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§ 4.4 The Simplex Method: Non-Standard Form
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§ 4.4 The Simplex Method: Non-StandardForm
Standard form and what can be relaxed
What were the conditions for standard form we have been adheringto?
Conditions for Standard Form1 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 relaxed1 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].
Standard form and what can be relaxed
What were the conditions for standard form we have been adheringto?
Conditions for Standard Form1 Object function is to be maximized.
2 Each variable must be constrained to be greater than or equal to0.
3 all other constraints must be of the form[linear expression] ≤ [non− negative constant].
What can be relaxed1 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].
Standard form and what can be relaxed
What were the conditions for standard form we have been adheringto?
Conditions for Standard Form1 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 relaxed1 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].
Standard form and what can be relaxed
What were the conditions for standard form we have been adheringto?
Conditions for Standard Form1 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 relaxed1 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].
Standard form and what can be relaxed
What were the conditions for standard form we have been adheringto?
Conditions for Standard Form1 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 relaxed1 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].
Standard form and what can be relaxed
What were the conditions for standard form we have been adheringto?
Conditions for Standard Form1 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 relaxed1 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].
Standard form and what can be relaxed
What were the conditions for standard form we have been adheringto?
Conditions for Standard Form1 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 relaxed1 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
ExampleMaximize x + y subject to the following constraints.
x + y ≥ 1x + y ≤ −1x ≥ 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 essentiallymeans we are adding a new set of steps to the beginning of the oneswe established in the last section.
Not really an example
ExampleMaximize x + y subject to the following constraints.
x + y ≥ 1x + y ≤ −1x ≥ 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 essentiallymeans we are adding a new set of steps to the beginning of the oneswe established in the last section.
Not really an example
ExampleMaximize x + y subject to the following constraints.
x + y ≥ 1x + y ≤ −1x ≥ 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 essentiallymeans we are adding a new set of steps to the beginning of the oneswe established in the last section.
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, removeby pivoting.
1 Select one of the negatives in this row. This will be the pivotcolumn.
2 Take ratios as before. The least positive one will indicate thepivot element.
3 Pivot.3 If there is another negative at this point in the upper part of the
last column, repeat. Otherwise revert to the original steps weused for standard form problems.
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, removeby pivoting.
1 Select one of the negatives in this row. This will be the pivotcolumn.
2 Take ratios as before. The least positive one will indicate thepivot element.
3 Pivot.3 If there is another negative at this point in the upper part of the
last column, repeat. Otherwise revert to the original steps weused for standard form problems.
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, removeby pivoting.
1 Select one of the negatives in this row. This will be the pivotcolumn.
2 Take ratios as before. The least positive one will indicate thepivot element.
3 Pivot.3 If there is another negative at this point in the upper part of the
last column, repeat. Otherwise revert to the original steps weused for standard form problems.
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, removeby pivoting.
1 Select one of the negatives in this row. This will be the pivotcolumn.
2 Take ratios as before. The least positive one will indicate thepivot element.
3 Pivot.
3 If there is another negative at this point in the upper part of thelast column, repeat. Otherwise revert to the original steps weused for standard form problems.
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, removeby pivoting.
1 Select one of the negatives in this row. This will be the pivotcolumn.
2 Take ratios as before. The least positive one will indicate thepivot element.
3 Pivot.3 If there is another negative at this point in the upper part of the
last column, repeat. Otherwise revert to the original steps weused for standard form problems.
First Example
ExampleMaximize 40x + 30y subject to
x + y ≤ 5−2x + 3y ≥ 12x ≥ 0, y ≥ 0
Is this in standard form?
To remove the inequality issue, we multiply both sides of thatinequality to get this to be a ≤ problem.
−2x + 3y ≥ 12⇒ 2x− 3y ≤ −12
First Example
ExampleMaximize 40x + 30y subject to
x + y ≤ 5−2x + 3y ≥ 12x ≥ 0, y ≥ 0
Is this in standard form?
To remove the inequality issue, we multiply both sides of thatinequality to get this to be a ≤ problem.
−2x + 3y ≥ 12⇒ 2x− 3y ≤ −12
First Example
ExampleMaximize 40x + 30y subject to
x + y ≤ 5−2x + 3y ≥ 12x ≥ 0, y ≥ 0
Is this in standard form?
To remove the inequality issue, we multiply both sides of thatinequality to get this to be a ≤ problem.
−2x + 3y ≥ 12⇒ 2x− 3y ≤ −12
First Example
Now we rewrite the system by introducing slack variables.
x + y + u = 52x− 3y + v = −12−40x− 30y + M = 0x ≥ 0, y ≥ 0
Now the initial tableau. 1 1 1 0 0 52 -3 0 1 0 -12
-40 -30 0 0 1 0
First Example
Now we rewrite the system by introducing slack variables.x + y + u = 52x− 3y + v = −12−40x− 30y + M = 0x ≥ 0, y ≥ 0
Now the initial tableau. 1 1 1 0 0 52 -3 0 1 0 -12
-40 -30 0 0 1 0
First Example
Now we rewrite the system by introducing slack variables.x + y + u = 52x− 3y + v = −12−40x− 30y + M = 0x ≥ 0, y ≥ 0
Now the initial tableau.
1 1 1 0 0 52 -3 0 1 0 -12
-40 -30 0 0 1 0
First Example
Now we rewrite the system by introducing slack variables.x + y + u = 52x− 3y + v = −12−40x− 30y + M = 0x ≥ 0, y ≥ 0
Now the initial tableau. 1 1 1 0 0 52 -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 ofproblem spots in the last column. 1 1 1 0 0 5
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 −12to be the pivot column and then we take ratios like before to seewhich gives the smallest positive ratio. 1 1 1 0 0 5
2 -3 0 1 0 -12-40 -30 0 0 1 0
51 = 5−12−3 = 4
which gives a pivot element of ...
First Example
Before we even look at the bottom row, we have to take care ofproblem spots in the last column. 1 1 1 0 0 5
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 −12to be the pivot column and then we take ratios like before to seewhich gives the smallest positive ratio. 1 1 1 0 0 5
2 -3 0 1 0 -12-40 -30 0 0 1 0
51 = 5−12−3 = 4
which gives a pivot element of ...
First Example
Before we even look at the bottom row, we have to take care ofproblem spots in the last column. 1 1 1 0 0 5
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 −12to be the pivot column and then we take ratios like before to seewhich gives the smallest positive ratio. 1 1 1 0 0 5
2 -3 0 1 0 -12-40 -30 0 0 1 0
51 = 5−12−3 = 4
which gives a pivot element of ...
First Example
1 1 1 0 0 5
2 -3 0 1 0 -12-40 -30 0 0 1 0
After we pivot, we have 1 1 1 0 0 5−2
3 1 0 − 13 0 4
-40 -30 0 0 1 0
∼
53 0 1 1
3 0 1− 2
3 1 0 −13 0 4
-60 0 0 -10 1 120
First Example
1 1 1 0 0 5
2 -3 0 1 0 -12-40 -30 0 0 1 0
After we pivot, we have 1 1 1 0 0 5
−23 1 0 − 1
3 0 4-40 -30 0 0 1 0
∼
53 0 1 1
3 0 1− 2
3 1 0 −13 0 4
-60 0 0 -10 1 120
First Example
1 1 1 0 0 5
2 -3 0 1 0 -12-40 -30 0 0 1 0
After we pivot, we have 1 1 1 0 0 5
−23 1 0 − 1
3 0 4-40 -30 0 0 1 0
∼
53 0 1 1
3 0 1− 2
3 1 0 −13 0 4
-60 0 0 -10 1 120
First Example
Since we have no more negatives in the upper portion of the rightcolumn, this is now a standard form problem and we proceed as usual. 5
3 0 1 13 0 1
−23 1 0 − 1
3 0 4-60 0 0 -10 1 120
153= 3
54
− 23= −6
And so, our pivot element is
53 0 1 1
3 0 1
−23 1 0 −1
3 0 4-60 0 0 -10 1 120
First Example
Since we have no more negatives in the upper portion of the rightcolumn, this is now a standard form problem and we proceed as usual. 5
3 0 1 13 0 1
−23 1 0 − 1
3 0 4-60 0 0 -10 1 120
153= 3
54
− 23= −6
And so, our pivot element is
53 0 1 1
3 0 1
−23 1 0 −1
3 0 4-60 0 0 -10 1 120
First Example
Since we have no more negatives in the upper portion of the rightcolumn, this is now a standard form problem and we proceed as usual. 5
3 0 1 13 0 1
−23 1 0 − 1
3 0 4-60 0 0 -10 1 120
153= 3
54
− 23= −6
And so, our pivot element is
53 0 1 1
3 0 1
−23 1 0 −1
3 0 4-60 0 0 -10 1 120
First Example
Since we have no more negatives in the upper portion of the rightcolumn, this is now a standard form problem and we proceed as usual. 5
3 0 1 13 0 1
−23 1 0 − 1
3 0 4-60 0 0 -10 1 120
153= 3
54
− 23= −6
And so, our pivot element is
53 0 1 1
3 0 1
−23 1 0 −1
3 0 4-60 0 0 -10 1 120
First Example
And now we pivot.
1 0 35
15 0 3
5−2
3 1 0 − 13 0 4
-60 0 0 -10 1 120
∼
1 0 35
15 0 3
50 1 2
5 − 15 0 22
50 0 36 2 1 156
Since we took care of all of the negatives in the bottom row, we aredone. 1 0 3
515 0 3
50 1 2
5 − 15 0 22
50 0 36 2 1 156
The maximum of 156 occurs at (3
5 ,225 ).
