4.3 the simplex method: standard form -...
TRANSCRIPT
§ 4.3 The Simplex Method: Standard Form
The Process
Let’s explain this process through an example.
ExampleMaximize 7x + 4y subject to the following constraints:
3x + 2y ≤ 36x + 4y ≤ 32x ≥ 0, y ≥ 0
The Process
Let’s explain this process through an example.
ExampleMaximize 7x + 4y subject to the following constraints:
3x + 2y ≤ 36x + 4y ≤ 32x ≥ 0, y ≥ 0
The Process
The process we wil use to solve this problem (and the others in thischapter) is called the Simplex Method.
In order to solve a problem using this method (at least at this point),we need the problem to be in 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].
We will later be able to relax conditions 1 and 3, but never the secondone.
The Process
The process we wil use to solve this problem (and the others in thischapter) is called the Simplex Method.
In order to solve a problem using this method (at least at this point),we need the problem to be in 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].
We will later be able to relax conditions 1 and 3, but never the secondone.
The Process
The process we wil use to solve this problem (and the others in thischapter) is called the Simplex Method.
In order to solve a problem using this method (at least at this point),we need the problem to be in standard form.
1 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].
We will later be able to relax conditions 1 and 3, but never the secondone.
The Process
The process we wil use to solve this problem (and the others in thischapter) is called the Simplex Method.
In order to solve a problem using this method (at least at this point),we need the problem to be in 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].
We will later be able to relax conditions 1 and 3, but never the secondone.
The Process
The process we wil use to solve this problem (and the others in thischapter) is called the Simplex Method.
In order to solve a problem using this method (at least at this point),we need the problem to be in 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].
We will later be able to relax conditions 1 and 3, but never the secondone.
The Process
The process we wil use to solve this problem (and the others in thischapter) is called the Simplex Method.
In order to solve a problem using this method (at least at this point),we need the problem to be in 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].
We will later be able to relax conditions 1 and 3, but never the secondone.
Slack Variables
Once we check to see if we have the problem in standard form, weintroduce slack variables. In order to use this process, we must writeall of the inequalities as equalities.
3x + 2y ≤ 36⇒ 3x + 2y + u = 36
u is a slack variable and u ≥ 0.
Similarly, for v ≥ 0,
x + 4y ≤ 32⇒ x + 4y + v = 32
Slack Variables
Once we check to see if we have the problem in standard form, weintroduce slack variables. In order to use this process, we must writeall of the inequalities as equalities.
3x + 2y ≤ 36⇒ 3x + 2y + u = 36
u is a slack variable and u ≥ 0.
Similarly, for v ≥ 0,
x + 4y ≤ 32⇒ x + 4y + v = 32
Slack Variables
Once we check to see if we have the problem in standard form, weintroduce slack variables. In order to use this process, we must writeall of the inequalities as equalities.
3x + 2y ≤ 36⇒ 3x + 2y + u = 36
u is a slack variable and u ≥ 0.
Similarly, for v ≥ 0,
x + 4y ≤ 32⇒ x + 4y + v = 32
Slack Variables
Once we check to see if we have the problem in standard form, weintroduce slack variables. In order to use this process, we must writeall of the inequalities as equalities.
3x + 2y ≤ 36⇒ 3x + 2y + u = 36
u is a slack variable and u ≥ 0.
Similarly, for v ≥ 0,
x + 4y ≤ 32⇒ x + 4y + v = 32
The Object Function
We also have to rewrite the object function M = 7x + 4y.
M = 7x + 4y⇒ −7x− 4y + M = 0
Notice that we rewrote this in a way to keep M positive.
So, our new system of equations will be3x+2y+u=36x+4y+v=32-7x-4y+M=0x ≥ 0, y ≥ 0
The Object Function
We also have to rewrite the object function M = 7x + 4y.
M = 7x + 4y⇒ −7x− 4y + M = 0
Notice that we rewrote this in a way to keep M positive.
So, our new system of equations will be3x+2y+u=36x+4y+v=32-7x-4y+M=0x ≥ 0, y ≥ 0
The Object Function
We also have to rewrite the object function M = 7x + 4y.
M = 7x + 4y⇒ −7x− 4y + M = 0
Notice that we rewrote this in a way to keep M positive.
So, our new system of equations will be3x+2y+u=36x+4y+v=32-7x-4y+M=0x ≥ 0, y ≥ 0
The Object Function
We also have to rewrite the object function M = 7x + 4y.
M = 7x + 4y⇒ −7x− 4y + M = 0
Notice that we rewrote this in a way to keep M positive.
So, our new system of equations will be3x+2y+u=36x+4y+v=32-7x-4y+M=0x ≥ 0, y ≥ 0
The Matrix and the Initial Simplex Tableau
At this point, we can write the following matrix. 3 2 1 0 0 361 4 0 1 0 32-7 -4 0 0 1 0
What do the boxed columns represent?
The boxed columns are those of the identity matrix. These columnswill always appear in our Simplex tableau, whether it is the initialtableau or we have performed a pivot.
The Matrix and the Initial Simplex Tableau
At this point, we can write the following matrix. 3 2 1 0 0 361 4 0 1 0 32-7 -4 0 0 1 0
What do the boxed columns represent?
The boxed columns are those of the identity matrix. These columnswill always appear in our Simplex tableau, whether it is the initialtableau or we have performed a pivot.
The Matrix and the Initial Simplex Tableau
At this point, we can write the following matrix. 3 2 1 0 0 361 4 0 1 0 32-7 -4 0 0 1 0
What do the boxed columns represent?
The boxed columns are those of the identity matrix. These columnswill always appear in our Simplex tableau, whether it is the initialtableau or we have performed a pivot.
Variable Types
In fact, we can read our answer directly from this, once we define thetwo types of variables.
Group I variables are those we will set equal to 0.These are the ones that do not correspond to the identitymatrix. We set these equal to 0. Here, the group Ivariables are x and y.
Group II variables are the ones that correspond to theidentity matrix. Here those are u, v and M. We read thevalues just as we did when solving systems ofequations.
Variable Types
In fact, we can read our answer directly from this, once we define thetwo types of variables.
Group I variables are those we will set equal to 0.These are the ones that do not correspond to the identitymatrix. We set these equal to 0. Here, the group Ivariables are x and y.
Group II variables are the ones that correspond to theidentity matrix. Here those are u, v and M. We read thevalues just as we did when solving systems ofequations.
Variable Types
In fact, we can read our answer directly from this, once we define thetwo types of variables.
Group I variables are those we will set equal to 0.These are the ones that do not correspond to the identitymatrix. We set these equal to 0. Here, the group Ivariables are x and y.
Group II variables are the ones that correspond to theidentity matrix. Here those are u, v and M. We read thevalues just as we did when solving systems ofequations.
