§ 4.4 The Simplex Method: Non-Standard
Form
Standard form and what can be relaxed
What were the conditions for standard form we have been adhering
to?
Standard form and what can be relaxed
What were the conditions for standard form we have been adhering
to?
Conditions for Standard Form
1 Object function is to be maximized.
Standard form and what can be relaxed
What were the conditions for standard form we have been adhering
to?
Conditions for Standard Form
1 Object function is to be maximized.
2
Each variable must be constrained to be greater than or equal to
0.
Standard form and what can be relaxed
What were the conditions for standard form we have been adhering
to?
Conditions for Standard Form
1 Object function is to be maximized.
2
Each variable must be constrained to be greater than or equal to
0.
3
all other constraints must be of the form
[linear expression] ≤ [non − negative constant].
Standard form and what can be relaxed
What were the conditions for standard form we have been adhering
to?
Conditions for Standard Form
1 Object function is to be maximized.
2
Each variable must be constrained to be greater than or equal to
0.
3
all other constraints must be of the form
[linear expression] ≤ [non − negative constant].
What can be relaxed
1 We can do minimization problems.
Standard form and what can be relaxed
What were the conditions for standard form we have been adhering
to?
Conditions for Standard Form
1 Object function is to be maximized.
2
Each variable must be constrained to be greater than or equal to
0.
3
all other constraints must be of the form
[linear expression] ≤ [non − negative constant].
What can be relaxed
1 We can do minimization problems.
2
We can handle constraints of the form
[linear expression] ≥ [non − negative constant].
Standard form and what can be relaxed
What were the conditions for standard form we have been adhering
to?
Conditions for Standard Form
1 Object function is to be maximized.
2
Each variable must be constrained to be greater than or equal to
0.
3
all other constraints must be of the form
[linear expression] ≤ [non − negative constant].
What can be relaxed
1 We can do minimization problems.
2
We can handle constraints of the form
[linear expression] ≥ [non − negative constant].
3
We can handle constraints of the form
[linear expression] ≤ [negative constant].
Not really an example
Example
Maximize x + y subject to the following constraints.

 x+y≥1
x + y ≤ −1

x ≥ 0, y ≥ 0
Not really an example
Example
Maximize x + y subject to the following constraints.

 x+y≥1
x + y ≤ −1

x ≥ 0, y ≥ 0
To take care of the first constraint, we could multiply both sides by
−1 which would invert the inequality.
Not really an example
Example
Maximize x + y subject to the following constraints.

 x+y≥1
x + y ≤ −1

x ≥ 0, y ≥ 0
To take care of the first constraint, we could multiply both sides by
−1 which would invert the inequality.
x + y ≥ 1 ⇒ −x − y ≤ −1
To take care any negative on the right, we will pivot. This essentially
means we are adding a new set of steps to the beginning of the ones
we established in the last section.
The New Steps
1
Rewrite all inequalities in the form
[linear expression] ≤ [constant]
The New Steps
1
Rewrite all inequalities in the form
[linear expression] ≤ [constant]
2
If a negative appears in the upper part of the last column, remove
by pivoting.
1
Select one of the negatives in this row. This will be the pivot
column.
The New Steps
1
Rewrite all inequalities in the form
[linear expression] ≤ [constant]
2
If a negative appears in the upper part of the last column, remove
by pivoting.
1
2
Select one of the negatives in this row. This will be the pivot
column.
Take ratios as before. The least positive one will indicate the
pivot element.
The New Steps
1
Rewrite all inequalities in the form
[linear expression] ≤ [constant]
2
If a negative appears in the upper part of the last column, remove
by pivoting.
1
2
3
Select one of the negatives in this row. This will be the pivot
column.
Take ratios as before. The least positive one will indicate the
pivot element.
Pivot.
The New Steps
1
Rewrite all inequalities in the form
[linear expression] ≤ [constant]
2
If a negative appears in the upper part of the last column, remove
by pivoting.
1
2
3
3
Select one of the negatives in this row. This will be the pivot
column.
Take ratios as before. The least positive one will indicate the
pivot element.
Pivot.
If there is another negative at this point in the upper part of the
last column, repeat. Otherwise revert to the original steps we
used for standard form problems.
First Example
Example
Maximize 40x + 30y subject to

 x+y≤5
−2x + 3y ≥ 12

x ≥ 0, y ≥ 0
Is this in standard form?
First Example
Example
Maximize 40x + 30y subject to

 x+y≤5
−2x + 3y ≥ 12

x ≥ 0, y ≥ 0
Is this in standard form?
To remove the inequality issue, we multiply both sides of that
inequality to get this to be a ≤ problem.
First Example
Example
Maximize 40x + 30y subject to

 x+y≤5
−2x + 3y ≥ 12

x ≥ 0, y ≥ 0
Is this in standard form?
To remove the inequality issue, we multiply both sides of that
inequality to get this to be a ≤ problem.
−2x + 3y ≥ 12 ⇒ 2x − 3y ≤ −12
First Example
Now we rewrite the system by introducing slack variables.
First Example
Now we rewrite the system by introducing slack variables.

x+y+u=5



2x − 3y + v = −12
−40x − 30y + M = 0



x ≥ 0, y ≥ 0
First Example
Now we rewrite the system by introducing slack variables.

x+y+u=5



2x − 3y + v = −12
−40x − 30y + M = 0



x ≥ 0, y ≥ 0
Now the initial tableau.
First Example
Now we rewrite the system by introducing slack variables.

x+y+u=5



2x − 3y + v = −12
−40x − 30y + M = 0



x ≥ 0, y ≥ 0
Now the initial tableau.

1
 2
-40
1
-3
-30
1
0
0
0
1
0
0
0
1

5
-12 
0
First Example
Before we even look at the bottom row, we have to take care of
problem spots in the last column.


5
1
1
1 0 0
 2
-3 0 1 0 -12 
-40 -30 0 0 1
0
First Example
Before we even look at the bottom row, we have to take care of
problem spots in the last column.


5
1
1
1 0 0
 2
-3 0 1 0 -12 
-40 -30 0 0 1
0
We select any column that has a negative in the same row as the −12
to be the pivot column and then we take ratios like before to see
which gives the smallest positive ratio.

1
 2
-40
1
-3
-30
1
0
0
0
1
0
0
0
1

5
-12 
0
First Example
Before we even look at the bottom row, we have to take care of
problem spots in the last column.


5
1
1
1 0 0
 2
-3 0 1 0 -12 
-40 -30 0 0 1
0
We select any column that has a negative in the same row as the −12
to be the pivot column and then we take ratios like before to see
which gives the smallest positive ratio.

1
1
1 0 0
5
 2
-3 0 1 0 -12 
-40 -30 0 0 1
0
which gives a pivot element of ...

5
1 =5
−12
−3 =
4
First Example

1

 2
-40
1
1
0
0
-3
-30
0
0
1
0
0
1
5


-12 
0
First Example

1

 2
-40
1
1
0
0
-3
-30
0
0
1
0
0
1
1
1
-30
1
0
0
0
− 13
0
5


-12 
0
After we pivot, we have

1
 −2
3
-40
0
0
1

5
4 
0
First Example

1

 2
-40
1
1
0
0
-3
-30
0
0
1
0
0
1
1
1
-30
1
0
0
0
− 13
0
5


-12 
0
After we pivot, we have

1
 −2
3
-40
 5
3
∼  − 32
-60
0
1
0
1
0
0
1
3
− 13
-10
0
0
1
0
0
1

5
4 
0

1
4 
120
First Example
Since we have no more negatives in the upper portion of the right
column, this is now a standard form problem and we proceed as usual.

5
3
 −2
3
-60
0
1
0
1
0
0
1
3
− 13
-10
0
0
1

1
4 
120
First Example
Since we have no more negatives in the upper portion of the right
column, this is now a standard form problem and we proceed as usual.

5
3
 −2
3
-60
0
1
0
1
0
0
1
3
− 13
-10
0
0
1

1
4 
120
1
5
3
=
4
− 23
3
5
= −6
First Example
Since we have no more negatives in the upper portion of the right
column, this is now a standard form problem and we proceed as usual.