First Example
And now we pivot.
1 0 35
15 0 3
5−2
3 1 0 − 13 0 4
-60 0 0 -10 1 120
∼ 1 0 3
515 0 3
50 1 2
5 − 15 0 22
50 0 36 2 1 156
Since we took care of all of the negatives in the bottom row, we aredone. 1 0 3
515 0 3
50 1 2
5 − 15 0 22
50 0 36 2 1 156
The maximum of 156 occurs at (3
5 ,225 ).
First Example
And now we pivot.
1 0 35
15 0 3
5−2
3 1 0 − 13 0 4
-60 0 0 -10 1 120
∼ 1 0 3
515 0 3
50 1 2
5 − 15 0 22
50 0 36 2 1 156
Since we took care of all of the negatives in the bottom row, we aredone. 1 0 3
515 0 3
50 1 2
5 − 15 0 22
50 0 36 2 1 156
The maximum of 156 occurs at (35 ,
225 ).
First Example
And now we pivot.
1 0 35
15 0 3
5−2
3 1 0 − 13 0 4
-60 0 0 -10 1 120
∼ 1 0 3
515 0 3
50 1 2
5 − 15 0 22
50 0 36 2 1 156
Since we took care of all of the negatives in the bottom row, we aredone. 1 0 3
515 0 3
50 1 2
5 − 15 0 22
50 0 36 2 1 156
The maximum of 156 occurs at (3
5 ,225 ).
How to handle minimization problems
The only thing we need to do is write the object function slightlydifferent and then solve in the usual manner.
If we wanted to minimize m = ab + by, this is the same asmaximizing M = −ax− by.
That is, the maximum value attained at a point is the negative of theminimum value attained at the same point.
How to handle minimization problems
The only thing we need to do is write the object function slightlydifferent and then solve in the usual manner.
If we wanted to minimize m = ab + by, this is the same asmaximizing M = −ax− by.
That is, the maximum value attained at a point is the negative of theminimum value attained at the same point.
How to handle minimization problems
The only thing we need to do is write the object function slightlydifferent and then solve in the usual manner.
If we wanted to minimize m = ab + by, this is the same asmaximizing M = −ax− by.
That is, the maximum value attained at a point is the negative of theminimum value attained at the same point.
First Minimization Example
ExampleMinimize −x + 2y subject to the constraints
2x + 3y ≤ 62x + y ≤ 14x ≥ 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
ExampleMinimize −x + 2y subject to the constraints
2x + 3y ≤ 62x + y ≤ 14x ≥ 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
M = x− 2y.
When we rewrite in the correct form, we get −x + 2y + M = 0
Now the initial tableau.
2 3 1 0 0 62 1 0 1 0 14-1 2 0 0 1 0
What is the pivot column?
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 3 1 0 0 62 1 0 1 0 14-1 2 0 0 1 0
What is the pivot column?
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 3 1 0 0 62 1 0 1 0 14-1 2 0 0 1 0
What is the pivot column?
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 3 1 0 0 62 1 0 1 0 14-1 2 0 0 1 0
What is the pivot column?
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 3 1 0 0 62 1 0 1 0 14-1 2 0 0 1 0
What is the pivot column?
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 3 1 0 0 62 1 0 1 0 14-1 2 0 0 1 0
What is the pivot column?
First Minimization Example
2 3 1 0 0 62 1 0 1 0 14-1 2 0 0 1 0
62 = 3142 = 7
So, the first pivot element is
2 3 1 0 0 62 1 0 1 0 14-1 2 0 0 1 0
First Minimization Example
2 3 1 0 0 62 1 0 1 0 14-1 2 0 0 1 0
62 = 3142 = 7
So, the first pivot element is
2 3 1 0 0 62 1 0 1 0 14-1 2 0 0 1 0
First Minimization Example
2 3 1 0 0 62 1 0 1 0 14-1 2 0 0 1 0
62 = 3142 = 7
So, the first pivot element is
2 3 1 0 0 62 1 0 1 0 14-1 2 0 0 1 0
First Minimization Example
2 3 1 0 0 62 1 0 1 0 14-1 2 0 0 1 0
62 = 3142 = 7
So, the first pivot element is
2 3 1 0 0 62 1 0 1 0 14-1 2 0 0 1 0
First Minimization Example
Now we execute the first pivot.
2 3 1 0 0 62 1 0 1 0 14-1 2 0 0 1 0
∼ 1 3
212 0 0 3
2 1 0 1 0 14-1 2 0 0 1 0
∼
1 32
12 0 0 3
0 -2 -1 1 0 80 7
212 0 1 3
Are we done?
First Minimization Example
Now we execute the first pivot.
2 3 1 0 0 62 1 0 1 0 14-1 2 0 0 1 0
∼
1 32
12 0 0 3
2 1 0 1 0 14-1 2 0 0 1 0
∼
1 32
12 0 0 3
0 -2 -1 1 0 80 7
212 0 1 3
Are we done?
First Minimization Example
Now we execute the first pivot.
2 3 1 0 0 62 1 0 1 0 14-1 2 0 0 1 0
∼ 1 3
212 0 0 3
2 1 0 1 0 14-1 2 0 0 1 0
∼
1 32
12 0 0 3
0 -2 -1 1 0 80 7
212 0 1 3
Are we done?
First Minimization Example
Now we execute the first pivot.
2 3 1 0 0 62 1 0 1 0 14-1 2 0 0 1 0
∼ 1 3
212 0 0 3
2 1 0 1 0 14-1 2 0 0 1 0
∼
1 32
12 0 0 3
0 -2 -1 1 0 80 7
212 0 1 3
Are we done?
First Minimization Example
Now we execute the first pivot.
2 3 1 0 0 62 1 0 1 0 14-1 2 0 0 1 0
∼ 1 3
212 0 0 3
2 1 0 1 0 14-1 2 0 0 1 0
∼
1 32
12 0 0 3
0 -2 -1 1 0 80 7
212 0 1 3
Are we done?
First Minimization Example
1 32
12 0 0 3
0 -2 -1 1 0 80 7
212 0 1 3
When we read this solution, we see that we get a maximum of M = 3at the point (3, 0). Problem ...
Since this is a minimization problem, we have to ‘undo’ the work wedid at the beginning by multiplying this maximization by -1 on bothsides to change it back to a minimization.
Our solution, therefore, is that we have a minimum of m = −3 at thepoint (3, 0).
First Minimization Example
1 32
12 0 0 3
0 -2 -1 1 0 80 7
212 0 1 3
When we read this solution, we see that we get a maximum of M = 3at the point (3, 0). Problem ...
Since this is a minimization problem, we have to ‘undo’ the work wedid at the beginning by multiplying this maximization by -1 on bothsides to change it back to a minimization.
Our solution, therefore, is that we have a minimum of m = −3 at thepoint (3, 0).
First Minimization Example
1 32
12 0 0 3
0 -2 -1 1 0 80 7
212 0 1 3
When we read this solution, we see that we get a maximum of M = 3at the point (3, 0). Problem ...
Since this is a minimization problem, we have to ‘undo’ the work wedid at the beginning by multiplying this maximization by -1 on bothsides to change it back to a minimization.
Our solution, therefore, is that we have a minimum of m = −3 at thepoint (3, 0).
First Minimization Example
1 32
12 0 0 3
0 -2 -1 1 0 80 7
212 0 1 3
When we read this solution, we see that we get a maximum of M = 3at the point (3, 0). Problem ...
Since this is a minimization problem, we have to ‘undo’ the work wedid at the beginning by multiplying this maximization by -1 on bothsides to change it back to a minimization.
Our solution, therefore, is that we have a minimum of m = −3 at thepoint (3, 0).
Next Example
ExampleMinimize 3x + 2y subject to the constraints
x + 2y ≥ 203x + y ≥ 25x ≥ 0, y ≥ 0
Is this problem in standard form?−x− 2y ≤ −20−3x− y ≤ −25x ≥ 0, y ≥ 0
What do we need to do to put this into the form we need?
Next Example
ExampleMinimize 3x + 2y subject to the constraints
x + 2y ≥ 203x + y ≥ 25x ≥ 0, y ≥ 0
Is this problem in standard form?−x− 2y ≤ −20−3x− y ≤ −25x ≥ 0, y ≥ 0
What do we need to do to put this into the form we need?
Next Example
ExampleMinimize 3x + 2y subject to the constraints
x + 2y ≥ 203x + y ≥ 25x ≥ 0, y ≥ 0
Is this problem in standard form?−x− 2y ≤ −20−3x− y ≤ −25x ≥ 0, y ≥ 0
What do we need to do to put this into the form we need?
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
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
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
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
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
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
−x− 2y + u = −20−3x− y + v = −253x + 2y + M = 0x ≥ 0, y ≥ 0
And when we write the initial tableau, we have
-1 -2 1 0 0 -20-3 -1 0 1 0 -253 2 0 0 1 0
Next Example
So, the system we need to work with is
−x− 2y + u = −20−3x− y + v = −253x + 2y + M = 0x ≥ 0, y ≥ 0
And when we write the initial tableau, we have
-1 -2 1 0 0 -20-3 -1 0 1 0 -253 2 0 0 1 0
Next Example
So, the system we need to work with is
−x− 2y + u = −20−3x− y + v = −253x + 2y + M = 0x ≥ 0, y ≥ 0
And when we write the initial tableau, we have
-1 -2 1 0 0 -20-3 -1 0 1 0 -253 2 0 0 1 0
Next Example
So, the system we need to work with is
−x− 2y + u = −20−3x− y + v = −253x + 2y + M = 0x ≥ 0, y ≥ 0
And when we write the initial tableau, we have
-1 -2 1 0 0 -20-3 -1 0 1 0 -253 2 0 0 1 0
Next Example
Remember, we don’t care whether or not there are negatives in thebottom row yet. Our first concern is the rightmost column.