The Initial Tableau
3 2 1 0 0 361 4 0 1 0 32-7 -4 0 0 1 0
So, our particular solution here would be x = 0 ,y = 0, u = 36,v = 32, M = 0 and we know this can’t be the maximum.
We need to pivot to attain other solutions. But the question is, wheredo we pivot and when do we stop?
The Initial Tableau
3 2 1 0 0 361 4 0 1 0 32-7 -4 0 0 1 0
So, our particular solution here would be
x = 0 ,y = 0, u = 36,v = 32, M = 0 and we know this can’t be the maximum.
We need to pivot to attain other solutions. But the question is, wheredo we pivot and when do we stop?
The Initial Tableau
3 2 1 0 0 361 4 0 1 0 32-7 -4 0 0 1 0
So, our particular solution here would be x = 0 ,y = 0, u = 36,v = 32, M = 0 and we know this can’t be the maximum.
We need to pivot to attain other solutions. But the question is, wheredo we pivot and when do we stop?
The Initial Tableau
3 2 1 0 0 361 4 0 1 0 32-7 -4 0 0 1 0
So, our particular solution here would be x = 0 ,y = 0, u = 36,v = 32, M = 0 and we know this can’t be the maximum.
We need to pivot to attain other solutions. But the question is, wheredo we pivot and when do we stop?
Choosing the Pivot Element
The goal is to pivot until we have no negatives in the bottom row. Thebottom row corresponds to the object function, so the pivot elementcannot come from that row.
To choose the pivot element, we begin with a column that has anegative in the bottom row. It doesn’t matter which one, and there isno defined exact way to choose which one to choose, but in myexperience it seems like less pivots are required when we select thecolumn that has the largest negative in it. 3 2 1 0 0 36
1 4 0 1 0 32-7 -4 0 0 1 0
Choosing the Pivot Element
The goal is to pivot until we have no negatives in the bottom row. Thebottom row corresponds to the object function, so the pivot elementcannot come from that row.
To choose the pivot element, we begin with a column that has anegative in the bottom row. It doesn’t matter which one, and there isno defined exact way to choose which one to choose, but in myexperience it seems like less pivots are required when we select thecolumn that has the largest negative in it. 3 2 1 0 0 36
1 4 0 1 0 32-7 -4 0 0 1 0
Choosing the Pivot Element
Once we select the column, we take ratios of the element in thatcolumn and the corresponding element in the answer column. 3 2 1 0 0 36
1 4 0 1 0 32-7 -4 0 0 1 0
363 = 12321 = 32
Since the smallest positive ratio is 12, we select the pivot element tothe 3 in the 1, 1 position.
Choosing the Pivot Element
Once we select the column, we take ratios of the element in thatcolumn and the corresponding element in the answer column. 3 2 1 0 0 36
1 4 0 1 0 32-7 -4 0 0 1 0
363 = 12321 = 32
Since the smallest positive ratio is 12, we select the pivot element tothe 3 in the 1, 1 position.
Pivoting
3 2 1 0 0 361 4 0 1 0 32-7 -4 0 0 1 0
After we pivot, we have 1 23
13 0 0 12
0 103 −1
3 1 0 200 2
373 0 1 84
Pivoting
3 2 1 0 0 361 4 0 1 0 32-7 -4 0 0 1 0
After we pivot, we have 1 2
313 0 0 12
0 103 −1
3 1 0 200 2
373 0 1 84
Are We Done?
Since there are no negatives left in the bottom row, we have optimizedthe problem.
When we read the solution, we see that x = 12, v = 20 and M = 84.We assign y = 0 and u = 0 because those don’t correspond to thecolumns of the identity matrix.
But, when we express the answer, we only give it in terms of the realvariables - not the slack variables, so we would say that we attain themaximum value of M = 84 at the point (12, 0).
Are We Done?
Since there are no negatives left in the bottom row, we have optimizedthe problem.
When we read the solution, we see that x = 12, v = 20 and M = 84.
We assign y = 0 and u = 0 because those don’t correspond to thecolumns of the identity matrix.
But, when we express the answer, we only give it in terms of the realvariables - not the slack variables, so we would say that we attain themaximum value of M = 84 at the point (12, 0).
Are We Done?
Since there are no negatives left in the bottom row, we have optimizedthe problem.
When we read the solution, we see that x = 12, v = 20 and M = 84.We assign y = 0 and u = 0 because those don’t correspond to thecolumns of the identity matrix.
But, when we express the answer, we only give it in terms of the realvariables - not the slack variables, so we would say that we attain themaximum value of M = 84 at the point (12, 0).
Are We Done?
Since there are no negatives left in the bottom row, we have optimizedthe problem.
When we read the solution, we see that x = 12, v = 20 and M = 84.We assign y = 0 and u = 0 because those don’t correspond to thecolumns of the identity matrix.
But, when we express the answer, we only give it in terms of the realvariables - not the slack variables, so we would say that we attain themaximum value of M = 84 at the point (12, 0).
Another Example
ExampleMaximize 2x + 6y subject to the following constraints:
3x + 6y ≤ 45x + y ≤ 10x ≥ 0, y ≥ 0
Step 1: What are the 3 properties for being in standard form?
1 Maximization problem2 Variables constrained to be nonnegative3 [linear constraint] ≤ [nonnegative constant]
Another Example
ExampleMaximize 2x + 6y subject to the following constraints:
3x + 6y ≤ 45x + y ≤ 10x ≥ 0, y ≥ 0
Step 1:
What are the 3 properties for being in standard form?
1 Maximization problem2 Variables constrained to be nonnegative3 [linear constraint] ≤ [nonnegative constant]
Another Example
ExampleMaximize 2x + 6y subject to the following constraints:
3x + 6y ≤ 45x + y ≤ 10x ≥ 0, y ≥ 0
Step 1: What are the 3 properties for being in standard form?
1 Maximization problem2 Variables constrained to be nonnegative3 [linear constraint] ≤ [nonnegative constant]
Another Example
ExampleMaximize 2x + 6y subject to the following constraints:
3x + 6y ≤ 45x + y ≤ 10x ≥ 0, y ≥ 0
Step 1: What are the 3 properties for being in standard form?
1 Maximization problem
2 Variables constrained to be nonnegative3 [linear constraint] ≤ [nonnegative constant]
Another Example
ExampleMaximize 2x + 6y subject to the following constraints:
3x + 6y ≤ 45x + y ≤ 10x ≥ 0, y ≥ 0
Step 1: What are the 3 properties for being in standard form?
1 Maximization problem2 Variables constrained to be nonnegative
3 [linear constraint] ≤ [nonnegative constant]
Another Example
ExampleMaximize 2x + 6y subject to the following constraints:
3x + 6y ≤ 45x + y ≤ 10x ≥ 0, y ≥ 0
Step 1: What are the 3 properties for being in standard form?