5
3
 −2
3
-60
0
1
0
1
0
0
1
3
− 13
-10
0
0
1
And so, our pivot element is

1
4 
120
1
5
3
=
4
− 23
3
5
= −6
First Example
Since we have no more negatives in the upper portion of the right
column, this is now a standard form problem and we proceed as usual.

5
3
 −2
3
-60
0
1
0
1
0
0
1
3
− 13
-10
0
0
1

1
4 
120
1
5
3
=
4
− 23
3
5
= −6
And so, our pivot element is

5
3

 −2
3
-60
0
1
1
3
0
1
0
0
0
− 13
-10
0
1
1


4 
120
First Example
And now we pivot.

1
 −2
3
-60
0
1
0
3
5
0
0
1
5
− 13
-10
0
0
1
3
5

4 ∼
120
First Example
And now we pivot.

1
 −2
3
-60
0
1
0
3
5
0
0
1
5
− 13
-10
0
0
1
3
5


1
4 ∼  0
0
120
0
1
0
3
5
2
5
36
1
5
− 15
2
0
0
1
3
5
22
5
156


First Example
And now we pivot.

1
 −2
3
-60
0
1
0
3
5
0
0
1
5
− 13
-10
0
0
1
3
5


1
4 ∼  0
0
120
0
1
0
3
5
2
5
36
1
5
− 15
2
0
0
1
3
5
22
5
156
Since we took care of all of the negatives in the bottom row, we are
done.


1
3
1 0 35
0
5
5
 0 1 2 − 1 0 22 
5
5
5
0 0 36
2
1 156


First Example
And now we pivot.

1
 −2
3
-60
0
1
0
3
5
0
0
1
5
− 13
-10
0
0
1
3
5


1
4 ∼  0
0
120
0
1
0
3
5
2
5
36
1
5
− 15
2
0
0
1
3
5
22
5
156
Since we took care of all of the negatives in the bottom row, we are
done.


1
3
1 0 35
0
5
5
 0 1 2 − 1 0 22 
5
5
5
0 0 36
2
1 156
The maximum of 156 occurs at ( 35 , 22
5 ).


How to handle minimization problems
The only thing we need to do is write the object function slightly
different and then solve in the usual manner.
How to handle minimization problems
The only thing we need to do is write the object function slightly
different and then solve in the usual manner.
If we wanted to minimize m = ab + by, this is the same as
maximizing M = −ax − by.
How to handle minimization problems
The only thing we need to do is write the object function slightly
different and then solve in the usual manner.
If we wanted to minimize m = ab + by, this is the same as
maximizing M = −ax − by.
That is, the maximum value attained at a point is the negative of the
minimum value attained at the same point.
First Minimization Example
Example
Minimize −x + 2y subject to the constraints

 2x + 3y ≤ 6
2x + y ≤ 14

x ≥ 0, y ≥ 0
First Minimization Example
Example
Minimize −x + 2y subject to the constraints

 2x + 3y ≤ 6
2x + y ≤ 14

x ≥ 0, y ≥ 0
We start by introducing slack variables as before.
2x + 3y ≤ 6 ⇒ 2x + 3y + u = 6
2x + y ≤ 14 ⇒ 2x + y + v = 14
First Minimization Example
Now the object function ... m = −x + 2y, so
First Minimization Example
Now the object function ... m = −x + 2y, so M = x − 2y.
First Minimization Example
Now the object function ... m = −x + 2y, so M = x − 2y.
When we rewrite in the correct form, we get −x + 2y + M = 0
First Minimization Example
Now the object function ... m = −x + 2y, so M = x − 2y.
When we rewrite in the correct form, we get −x + 2y + M = 0
Now the initial tableau.
First Minimization Example
Now the object function ... m = −x + 2y, so M = x − 2y.
When we rewrite in the correct form, we get −x + 2y + M = 0
Now the initial tableau.

2
 2
-1
3
1
2
1
0
0
0
1
0
0
0
1

6
14 
0
First Minimization Example
Now the object function ... m = −x + 2y, so M = x − 2y.
When we rewrite in the correct form, we get −x + 2y + M = 0
Now the initial tableau.

2
 2
-1
What is the pivot column?
3
1
2
1
0
0
0
1
0
0
0
1

6
14 
0
First Minimization Example

2
 2
-1
3
1
2
1
0
0
0
1
0
0
0
1

6
14 
0
First Minimization Example

2
 2
-1
3
1
2
1
0
0
0
1
0
0
0
1

6
14 
0
6
2 =3
14
2 =7
First Minimization Example

2
 2
-1
3
1
2
1
0
0
0
1
0
0
0
1
So, the first pivot element is

6
14 
0
6
2 =3
14
2 =7
First Minimization Example

2
 2
-1
3
1
2
1
0
0
0
1
0
0
0
1

6
14 
0
6
2 =3
14
2 =7
So, the first pivot element is

2
 2
-1
3
1
2
1
0
0
0
1
0
0
0
1

6
14 
0
First Minimization Example
Now we execute the first pivot.
First Minimization Example
Now we execute the first pivot.

2
 2
-1
3
1
2
1
0
0
0
1
0
0
0
1

6
14  ∼
0
First Minimization Example
Now we execute the first pivot.

2
 2
-1
3
1
2
1
0
0
0
1
0
0
0
1
 
1
6


2
14 ∼
-1
0
∼
3
2
1
2
1
2
0
0
0
1
0
0
0
1

3
14 
0
First Minimization Example
Now we execute the first pivot.

2
 2
-1
3
1
2
1
0
0
0
1
0
0
0
1

 
3
1
1
0 0 3
6
2
2
1 0 1 0 14 
14  ∼  2
-1 2 0 0 1 0
0


1
1 23
0 0 3
2
∼  0 -2 -1 1 0 8 
1
0 27
0 1 3
2
First Minimization Example
Now we execute the first pivot.

2
 2
-1
3
1
2
1
0
0
Are we done?
0
1
0
0
0
1

 
3
1
1
0 0 3
6
2
2
1 0 1 0 14 
14  ∼  2
-1 2 0 0 1 0
0


1
1 23
0 0 3
2
∼  0 -2 -1 1 0 8 
1
0 27
0 1 3
2
First Minimization Example

1
 0
0
3
2
1
2
-2
-1
7
2
1
2
0
1
0
0
0
1

3
8 
3
First Minimization Example

1
 0
0
3
2
1
2
-2
-1
7
2
1
2
0
1
0
0
0
1

3
8 
3
When we read this solution, we see that we get a maximum of M = 3
at the point (3, 0). Problem ...
First Minimization Example

1
 0
0
3
2
1
2
-2
-1
7
2
1
2
0
1
0
0
0
1

3
8 
3
When we read this solution, we see that we get a maximum of M = 3
at the point (3, 0). Problem ...
Since this is a minimization problem, we have to ‘undo’ the work we
did at the beginning by multiplying this maximization by -1 on both
sides to change it back to a minimization.
First Minimization Example

1
 0
0
3
2
1
2
-2
-1
7
2
1
2
0
1
0
0
0
1

3
8 
3
When we read this solution, we see that we get a maximum of M = 3
at the point (3, 0). Problem ...
Since this is a minimization problem, we have to ‘undo’ the work we
did at the beginning by multiplying this maximization by -1 on both
sides to change it back to a minimization.
Our solution, therefore, is that we have a minimum of m = −3 at the
point (3, 0).
Next Example
Example
Minimize 3x + 2y subject to the constraints

 x + 2y ≥ 20
3x + y ≥ 25

x ≥ 0, y ≥ 0
Next Example
Example
Minimize 3x + 2y subject to the constraints

 x + 2y ≥ 20
3x + y ≥ 25

x ≥ 0, y ≥ 0
Is this problem in standard form?

 −x − 2y ≤ −20
−3x − y ≤ −25

x ≥ 0, y ≥ 0
Next Example
Example
Minimize 3x + 2y subject to the constraints

 x + 2y ≥ 20
3x + y ≥ 25

x ≥ 0, y ≥ 0
Is this problem in standard form?