We choose either negative in the right column and then any columnwith a negative in the same row as the chosen negative to select apivot column. -1 -2 1 0 0 -20
-3 -1 0 1 0 -253 2 0 0 1 0
−20−2 = 10−25−1 = 25
Next Example
Remember, we don’t care whether or not there are negatives in thebottom row yet. Our first concern is the rightmost column.
We choose either negative in the right column and then any columnwith a negative in the same row as the chosen negative to select apivot column.
-1 -2 1 0 0 -20-3 -1 0 1 0 -253 2 0 0 1 0
−20−2 = 10−25−1 = 25
Next Example
Remember, we don’t care whether or not there are negatives in thebottom row yet. Our first concern is the rightmost column.
We choose either negative in the right column and then any columnwith a negative in the same row as the chosen negative to select apivot column. -1 -2 1 0 0 -20
-3 -1 0 1 0 -253 2 0 0 1 0
−20−2 = 10−25−1 = 25
Next Example
Remember, we don’t care whether or not there are negatives in thebottom row yet. Our first concern is the rightmost column.
We choose either negative in the right column and then any columnwith a negative in the same row as the chosen negative to select apivot column. -1 -2 1 0 0 -20
-3 -1 0 1 0 -253 2 0 0 1 0
−20−2 = 10−25−1 = 25
Next Example
Now we pivot.
-1 -2 1 0 0 -20-3 -1 0 1 0 -253 2 0 0 1 0
∼ 1
2 1 −12 0 0 10
-3 -1 0 1 0 -253 2 0 0 1 0
∼
12 1 −1
2 0 0 10− 5
2 0 −12 1 0 -15
2 0 1 0 1 -20
Still a negative in the upper part of the rightmost column, so we needto keep going.
Choice of column again doesn’t matter as long as there is a negativein the same row as the −15.
Next Example
Now we pivot.
-1 -2 1 0 0 -20-3 -1 0 1 0 -253 2 0 0 1 0
∼
12 1 −1
2 0 0 10-3 -1 0 1 0 -253 2 0 0 1 0
∼
12 1 −1
2 0 0 10− 5
2 0 −12 1 0 -15
2 0 1 0 1 -20
Still a negative in the upper part of the rightmost column, so we needto keep going.
Choice of column again doesn’t matter as long as there is a negativein the same row as the −15.
Next Example
Now we pivot.
-1 -2 1 0 0 -20-3 -1 0 1 0 -253 2 0 0 1 0
∼ 1
2 1 −12 0 0 10
-3 -1 0 1 0 -253 2 0 0 1 0
∼
12 1 −1
2 0 0 10− 5
2 0 −12 1 0 -15
2 0 1 0 1 -20
Still a negative in the upper part of the rightmost column, so we needto keep going.
Choice of column again doesn’t matter as long as there is a negativein the same row as the −15.
Next Example
Now we pivot.
-1 -2 1 0 0 -20-3 -1 0 1 0 -253 2 0 0 1 0
∼ 1
2 1 −12 0 0 10
-3 -1 0 1 0 -253 2 0 0 1 0
∼
12 1 −1
2 0 0 10− 5
2 0 −12 1 0 -15
2 0 1 0 1 -20
Still a negative in the upper part of the rightmost column, so we needto keep going.
Choice of column again doesn’t matter as long as there is a negativein the same row as the −15.
Next Example
Now we pivot.
-1 -2 1 0 0 -20-3 -1 0 1 0 -253 2 0 0 1 0
∼ 1
2 1 −12 0 0 10
-3 -1 0 1 0 -253 2 0 0 1 0
∼
12 1 −1
2 0 0 10− 5
2 0 −12 1 0 -15
2 0 1 0 1 -20
Still a negative in the upper part of the rightmost column, so we needto keep going.
Choice of column again doesn’t matter as long as there is a negativein the same row as the −15.
Next Example
Now we pivot.
-1 -2 1 0 0 -20-3 -1 0 1 0 -253 2 0 0 1 0
∼ 1
2 1 −12 0 0 10
-3 -1 0 1 0 -253 2 0 0 1 0
∼
12 1 −1
2 0 0 10− 5
2 0 −12 1 0 -15
2 0 1 0 1 -20
Still a negative in the upper part of the rightmost column, so we needto keep going.
Choice of column again doesn’t matter as long as there is a negativein the same row as the −15.
Next Example
12 1 −1
2 0 0 10−5
2 0 −12 1 0 -15
2 0 1 0 1 -20
1012= 20
−15− 5
2= 6
12 1 −1
2 0 0 10
−52 0 −1
2 1 0 -15
2 0 1 0 1 -20
Next Example
12 1 −1
2 0 0 10−5
2 0 −12 1 0 -15
2 0 1 0 1 -20
1012= 20
−15− 5
2= 6
12 1 −1
2 0 0 10
−52 0 −1
2 1 0 -15
2 0 1 0 1 -20
Next Example
12 1 −1
2 0 0 10−5
2 0 −12 1 0 -15
2 0 1 0 1 -20
1012= 20
−15− 5
2= 6
12 1 −1
2 0 0 10
−52 0 −1
2 1 0 -15
2 0 1 0 1 -20
Next Example
And now we pivot ...
12 1 −1
2 0 0 10
−52 0 −1
2 1 0 -15
2 0 1 0 1 -20
∼
12 1 −1
2 0 0 101 0 1
5 −25 0 6
2 0 1 0 1 -20
∼
0 1 −35
15 0 7
1 0 15 −2
5 0 60 0 3
545 1 -32
Are we done?
Next Example
And now we pivot ...
12 1 −1
2 0 0 10
−52 0 −1
2 1 0 -15
2 0 1 0 1 -20
∼
12 1 −1
2 0 0 101 0 1
5 −25 0 6
2 0 1 0 1 -20
∼
0 1 −35
15 0 7
1 0 15 −2
5 0 60 0 3
545 1 -32
Are we done?
Next Example
And now we pivot ...
12 1 −1
2 0 0 10
−52 0 −1
2 1 0 -15
2 0 1 0 1 -20
∼
12 1 −1
2 0 0 101 0 1
5 −25 0 6
2 0 1 0 1 -20
∼
0 1 −35
15 0 7
1 0 15 −2
5 0 60 0 3
545 1 -32
Are we done?
Next Example
And now we pivot ...
12 1 −1
2 0 0 10
−52 0 −1
2 1 0 -15
2 0 1 0 1 -20
∼
12 1 −1
2 0 0 101 0 1
5 −25 0 6
2 0 1 0 1 -20
∼
0 1 −35
15 0 7
1 0 15 −2
5 0 60 0 3
545 1 -32
Are we done?
Next Example
0 1 − 35
15 0 7
1 0 15 − 2
5 0 60 0 3
545 1 -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).
Next Example
0 1 − 35
15 0 7
1 0 15 − 2
5 0 60 0 3
545 1 -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).
Next Example
0 1 − 35
15 0 7
1 0 15 − 2
5 0 60 0 3
545 1 -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).
Next Example
0 1 − 35
15 0 7
1 0 15 − 2
5 0 60 0 3
545 1 -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).
Next Example
0 1 − 35
15 0 7
1 0 15 − 2
5 0 60 0 3
545 1 -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
ExampleMinimize 3x + 5y + z subject to the constraints
x + y + z ≥ 20y + 2z ≥ 10x ≥ 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
ExampleMinimize 3x + 5y + z subject to the constraints
x + y + z ≥ 20y + 2z ≥ 10x ≥ 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
ExampleMinimize 3x + 5y + z subject to the constraints
x + y + z ≥ 20y + 2z ≥ 10x ≥ 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:
Maximize −3x− 5y− z subject to the constraints−x− y− z ≤ −20−y− 2z ≤ −10x ≥ 0, y ≥ 0, z ≥ 0
After introducing slack variables, we have−x− y− z + u = −20−y− 2z + v = −103x + 5y + z + M = 0x ≥ 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 ≤ −10x ≥ 0, y ≥ 0, z ≥ 0
After introducing slack variables, we have−x− y− z + u = −20−y− 2z + v = −103x + 5y + z + M = 0x ≥ 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 ≤ −10x ≥ 0, y ≥ 0, z ≥ 0
After introducing slack variables, we have−x− y− z + u = −20−y− 2z + v = −103x + 5y + z + M = 0x ≥ 0, y ≥ 0, z ≥ 0
Another Example
Our initial tableau, therefore is -1 -1 -1 1 0 0 -200 -1 -2 0 1 0 -103 5 1 0 0 1 0
Which column do we want to select as the pivot column?
-1 -1 -1 1 0 0 -200 -1 -2 0 1 0 -103 5 1 0 0 1 0
−20−1 = 20−10−1 = 10
This makes our choice of pivot ...
Another Example
Our initial tableau, therefore is -1 -1 -1 1 0 0 -200 -1 -2 0 1 0 -103 5 1 0 0 1 0
Which column do we want to select as the pivot column?
-1 -1 -1 1 0 0 -200 -1 -2 0 1 0 -103 5 1 0 0 1 0
−20−1 = 20−10−1 = 10
This makes our choice of pivot ...