1 Maximization problem2 Variables constrained to be nonnegative3 [linear constraint] ≤ [nonnegative constant]
Another Example
Step 2:
Introduce slack variables
3x + 6y + u = 45x + y + v = 10x ≥ 0, y ≥ 0
Step 3: Rewrite object function.
M = 2x + 6y⇒ −2x− 6y + M = 0
Another Example
Step 2: Introduce slack variables
3x + 6y + u = 45x + y + v = 10x ≥ 0, y ≥ 0
Step 3: Rewrite object function.
M = 2x + 6y⇒ −2x− 6y + M = 0
Another Example
Step 2: Introduce slack variables
3x + 6y + u = 45x + y + v = 10x ≥ 0, y ≥ 0
Step 3: Rewrite object function.
M = 2x + 6y⇒ −2x− 6y + M = 0
Another Example
Step 2: Introduce slack variables
3x + 6y + u = 45x + y + v = 10x ≥ 0, y ≥ 0
Step 3:
Rewrite object function.
M = 2x + 6y⇒ −2x− 6y + M = 0
Another Example
Step 2: Introduce slack variables
3x + 6y + u = 45x + y + v = 10x ≥ 0, y ≥ 0
Step 3: Rewrite object function.
M = 2x + 6y⇒ −2x− 6y + M = 0
Another Example
Step 2: Introduce slack variables
3x + 6y + u = 45x + y + v = 10x ≥ 0, y ≥ 0
Step 3: Rewrite object function.
M = 2x + 6y⇒ −2x− 6y + M = 0
Another Example
Step 4:
Set up the initial Simplex tableau 3 6 1 0 0 451 1 0 1 0 10-2 -6 0 0 1 0
Which is our pivot column? 3 6 1 0 0 45
1 1 0 1 0 10-2 -6 0 0 1 0
Another Example
Step 4: Set up the initial Simplex tableau
3 6 1 0 0 451 1 0 1 0 10-2 -6 0 0 1 0
Which is our pivot column? 3 6 1 0 0 45
1 1 0 1 0 10-2 -6 0 0 1 0
Another Example
Step 4: Set up the initial Simplex tableau 3 6 1 0 0 451 1 0 1 0 10-2 -6 0 0 1 0
Which is our pivot column? 3 6 1 0 0 451 1 0 1 0 10-2 -6 0 0 1 0
Another Example
Step 4: Set up the initial Simplex tableau 3 6 1 0 0 451 1 0 1 0 10-2 -6 0 0 1 0
Which is our pivot column?
3 6 1 0 0 451 1 0 1 0 10-2 -6 0 0 1 0
Another Example
Step 4: Set up the initial Simplex tableau 3 6 1 0 0 451 1 0 1 0 10-2 -6 0 0 1 0
Which is our pivot column? 3 6 1 0 0 45
1 1 0 1 0 10-2 -6 0 0 1 0
Another Example
Next we ...
determine the pivot element by ... taking ratios. 3 6 1 0 0 451 1 0 1 0 10-2 -6 0 0 1 0
456 = 15
2101 = 10
So, the pivot element is 3 6 1 0 0 451 1 0 1 0 10-2 -6 0 0 1 0
Another Example
Next we ... determine the pivot element by ...
taking ratios. 3 6 1 0 0 451 1 0 1 0 10-2 -6 0 0 1 0
456 = 15
2101 = 10
So, the pivot element is 3 6 1 0 0 451 1 0 1 0 10-2 -6 0 0 1 0
Another Example
Next we ... determine the pivot element by ... taking ratios. 3 6 1 0 0 451 1 0 1 0 10-2 -6 0 0 1 0
456 = 15
2101 = 10
So, the pivot element is 3 6 1 0 0 451 1 0 1 0 10-2 -6 0 0 1 0
Another Example
Next we ... determine the pivot element by ... taking ratios. 3 6 1 0 0 451 1 0 1 0 10-2 -6 0 0 1 0
456 = 15
2101 = 10
So, the pivot element is 3 6 1 0 0 451 1 0 1 0 10-2 -6 0 0 1 0
Another Example
Next we ... determine the pivot element by ... taking ratios. 3 6 1 0 0 451 1 0 1 0 10-2 -6 0 0 1 0
456 = 15
2101 = 10
So, the pivot element is
3 6 1 0 0 451 1 0 1 0 10-2 -6 0 0 1 0
Another Example
Next we ... determine the pivot element by ... taking ratios. 3 6 1 0 0 451 1 0 1 0 10-2 -6 0 0 1 0
456 = 15
2101 = 10
So, the pivot element is 3 6 1 0 0 451 1 0 1 0 10-2 -6 0 0 1 0
Another Example
After we pivot, we should have
12 1 1
6 0 0 152
− 12 0 −1
6 1 0 52
1 0 1 0 1 45
Are we done? 1
2 1 16 0 0 15
2− 1
2 0 −16 1 0 5
21 0 1 0 1 45
The maximum value of M = 45 occurs at the point (0, 15
2 ).
Another Example
After we pivot, we should have 12 1 1
6 0 0 152
− 12 0 −1
6 1 0 52
1 0 1 0 1 45
Are we done? 12 1 1
6 0 0 152
− 12 0 −1
6 1 0 52
1 0 1 0 1 45
The maximum value of M = 45 occurs at the point (0, 15
2 ).
Another Example
After we pivot, we should have 12 1 1
6 0 0 152
− 12 0 −1
6 1 0 52
1 0 1 0 1 45
Are we done?
12 1 1
6 0 0 152
− 12 0 −1
6 1 0 52
1 0 1 0 1 45
The maximum value of M = 45 occurs at the point (0, 15
2 ).
Another Example
After we pivot, we should have 12 1 1
6 0 0 152
− 12 0 −1
6 1 0 52
1 0 1 0 1 45
Are we done? 1
2 1 16 0 0 15
2− 1
2 0 −16 1 0 5
21 0 1 0 1 45
The maximum value of M = 45 occurs at the point (0, 152 ).
Another Example
After we pivot, we should have 12 1 1
6 0 0 152
− 12 0 −1
6 1 0 52
1 0 1 0 1 45
Are we done? 1
2 1 16 0 0 15
2− 1
2 0 −16 1 0 5
21 0 1 0 1 45
The maximum value of M = 45 occurs at the point (0, 15
2 ).
