simplex iterative method
simplex iterative method until a
until a final
final
tableau
tableau is obtained with the
is obtained with the solution
solution for the problem:
for the problem:
 The
The fundamental theorem of duality
fundamental theorem of duality tells us that the
tells us that the
solution
solution to the
to the primal problem
primal problem is
is x
x = 30
= 30 and
and y
y = 120
= 120, with a
, with a
minimum value
minimum value for
for C
C of
of 1140
1140.
.
u
u v
v w
w x
x y
y P
P Constant
Constant
1
1 0
0 –
–3/20
3/20 3/100
3/100 –
–1/50
1/50 0
0 1/50
1/50
0
0 1
1 11/10
11/10 –
–1/50
1/50 2/25
2/25 0
0 13/25
13/25
0
0 0
0 450
450 30
30 120
120 1
1 1140
1140
Solution
Solution for the
for the
primal problem
primal problem

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