Another Example
Our initial tableau, therefore is -1 -1 -1 1 0 0 -200 -1 -2 0 1 0 -103 5 1 0 0 1 0
Which column do we want to select as the pivot column?
-1 -1 -1 1 0 0 -200 -1 -2 0 1 0 -103 5 1 0 0 1 0
−20−1 = 20−10−1 = 10
This makes our choice of pivot ...
Another Example
Our initial tableau, therefore is -1 -1 -1 1 0 0 -200 -1 -2 0 1 0 -103 5 1 0 0 1 0
Which column do we want to select as the pivot column?
-1 -1 -1 1 0 0 -200 -1 -2 0 1 0 -103 5 1 0 0 1 0
−20−1 = 20−10−1 = 10
This makes our choice of pivot ...
Another Example
-1 -1 -1 1 0 0 -20
0 -1 -2 0 1 0 -103 5 1 0 0 1 0
∼
-1 -1 -1 1 0 0 -200 1 2 0 -1 0 103 5 1 0 0 1 0
∼
-1 0 1 1 -1 0 -100 1 2 0 -1 0 103 0 -9 0 5 1 -50
What is our next concern? Which column do we want to use?
Another Example
-1 -1 -1 1 0 0 -20
0 -1 -2 0 1 0 -103 5 1 0 0 1 0
∼
-1 -1 -1 1 0 0 -200 1 2 0 -1 0 103 5 1 0 0 1 0
∼
-1 0 1 1 -1 0 -100 1 2 0 -1 0 103 0 -9 0 5 1 -50
What is our next concern? Which column do we want to use?
Another Example
-1 -1 -1 1 0 0 -20
0 -1 -2 0 1 0 -103 5 1 0 0 1 0
∼
-1 -1 -1 1 0 0 -200 1 2 0 -1 0 103 5 1 0 0 1 0
∼
-1 0 1 1 -1 0 -100 1 2 0 -1 0 103 0 -9 0 5 1 -50
What is our next concern? Which column do we want to use?
Another Example
-1 -1 -1 1 0 0 -20
0 -1 -2 0 1 0 -103 5 1 0 0 1 0
∼
-1 -1 -1 1 0 0 -200 1 2 0 -1 0 103 5 1 0 0 1 0
∼
-1 0 1 1 -1 0 -100 1 2 0 -1 0 103 0 -9 0 5 1 -50
What is our next concern?
Which column do we want to use?
Another Example
-1 -1 -1 1 0 0 -20
0 -1 -2 0 1 0 -103 5 1 0 0 1 0
∼
-1 -1 -1 1 0 0 -200 1 2 0 -1 0 103 5 1 0 0 1 0
∼
-1 0 1 1 -1 0 -100 1 2 0 -1 0 103 0 -9 0 5 1 -50
What is our next concern? Which column do we want to use?
Another Example
-1 0 1 1 -1 0 -100 1 2 0 -1 0 103 0 -9 0 5 1 -50
−10−1 = 10
100 = und
So, our pivot element is ...
-1 0 1 1 -1 0 -100 1 2 0 -1 0 103 0 -9 0 5 1 -50
Another Example
-1 0 1 1 -1 0 -100 1 2 0 -1 0 103 0 -9 0 5 1 -50
−10−1 = 10
100 = und
So, our pivot element is ...
-1 0 1 1 -1 0 -100 1 2 0 -1 0 103 0 -9 0 5 1 -50
Another Example
-1 0 1 1 -1 0 -100 1 2 0 -1 0 103 0 -9 0 5 1 -50
−10−1 = 10
100 = und
So, our pivot element is ...
-1 0 1 1 -1 0 -100 1 2 0 -1 0 103 0 -9 0 5 1 -50
Another Example
-1 0 1 1 -1 0 -100 1 2 0 -1 0 103 0 -9 0 5 1 -50
−10−1 = 10
100 = und
So, our pivot element is ...
-1 0 1 1 -1 0 -100 1 2 0 -1 0 103 0 -9 0 5 1 -50
Another Example
-1 0 1 1 -1 0 -100 1 2 0 -1 0 103 0 -9 0 5 1 -50
∼
1 0 -1 -1 1 0 100 1 2 0 -1 0 103 0 -9 0 5 1 -50
∼
1 0 -1 -1 1 0 100 1 2 0 -1 0 100 0 -6 3 2 1 -80
Now what?
Another Example
-1 0 1 1 -1 0 -100 1 2 0 -1 0 103 0 -9 0 5 1 -50
∼
1 0 -1 -1 1 0 100 1 2 0 -1 0 103 0 -9 0 5 1 -50
∼
1 0 -1 -1 1 0 100 1 2 0 -1 0 100 0 -6 3 2 1 -80
Now what?
Another Example
-1 0 1 1 -1 0 -100 1 2 0 -1 0 103 0 -9 0 5 1 -50
∼
1 0 -1 -1 1 0 100 1 2 0 -1 0 103 0 -9 0 5 1 -50
∼
1 0 -1 -1 1 0 100 1 2 0 -1 0 100 0 -6 3 2 1 -80
Now what?
Another Example
-1 0 1 1 -1 0 -100 1 2 0 -1 0 103 0 -9 0 5 1 -50
∼
1 0 -1 -1 1 0 100 1 2 0 -1 0 103 0 -9 0 5 1 -50
∼
1 0 -1 -1 1 0 100 1 2 0 -1 0 100 0 -6 3 2 1 -80
Now what?
Another Example
1 0 -1 -1 1 0 100 1 2 0 -1 0 100 0 -6 3 2 1 -80
10−1 = −10102 = 5
So, the pivot element is ...
Another Example
1 0 -1 -1 1 0 100 1 2 0 -1 0 100 0 -6 3 2 1 -80
10−1 = −10102 = 5
So, the pivot element is ...
Another Example
1 0 -1 -1 1 0 100 1 2 0 -1 0 100 0 -6 3 2 1 -80
10−1 = −10102 = 5
So, the pivot element is ...
Another Example
1 0 -1 -1 1 0 100 1 2 0 -1 0 100 0 -6 3 2 1 -80
∼
1 0 -1 -1 1 0 100 1
2 1 0 − 12 0 5
0 0 -6 3 2 1 -80
∼
1 12 0 -1 1
2 0 150 1
2 1 0 − 12 0 5
0 3 0 3 -1 1 -50
Do we need to keep going?
Another Example
1 0 -1 -1 1 0 100 1 2 0 -1 0 100 0 -6 3 2 1 -80
∼
1 0 -1 -1 1 0 100 1
2 1 0 − 12 0 5
0 0 -6 3 2 1 -80
∼
1 12 0 -1 1
2 0 150 1
2 1 0 − 12 0 5
0 3 0 3 -1 1 -50
Do we need to keep going?
Another Example
1 0 -1 -1 1 0 100 1 2 0 -1 0 100 0 -6 3 2 1 -80
∼
1 0 -1 -1 1 0 100 1
2 1 0 − 12 0 5
0 0 -6 3 2 1 -80
∼
1 12 0 -1 1
2 0 150 1
2 1 0 − 12 0 5
0 3 0 3 -1 1 -50
Do we need to keep going?
Another Example
1 0 -1 -1 1 0 100 1 2 0 -1 0 100 0 -6 3 2 1 -80
∼
1 0 -1 -1 1 0 100 1
2 1 0 − 12 0 5
0 0 -6 3 2 1 -80
∼
1 12 0 -1 1
2 0 150 1
2 1 0 − 12 0 5
0 3 0 3 -1 1 -50
Do we need to keep going?
Another Example
1 12 0 -1 1
2 0 150 1
2 1 0 − 12 0 5
0 3 0 3 -1 1 -50
1512= 30
5− 1
2= −10
So, the pivot element is ...
Another Example
1 12 0 -1 1
2 0 150 1
2 1 0 − 12 0 5
0 3 0 3 -1 1 -50
1512= 30
5− 1
2= −10
So, the pivot element is ...
Another Example
1 12 0 -1 1
2 0 150 1
2 1 0 − 12 0 5
0 3 0 3 -1 1 -50
1512= 30
5− 1
2= −10
So, the pivot element is ...
Another Example
1 12 0 -1 1
2 0 15
0 12 1 0 − 1
2 0 50 3 0 3 -1 1 -50
∼
2 1 0 -2 1 0 300 1
2 1 0 −12 0 5
0 3 0 3 -1 1 -50
∼
2 1 0 -2 1 0 301 1 1 -1 0 0 202 4 0 1 0 1 -20
Are we done?
Another Example
1 12 0 -1 1
2 0 15
0 12 1 0 − 1
2 0 50 3 0 3 -1 1 -50
∼
2 1 0 -2 1 0 300 1
2 1 0 −12 0 5
0 3 0 3 -1 1 -50
∼
2 1 0 -2 1 0 301 1 1 -1 0 0 202 4 0 1 0 1 -20
Are we done?
Another Example
1 12 0 -1 1
2 0 15
0 12 1 0 − 1
2 0 50 3 0 3 -1 1 -50
∼
2 1 0 -2 1 0 300 1
2 1 0 −12 0 5
0 3 0 3 -1 1 -50
∼
2 1 0 -2 1 0 301 1 1 -1 0 0 202 4 0 1 0 1 -20
Are we done?
Another Example
1 12 0 -1 1
2 0 15
0 12 1 0 − 1
2 0 50 3 0 3 -1 1 -50
∼
2 1 0 -2 1 0 300 1
2 1 0 −12 0 5
0 3 0 3 -1 1 -50
∼
2 1 0 -2 1 0 301 1 1 -1 0 0 202 4 0 1 0 1 -20
Are we done?