The Next One
ExampleMaximize 5x + 6y subject to the following constraints:
x− y ≤ 223 x + y ≤ 8x ≥ 0, y ≥ 0
After introducing slack variables and rewriting, we havex− y + u = 223 x + y + v = 8−5x− 6y + M = 0x ≥ 0, y ≥ 0
The Next One
ExampleMaximize 5x + 6y subject to the following constraints:
x− y ≤ 223 x + y ≤ 8x ≥ 0, y ≥ 0
After introducing slack variables and rewriting, we have
x− y + u = 223 x + y + v = 8−5x− 6y + M = 0x ≥ 0, y ≥ 0
The Next One
ExampleMaximize 5x + 6y subject to the following constraints:
x− y ≤ 223 x + y ≤ 8x ≥ 0, y ≥ 0
After introducing slack variables and rewriting, we havex− y + u = 223 x + y + v = 8−5x− 6y + M = 0x ≥ 0, y ≥ 0
The Next One
Our initial tableau is
1 -1 1 0 0 223 1 0 1 0 8-5 -6 0 0 1 0
Notice that the largest negative in the bottom row is in the secondcolumn, but then we have no choice for pivot element as only oneratio will be positive. 1 -1 1 0 0 2
23 1 0 1 0 8-5 -6 0 0 1 0
The Next One
Our initial tableau is 1 -1 1 0 0 223 1 0 1 0 8-5 -6 0 0 1 0
Notice that the largest negative in the bottom row is in the secondcolumn, but then we have no choice for pivot element as only oneratio will be positive. 1 -1 1 0 0 2
23 1 0 1 0 8-5 -6 0 0 1 0
The Next One
Our initial tableau is 1 -1 1 0 0 223 1 0 1 0 8-5 -6 0 0 1 0
Notice that the largest negative in the bottom row is in the secondcolumn, but then we have no choice for pivot element as only oneratio will be positive.
1 -1 1 0 0 223 1 0 1 0 8-5 -6 0 0 1 0
The Next One
Our initial tableau is 1 -1 1 0 0 223 1 0 1 0 8-5 -6 0 0 1 0
Notice that the largest negative in the bottom row is in the secondcolumn, but then we have no choice for pivot element as only oneratio will be positive. 1 -1 1 0 0 2
23 1 0 1 0 8-5 -6 0 0 1 0
The Next One
After the first pivot, we have
53 0 1 1 0 1023 1 0 1 0 8-1 0 0 6 1 48
Are we done? 5
3 0 1 1 0 1023 1 0 1 0 8-1 0 0 6 1 48
1053= 6
823= 12
The pivot element is in the 1, 1 position.
The Next One
After the first pivot, we have 53 0 1 1 0 1023 1 0 1 0 8-1 0 0 6 1 48
Are we done? 53 0 1 1 0 1023 1 0 1 0 8-1 0 0 6 1 48
1053= 6
823= 12
The pivot element is in the 1, 1 position.
The Next One
After the first pivot, we have 53 0 1 1 0 1023 1 0 1 0 8-1 0 0 6 1 48
Are we done?
53 0 1 1 0 1023 1 0 1 0 8-1 0 0 6 1 48
1053= 6
823= 12
The pivot element is in the 1, 1 position.
The Next One
After the first pivot, we have 53 0 1 1 0 1023 1 0 1 0 8-1 0 0 6 1 48
Are we done? 5
3 0 1 1 0 1023 1 0 1 0 8-1 0 0 6 1 48
1053= 6
823= 12
The pivot element is in the 1, 1 position.
The Next One
After the first pivot, we have 53 0 1 1 0 1023 1 0 1 0 8-1 0 0 6 1 48
Are we done? 5
3 0 1 1 0 1023 1 0 1 0 8-1 0 0 6 1 48
1053= 6
823= 12
The pivot element is in the 1, 1 position.
The Next One
After the first pivot, we have 53 0 1 1 0 1023 1 0 1 0 8-1 0 0 6 1 48
Are we done? 5
3 0 1 1 0 1023 1 0 1 0 8-1 0 0 6 1 48
1053= 6
823= 12
The pivot element is in the 1, 1 position.
The Next One
53 0 1 1 0 1023 1 0 1 0 8-1 0 0 6 1 48
And after the next pivot, we have 1 0 35
35 0 6
0 1 − 25
35 0 4
0 0 35
335 1 54
The maximum value of M = 54 is obtained at the point (6, 4).
The Next One
53 0 1 1 0 1023 1 0 1 0 8-1 0 0 6 1 48
And after the next pivot, we have
1 0 35
35 0 6
0 1 − 25
35 0 4
0 0 35
335 1 54
The maximum value of M = 54 is obtained at the point (6, 4).
The Next One
53 0 1 1 0 1023 1 0 1 0 8-1 0 0 6 1 48
And after the next pivot, we have 1 0 3
535 0 6
0 1 − 25
35 0 4
0 0 35
335 1 54
The maximum value of M = 54 is obtained at the point (6, 4).
The Next One
53 0 1 1 0 1023 1 0 1 0 8-1 0 0 6 1 48
And after the next pivot, we have 1 0 3
535 0 6
0 1 − 25
35 0 4
0 0 35
335 1 54
The maximum value of M = 54 is obtained at the point (6, 4).
Another of the Same Size
ExampleMaximize 2x + y subject to the following constraints:
32 x + y ≤ 385x + y ≤ 80x ≥ 0, y ≥ 0
After introducing slack variables and rewriting, we have32 x + y + u = 385x + y + v = 80−2x− y + M = 0x ≥ 0, y ≥ 0
Another of the Same Size
ExampleMaximize 2x + y subject to the following constraints:
32 x + y ≤ 385x + y ≤ 80x ≥ 0, y ≥ 0
After introducing slack variables and rewriting, we have
32 x + y + u = 385x + y + v = 80−2x− y + M = 0x ≥ 0, y ≥ 0
Another of the Same Size
ExampleMaximize 2x + y subject to the following constraints:
32 x + y ≤ 385x + y ≤ 80x ≥ 0, y ≥ 0
After introducing slack variables and rewriting, we have32 x + y + u = 385x + y + v = 80−2x− y + M = 0x ≥ 0, y ≥ 0
Another of the Same Size
Initial tableau is ...
32 1 1 0 0 385 1 0 1 0 80-2 -1 0 0 1 0
Where is the first pivot? 3
2 1 1 0 0 385 1 0 1 0 80-2 -1 0 0 1 0
3832= 76
3805 = 16
Another of the Same Size
Initial tableau is ...
32 1 1 0 0 385 1 0 1 0 80-2 -1 0 0 1 0
Where is the first pivot? 32 1 1 0 0 385 1 0 1 0 80-2 -1 0 0 1 0
3832= 76
3805 = 16
Another of the Same Size
Initial tableau is ...
32 1 1 0 0 385 1 0 1 0 80-2 -1 0 0 1 0
Where is the first pivot?
32 1 1 0 0 385 1 0 1 0 80-2 -1 0 0 1 0
3832= 76
3805 = 16
Another of the Same Size
Initial tableau is ...