 −x − 2y ≤ −20
−3x − y ≤ −25

x ≥ 0, y ≥ 0
What do we need to do to put this into the form we need?
Next Example
The inequalities ...
Next Example
The inequalities ...
x + 2y ≥ 20 ⇒ −x − 2y ≤ −20
Next Example
The inequalities ...
x + 2y ≥ 20 ⇒ −x − 2y ≤ −20
3x + y ≥ 25 ⇒ −3x − y ≤ −25
Next Example
The inequalities ...
x + 2y ≥ 20 ⇒ −x − 2y ≤ −20
3x + y ≥ 25 ⇒ −3x − y ≤ −25
And the object function?
Next Example
The inequalities ...
x + 2y ≥ 20 ⇒ −x − 2y ≤ −20
3x + y ≥ 25 ⇒ −3x − y ≤ −25
And the object function?
m = 3x + 2y ⇒ M = −3x − 2y
Next Example
The inequalities ...
x + 2y ≥ 20 ⇒ −x − 2y ≤ −20
3x + y ≥ 25 ⇒ −3x − y ≤ −25
And the object function?
m = 3x + 2y ⇒ M = −3x − 2y
3x + 2y + M = 0
Next Example
So, the system we need to work with is
Next Example
So, the system we need to work with is

−x − 2y + u = −20



−3x − y + v = −25
 3x + 2y + M = 0


x ≥ 0, y ≥ 0
Next Example
So, the system we need to work with is

−x − 2y + u = −20



−3x − y + v = −25
 3x + 2y + M = 0


x ≥ 0, y ≥ 0
And when we write the initial tableau, we have
Next Example
So, the system we need to work with is

−x − 2y + u = −20



−3x − y + v = −25
 3x + 2y + M = 0


x ≥ 0, y ≥ 0
And when we write the initial tableau, we have

-1
 -3
3
-2
-1
2
1
0
0
0
1
0
0
0
1

-20
-25 
0
Next Example
Remember, we don’t care whether or not there are negatives in the
bottom row yet. Our first concern is the rightmost column.
Next Example
Remember, we don’t care whether or not there are negatives in the
bottom row yet. Our first concern is the rightmost column.
We choose either negative in the right column and then any column
with a negative in the same row as the chosen negative to select a
pivot column.
Next Example
Remember, we don’t care whether or not there are negatives in the
bottom row yet. Our first concern is the rightmost column.
We choose either negative in the right column and then any column
with a negative in the same row as the chosen negative to select a
pivot column.

-1
 -3
3
-2
-1
2
1
0
0
0
1
0
0
0
1

-20
-25 
0
Next Example
Remember, we don’t care whether or not there are negatives in the
bottom row yet. Our first concern is the rightmost column.
We choose either negative in the right column and then any column
with a negative in the same row as the chosen negative to select a
pivot column.

-1
 -3
3
-2
-1
2
1
0
0
0
1
0
0
0
1

-20
-25 
0
−20
−2
−25
−1
= 10
= 25
Next Example
Now we pivot.
Next Example
Now we pivot.

-1

 -3
3
-2
-1
2
1
0
0
0
1
0
0
0
1

-20

-25  ∼
0
Next Example
Now we pivot.

-1

 -3
3
-2
-1
2
1
0
0
0
1
0
0
0
1
 
1
-20
  2
-3
-25  ∼
3
0
∼
1
-1
2
− 12
0
0
0
1
0
0
0
1

10
-25 
0
Next Example
Now we pivot.

-1

 -3
3
-2
-1
2
1
0
0
0
1
0
0
0
1
 
1
1
-20
  2
-3 -1
-25  ∼
3
2
0
 1
1
2
5

−2 0
∼
2
0
− 12
0
0
0
1
0
0
0
1
− 12
− 12
1
0
1
0
0
0
1

10
-25 
0

10
-15 
-20
Next Example
Now we pivot.

-1

 -3
3
-2
-1
2
1
0
0
0
1
0
0
0
1
 
1
1
-20
  2
-3 -1
-25  ∼
3
2
0
 1
1
2
5

−2 0
∼
2
0
− 12
0
0
0
1
0
0
0
1
− 12
− 12
1
0
1
0
0
0
1

10
-25 
0

10
-15 
-20
Still a negative in the upper part of the rightmost column, so we need
to keep going.
Next Example
Now we pivot.

-1

 -3
3
-2
-1
2
1
0
0
0
1
0
0
0
1
 
1
1
-20
  2
-3 -1
-25  ∼
3
2
0
 1
1
2
5

−2 0
∼
2
0
− 12
0
0
0
1
0
0
0
1
− 12
− 12
1
0
1
0
0
0
1

10
-25 
0

10
-15 
-20
Still a negative in the upper part of the rightmost column, so we need
to keep going.
Choice of column again doesn’t matter as long as there is a negative
in the same row as the −15.
Next Example

1
2
 −5
2
2
1
0
0
− 21
− 21
1
0
1
0
0
0
1

10
-15 
-20
Next Example


1
2
− 25
2
1
0
0
− 21
− 21
1
0
1
0
0
0
1

10
-15 
-20
10
1
2
= 20
−15
− 52
=6
Next Example


1
2
− 25
2
1
0
0
− 21
− 21
1
0
1
0
10

10
-15 
-20
0
0
1
1
2
= 20
−15
− 52
=6

1
2
1
− 12
0
0
10



− 52
0
− 12
1
0
2
0
1
0
1

-15 

-20

Next Example
And now we pivot ...
∼

1
2
1
− 21
0
0
10



− 52
0
− 21
1
0
2
0
1
0
1

-15 

-20

Next Example
And now we pivot ...

1
2
1
− 21
0
0
10



− 52
0
− 21
1
0
2
0
1
0
1
0
− 25
0
0
0
1

-15 

-20

10
6 
-20

1
2
∼  1
2
∼
1
0
0
− 12
1
5
1

Next Example
And now we pivot ...

1
2
1
− 21
0
0



− 52
0
− 21
1
2
0
1
0

-15 

1 -20

0 10
0
6 
1 -20

0
7
0
6 
1 -32

1
2
∼  1
2

0
∼  1
0
1
0
0
− 12
1
0
0
− 35
1
5
1
1
5
3
5
0
− 25
0
1
5
− 25
4
5
0
10

Next Example
And now we pivot ...

1
2
1
− 21
0
0



− 52
0
− 21
1
2
0
1
0

-15 

1 -20

0 10
0
6 
1 -20

0
7
0
6 
1 -32

1
2
∼  1
2

0
∼  1
0
Are we done?
1
0
0
− 12
1
0
0
− 35
1
5
1
1
5
3
5
0
− 25
0
1
5
− 25
4
5
0
10

Next Example

0
 1
0
1
0
0
− 53
1
5
3
5
1
5
− 25
4
5
0
0
1

7
6 
-32
Next Example

0
 1
0
1
0
0
− 53
1
5
3
5
1
5
− 25
4
5
0
0
1

7
6 
-32
We have what for a solution from this tableau?
Next Example

0
 1
0
1
0
0
− 53
1
5
3
5
1
5
− 25
4
5
0
0
1

7
6 
-32
We have what for a solution from this tableau?
We have a maximum of M = −32 at the point (6, 7).
Next Example

0
 1
0
1
0
0
− 53
1
5
3
5
1
5
− 25
4
5
0
0
1

7
6 
-32
We have what for a solution from this tableau?
We have a maximum of M = −32 at the point (6, 7).
Is this our answer?
Next Example

0
 1
0
1
0
0
− 53
1
5
3
5
1
5
− 25
4
5
0
0
1

7
6 
-32
We have what for a solution from this tableau?
We have a maximum of M = −32 at the point (6, 7).
Is this our answer?
The minimum of m = 32 occurs at (6, 7).
Another Example
Example
Minimize 3x + 5y + z subject to the constraints

 x + y + z ≥ 20
y + 2z ≥ 10

x ≥ 0, y ≥ 0, z ≥ 0
Another Example
Example
Minimize 3x + 5y + z subject to the constraints

 x + y + z ≥ 20
y + 2z ≥ 10

x ≥ 0, y ≥ 0, z ≥ 0
Is this problem in standard form?
Another Example
Example
Minimize 3x + 5y + z subject to the constraints

 x + y + z ≥ 20
y + 2z ≥ 10

x ≥ 0, y ≥ 0, z ≥ 0
Is this problem in standard form?
What must we do to put this in the form we need?
Another Example
Rewritten, we need to solve the following:
Another Example
Rewritten, we need to solve the following:
Maximize −3x − 5y − z subject to the constraints