Another Example
2 1 0 -2 1 0 301 1 1 -1 0 0 202 4 0 1 0 1 -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).
Another Example
2 1 0 -2 1 0 301 1 1 -1 0 0 202 4 0 1 0 1 -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).
Another Example
2 1 0 -2 1 0 301 1 1 -1 0 0 202 4 0 1 0 1 -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).
Another Example
2 1 0 -2 1 0 301 1 1 -1 0 0 202 4 0 1 0 1 -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
ExampleMinimize 5x + 6y subject to the constraints
x + y ≤ 10x + 2y ≥ 122x + y ≥ 12x ≥ 0, y ≥ 0
What makes this problem not be in standard form?
Example 7
ExampleMinimize 5x + 6y subject to the constraints
x + y ≤ 10x + 2y ≥ 122x + y ≥ 12x ≥ 0, y ≥ 0
What makes this problem not be in standard form?
Example 7
The system we will be working with is
x + y + u = 10−x− 2y + v = −12−2x− y + w = −125x + 6y + M = 0x ≥ 0, y ≥ 0
which gives initial tableau
1 1 1 0 0 0 10-1 -2 0 1 0 0 -12-2 -1 0 0 1 0 -125 6 0 0 0 1 0
Where to start?
Example 7
The system we will be working with is
x + y + u = 10−x− 2y + v = −12−2x− y + w = −125x + 6y + M = 0x ≥ 0, y ≥ 0
which gives initial tableau
1 1 1 0 0 0 10-1 -2 0 1 0 0 -12-2 -1 0 0 1 0 -125 6 0 0 0 1 0
Where to start?
Example 7
The system we will be working with is
x + y + u = 10−x− 2y + v = −12−2x− y + w = −125x + 6y + M = 0x ≥ 0, y ≥ 0
which gives initial tableau
1 1 1 0 0 0 10-1 -2 0 1 0 0 -12-2 -1 0 0 1 0 -125 6 0 0 0 1 0
Where to start?
Example 7
The system we will be working with is
x + y + u = 10−x− 2y + v = −12−2x− y + w = −125x + 6y + M = 0x ≥ 0, y ≥ 0
which gives initial tableau
1 1 1 0 0 0 10-1 -2 0 1 0 0 -12-2 -1 0 0 1 0 -125 6 0 0 0 1 0
Where to start?
Example 7
The system we will be working with is
x + y + u = 10−x− 2y + v = −12−2x− y + w = −125x + 6y + M = 0x ≥ 0, y ≥ 0
which gives initial tableau
1 1 1 0 0 0 10-1 -2 0 1 0 0 -12-2 -1 0 0 1 0 -125 6 0 0 0 1 0
Where to start?
Example 7
1 1 1 0 0 0 10-1 -2 0 1 0 0 -12-2 -1 0 0 1 0 -125 6 0 0 0 1 0
101 = 10−12−2 = 6−12−1 = 12
And we choose ...
Example 7
1 1 1 0 0 0 10-1 -2 0 1 0 0 -12-2 -1 0 0 1 0 -125 6 0 0 0 1 0
101 = 10−12−2 = 6−12−1 = 12
And we choose ...
Example 7
1 1 1 0 0 0 10-1 -2 0 1 0 0 -12-2 -1 0 0 1 0 -125 6 0 0 0 1 0
101 = 10−12−2 = 6−12−1 = 12
And we choose ...
Example 7
1 1 1 0 0 0 10
-1 -2 0 1 0 0 -12-2 -1 0 0 1 0 -125 6 0 0 0 1 0
∼
1 1 1 0 0 0 1012 1 0 −1
2 0 0 6-2 -1 0 0 1 0 -125 6 0 0 0 1 0
∼
12 0 1 1
2 0 0 412 1 0 −1
2 0 0 6−3
2 0 0 −12 1 0 -6
2 0 0 3 0 1 -36
It would be too easy if we were done. What’s next?
Example 7
1 1 1 0 0 0 10
-1 -2 0 1 0 0 -12-2 -1 0 0 1 0 -125 6 0 0 0 1 0
∼
1 1 1 0 0 0 1012 1 0 −1
2 0 0 6-2 -1 0 0 1 0 -125 6 0 0 0 1 0
∼
12 0 1 1
2 0 0 412 1 0 −1
2 0 0 6−3
2 0 0 −12 1 0 -6
2 0 0 3 0 1 -36
It would be too easy if we were done. What’s next?
Example 7
1 1 1 0 0 0 10
-1 -2 0 1 0 0 -12-2 -1 0 0 1 0 -125 6 0 0 0 1 0
∼
1 1 1 0 0 0 1012 1 0 −1
2 0 0 6-2 -1 0 0 1 0 -125 6 0 0 0 1 0
∼
12 0 1 1
2 0 0 412 1 0 −1
2 0 0 6−3
2 0 0 −12 1 0 -6
2 0 0 3 0 1 -36
It would be too easy if we were done. What’s next?
Example 7
1 1 1 0 0 0 10
-1 -2 0 1 0 0 -12-2 -1 0 0 1 0 -125 6 0 0 0 1 0
∼
1 1 1 0 0 0 1012 1 0 −1
2 0 0 6-2 -1 0 0 1 0 -125 6 0 0 0 1 0
∼
12 0 1 1
2 0 0 412 1 0 −1
2 0 0 6−3
2 0 0 −12 1 0 -6
2 0 0 3 0 1 -36
It would be too easy if we were done. What’s next?
Example 7
12 0 1 1
2 0 0 412 1 0 −1
2 0 0 6−3
2 0 0 −12 1 0 -6
2 0 0 3 0 1 -36
412= 8
612= 12
−6− 3
2= 4
So, our pivot element is ...
Example 7
12 0 1 1
2 0 0 412 1 0 −1
2 0 0 6−3
2 0 0 −12 1 0 -6
2 0 0 3 0 1 -36
412= 8
612= 12
−6− 3
2= 4
So, our pivot element is ...
Example 7
12 0 1 1
2 0 0 412 1 0 −1
2 0 0 6−3
2 0 0 −12 1 0 -6
2 0 0 3 0 1 -36
412= 8
612= 12
−6− 3
2= 4
So, our pivot element is ...
Example 7
12 0 1 1
2 0 0 412 1 0 −1
2 0 0 6
−32 0 0 −1
2 1 0 -6
2 0 0 3 0 1 -36
∼
12 0 1 1
2 0 0 412 1 0 − 1
2 0 0 61 0 0 1
3 −23 0 4
2 0 0 3 0 1 -36
∼
0 0 1 1
313 0 2
0 1 0 −23
13 0 4
1 0 0 13 −2
3 0 40 0 0 7
343 1 -44
Example 7
12 0 1 1
2 0 0 412 1 0 −1
2 0 0 6
−32 0 0 −1
2 1 0 -6
2 0 0 3 0 1 -36
∼
12 0 1 1
2 0 0 412 1 0 − 1
2 0 0 61 0 0 1
3 −23 0 4
2 0 0 3 0 1 -36
∼
0 0 1 1
313 0 2
0 1 0 −23
13 0 4
1 0 0 13 −2
3 0 40 0 0 7
343 1 -44
Example 7
12 0 1 1
2 0 0 412 1 0 −1
2 0 0 6
−32 0 0 −1
2 1 0 -6
2 0 0 3 0 1 -36
∼
12 0 1 1
2 0 0 412 1 0 − 1
2 0 0 61 0 0 1
3 −23 0 4
2 0 0 3 0 1 -36
∼
0 0 1 1
313 0 2
0 1 0 −23
13 0 4
1 0 0 13 −2
3 0 40 0 0 7
343 1 -44
Example 7
Are we done?
0 0 1 1
313 0 2
0 1 0 −23
13 0 4
1 0 0 13 − 2
3 0 40 0 0 7
343 1 -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 7
Are we done?
0 0 1 1
313 0 2
0 1 0 −23
13 0 4
1 0 0 13 − 2
3 0 40 0 0 7
343 1 -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 7
Are we done?
0 0 1 1
313 0 2
0 1 0 −23
13 0 4
1 0 0 13 − 2
3 0 40 0 0 7
343 1 -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 7
Are we done?
0 0 1 1
313 0 2
0 1 0 −23
13 0 4
1 0 0 13 − 2
3 0 40 0 0 7
343 1 -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 7
Are we done?