32 1 1 0 0 385 1 0 1 0 80-2 -1 0 0 1 0
Where is the first pivot? 3
2 1 1 0 0 385 1 0 1 0 80-2 -1 0 0 1 0
3832= 76
3805 = 16
Another of the Same Size
Initial tableau is ...
32 1 1 0 0 385 1 0 1 0 80-2 -1 0 0 1 0
Where is the first pivot? 3
2 1 1 0 0 385 1 0 1 0 80-2 -1 0 0 1 0
3832= 76
3805 = 16
Another of the Same Size
32 1 1 0 0 385 1 0 1 0 80-2 -1 0 0 1 0
After the first pivot is completed, we have 0 710 1 − 3
10 0 141 1
5 0 15 0 16
0 − 35 0 2
5 1 32
Are we done?
Another of the Same Size
32 1 1 0 0 385 1 0 1 0 80-2 -1 0 0 1 0
After the first pivot is completed, we have 0 7
10 1 − 310 0 14
1 15 0 1
5 0 160 − 3
5 0 25 1 32
Are we done?
Another of the Same Size
32 1 1 0 0 385 1 0 1 0 80-2 -1 0 0 1 0
After the first pivot is completed, we have 0 7
10 1 − 310 0 14
1 15 0 1
5 0 160 − 3
5 0 25 1 32
Are we done?
Another of the Same Size
0 710 1 − 3
10 0 141 1
5 0 15 0 16
0 − 35 0 2
5 1 32
147
10= 20
1615= 80
0 710 1 − 3
10 0 14
1 15 0 1
5 0 160 − 3
5 0 25 1 32
Another of the Same Size
0 710 1 − 3
10 0 141 1
5 0 15 0 16
0 − 35 0 2
5 1 32
147
10= 20
1615= 80
0 710 1 − 3
10 0 14
1 15 0 1
5 0 160 − 3
5 0 25 1 32
Another of the Same Size
0 710 1 − 3
10 0 141 1
5 0 15 0 16
0 − 35 0 2
5 1 32
147
10= 20
1615= 80
0 710 1 − 3
10 0 14
1 15 0 1
5 0 160 − 3
5 0 25 1 32
Another of the Same Size
After the second pivot is completed, we have 0 1 107 −3
7 0 201 0 −2
74
35 0 120 0 6
717 1 44
Are we done yet?
0 1 107 −3
7 0 201 0 −2
74
35 0 120 0 6
717 1 44
The maximum value of M = 44 is obtained at the point (12, 20).
Another of the Same Size
After the second pivot is completed, we have 0 1 107 −3
7 0 201 0 −2
74
35 0 120 0 6
717 1 44
Are we done yet?
0 1 107 −3
7 0 201 0 −2
74
35 0 120 0 6
717 1 44
The maximum value of M = 44 is obtained at the point (12, 20).
Another of the Same Size
After the second pivot is completed, we have 0 1 107 −3
7 0 201 0 −2
74
35 0 120 0 6
717 1 44
Are we done yet?
0 1 107 −3
7 0 201 0 −2
74
35 0 120 0 6
717 1 44
The maximum value of M = 44 is obtained at the point (12, 20).
Another of the Same Size
After the second pivot is completed, we have 0 1 107 −3
7 0 201 0 −2
74
35 0 120 0 6
717 1 44
Are we done yet?
0 1 107 −3
7 0 201 0 −2
74
35 0 120 0 6
717 1 44
The maximum value of M = 44 is obtained at the point (12, 20).
One More
ExampleMaximize 6x + 8y subject to the following constraints:
2x + 9y ≤ 364x + 3y ≤ 42x ≥ 0, y ≥ 0
After introducing slack variables, we have2x + 9y + u = 364x + 3y + v = 42−6x− 8y + M = 0x ≥ 0, y ≥ 0
One More
ExampleMaximize 6x + 8y subject to the following constraints:
2x + 9y ≤ 364x + 3y ≤ 42x ≥ 0, y ≥ 0
After introducing slack variables, we have2x + 9y + u = 364x + 3y + v = 42−6x− 8y + M = 0x ≥ 0, y ≥ 0
One More
The initial simplex tableau is ...
2 9 1 0 0 364 3 0 1 0 42-6 -8 0 0 1 0
First pivot column? Pivot element? 2 9 1 0 0 36
4 3 0 1 0 42-6 -8 0 0 1 0
369 = 4
423 = 14
One More
The initial simplex tableau is ...
2 9 1 0 0 364 3 0 1 0 42-6 -8 0 0 1 0
First pivot column? Pivot element? 2 9 1 0 0 364 3 0 1 0 42-6 -8 0 0 1 0
369 = 4
423 = 14
One More
The initial simplex tableau is ...
2 9 1 0 0 364 3 0 1 0 42-6 -8 0 0 1 0
First pivot column? Pivot element? 2 9 1 0 0 36
4 3 0 1 0 42-6 -8 0 0 1 0
369 = 4
423 = 14
One More
The initial simplex tableau is ...
2 9 1 0 0 364 3 0 1 0 42-6 -8 0 0 1 0
First pivot column? Pivot element? 2 9 1 0 0 36
4 3 0 1 0 42-6 -8 0 0 1 0
369 = 4
423 = 14
One More
2 9 1 0 0 364 3 0 1 0 42-6 -8 0 0 1 0
After the pivot is complete, we have
29 1 1
9 0 0 4103 0 −1
3 1 0 30−38
9 0 89 0 1 32
Are we done? What is the pivot column?
One More
2 9 1 0 0 364 3 0 1 0 42-6 -8 0 0 1 0
After the pivot is complete, we have
29 1 1
9 0 0 4103 0 −1
3 1 0 30−38
9 0 89 0 1 32
Are we done? What is the pivot column?
One More
2 9 1 0 0 364 3 0 1 0 42-6 -8 0 0 1 0
After the pivot is complete, we have
29 1 1
9 0 0 4103 0 −1
3 1 0 30−38
9 0 89 0 1 32
Are we done? What is the pivot column?
One More
2 9 1 0 0 364 3 0 1 0 42-6 -8 0 0 1 0
After the pivot is complete, we have
29 1 1
9 0 0 4103 0 −1
3 1 0 30−38
9 0 89 0 1 32
Are we done?
What is the pivot column?
One More
2 9 1 0 0 364 3 0 1 0 42-6 -8 0 0 1 0
After the pivot is complete, we have
29 1 1
9 0 0 4103 0 −1
3 1 0 30−38
9 0 89 0 1 32
Are we done? What is the pivot column?