 −x − y − z ≤ −20
−y − 2z ≤ −10

x ≥ 0, y ≥ 0, z ≥ 0
Another Example
Rewritten, we need to solve the following:
Maximize −3x − 5y − z subject to the constraints

 −x − y − z ≤ −20
−y − 2z ≤ −10

x ≥ 0, y ≥ 0, z ≥ 0
After introducing slack variables, we have

−x − y − z + u = −20



−y − 2z + v = −10
3x + 5y + z + M = 0



x ≥ 0, y ≥ 0, z ≥ 0
Another Example
Our initial tableau, therefore is

-1 -1 -1
 0 -1 -2
3 5 1
1
0
0
0
1
0
0
0
1

-20
-10 
0
Which column do we want to select as the pivot column?
Another Example
Our initial tableau, therefore is

-1 -1 -1
 0 -1 -2
3 5 1
1
0
0
0
1
0
0
0
1

-20
-10 
0
Which column do we want to select as the pivot column?

-1
 0
3
-1
-1
5
-1
-2
1
1
0
0
0
1
0
0
0
1

-20
-10 
0
Another Example
Our initial tableau, therefore is

-1 -1 -1
 0 -1 -2
3 5 1
1
0
0
0
1
0
0
0
1

-20
-10 
0
Which column do we want to select as the pivot column?

-1
 0
3
-1
-1
5
-1
-2
1
1
0
0
0
1
0
0
0
1

-20
-10 
0
−20
−1
−10
−1
= 20
= 10
Another Example
Our initial tableau, therefore is

-1 -1 -1
 0 -1 -2
3 5 1
1
0
0
0
1
0
0
0
1

-20
-10 
0
Which column do we want to select as the pivot column?

-1
 0
3
-1
-1
5
-1
-2
1
1
0
0
0
1
0
0
0
1

-20
-10 
0
This makes our choice of pivot ...
−20
−1
−10
−1
= 20
= 10
Another Example


-1
-1
-1
1
0
0
-20

 0
3
-1
5
-2
1
0
0
1
0
0
1

-10 
0
Another Example


-1
-1
-1
1
0
0
-20

 0
3

-1
∼ 0
3
-1
5
-2
1
0
0
1
0
0
1
0
-1
0
0
0
1

-10 
0

-20
10 
0
-1
1
5
-1
2
1
1
0
0
Another Example


-1
-1
-1
1
0
0
-20

 0
3

-1
∼ 0
3

-1
5
-2
1
0
0
1
0
0
1

-10 
0

-20
10 
0

-10
10 
-50
-1

0
∼
3
-1
1
5
-1
2
1
1
0
0
0
-1
0
0
0
1
0
1
0
1
2
-9
1
0
0
-1
-1
5
0
0
1
Another Example


-1
-1
-1
1
0
0
-20

 0
3

-1
∼ 0
3

-1
5
-2
1
0
0
1
0
0
1

-10 
0

-20
10 
0

-10
10 
-50
-1

0
∼
3
-1
1
5
-1
2
1
1
0
0
0
-1
0
0
0
1
0
1
0
1
2
-9
1
0
0
-1
-1
5
0
0
1
What is our next concern?
Another Example


-1
-1
-1
1
0
0
-20

 0
3

-1
∼ 0
3

-1
5
-2
1
0
0
1
0
0
1

-10 
0

-20
10 
0

-10
10 
-50
-1

0
∼
3
-1
1
5
-1
2
1
1
0
0
0
-1
0
0
0
1
0
1
0
1
2
-9
1
0
0
-1
-1
5
0
0
1
What is our next concern? Which column do we want to use?
Another Example

-1
 0
3
0
1
0
1
2
-9
1
0
0
-1
-1
5
0
0
1

-10
10 
-50
Another Example

-1
 0
3
0
1
0
1
2
-9
1
0
0
-1
-1
5
0
0
1

-10
10 
-50
−10
−1 = 10
10
0 = und
Another Example

-1
 0
3
0
1
0
1
2
-9
1
0
0
-1
-1
5
So, our pivot element is ...
0
0
1

-10
10 
-50
−10
−1 = 10
10
0 = und
Another Example

-1
 0
3
0
1
0
1
2
-9
1
0
0
-1
-1
5
−10
−1 = 10
10
0 = und

-10
10 
-50
0
0
1
So, our pivot element is ...

-1

 0
3
0
1
0
1
2
-9
1
0
0
-1
-1
5
0
0
1

-10

10 
-50
Another Example

-1

 0
3
∼
0
1
0
1
2
-9
1
0
0
-1
-1
5
0
0
1

-10

10 
-50
Another Example

-1

 0
3

1
∼  0
3
∼
0
1
0
0
1
0
1
2
-9
-1
2
-9
1
0
0
-1
-1
5
0
0
1
-1
0
0
1
-1
5
0
0
1

-10

10 
-50

10
10 
-50
Another Example

-1 0

 0
1
3
0

1
0
∼  0
1
3
0

1 0

0 1
∼
0 0
1
2
-9

-10

10 
-50

10
10 
1
0
0
-1
-1
5
0
0
1
-1
2
-9
-1
0
0
1
-1
5
0
0
1
-50
-1
2
-6
-1
0
3
1
-1
2
0
0
1

10
10 
-80
Another Example

-1 0

 0
1
3
0

1
0
∼  0
1
3
0

1 0

0 1
∼
0 0
Now what?
1
2
-9

-10

10 
-50

10
10 
1
0
0
-1
-1
5
0
0
1
-1
2
-9
-1
0
0
1
-1
5
0
0
1
-50
-1
2
-6
-1
0
3
1
-1
2
0
0
1

10
10 
-80
Another Example

1
 0
0
0
1
0
-1
2
-6
-1
0
3
1
-1
2
0
0
1

10
10 
-80
Another Example

1
 0
0
0
1
0
-1
2
-6
-1
0
3
1
-1
2
0
0
1

10
10 
-80
10
−1 = −10
10
2 =5
Another Example

1
 0
0
0
1
0
-1
2
-6
-1
0
3
1
-1
2
So, the pivot element is ...
0
0
1

10
10 
-80
10
−1 = −10
10
2 =5
Another Example

1
 0
0
∼
0
1
0
-1
2
-6
-1
0
3
1
-1
2
0
0
1

10
10 
-80
Another Example

1 0 -1 -1 1
 0 1
2
0 -1
0 0 -6
3 2

1 0 -1 -1
1
1

∼
0 2
0 − 21
1
0 0 -6
3
2
∼
0
0
1
0
0
1

10
10 
-80

10
5 
-80
Another Example

1 0 -1 -1
 0 1
2
0
0 0 -6
3

1 0 -1 -1

∼
0 12
0
1
0 0 -6
3

1 12 0 -1
∼  0 12 1 0
0 3 0 3
1
-1
2
0
0
1
1
− 21
2
0
0
1
1
2
0
0
1
− 12
-1

10
10 
-80

10
5 
-80

15
5 
-50
Another Example

1 0 -1 -1
 0 1
2
0
0 0 -6
3

1 0 -1 -1

∼
0 12
0
1
0 0 -6
3

1 12 0 -1
∼  0 12 1 0
0 3 0 3
Do we need to keep going?
1
-1
2
0
0
1
1
− 21
2
0
0
1
1
2
0
0
1
− 12
-1

10
10 
-80

10
5 
-80

15
5 
-50
Another Example

1
 0
0
1
2
1
2
3
0
1
0
-1
0
3
1
2
− 21
-1
0
0
1

15
5 
-50
Another Example

1
 0
0
1
2
1
2
3
0
1
0
-1
0
3
1
2
− 21
-1
0
0
1

15
5 
-50
15
1
2
5
− 12
= 30
= −10
Another Example

1
 0
0
1
2
1
2
3
0
1
0
-1
0
3
1
2
− 21
-1
So, the pivot element is ...
0
0
1

15
5 
-50
15
1
2
5
− 12
= 30
= −10
Another Example

1

 0
0
∼
1
2
1
2
3

0
-1
1
2
0
15
1
0
0
3
− 12
-1
0
1

5 
-50
Another Example

1

 0
0

2

∼
0
0
∼
1
2
1
2
3
1
1
2
3

0
-1
1
2
0
15
1
0
0
3
− 12
-1
0
1
1
− 21
-1
0
0
1

5 
-50

30
5 
-50
0
1
0
-2
0
3
Another Example

1

 0
0

2

∼
0
0

1
2
1
2
3
2
∼  1
2

0
-1
1
2
0
15
1
0
0
3
− 12
-1
0
1
1
− 21
-1
0
0
1
1
0
0
0
0
1

5 
-50

30
5 
-50

30
20 
-20
1
0
1
0
1
2
3
1
1
4
-2
0
3
0
1
0
-2
-1
1
Another Example

1

 0
0

2

∼
0
0

1
2
1
2
3
2
∼  1
2
Are we done?