0 0 1 1
313 0 2
0 1 0 −23
13 0 4
1 0 0 13 − 2
3 0 40 0 0 7
343 1 -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
ExampleMinimize 10x + 15y subject to the constraints
x + y ≥ 175x + 8y ≥ 42x ≥ 0, y ≥ 0
The system we need to solve is
−x− y + u = −17−5x− 8y + v = −4210x + 15y + M = 0x ≥ 0, y ≥ 0
Example 8
ExampleMinimize 10x + 15y subject to the constraints
x + y ≥ 175x + 8y ≥ 42x ≥ 0, y ≥ 0
The system we need to solve is
−x− y + u = −17−5x− 8y + v = −4210x + 15y + M = 0x ≥ 0, y ≥ 0
Example 8
ExampleMinimize 10x + 15y subject to the constraints
x + y ≥ 175x + 8y ≥ 42x ≥ 0, y ≥ 0
The system we need to solve is
−x− y + u = −17−5x− 8y + v = −4210x + 15y + M = 0x ≥ 0, y ≥ 0
Example 8
Our initial tableau is -1 -1 1 0 0 -17-5 -8 0 1 0 -4210 15 0 0 1 0
Where do we start? -1 -1 1 0 0 -17-5 -8 0 1 0 -4210 15 0 0 1 0
−17−1 = 17−42−5 = 42
5
Example 8
Our initial tableau is -1 -1 1 0 0 -17-5 -8 0 1 0 -4210 15 0 0 1 0
Where do we start? -1 -1 1 0 0 -17
-5 -8 0 1 0 -4210 15 0 0 1 0
−17−1 = 17−42−5 = 42
5
Example 8
Our initial tableau is -1 -1 1 0 0 -17-5 -8 0 1 0 -4210 15 0 0 1 0
Where do we start? -1 -1 1 0 0 -17
-5 -8 0 1 0 -4210 15 0 0 1 0
−17−1 = 17−42−5 = 42
5
Example 8
-1 -1 1 0 0 -17
-5 -8 0 1 0 -4210 15 0 0 1 0
∼
-1 -1 1 0 0 -171 8
5 0 −15 0 42
510 15 0 0 1 0
∼
0 35 1 −1
5 0 − 435
1 85 0 −1
5 0 425
0 -1 0 2 1 84
Done?
Example 8
-1 -1 1 0 0 -17
-5 -8 0 1 0 -4210 15 0 0 1 0
∼
-1 -1 1 0 0 -171 8
5 0 −15 0 42
510 15 0 0 1 0
∼
0 35 1 −1
5 0 − 435
1 85 0 −1
5 0 425
0 -1 0 2 1 84
Done?
Example 8
-1 -1 1 0 0 -17
-5 -8 0 1 0 -4210 15 0 0 1 0
∼
-1 -1 1 0 0 -171 8
5 0 −15 0 42
510 15 0 0 1 0
∼
0 35 1 −1
5 0 − 435
1 85 0 −1
5 0 425
0 -1 0 2 1 84
Done?
Example 8
-1 -1 1 0 0 -17
-5 -8 0 1 0 -4210 15 0 0 1 0
∼
-1 -1 1 0 0 -171 8
5 0 −15 0 42
510 15 0 0 1 0
∼
0 35 1 −1
5 0 − 435
1 85 0 −1
5 0 425
0 -1 0 2 1 84
Done?
Example 8
0 35 1 − 1
5 0 − 435
1 85 0 − 1
5 0 425
0 -1 0 2 1 84
− 435
− 15= 43
425
− 15= −42
So, our next pivot element is ...
Example 8
0 35 1 − 1
5 0 − 435
1 85 0 − 1
5 0 425
0 -1 0 2 1 84
− 43
5− 1
5= 43
425
− 15= −42
So, our next pivot element is ...
Example 8
0 35 1 − 1
5 0 − 435
1 85 0 − 1
5 0 425
0 -1 0 2 1 84
− 43
5− 1
5= 43
425
− 15= −42
So, our next pivot element is ...
Example 8
0 35 1 − 1
5 0 − 435
1 85 0 − 1
5 0 425
0 -1 0 2 1 -84
∼
0 -3 -5 1 0 431 8
5 -1 −15 0 42
50 -1 0 2 1 -84
∼
0 -3 -5 1 0 431 1 -1 0 0 170 5 10 0 1 -170
And ...
Example 8
0 35 1 − 1
5 0 − 435
1 85 0 − 1
5 0 425
0 -1 0 2 1 -84
∼
0 -3 -5 1 0 431 8
5 -1 −15 0 42
50 -1 0 2 1 -84
∼
0 -3 -5 1 0 431 1 -1 0 0 170 5 10 0 1 -170
And ...
Example 8
0 35 1 − 1
5 0 − 435
1 85 0 − 1
5 0 425
0 -1 0 2 1 -84
∼
0 -3 -5 1 0 431 8
5 -1 −15 0 42
50 -1 0 2 1 -84
∼
0 -3 -5 1 0 431 1 -1 0 0 170 5 10 0 1 -170
And ...
Example 8
0 35 1 − 1
5 0 − 435
1 85 0 − 1
5 0 425
0 -1 0 2 1 -84
∼
0 -3 -5 1 0 431 8
5 -1 −15 0 42
50 -1 0 2 1 -84
∼
0 -3 -5 1 0 431 1 -1 0 0 170 5 10 0 1 -170
And ...
Example 8
0 -3 -5 1 0 431 1 -1 0 0 170 5 10 0 1 -170
So, we have a maximum of m = −170 at the point (17, 0).
The minimum of m = 170 occurs at (17, 0).
Example 8
0 -3 -5 1 0 431 1 -1 0 0 170 5 10 0 1 -170
So, we have a maximum of m = −170 at the point (17, 0).
The minimum of m = 170 occurs at (17, 0).
Example 8
0 -3 -5 1 0 431 1 -1 0 0 170 5 10 0 1 -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
ExampleMinimize 3x + 4y subject to the constraints
2x + y ≥ 10x + 2y ≥ 14x ≥ 0, y ≥ 0
After rewriting, we have−2x− y + u = −10−x− 2y + v = −143x + 4y + M = 0x ≥ 0, y ≥ 0
Example 9
ExampleMinimize 3x + 4y subject to the constraints
2x + y ≥ 10x + 2y ≥ 14x ≥ 0, y ≥ 0
After rewriting, we have−2x− y + u = −10−x− 2y + v = −143x + 4y + M = 0x ≥ 0, y ≥ 0
Example 9
The initial tableau is
-2 -1 1 0 0 -10-1 -2 0 1 0 -143 4 0 0 1 0
Which column do we want to use? -2 -1 1 0 0 -10
-1 -2 0 1 0 -143 4 0 0 1 0
−10−1 = 10−14−2 = 7
Example 9
The initial tableau is
-2 -1 1 0 0 -10-1 -2 0 1 0 -143 4 0 0 1 0
Which column do we want to use? -2 -1 1 0 0 -10-1 -2 0 1 0 -143 4 0 0 1 0
−10−1 = 10−14−2 = 7
Example 9
The initial tableau is
-2 -1 1 0 0 -10-1 -2 0 1 0 -143 4 0 0 1 0
Which column do we want to use?
-2 -1 1 0 0 -10-1 -2 0 1 0 -143 4 0 0 1 0
−10−1 = 10−14−2 = 7
Example 9
The initial tableau is
-2 -1 1 0 0 -10-1 -2 0 1 0 -143 4 0 0 1 0
Which column do we want to use? -2 -1 1 0 0 -10
-1 -2 0 1 0 -143 4 0 0 1 0
−10−1 = 10−14−2 = 7
Example 9
The initial tableau is
-2 -1 1 0 0 -10-1 -2 0 1 0 -143 4 0 0 1 0
Which column do we want to use? -2 -1 1 0 0 -10
-1 -2 0 1 0 -143 4 0 0 1 0
−10−1 = 10−14−2 = 7
Example 9
So, our pivot element is ...
-2 -1 1 0 0 -10
-1 -2 0 1 0 -143 4 0 0 1 0
∼
-2 -1 1 0 0 -1012 1 0 −1
2 0 73 4 0 0 1 0
∼
− 32 0 1 −1
2 0 -312 1 0 −1
2 0 71 0 0 2 1 -28
Done? Which column?
Example 9
So, our pivot element is ...
-2 -1 1 0 0 -10
-1 -2 0 1 0 -143 4 0 0 1 0
∼
-2 -1 1 0 0 -1012 1 0 −1
2 0 73 4 0 0 1 0
∼
− 32 0 1 −1
2 0 -312 1 0 −1
2 0 71 0 0 2 1 -28
Done? Which column?
Example 9
So, our pivot element is ...
-2 -1 1 0 0 -10
-1 -2 0 1 0 -143 4 0 0 1 0
∼
-2 -1 1 0 0 -1012 1 0 −1
2 0 73 4 0 0 1 0
∼
− 32 0 1 −1
2 0 -312 1 0 −1
2 0 71 0 0 2 1 -28
Done? Which column?
Example 9
So, our pivot element is ...
-2 -1 1 0 0 -10
-1 -2 0 1 0 -143 4 0 0 1 0
∼
-2 -1 1 0 0 -1012 1 0 −1
2 0 73 4 0 0 1 0
∼
− 32 0 1 −1
2 0 -312 1 0 −1
2 0 71 0 0 2 1 -28
Done? Which column?
Example 9
So, our pivot element is ...
-2 -1 1 0 0 -10
-1 -2 0 1 0 -143 4 0 0 1 0
∼
-2 -1 1 0 0 -1012 1 0 −1
2 0 73 4 0 0 1 0
∼
− 32 0 1 −1
2 0 -312 1 0 −1
2 0 71 0 0 2 1 -28
Done?
Which column?
Example 9
So, our pivot element is ...
-2 -1 1 0 0 -10
-1 -2 0 1 0 -143 4 0 0 1 0
∼
-2 -1 1 0 0 -1012 1 0 −1
2 0 73 4 0 0 1 0
∼
− 32 0 1 −1
2 0 -312 1 0 −1
2 0 71 0 0 2 1 -28
Done? Which column?