One More
29 1 1
9 0 0 4103 0 −1
3 1 0 30−38
9 0 89 0 1 32
429= 18
30103= 9
So the pivot element is
29 1 1
9 0 0 4103 0 − 1
3 1 0 30
−389 0 8
9 0 1 32
One More
29 1 1
9 0 0 4103 0 −1
3 1 0 30−38
9 0 89 0 1 32
429= 18
30103= 9
So the pivot element is
29 1 1
9 0 0 4103 0 − 1
3 1 0 30
−389 0 8
9 0 1 32
One More
29 1 1
9 0 0 4103 0 −1
3 1 0 30−38
9 0 89 0 1 32
429= 18
30103= 9
So the pivot element is
29 1 1
9 0 0 4103 0 − 1
3 1 0 30
−389 0 8
9 0 1 32
One More
29 1 1
9 0 0 4103 0 −1
3 1 0 30−38
9 0 89 0 1 32
429= 18
30103= 9
So the pivot element is
29 1 1
9 0 0 4103 0 − 1
3 1 0 30
−389 0 8
9 0 1 32
One More
After the pivot is complete, we have
0 1 215 − 1
15 0 21 0 − 1
10310 0 9
0 0 2645
1915 1 70
Are we done?
0 1 215 − 1
15 0 21 0 − 1
10310 0 9
0 0 2645
1915 1 70
The maximum value of M = 70 is obtained at the point (9, 2).
One More
After the pivot is complete, we have
0 1 215 − 1
15 0 21 0 − 1
10310 0 9
0 0 2645
1915 1 70
Are we done?
0 1 215 − 1
15 0 21 0 − 1
10310 0 9
0 0 2645
1915 1 70
The maximum value of M = 70 is obtained at the point (9, 2).
One More
After the pivot is complete, we have
0 1 215 − 1
15 0 21 0 − 1
10310 0 9
0 0 2645
1915 1 70
Are we done?
0 1 215 − 1
15 0 21 0 − 1
10310 0 9
0 0 2645
1915 1 70
The maximum value of M = 70 is obtained at the point (9, 2).
One More
After the pivot is complete, we have
0 1 215 − 1
15 0 21 0 − 1
10310 0 9
0 0 2645
1915 1 70
Are we done?
0 1 215 − 1
15 0 21 0 − 1
10310 0 9
0 0 2645
1915 1 70
The maximum value of M = 70 is obtained at the point (9, 2).
One More
After the pivot is complete, we have
0 1 215 − 1
15 0 21 0 − 1
10310 0 9
0 0 2645
1915 1 70
Are we done?
0 1 215 − 1
15 0 21 0 − 1
10310 0 9
0 0 2645
1915 1 70
The maximum value of M = 70 is obtained at the point (9, 2).
An Additional Constraint
ExampleMaximize 10x + 15y subject to the following constraints:
x ≤ 2y ≤ 3x + y ≤ 4x ≥ 0, y ≥ 0
After introducing slack variables, we have ...
x + u = 2y + v = 3x + y + w = 4−10x− 15y + M = 0x ≥ 0, y ≥ 0
Dimensions of the matrix?
An Additional Constraint
ExampleMaximize 10x + 15y subject to the following constraints:
x ≤ 2y ≤ 3x + y ≤ 4x ≥ 0, y ≥ 0
After introducing slack variables, we have ...
x + u = 2y + v = 3x + y + w = 4−10x− 15y + M = 0x ≥ 0, y ≥ 0
Dimensions of the matrix?
An Additional Constraint
ExampleMaximize 10x + 15y subject to the following constraints:
x ≤ 2y ≤ 3x + y ≤ 4x ≥ 0, y ≥ 0
After introducing slack variables, we have ...
x + u = 2y + v = 3x + y + w = 4−10x− 15y + M = 0x ≥ 0, y ≥ 0
Dimensions of the matrix?
An Additional Constraint
ExampleMaximize 10x + 15y subject to the following constraints:
x ≤ 2y ≤ 3x + y ≤ 4x ≥ 0, y ≥ 0
After introducing slack variables, we have ...
x + u = 2y + v = 3x + y + w = 4−10x− 15y + M = 0x ≥ 0, y ≥ 0
Dimensions of the matrix?
An Additional Constraint
1 0 1 0 0 0 20 1 0 1 0 0 31 1 0 0 1 0 4
-10 -15 0 0 0 1 0
Pivot column?
1 0 1 0 0 0 20 1 0 1 0 0 31 1 0 0 1 0 4
-10 -15 0 0 0 1 0
An Additional Constraint
1 0 1 0 0 0 20 1 0 1 0 0 31 1 0 0 1 0 4
-10 -15 0 0 0 1 0
Pivot column?
1 0 1 0 0 0 20 1 0 1 0 0 31 1 0 0 1 0 4
-10 -15 0 0 0 1 0
An Additional Constraint
1 0 1 0 0 0 20 1 0 1 0 0 31 1 0 0 1 0 4
-10 -15 0 0 0 1 0
Pivot column?
1 0 1 0 0 0 20 1 0 1 0 0 31 1 0 0 1 0 4
-10 -15 0 0 0 1 0
An Additional Constraint
1 0 1 0 0 0 20 1 0 1 0 0 31 1 0 0 1 0 4
-10 -15 0 0 0 1 0
20 = und31 = 341 = 4
1 0 1 0 0 0 20 1 0 1 0 0 31 1 0 0 1 0 4
-10 -15 0 0 0 1 0
An Additional Constraint
1 0 1 0 0 0 20 1 0 1 0 0 31 1 0 0 1 0 4
-10 -15 0 0 0 1 0
20 = und31 = 341 = 4
1 0 1 0 0 0 20 1 0 1 0 0 31 1 0 0 1 0 4
-10 -15 0 0 0 1 0
An Additional Constraint
1 0 1 0 0 0 20 1 0 1 0 0 31 1 0 0 1 0 4
-10 -15 0 0 0 1 0
20 = und31 = 341 = 4
1 0 1 0 0 0 20 1 0 1 0 0 31 1 0 0 1 0 4
-10 -15 0 0 0 1 0
An Additional Constraint
After we complete the pivot, we have „,
1 0 1 0 0 0 20 1 0 1 0 0 31 0 0 -1 1 0 1
-10 0 0 15 0 1 45
Are we done? What is the pivot column?
1 0 1 0 0 0 20 1 0 1 0 0 31 0 0 -1 1 0 1
-10 0 0 15 0 1 45
An Additional Constraint
After we complete the pivot, we have „,
1 0 1 0 0 0 20 1 0 1 0 0 31 0 0 -1 1 0 1
-10 0 0 15 0 1 45
Are we done? What is the pivot column?
1 0 1 0 0 0 20 1 0 1 0 0 31 0 0 -1 1 0 1
-10 0 0 15 0 1 45
An Additional Constraint
After we complete the pivot, we have „,
1 0 1 0 0 0 20 1 0 1 0 0 31 0 0 -1 1 0 1
-10 0 0 15 0 1 45
Are we done?
What is the pivot column?