0
-1
1
2
0
15
1
0
0
3
− 12
-1
0
1
1
− 21
-1
0
0
1
1
0
0
0
0
1

5 
-50

30
5 
-50

30
20 
-20
1
0
1
0
1
2
3
1
1
4
-2
0
3
0
1
0
-2
-1
1
Another Example

2
 1
2
1
1
4
0
1
0
-2
-1
1
1
0
0
0
0
1

30
20 
-20
Another Example

2
 1
2
1
1
4
0
1
0
-2
-1
1
1
0
0
0
0
1

30
20 
-20
We have a maximum of M = −20 at the point (0, 0, 20).
Another Example

2
 1
2
1
1
4
0
1
0
-2
-1
1
1
0
0
0
0
1

30
20 
-20
We have a maximum of M = −20 at the point (0, 0, 20).
Is this our solution?
Another Example

2
 1
2
1
1
4
0
1
0
-2
-1
1
1
0
0
0
0
1

30
20 
-20
We have a maximum of M = −20 at the point (0, 0, 20).
Is this our solution?
The minimum of m = 20 occurs at (0, 0, 20).
Example 7
Example
Minimize 5x + 6y subject to the constraints

x + y ≤ 10



x + 2y ≥ 12
 2x + y ≥ 12


x ≥ 0, y ≥ 0
Example 7
Example
Minimize 5x + 6y subject to the constraints

x + y ≤ 10



x + 2y ≥ 12
 2x + y ≥ 12


x ≥ 0, y ≥ 0
What makes this problem not be in standard form?
Example 7
The system we will be working with is
Example 7
The system we will be working with is

x + y + u = 10




 −x − 2y + v = −12
−2x − y + w = −12


5x + 6y + M = 0



x ≥ 0, y ≥ 0
Example 7
The system we will be working with is

x + y + u = 10




 −x − 2y + v = −12
−2x − y + w = −12


5x + 6y + M = 0



x ≥ 0, y ≥ 0
which gives initial tableau
Example 7
The system we will be working with is

x + y + u = 10




 −x − 2y + v = −12
−2x − y + w = −12


5x + 6y + M = 0



x ≥ 0, y ≥ 0
which gives initial tableau

1
 -1

 -2
5
1
-2
-1
6
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1

10
-12 

-12 
0
Example 7
The system we will be working with is

x + y + u = 10




 −x − 2y + v = −12
−2x − y + w = −12


5x + 6y + M = 0



x ≥ 0, y ≥ 0
which gives initial tableau

1
 -1

 -2
5
Where to start?
1
-2
-1
6
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1

10
-12 

-12 
0
Example 7

1
 -1

 -2
5
1
-2
-1
6
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1

10
-12 

-12 
0
Example 7

1
 -1

 -2
5
1
-2
-1
6
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1

10
-12 

-12 
0
10
1 = 10
−12
−2 = 6
−12
−1 = 12
Example 7

1
 -1

 -2
5
1
-2
-1
6
1
0
0
0
And we choose ...
0
1
0
0
0
0
1
0
0
0
0
1

10
-12 

-12 
0
10
1 = 10
−12
−2 = 6
−12
−1 = 12
Example 7

∼

1
1
1
0
0
0
10

 -1

 -2
5
-2
-1
6
0
0
0
1
0
0
0
1
0
0
0
1

-12 

-12 
0
Example 7


1
1
0
0
0
10

 -1 -2

 -2 -1
5
6

1
1
 1
1
2
∼ 
 -2 -1
5
6
0
0
0
1
0
0
0
1
0
0
0
1
0
− 12
0
0
0
0
1
0
0
0
0
1

-12 

-12 
0

10
6 

-12 
∼
1
1
0
0
0
0
Example 7

1

1
1
0
0
0
10

 -1 -2

 -2 -1
5
6

1
1
 1
1
2
∼ 
 -2 -1
5
6
 1
0
2
 1
1
2
∼ 
 −3 0
2
2
0
0
0
0
1
0
0
0
1
0
0
0
1
0
− 12
0
0
0
0
1
0
0
0
0
1

-12 

-12 
0

10
6 

-12 
0
0
1
0
0
0
0
1
1
0
0
0
1
0
0
0
1
2
− 12
− 12
3
0

4
6 

-6 
-36
Example 7

1

1
1
0
0
0
10

 -1 -2

 -2 -1
5
6

1
1
 1
1
2
∼ 
 -2 -1
5
6
 1
0
2
 1
1
2
∼ 
 −3 0
2
2
0
0
0
0
1
0
0
0
1
0
0
0
1
0
− 12
0
0
0
0
1
0
0
0
0
1

-12 

-12 
0

10
6 

-12 
0
0
1
0
0
0
0
1
1
0
0
0
1
0
0
0
1
2
− 12
− 12
3
0

4
6 

-6 
-36
It would be too easy if we were done. What’s next?
Example 7

1
2
1
2

 3
 −
2
2
0
1
0
0
1
0
0
0
1
2
− 12
− 12
3
0
0
1
0
0
0
0
1

4
6 

-6 
-36
Example 7

1
2
1
2

 3
 −
2
2
0
1
0
0
1
0
0
0
1
2
− 12
− 12
3
4
0
0
1
0
0
0
0
1

4
6 

-6 
-36
1
2
6
1
2
=8
= 12
−6
− 32
=4
Example 7

1
2
1
2

 3
 −
2
2
0
1
0
0
1
0
0
0
1
2
− 12
− 12
3
4
0
0
1
0
So, our pivot element is ...
0
0
0
1

4
6 

-6 
-36
1
2
6
1
2
=8
= 12
−6
− 32
=4
Example 7






∼
1
2
1
2
0
1
1
0
1
2
− 12
0
0
0
0
− 32
0
0
− 12
1
0
2
0
0
3
0
1

4
6 


-6 

-36
Example 7
1
2
1
2
0
1
1
0
1
2
− 12
0
0
0
0
− 32
0
0
− 12
1
0
2
0
0
3
0
1
0
0
− 23
0
0
0
0
1







1
2
1
2

∼ 
 1
2
∼
0
1
0
0
1
0
0
0
1
2
− 12
1
3
3

4
6 


-6 

-36

4
6 

4 
-36
Example 7
1
2
1
2
0
1
1
0
1
2
− 12
0
0
0
0
− 32
0
0
− 12
1
0
2
0
0
3
0
1
0
0
− 23
0
0
0
0
1
-36
0
0
0
1

2
4 

4 
-44







1
2
1
2

∼ 
 1
2

0
 0
∼ 
 1
0
0
1
0
0
0
1
0
0
1
0
0
0
1
0
0
0
1
2
− 12
1
3
3
1
3
− 23
1
3
7
3
1
3
1
3
− 32
4
3

4
6 


-6 

-36

4
6 

4 
Example 7
Are we done?
Example 7
Are we done?

0
 0

 1
0
Solution?
0
1
0
0
1
0
0
0
1
3
− 23
1
3
7
3
1
3
1
3
− 23
4
3
0
0
0
1

2
4 

4 
-44
Example 7
Are we done?