Example 9
−32 0 1 − 1
2 0 -312 1 0 − 1
2 0 71 0 0 2 1 -28
−3− 3
2= 2
712= 14
So, our pivot element is ... −32 0 1 −1
2 0 -312 1 0 −1
2 0 71 0 0 2 1 -28
Example 9
−32 0 1 − 1
2 0 -312 1 0 − 1
2 0 71 0 0 2 1 -28
−3− 3
2= 2
712= 14
So, our pivot element is ... −32 0 1 −1
2 0 -312 1 0 −1
2 0 71 0 0 2 1 -28
Example 9
−32 0 1 − 1
2 0 -312 1 0 − 1
2 0 71 0 0 2 1 -28
−3− 3
2= 2
712= 14
So, our pivot element is ... −32 0 1 −1
2 0 -312 1 0 −1
2 0 71 0 0 2 1 -28
Example 9
−32 0 1 −1
2 0 -312 1 0 −1
2 0 71 0 0 2 1 -28
∼
1 0 −23
13 0 2
12 1 0 −1
2 0 71 0 0 2 1 -28
∼
1 0 − 23
13 0 2
0 1 13
23 0 6
0 0 23
53 1 -30
Done?
Example 9
−32 0 1 −1
2 0 -312 1 0 −1
2 0 71 0 0 2 1 -28
∼
1 0 −23
13 0 2
12 1 0 −1
2 0 71 0 0 2 1 -28
∼
1 0 − 23
13 0 2
0 1 13
23 0 6
0 0 23
53 1 -30
Done?
Example 9
−32 0 1 −1
2 0 -312 1 0 −1
2 0 71 0 0 2 1 -28
∼
1 0 −23
13 0 2
12 1 0 −1
2 0 71 0 0 2 1 -28
∼
1 0 − 23
13 0 2
0 1 13
23 0 6
0 0 23
53 1 -30
Done?
Example 9
−32 0 1 −1
2 0 -312 1 0 −1
2 0 71 0 0 2 1 -28
∼
1 0 −23
13 0 2
12 1 0 −1
2 0 71 0 0 2 1 -28
∼
1 0 − 23
13 0 2
0 1 13
23 0 6
0 0 23
53 1 -30
Done?
Example 9
1 0 − 23
13 0 2
0 1 13 − 2
3 0 60 0 2
353 1 -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 9
1 0 − 23
13 0 2
0 1 13 − 2
3 0 60 0 2
353 1 -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 9
1 0 − 23
13 0 2
0 1 13 − 2
3 0 60 0 2
353 1 -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 9
1 0 − 23
13 0 2
0 1 13 − 2
3 0 60 0 2
353 1 -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
ExampleMinimize 2x + y + 2z subject to the constraints
x + 5y + z ≤ 100x + 2y + z ≥ 502x + 4y + z ≥ 80x ≥ 0, y ≥ 0, z ≥ 0
The system of equations we want to work with isx + 5y + z + u = 100−x− 2y− z + v = −50−2x− 4y− z + w = −802x + y + 2z + M = 0x ≥ 0, y ≥ 0, z ≥ 0
Example 10
ExampleMinimize 2x + y + 2z subject to the constraints
x + 5y + z ≤ 100x + 2y + z ≥ 502x + 4y + z ≥ 80x ≥ 0, y ≥ 0, z ≥ 0
The system of equations we want to work with isx + 5y + z + u = 100−x− 2y− z + v = −50−2x− 4y− z + w = −802x + y + 2z + M = 0x ≥ 0, y ≥ 0, z ≥ 0
Example 10
The initial tableau therefore is
1 5 1 1 0 0 0 100-1 -2 -1 0 1 0 0 -50-2 -4 -1 0 0 1 0 -802 1 2 0 0 0 1 0
Pivot column? Pivot element?
1 5 1 1 0 0 0 100-1 -2 -1 0 1 0 0 -50-2 -4 -1 0 0 1 0 -802 1 2 0 0 0 1 0
1001 = 100−50−1 = 50−80−1 = 80
Example 10
The initial tableau therefore is
1 5 1 1 0 0 0 100-1 -2 -1 0 1 0 0 -50-2 -4 -1 0 0 1 0 -802 1 2 0 0 0 1 0
Pivot column? Pivot element?1 5 1 1 0 0 0 100-1 -2 -1 0 1 0 0 -50-2 -4 -1 0 0 1 0 -802 1 2 0 0 0 1 0
1001 = 100−50−1 = 50−80−1 = 80
Example 10
The initial tableau therefore is
1 5 1 1 0 0 0 100-1 -2 -1 0 1 0 0 -50-2 -4 -1 0 0 1 0 -802 1 2 0 0 0 1 0
Pivot column? Pivot element?
1 5 1 1 0 0 0 100-1 -2 -1 0 1 0 0 -50-2 -4 -1 0 0 1 0 -802 1 2 0 0 0 1 0
1001 = 100−50−1 = 50−80−1 = 80
Example 10
The initial tableau therefore is
1 5 1 1 0 0 0 100-1 -2 -1 0 1 0 0 -50-2 -4 -1 0 0 1 0 -802 1 2 0 0 0 1 0
Pivot column? Pivot element?
1 5 1 1 0 0 0 100-1 -2 -1 0 1 0 0 -50-2 -4 -1 0 0 1 0 -802 1 2 0 0 0 1 0
100
1 = 100−50−1 = 50−80−1 = 80
Example 10
1 5 1 1 0 0 0 100
-1 -2 -1 0 1 0 0 -50-2 -4 -1 0 0 1 0 -802 1 2 0 0 0 1 0
∼
1 5 1 1 0 0 0 1001 2 1 0 -1 0 0 50-2 -4 -1 0 0 1 0 -802 1 2 0 0 0 1 0
∼
0 3 0 1 1 0 0 501 2 1 0 -1 0 0 50-1 -2 0 0 -1 1 0 -300 -3 0 0 2 0 1 -100
Example 10
1 5 1 1 0 0 0 100
-1 -2 -1 0 1 0 0 -50-2 -4 -1 0 0 1 0 -802 1 2 0 0 0 1 0
∼
1 5 1 1 0 0 0 1001 2 1 0 -1 0 0 50-2 -4 -1 0 0 1 0 -802 1 2 0 0 0 1 0
∼
0 3 0 1 1 0 0 501 2 1 0 -1 0 0 50-1 -2 0 0 -1 1 0 -300 -3 0 0 2 0 1 -100
Example 10
1 5 1 1 0 0 0 100
-1 -2 -1 0 1 0 0 -50-2 -4 -1 0 0 1 0 -802 1 2 0 0 0 1 0
∼
1 5 1 1 0 0 0 1001 2 1 0 -1 0 0 50-2 -4 -1 0 0 1 0 -802 1 2 0 0 0 1 0
∼
0 3 0 1 1 0 0 501 2 1 0 -1 0 0 50-1 -2 0 0 -1 1 0 -300 -3 0 0 2 0 1 -100
Example 10
Since we are not done, we need to choose a new pivot column andelement.
0 3 0 1 1 0 0 501 2 1 0 -1 0 0 50-1 -2 0 0 -1 1 0 -300 -3 0 0 2 0 1 -100
501 = 5050−1 = −50−30−1 = 30
So, the pivot element is ...
Example 10
Since we are not done, we need to choose a new pivot column andelement.
0 3 0 1 1 0 0 501 2 1 0 -1 0 0 50-1 -2 0 0 -1 1 0 -300 -3 0 0 2 0 1 -100
501 = 5050−1 = −50−30−1 = 30
So, the pivot element is ...
Example 10
Since we are not done, we need to choose a new pivot column andelement.
0 3 0 1 1 0 0 501 2 1 0 -1 0 0 50-1 -2 0 0 -1 1 0 -300 -3 0 0 2 0 1 -100
501 = 5050−1 = −50−30−1 = 30
So, the pivot element is ...
Example 10
Since we are not done, we need to choose a new pivot column andelement.
0 3 0 1 1 0 0 501 2 1 0 -1 0 0 50-1 -2 0 0 -1 1 0 -300 -3 0 0 2 0 1 -100
501 = 5050−1 = −50−30−1 = 30
So, the pivot element is ...
Example 10
0 3 0 1 1 0 0 501 2 1 0 -1 0 0 50
-1 -2 0 0 -1 1 0 -300 -3 0 0 2 0 1 -100
∼
0 3 0 1 1 0 0 501 2 1 0 -1 0 0 501 2 0 0 1 -1 0 300 -3 0 0 2 0 1 -100
∼
-1 1 0 1 0 1 0 202 4 1 0 0 -1 0 801 2 0 0 1 -1 0 30-2 -7 0 0 0 2 1 -160
Example 10
0 3 0 1 1 0 0 501 2 1 0 -1 0 0 50
-1 -2 0 0 -1 1 0 -300 -3 0 0 2 0 1 -100
∼
0 3 0 1 1 0 0 501 2 1 0 -1 0 0 501 2 0 0 1 -1 0 300 -3 0 0 2 0 1 -100
∼
-1 1 0 1 0 1 0 202 4 1 0 0 -1 0 801 2 0 0 1 -1 0 30-2 -7 0 0 0 2 1 -160
Example 10
0 3 0 1 1 0 0 501 2 1 0 -1 0 0 50
-1 -2 0 0 -1 1 0 -300 -3 0 0 2 0 1 -100
∼
0 3 0 1 1 0 0 501 2 1 0 -1 0 0 501 2 0 0 1 -1 0 300 -3 0 0 2 0 1 -100
∼
-1 1 0 1 0 1 0 202 4 1 0 0 -1 0 801 2 0 0 1 -1 0 30-2 -7 0 0 0 2 1 -160
Example 10
Now that it is in standard form, we need to take care of the bottomrow.
-1 1 0 1 0 1 0 202 4 1 0 0 -1 0 801 2 0 0 1 -1 0 30-2 -7 0 0 0 2 1 -160
201 = 20
804 = 20
302 = 15
So the pivot element is ...