1 0 1 0 0 0 20 1 0 1 0 0 31 0 0 -1 1 0 1
-10 0 0 15 0 1 45
An Additional Constraint
After we complete the pivot, we have „,
1 0 1 0 0 0 20 1 0 1 0 0 31 0 0 -1 1 0 1
-10 0 0 15 0 1 45
Are we done? What is the pivot column?
1 0 1 0 0 0 20 1 0 1 0 0 31 0 0 -1 1 0 1
-10 0 0 15 0 1 45
An Additional Constraint
After we complete the pivot, we have „,
1 0 1 0 0 0 20 1 0 1 0 0 31 0 0 -1 1 0 1
-10 0 0 15 0 1 45
Are we done? What is the pivot column?
1 0 1 0 0 0 20 1 0 1 0 0 31 0 0 -1 1 0 1
-10 0 0 15 0 1 45
An Additional Constraint
Which is the pivot element?
1 0 1 0 0 0 20 1 0 1 0 0 31 0 0 -1 1 0 1
-10 0 0 15 0 1 45
21 = 230 = und11 = 1
1 0 1 0 0 0 20 1 0 1 0 0 31 0 0 -1 1 0 1
-10 0 0 15 0 1 45
An Additional Constraint
Which is the pivot element?
1 0 1 0 0 0 20 1 0 1 0 0 31 0 0 -1 1 0 1
-10 0 0 15 0 1 45
21 = 230 = und11 = 1
1 0 1 0 0 0 20 1 0 1 0 0 31 0 0 -1 1 0 1
-10 0 0 15 0 1 45
An Additional Constraint
Which is the pivot element?
1 0 1 0 0 0 20 1 0 1 0 0 31 0 0 -1 1 0 1
-10 0 0 15 0 1 45
21 = 230 = und11 = 1
1 0 1 0 0 0 20 1 0 1 0 0 31 0 0 -1 1 0 1
-10 0 0 15 0 1 45
An Additional Constraint
After we complete the pivot, we have
0 0 1 1 -1 0 10 1 0 1 0 0 31 0 0 -1 1 0 10 0 0 5 10 1 55
Are we done?
0 0 1 1 -1 0 10 1 0 1 0 0 31 0 0 -1 1 0 10 0 0 5 10 1 55
The maximum value of M = 55 is obtained at the point (1, 3).
An Additional Constraint
After we complete the pivot, we have
0 0 1 1 -1 0 10 1 0 1 0 0 31 0 0 -1 1 0 10 0 0 5 10 1 55
Are we done?
0 0 1 1 -1 0 10 1 0 1 0 0 31 0 0 -1 1 0 10 0 0 5 10 1 55
The maximum value of M = 55 is obtained at the point (1, 3).
An Additional Constraint
After we complete the pivot, we have
0 0 1 1 -1 0 10 1 0 1 0 0 31 0 0 -1 1 0 10 0 0 5 10 1 55
Are we done?
0 0 1 1 -1 0 10 1 0 1 0 0 31 0 0 -1 1 0 10 0 0 5 10 1 55
The maximum value of M = 55 is obtained at the point (1, 3).
An Additional Constraint
After we complete the pivot, we have
0 0 1 1 -1 0 10 1 0 1 0 0 31 0 0 -1 1 0 10 0 0 5 10 1 55
Are we done?
0 0 1 1 -1 0 10 1 0 1 0 0 31 0 0 -1 1 0 10 0 0 5 10 1 55
The maximum value of M = 55 is obtained at the point (1, 3).
An Additional Constraint
After we complete the pivot, we have
0 0 1 1 -1 0 10 1 0 1 0 0 31 0 0 -1 1 0 10 0 0 5 10 1 55
Are we done?
0 0 1 1 -1 0 10 1 0 1 0 0 31 0 0 -1 1 0 10 0 0 5 10 1 55
The maximum value of M = 55 is obtained at the point (1, 3).
A Little More Complicated
ExampleMaximize 3x + 2y subject to the following constraints:
2x + y ≤ 183x + 3y ≤ 423x + y ≤ 24x ≥ 0, y ≥ 0
What do we do first?
2x + y + u = 183x + 3y + v = 423x + y + w = 24−3x− 2y + M = 0x ≥ 0, y ≥ 0
A Little More Complicated
ExampleMaximize 3x + 2y subject to the following constraints:
2x + y ≤ 183x + 3y ≤ 423x + y ≤ 24x ≥ 0, y ≥ 0
What do we do first?
2x + y + u = 183x + 3y + v = 423x + y + w = 24−3x− 2y + M = 0x ≥ 0, y ≥ 0
A Little More Complicated
ExampleMaximize 3x + 2y subject to the following constraints:
2x + y ≤ 183x + 3y ≤ 423x + y ≤ 24x ≥ 0, y ≥ 0
What do we do first?
2x + y + u = 183x + 3y + v = 423x + y + w = 24−3x− 2y + M = 0x ≥ 0, y ≥ 0
A Little More Complicated
The initial simplex tableau is
2 1 1 0 0 0 183 3 0 1 0 0 423 1 0 0 1 0 24-3 -2 0 0 0 1 0
Which is the pivot column? The pivot element?
2 1 1 0 0 0 183 3 0 1 0 0 423 1 0 0 1 0 24-3 -2 0 0 0 1 0
182 = 9423 = 14243 = 8
So the pivot element is ...
A Little More Complicated
The initial simplex tableau is
2 1 1 0 0 0 183 3 0 1 0 0 423 1 0 0 1 0 24-3 -2 0 0 0 1 0
Which is the pivot column? The pivot element?
2 1 1 0 0 0 183 3 0 1 0 0 423 1 0 0 1 0 24-3 -2 0 0 0 1 0
182 = 9423 = 14243 = 8
So the pivot element is ...
A Little More Complicated
The initial simplex tableau is
2 1 1 0 0 0 183 3 0 1 0 0 423 1 0 0 1 0 24-3 -2 0 0 0 1 0
Which is the pivot column? The pivot element?
2 1 1 0 0 0 183 3 0 1 0 0 423 1 0 0 1 0 24-3 -2 0 0 0 1 0
182 = 9423 = 14243 = 8
So the pivot element is ...
A Little More Complicated
The initial simplex tableau is
2 1 1 0 0 0 183 3 0 1 0 0 423 1 0 0 1 0 24-3 -2 0 0 0 1 0
Which is the pivot column? The pivot element?
2 1 1 0 0 0 183 3 0 1 0 0 423 1 0 0 1 0 24-3 -2 0 0 0 1 0
182 = 9423 = 14243 = 8
So the pivot element is ...
A Little More Complicated
The initial simplex tableau is
2 1 1 0 0 0 183 3 0 1 0 0 423 1 0 0 1 0 24-3 -2 0 0 0 1 0
Which is the pivot column? The pivot element?