0
 0

 1
0
0
1
0
0
1
0
0
0
1
3
− 23
1
3
7
3
1
3
1
3
− 23
4
3
0
0
0
1

2
4 

4 
-44
Solution? The maximum value M = −44 occurs at (4, 4).
Example 7
Are we done?

0
 0

 1
0
0
1
0
0
1
0
0
0
1
3
− 23
1
3
7
3
1
3
1
3
− 23
4
3
0
0
0
1

2
4 

4 
-44
Solution? The maximum value M = −44 occurs at (4, 4).
Is this our answer?
Example 7
Are we done?

0
 0

 1
0
0
1
0
0
1
0
0
0
1
3
− 23
1
3
7
3
1
3
1
3
− 23
4
3
0
0
0
1

2
4 

4 
-44
Solution? The maximum value M = −44 occurs at (4, 4).
Is this our answer?
The minimum value m = 44 occurs at (4, 4).
Example 8
Example
Minimize 10x + 15y subject to the constraints

 x + y ≥ 17
5x + 8y ≥ 42

x ≥ 0, y ≥ 0
Example 8
Example
Minimize 10x + 15y subject to the constraints

 x + y ≥ 17
5x + 8y ≥ 42

x ≥ 0, y ≥ 0
The system we need to solve is
Example 8
Example
Minimize 10x + 15y subject to the constraints

 x + y ≥ 17
5x + 8y ≥ 42

x ≥ 0, y ≥ 0
The system we need to solve is

−x − y + u = −17



−5x − 8y + v = −42
10x + 15y + M = 0



x ≥ 0, y ≥ 0
Example 8
Our initial tableau is

-1
 -5
10
-1
-8
15
1
0
0
0
1
0
0
0
1

-17
-42 
0
Example 8
Our initial tableau is

-1
 -5
10
Where do we start?

-1 -1 1
 -5 -8 0
10 15 0
0
1
0
-1
-8
15
0
0
1
1
0
0

-17
-42 
0
0
1
0
0
0
1

-17
-42 
0
Example 8
Our initial tableau is

-1
 -5
10
Where do we start?

-1 -1 1
 -5 -8 0
10 15 0
0
1
0
-1
-8
15
0
0
1
1
0
0

-17
-42 
0
0
1
0
0
0
1

-17
-42 
0
−17
−1
−42
−5
= 17
= 42
5
Example 8

∼

-1
-1
1
0
0
-17

 -5
10
-8
15
0
0
1
0
0
1

-42 
0
Example 8

-1
1

0
0
-17

-8 0 1
 -5
10 15 0 0

-1 -1 1
0
8
1
∼  1
0
−
5
5
10 15 0
0
0
1

-42 
0

-17
42 
∼
-1
0
0
1
5
0
Example 8

-1
-1

-8
 -5
10 15

-1 -1 1
8
∼  1
0
5
10 15 0

0 53 1
∼  1 58 0
0 -1 0

1
0
0
-17
0
0
1
0
0
1
0
− 15
0
0
0
1

-42 
0

-17
42 
− 15
− 15
2
0
0
1
5
0
− 43
5
42
5
84


Example 8

-1
-1

-8
 -5
10 15

-1 -1 1
8
∼  1
0
5
10 15 0

0 53 1
∼  1 58 0
0 -1 0
Done?

1
0
0
-17
0
0
1
0
0
1
0
− 15
0
0
0
1

-42 
0

-17
42 
− 15
− 15
2
0
0
1
5
0
− 43
5
42
5
84


Example 8

0
 1
0
3
5
8
5
-1
1
0
0
− 51
− 51
2
0
0
1
− 43
5
42
5
84


Example 8

0
 1
0
3
5
8
5
-1
1
0
0
− 51
− 51
2
0
0
1
− 43
5
42
5
84


− 43
5
− 15
42
5
− 15
= 43
= −42
Example 8

0
 1
0
3
5
8
5
-1
1
0
0
− 51
− 51
2
0
0
1
− 43
5
42
5
84
So, our next pivot element is ...


− 43
5
− 15
42
5
− 15
= 43
= −42
Example 8

 0

 1
0
∼
3
5
8
5
-1
1
0
0
− 51
− 51
2
0
0
1
− 43
5
42
5
-84




Example 8

 0

 1
0

0
∼  1
0
∼
3
5
8
5
− 51
− 51
1
0
0
-1
-3
8
5
-1
2
-5
-1
0
1
− 15
2
0
0
1
0
0
1
− 43
5
42
5




-84
43

42
5

-84
Example 8

 0

 1
0

3
5
8
5
-1
1
0
0
− 51
− 51
2
0
0
1
− 43
5
42
5




-84

0 -3 -5
1
0 43

∼  1 85 -1 − 15 0 42
5
0 -1 0
2
1 -84


0 -3 -5 1 0
43
17 
∼  1 1 -1 0 0
0 5 10 0 1 -170
Example 8

 0

 1
0

3
5
8
5
-1
1
0
0
− 51
− 51
2
0
0
1
− 43
5
42
5




-84

0 -3 -5
1
0 43

∼  1 85 -1 − 15 0 42
5
0 -1 0
2
1 -84


0 -3 -5 1 0
43
17 
∼  1 1 -1 0 0
0 5 10 0 1 -170
And ...
Example 8

0
 1
0
-3
1
5
-5
-1
10
1
0
0
0
0
1

43
17 
-170
Example 8

0
 1
0
-3
1
5
-5
-1
10
1
0
0
0
0
1

43
17 
-170
So, we have a maximum of m = −170 at the point (17, 0).
Example 8

0
 1
0
-3
1
5
-5
-1
10
1
0
0
0
0
1

43
17 
-170
So, we have a maximum of m = −170 at the point (17, 0).
The minimum of m = 170 occurs at (17, 0).
Example 9
Example
Minimize 3x + 4y subject to the constraints

 2x + y ≥ 10
x + 2y ≥ 14

x ≥ 0, y ≥ 0
Example 9
Example
Minimize 3x + 4y subject to the constraints

 2x + y ≥ 10
x + 2y ≥ 14

x ≥ 0, y ≥ 0
After rewriting, we have

−2x − y + u = −10



−x − 2y + v = −14
3x + 4y + M = 0



x ≥ 0, y ≥ 0
Example 9
The initial tableau is
Example 9
The initial tableau is

-2
 -1
3
-1
-2
4
1
0
0
0
1
0
0
0
1

-10
-14 
0
Example 9
The initial tableau is

-2
 -1
3
-1
-2
4
1
0
0
Which column do we want to use?
0
1
0
0
0
1

-10
-14 
0
Example 9
The initial tableau is

-2
 -1
3
-1
-2
4
1
0
0
Which column do we want to use?


-2 -1 1 0 0 -10
 -1 -2 0 1 0 -14 
3 4 0 0 1
0
0
1
0
0
0
1

-10
-14 
0
Example 9
The initial tableau is

-2
 -1
3
-1
-2
4
1
0
0
Which column do we want to use?


-2 -1 1 0 0 -10
 -1 -2 0 1 0 -14 
3 4 0 0 1
0
0
1
0
0
0
1

-10
-14 
0
−10
−1
−14
−2
= 10
=7
Example 9
So, our pivot element is ...
Example 9
So, our pivot element is ...

∼

-2
-1
1
0
0
-10

 -1
3
-2
4
0
0
1
0
0
1

-14 
0
Example 9
So, our pivot element is ...

-1
1

0
0
-10

 -1 -2 0 1
3
4
0 0

-2 -1 1
0
1

∼
1
0 − 12
2
3
4
0
0
0
1

-14 
0

-10
7 
0
∼
-2
0
0
1
Example 9
So, our pivot element is ...

-2
-1
1
0
0

 -1 -2
3
4

-2 -1
1

∼
1
2
3
4
 3
−2 0
1

1
∼
2
1
0
0
0
1
0
0
1
1
0
0
0
− 12
0
1
0
0
− 12
− 12
2
-10


-14 
0

0 -10
0
7 
1
0

0 -3
7 
0
1 -28
Example 9
So, our pivot element is ...