Example 10
Now that it is in standard form, we need to take care of the bottomrow.
-1 1 0 1 0 1 0 202 4 1 0 0 -1 0 801 2 0 0 1 -1 0 30-2 -7 0 0 0 2 1 -160
201 = 20
804 = 20
302 = 15
So the pivot element is ...
Example 10
Now that it is in standard form, we need to take care of the bottomrow.
-1 1 0 1 0 1 0 202 4 1 0 0 -1 0 801 2 0 0 1 -1 0 30-2 -7 0 0 0 2 1 -160
201 = 20
804 = 20
302 = 15
So the pivot element is ...
Example 10
Now that it is in standard form, we need to take care of the bottomrow.
-1 1 0 1 0 1 0 202 4 1 0 0 -1 0 801 2 0 0 1 -1 0 30-2 -7 0 0 0 2 1 -160
201 = 20
804 = 20
302 = 15
So the pivot element is ...
Example 10
-1 1 0 1 0 1 0 202 4 1 0 0 -1 0 801 2 0 0 1 -1 0 30-2 -7 0 0 0 2 1 -160
∼
-1 1 0 1 0 1 0 202 4 1 0 0 -1 0 8012 1 0 0 1
2 −12 0 15
-2 -7 0 0 0 2 1 -160
∼
− 3
2 0 0 1 −12
32 0 5
0 0 1 0 -2 1 0 2012 1 0 0 1
2 −12 0 15
32 0 0 0 7
2 −32 1 -55
Example 10
-1 1 0 1 0 1 0 202 4 1 0 0 -1 0 801 2 0 0 1 -1 0 30-2 -7 0 0 0 2 1 -160
∼
-1 1 0 1 0 1 0 202 4 1 0 0 -1 0 8012 1 0 0 1
2 −12 0 15
-2 -7 0 0 0 2 1 -160
∼
− 3
2 0 0 1 −12
32 0 5
0 0 1 0 -2 1 0 2012 1 0 0 1
2 −12 0 15
32 0 0 0 7
2 −32 1 -55
Example 10
-1 1 0 1 0 1 0 202 4 1 0 0 -1 0 801 2 0 0 1 -1 0 30-2 -7 0 0 0 2 1 -160
∼
-1 1 0 1 0 1 0 202 4 1 0 0 -1 0 8012 1 0 0 1
2 −12 0 15
-2 -7 0 0 0 2 1 -160
∼
− 3
2 0 0 1 −12
32 0 5
0 0 1 0 -2 1 0 2012 1 0 0 1
2 −12 0 15
32 0 0 0 7
2 −32 1 -55
Example 10
Still a negative in the bottom row, so ...
−3
2 0 0 1 −12
32 0 5
0 0 1 0 -2 1 0 2012 1 0 0 1
2 − 12 0 15
32 0 0 0 7
2 − 32 1 -55
532= 10
3201 = 2015− 1
2= −30
And so, the (hopefully) last pivot is ...
Example 10
Still a negative in the bottom row, so ...
−3
2 0 0 1 −12
32 0 5
0 0 1 0 -2 1 0 2012 1 0 0 1
2 − 12 0 15
32 0 0 0 7
2 − 32 1 -55
532= 10
3201 = 2015− 1
2= −30
And so, the (hopefully) last pivot is ...
Example 10
Still a negative in the bottom row, so ...
−3
2 0 0 1 −12
32 0 5
0 0 1 0 -2 1 0 2012 1 0 0 1
2 − 12 0 15
32 0 0 0 7
2 − 32 1 -55
532= 10
3201 = 2015− 1
2= −30
And so, the (hopefully) last pivot is ...
Example 10
Still a negative in the bottom row, so ...
−3
2 0 0 1 −12
32 0 5
0 0 1 0 -2 1 0 2012 1 0 0 1
2 − 12 0 15
32 0 0 0 7
2 − 32 1 -55
532= 10
3201 = 2015− 1
2= −30
And so, the (hopefully) last pivot is ...
Example 10
−3
2 0 0 1 − 12
32 0 5
0 0 1 0 -2 1 0 2012 1 0 0 1
2 − 12 0 15
32 0 0 0 7
2 − 32 1 -55
∼
-1 0 0 2
3 −13 1 0 10
30 0 1 0 -2 1 0 2012 1 0 0 1
2 −12 0 15
32 0 0 0 7
2 −32 1 -55
∼
-1 0 0 2
3 −13 1 0 10
31 0 1 −2
3 −53 0 0 50
30 1 0 1
313 0 0 50
30 0 0 1 0 0 1 -50
Example 10
−3
2 0 0 1 − 12
32 0 5
0 0 1 0 -2 1 0 2012 1 0 0 1
2 − 12 0 15
32 0 0 0 7
2 − 32 1 -55
∼
-1 0 0 2
3 −13 1 0 10
30 0 1 0 -2 1 0 2012 1 0 0 1
2 −12 0 15
32 0 0 0 7
2 −32 1 -55
∼
-1 0 0 2
3 −13 1 0 10
31 0 1 −2
3 −53 0 0 50
30 1 0 1
313 0 0 50
30 0 0 1 0 0 1 -50
Example 10
−3
2 0 0 1 − 12
32 0 5
0 0 1 0 -2 1 0 2012 1 0 0 1
2 − 12 0 15
32 0 0 0 7
2 − 32 1 -55
∼
-1 0 0 2
3 −13 1 0 10
30 0 1 0 -2 1 0 2012 1 0 0 1
2 −12 0 15
32 0 0 0 7
2 −32 1 -55
∼
-1 0 0 2
3 −13 1 0 10
31 0 1 −2
3 −53 0 0 50
30 1 0 1
313 0 0 50
30 0 0 1 0 0 1 -50
Example 10
We took care of the last column and bottom row, so we havecompleted all of the required pivots.
-1 0 0 2
3 −13 1 0 10
31 0 1 −2
3 −53 0 0 50
30 1 0 1
313 0 0 50
30 0 0 1 0 0 1 -50
The table gives a maximum value of M = −50 at the point (0, 50
3 ,503 ).
Was this a minimization or maximization problem?
The minimum of m = 50 occurs at (0, 503 ,
503 ).
Example 10
We took care of the last column and bottom row, so we havecompleted all of the required pivots.
-1 0 0 2
3 −13 1 0 10
31 0 1 −2
3 −53 0 0 50
30 1 0 1
313 0 0 50
30 0 0 1 0 0 1 -50
The table gives a maximum value of M = −50 at the point (0, 503 ,
503 ).
Was this a minimization or maximization problem?
The minimum of m = 50 occurs at (0, 503 ,
503 ).
Example 10
We took care of the last column and bottom row, so we havecompleted all of the required pivots.
-1 0 0 2
3 −13 1 0 10
31 0 1 −2
3 −53 0 0 50
30 1 0 1
313 0 0 50
30 0 0 1 0 0 1 -50
The table gives a maximum value of M = −50 at the point (0, 50
3 ,503 ).
Was this a minimization or maximization problem?
The minimum of m = 50 occurs at (0, 503 ,
503 ).
Example 10
We took care of the last column and bottom row, so we havecompleted all of the required pivots.
-1 0 0 2
3 −13 1 0 10
31 0 1 −2
3 −53 0 0 50
30 1 0 1
313 0 0 50
30 0 0 1 0 0 1 -50
The table gives a maximum value of M = −50 at the point (0, 50
3 ,503 ).
Was this a minimization or maximization problem?
The minimum of m = 50 occurs at (0, 503 ,
503 ).
Example 10
We took care of the last column and bottom row, so we havecompleted all of the required pivots.
-1 0 0 2
3 −13 1 0 10
31 0 1 −2
3 −53 0 0 50
30 1 0 1
313 0 0 50
30 0 0 1 0 0 1 -50
The table gives a maximum value of M = −50 at the point (0, 50
3 ,503 ).
Was this a minimization or maximization problem?
The minimum of m = 50 occurs at (0, 503 ,
503 ).
Example 11
ExampleMaximize 12x + 10y subject to the constraints
x + 2y ≥ 24x + y ≤ 40x ≥ 0, y ≥ 0
Solution The maximum of m = 480 occurs at (40, 0).
Example 11
ExampleMaximize 12x + 10y subject to the constraints
x + 2y ≥ 24x + y ≤ 40x ≥ 0, y ≥ 0
Solution The maximum of m = 480 occurs at (40, 0).
Example 12
ExampleMaximize 4x + 3y subject to the constraints
2x + 3y ≤ 11x + 2y ≤ 6x ≥ 0, y ≥ 0
Solution The maximum M = 22 occurs at (112 , 0).
Example 12
ExampleMaximize 4x + 3y subject to the constraints
2x + 3y ≤ 11x + 2y ≤ 6x ≥ 0, y ≥ 0
Solution The maximum M = 22 occurs at (112 , 0).
Example 13
ExampleMinimize 2x + y + z subject to the constraints
3x− y− 4z ≤ −12x + 3y + 2z ≥ 10x− y + z ≤ 8x ≥ 0, y ≥ 0, z ≥ 0
Solution The minimum of m = 215 occurs at (0, 8
5 ,135 ).
Example 13
ExampleMinimize 2x + y + z subject to the constraints
3x− y− 4z ≤ −12x + 3y + 2z ≥ 10x− y + z ≤ 8x ≥ 0, y ≥ 0, z ≥ 0
Solution The minimum of m = 215 occurs at (0, 8
5 ,135 ).