Done?
-2
-1
1
0
0

 -1 -2
3
4

-2 -1
1

∼
1
2
3
4
 3
−2 0
1

1
∼
2
1
0
0
0
1
0
0
1
1
0
0
0
− 12
0
1
0
0
− 12
− 12
2
-10


-14 
0

0 -10
0
7 
1
0

0 -3
7 
0
1 -28
Example 9
So, our pivot element is ...

-2
-1
1
0
0

 -1 -2
3
4

-2 -1
1

∼
1
2
3
4
 3
−2 0
1

1
∼
2
1
0
0
0
1
0
0
1
Done? Which column?
1
0
0
0
− 12
0
1
0
0
− 12
− 12
2
-10


-14 
0

0 -10
0
7 
1
0

0 -3
7 
0
1 -28
Example 9


− 23
1
2
1
0
1
0
1
0
0
− 12
− 12
2
0
0
1

-3
7 
-28
Example 9


− 23
1
2
1
0
1
0
1
0
0
− 12
− 12
2
0
0
1

-3
7 
-28
−3
− 32
7
1
2
=2
= 14
Example 9


− 23
1
2
1
0
1
0
1
0
0
− 12
− 12
2
0
0
1
So, our pivot element is ...

3
0
 −2

1

1
2
1
0

-3
7 
-28
−3
− 32
7
1
2
1
− 12
0
0
0
− 12
2
0
1
=2
= 14

-3 

7 
-28
Example 9




− 32
1
2
1
∼
0
1
1
0
0
0
− 12
− 12
2

0
0
1
-3 

7 
-28
Example 9




− 32
1
2
1

1
∼ 
1
2
1
∼
0
1
0
0
1
1
0
0
0
− 23
0
0
− 12
− 12
2
1
3
− 12
2

0
0
1
0
0
1
-3 

7 
-28

2
7 
-28
Example 9




− 32
1
2
1

1
∼ 
1
2
1

1
∼  0
0
0
1
1
0
0
0
0
1
0
2
1
3
− 23
0
0
0
1
0
− 12
− 12
− 12
2
− 23
1
3
2
3
1
3
2
3
5
3

0
0
1
0
0
1
0
0
1
-3 

7 
-28

2
7 
-28

2
6 
-30
Example 9




− 32
1
2
1

1
∼ 
1
2
1

1
∼  0
0
Done?
0
1
1
0
0
0
0
1
0
2
1
3
− 23
0
0
0
1
0
− 12
− 12
− 12
2
− 23
1
3
2
3
1
3
2
3
5
3

0
0
1
0
0
1
0
0
1
-3 

7 
-28

2
7 
-28

2
6 
-30
Example 9

1
 0
0
0
1
0
− 32
1
3
2
3
1
3
− 23
5
3
0
0
1

2
6 
-30
Example 9

1
 0
0
0
1
0
− 32
1
3
2
3
1
3
− 23
5
3
0
0
1

2
6 
-30
So, we get a maximum of M = −30 at the point (2, 6).
Example 9

1
 0
0
0
1
0
− 32
1
3
2
3
1
3
− 23
5
3
0
0
1

2
6 
-30
So, we get a maximum of M = −30 at the point (2, 6).
Is this the answer?
Example 9

1
 0
0
0
1
0
− 32
1
3
2
3
1
3
− 23
5
3
0
0
1

2
6 
-30
So, we get a maximum of M = −30 at the point (2, 6).
Is this the answer?
The minimum of m = 30 occurs at (2, 6).
Example 10
Example
Minimize 2x + y + 2z subject to the constraints

x + 5y + z ≤ 100



x + 2y + z ≥ 50
2x + 4y + z ≥ 80



x ≥ 0, y ≥ 0, z ≥ 0
Example 10
Example
Minimize 2x + y + 2z subject to the constraints

x + 5y + z ≤ 100



x + 2y + z ≥ 50
2x + 4y + z ≥ 80



x ≥ 0, y ≥ 0, z ≥ 0
The system of equations we want to work with is

x + 5y + z + u = 100




 −x − 2y − z + v = −50
−2x − 4y − z + w = −80


2x + y + 2z + M = 0



x ≥ 0, y ≥ 0, z ≥ 0
Example 10
The initial tableau therefore is
Example 10
The initial tableau therefore is

1
 -1

 -2
2
5
-2
-4
1
1
-1
-1
2
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1

100
-50 

-80 
0
Example 10
The initial tableau therefore is

1
 -1

 -2
2
5
-2
-4
1
1
-1
-1
2
Pivot column? Pivot element?

1 5 1 1 0 0 0
 -1 -2 -1 0 1 0 0

 -2 -4 -1 0 0 1 0
2 1 2 0 0 0 1
1
0
0
0
0
1
0
0
0
0
1
0

100
-50 

-80 
0
0
0
0
1

100
-50 

-80 
0
Example 10
The initial tableau therefore is

1
 -1

 -2
2
5
-2
-4
1
1
-1
-1
2
Pivot column? Pivot element?

1 5 1 1 0 0 0
 -1 -2 -1 0 1 0 0

 -2 -4 -1 0 0 1 0
2 1 2 0 0 0 1
1
0
0
0
0
1
0
0
0
0
1
0

100
-50 

-80 
0
0
0
0
1

100
-50 

-80 
0
100
1 = 100
−50
−1 = 50
−80
−1 = 80
Example 10

∼

1
5
1
1
0
0
0
100

 -1

 -2
2
-2
-4
1
-1
-1
2
0
0
0
1
0
0
0
1
0
0
0
1

-50 

-80 
0
Example 10


1
5
1
1
0
0
0
100

 -1

 -2
2

1
 1
∼ 
 -2
2
-2
-4
1
-1
-1
2
0
0
0
1
0
0
0
1
0
0
0
1
5
2
-4
1
1
1
-1
2
0
-1
0
0
0
0
1
0
0
0
0
1

-50 

-80 
0

100
50 

-80 
0
∼
1
0
0
0
Example 10

1

 -1

 -2
2

1
 1
∼ 
 -2
2

0
 1
∼ 
 -1
0

5
1
1
0
0
0
100
-2
-4
1
-1
-1
2
0
0
0
1
0
0
0
1
0
0
0
1
5
2
-4
1
1
1
-1
2
1
0
0
0
0
-1
0
0
0
0
1
0
0
0
0
1
0
1
0
0
1
0
0
0
1
-1
-1
2
0
0
1
0
0
0
0
1

-50 

-80 
0

100
50 

-80 
0

50
50 

-30 
-100
3
2
-2
-3
Example 10
Since we are not done, we need to choose a new pivot column and
element.
Example 10
Since we are not done, we need to choose a new pivot column and
element.

0
 1

 -1
0
3
2
-2
-3
0
1
0
0
1
0
0
0
1
-1
-1
2
0
0
1
0
0
0
0
1

50
50 

-30 
-100
Example 10
Since we are not done, we need to choose a new pivot column and
element.

0
 1

 -1
0
3
2
-2
-3
0
1
0
0
1
0
0
0
1
-1
-1
2
0
0
1
0
0
0
0
1

50
50 

-30 
-100
50
1 = 50
50
−1 = −50
−30
−1 = 30
Example 10
Since we are not done, we need to choose a new pivot column and
element.

0
 1

 -1
0
3
2
-2
-3
0
1
0
0
1
0
0
0
1
-1
-1
2
0
0
1
0
So, the pivot element is ...
0
0
0
1

50
50 

-30 
-100
50
1 = 50
50
−1 = −50
−30
−1 = 30
Example 10

0
 1


 -1
0
∼
3
2
0
1
1
0
1
-1
0
0
0
0
-2
-3
0
0
0
0
-1
2
1
0
0
1

50
50 


-30 
-100
Example 10

0
 1


 -1
0

0
 1
∼ 
 1
0
∼
3
2
0
1
1
0
1
-1
0
0
0
0
-2
-3
0
0
0
0
-1
2
1
0
0
1
3
2
2
-3
0
1
0
0
1
0
0
0
1
-1
1
2
0
0
-1
0
0
0
0
1

50
50 


-30 
-100

50
50 

30 
-100
Example 10

0
 1


 -1
0

0
 1
∼ 
 1
0

-1
 2
∼ 
 1
-2
3
2
0
1
1
0
1
-1
0
0
0
0
-2
-3
0
0
0
0
-1
2
1
0
0
1
3
2
2
-3
0
1
0
0
1
0
0
0
1
-1
1
2
0
0
-1
0
0
0
0
1
1
-1
-1
2
0
0
0
1
1
4
2
-7
0
1
0
0
1
0
0
0
0
0
1
0

50
50 


-30 
-100

50
50 

30 
-100

20
80 

30 
-160
Example 10
Now that it is in standard form, we need to take care of the bottom
row.
Example 10
Now that it is in standard form, we need to take care of the bottom
row.

-1
 2

 1
-2
1
4
2
-7
0
1
0
0
1
0
0
0
0
0
1
0
1
-1
-1
2
0
0
0
1

20
80 

30 
-160
Example 10
Now that it is in standard form, we need to take care of the bottom
row.

-1
 2

 1
-2
1
4
2
-7
0
1
0
0
1
0
0
0
0
0
1
0
1
-1
-1
2
0
0
0
1

20
80 

30 
-160
20
1
80
4
30
2
= 20
= 20
= 15
Example 10
Now that it is in standard form, we need to take care of the bottom
row.

-1
 2

 1
-2
1
4
2
-7
0
1
0
0
1
0
0
0
0
0
1
0
1
-1
-1
2
So the pivot element is ...
0
0
0
1

20
80 

30 
-160
20
1
80
4
30
2
= 20
= 20
= 15
Example 10

-1
 2

 1
-2
∼
1
4
2
-7
0
1
0
0
1
0
0
0
0
0
1
0
1
-1
-1
2
0
0
0
1

20
80 

30 
-160
Example 10

-1
1
 2
4

 1
2
-2 -7

-1
1
 2
4
∼ 
 1
1
2
-2 -7
∼
0
1
0
0
0
1
0
0
1
0
0
0
1
0
0
0
0
0
1
0
0
0
1
2
0
1
-1
-1
2
0
0
0
1
1
-1
− 12
2
0
0
0
1

20
80 

30 
-160

20
80 

15 
-160
Example 10

-1
1
 2
4

 1
2
-2 -7

-1
1
 2
4
∼ 
 1
1
2
-2 -7
 3
−2 0
 0
0
∼ 
 1
1
2
3
0
2
0
1
0
0
1
0
0
0
0
1
0
0
1
0
0
0
0
1
0
0
1
0
0
0

20
80 

30 
-160

0
1
0
20
0 -1 0
80 

1
1

15
−
0
2
2
0
2
1 -160

3
− 21
0
5
2
-2
1
0 20 

1
1
− 2 0 15 
2
7
− 32 1 -55
2
0
0
1
0
1
-1
-1
2
0
0
0
1
Example 10
Still a negative in the bottom row, so ...
Example 10
Still a negative in the bottom row, so ...
− 32
 0
 1


2
3
2
0
0
1
0
0
1
0
0
1
0
0
0
− 12
-2
1
2
7
2
3
2
1
− 21
− 23
0
0
0
1

5
20 

15 
-55
Example 10
Still a negative in the bottom row, so ...
− 32
 0
 1


2
3
2
0
0
1
0
0
1
0
0
1
0
0
0
− 12
-2
1
2
7
2
3
2
1
− 21
− 23
0
0
0
1
5
10

5 32 = 3
20
20 
1 = 20
15 151 = −30
−
-55 2
Example 10
Still a negative in the bottom row, so ...
− 32
 0
 1


2
3
2
0
0
1
0
0
1
0
0
1
0
0
0
− 12
-2
1
2
7
2
3
2
1
− 21
− 23
0
0
0
1
5
10

5 32 = 3
20
20 
1 = 20
15 151 = −30
−
-55 2
And so, the (hopefully) last pivot is ...
Example 10

− 32

 0
 1

2
3
2
∼
0
0
1
− 21
3
2
0
0
1
0
1
0
0
0
0
0
-2
1
− 21
− 23
0
0
1
1
2
7
2
5


20 

15 
-55
Example 10

− 32

 0
 1

2
3
2

-1
 0
∼ 
 1
2
3
2
∼
0
0
1
− 21
3
2
0
0
1
0
1
0
0
0
0
0
-2
1
− 21
− 23
0
0
1
1
1
− 21
− 23
0
0
0
1
0
0
1
0
0
1
0
0
1
2
7
2
2
3
0
0
0
− 13
-2
1
2
7
2
5


20 

15 
-55

10
3
20 

15 
-55
Example 10

− 32

 0
 1

2
3
2

0
0
1
− 21
3
2
0
0
1
0
1
0
0
0
0
0
-2
1
− 21
− 23
0
0
1
1
1
− 21
− 23
0
0
0
1
1
0
0
0
0
0
0
1
-1
 0
∼ 
 1
0
0
1
0
0
1
0
0

0
0
1
0
0
1
0
0
2
3
2
-1
 1
∼ 
 0
0
1
2
7
2
2
3
− 13
-2
0
0
0
1
2
7
2
2
3
− 32
1
3
1
− 13
− 53
1
3
0
5


20 

15 
-55

10
3
20 

15 
-55
10 
3
50
3
50
3
-50



Example 10
We took care of the last column and bottom row, so we have
completed all of the required pivots.
Example 10
We took care of the last column and bottom row, so we have
completed all of the required pivots.

-1
 1

 0
0
0
0
1
0
0
1
0
0
2
3
− 32
− 13
− 53
1
0
1
3
1
3
1
0
0
0
0
0
0
1
10
3
50
3
50
3
-50




Example 10
We took care of the last column and bottom row, so we have
completed all of the required pivots.

-1
 1

 0
0
0
0
1
0
0
1
0
0
2
3
− 32
− 13
− 53
1
0
1
3
1
3
1
0
0
0
0
0
0
1
10
3
50
3
50
3




-50
50
The table gives a maximum value of M = −50 at the point (0, 50
3 , 3 ).
Example 10
We took care of the last column and bottom row, so we have
completed all of the required pivots.

-1
 1

 0
0
0
0
1
0
0
1
0
0
2
3
− 32
− 13
− 53
1
0
1
3
1
3
1
0
0
0
0
0
0
1
10
3
50
3
50
3




-50
50
The table gives a maximum value of M = −50 at the point (0, 50
3 , 3 ).
Was this a minimization or maximization problem?
Example 10
We took care of the last column and bottom row, so we have
completed all of the required pivots.

-1
 1

 0
0
0
0
1
0
0
1
0
0
2
3
− 32
− 13
− 53
1
0
1
3
1
3
1
0
0
0
0
0
0
1
10
3
50
3
50
3




-50
50
The table gives a maximum value of M = −50 at the point (0, 50
3 , 3 ).
Was this a minimization or maximization problem?
50
The minimum of m = 50 occurs at (0, 50
3 , 3 ).
Example 11
Example
Maximize 12x + 10y subject to the constraints

 x + 2y ≥ 24
x + y ≤ 40

x ≥ 0, y ≥ 0
Example 11
Example
Maximize 12x + 10y subject to the constraints

 x + 2y ≥ 24
x + y ≤ 40

x ≥ 0, y ≥ 0
Solution The maximum of m = 480 occurs at (40, 0).
Example 12
Example
Maximize 4x + 3y subject to the constraints

 2x + 3y ≤ 11
x + 2y ≤ 6

x ≥ 0, y ≥ 0
Example 12
Example
Maximize 4x + 3y subject to the constraints

 2x + 3y ≤ 11
x + 2y ≤ 6

x ≥ 0, y ≥ 0
Solution The maximum M = 22 occurs at ( 11
2 , 0).
Example 13
Example
Minimize 2x + y + z subject to the constraints

3x − y − 4z ≤ −12



x + 3y + 2z ≥ 10
x−y+z≤8



x ≥ 0, y ≥ 0, z ≥ 0
Example 13
Example
Minimize 2x + y + z subject to the constraints

3x − y − 4z ≤ −12



x + 3y + 2z ≥ 10
x−y+z≤8



x ≥ 0, y ≥ 0, z ≥ 0
Solution The minimum of m =
21
5
occurs at (0, 58 , 13
5 ).