2 1 1 0 0 0 183 3 0 1 0 0 423 1 0 0 1 0 24-3 -2 0 0 0 1 0
182 = 9423 = 14243 = 8
So the pivot element is ...
A Little More Complicated
The initial simplex tableau is
2 1 1 0 0 0 183 3 0 1 0 0 423 1 0 0 1 0 24-3 -2 0 0 0 1 0
Which is the pivot column? The pivot element?
2 1 1 0 0 0 183 3 0 1 0 0 423 1 0 0 1 0 24-3 -2 0 0 0 1 0
182 = 9423 = 14243 = 8
So the pivot element is ...
A Little More Complicated
2 1 1 0 0 0 183 3 0 1 0 0 423 1 0 0 1 0 24-3 -2 0 0 0 1 0
And after completing this pivot, we have
0 1
3 1 0 −23 0 2
0 2 0 1 -1 0 181 1
3 0 0 13 0 8
0 -1 0 0 1 1 24
Are we done yet? Which is the new pivot column?
A Little More Complicated
2 1 1 0 0 0 183 3 0 1 0 0 423 1 0 0 1 0 24-3 -2 0 0 0 1 0
And after completing this pivot, we have
0 1
3 1 0 −23 0 2
0 2 0 1 -1 0 181 1
3 0 0 13 0 8
0 -1 0 0 1 1 24
Are we done yet? Which is the new pivot column?
A Little More Complicated
2 1 1 0 0 0 183 3 0 1 0 0 423 1 0 0 1 0 24-3 -2 0 0 0 1 0
And after completing this pivot, we have
0 1
3 1 0 −23 0 2
0 2 0 1 -1 0 181 1
3 0 0 13 0 8
0 -1 0 0 1 1 24
Are we done yet? Which is the new pivot column?
A Little More Complicated
2 1 1 0 0 0 183 3 0 1 0 0 423 1 0 0 1 0 24-3 -2 0 0 0 1 0
And after completing this pivot, we have
0 1
3 1 0 −23 0 2
0 2 0 1 -1 0 181 1
3 0 0 13 0 8
0 -1 0 0 1 1 24
Are we done yet?
Which is the new pivot column?
A Little More Complicated
2 1 1 0 0 0 183 3 0 1 0 0 423 1 0 0 1 0 24-3 -2 0 0 0 1 0
And after completing this pivot, we have
0 1
3 1 0 −23 0 2
0 2 0 1 -1 0 181 1
3 0 0 13 0 8
0 -1 0 0 1 1 24
Are we done yet? Which is the new pivot column?
A Little More Complicated
0 1
3 1 0 −23 0 2
0 2 0 1 -1 0 181 1
3 0 0 13 0 8
0 -1 0 0 1 1 24
213= 6
182 = 9813= 24
So the next pivot element is
0 1
3 1 0 −23 0 2
0 2 0 1 -1 0 181 1
3 0 0 13 0 8
0 -1 0 0 1 1 24
A Little More Complicated
0 1
3 1 0 −23 0 2
0 2 0 1 -1 0 181 1
3 0 0 13 0 8
0 -1 0 0 1 1 24
213= 6
182 = 9813= 24
So the next pivot element is
0 1
3 1 0 −23 0 2
0 2 0 1 -1 0 181 1
3 0 0 13 0 8
0 -1 0 0 1 1 24
A Little More Complicated
0 1
3 1 0 −23 0 2
0 2 0 1 -1 0 181 1
3 0 0 13 0 8
0 -1 0 0 1 1 24
213= 6
182 = 9813= 24
So the next pivot element is
0 1
3 1 0 −23 0 2
0 2 0 1 -1 0 181 1
3 0 0 13 0 8
0 -1 0 0 1 1 24
A Little More Complicated
0 1
3 1 0 −23 0 2
0 2 0 1 -1 0 181 1
3 0 0 13 0 8
0 -1 0 0 1 1 24
213= 6
182 = 9813= 24
So the next pivot element is
0 1
3 1 0 −23 0 2
0 2 0 1 -1 0 181 1
3 0 0 13 0 8
0 -1 0 0 1 1 24
A Little More Complicated
After completing this pivot, we have0 1 3 0 -2 0 60 0 -6 1 3 0 61 0 -1 0 1 0 60 0 3 0 -1 1 30
Done yet? What is the pivot column?
0 1 3 0 -2 0 60 0 -6 1 3 0 61 0 -1 0 1 0 60 0 3 0 -1 1 30
A Little More Complicated
After completing this pivot, we have0 1 3 0 -2 0 60 0 -6 1 3 0 61 0 -1 0 1 0 60 0 3 0 -1 1 30
Done yet?
What is the pivot column?
0 1 3 0 -2 0 60 0 -6 1 3 0 61 0 -1 0 1 0 60 0 3 0 -1 1 30
A Little More Complicated
After completing this pivot, we have0 1 3 0 -2 0 60 0 -6 1 3 0 61 0 -1 0 1 0 60 0 3 0 -1 1 30
Done yet? What is the pivot column?
0 1 3 0 -2 0 60 0 -6 1 3 0 61 0 -1 0 1 0 60 0 3 0 -1 1 30
A Little More Complicated
After completing this pivot, we have0 1 3 0 -2 0 60 0 -6 1 3 0 61 0 -1 0 1 0 60 0 3 0 -1 1 30
Done yet? What is the pivot column?
0 1 3 0 -2 0 60 0 -6 1 3 0 61 0 -1 0 1 0 60 0 3 0 -1 1 30
A Little More Complicated
0 1 3 0 -2 0 60 0 -6 1 3 0 61 0 -1 0 1 0 60 0 3 0 -1 1 30
6−2 = −363 = 261 = 6
So, the pivot element is ...
0 1 3 0 -2 0 60 0 -6 1 3 0 61 0 -1 0 1 0 60 0 3 0 -1 1 30
A Little More Complicated
0 1 3 0 -2 0 60 0 -6 1 3 0 61 0 -1 0 1 0 60 0 3 0 -1 1 30
6−2 = −363 = 261 = 6
So, the pivot element is ...
0 1 3 0 -2 0 60 0 -6 1 3 0 61 0 -1 0 1 0 60 0 3 0 -1 1 30
A Little More Complicated
0 1 3 0 -2 0 60 0 -6 1 3 0 61 0 -1 0 1 0 60 0 3 0 -1 1 30
6−2 = −363 = 261 = 6
So, the pivot element is ...
0 1 3 0 -2 0 60 0 -6 1 3 0 61 0 -1 0 1 0 60 0 3 0 -1 1 30
A Little More Complicated
0 1 3 0 -2 0 60 0 -6 1 3 0 61 0 -1 0 1 0 60 0 3 0 -1 1 30
6−2 = −363 = 261 = 6
So, the pivot element is ...
0 1 3 0 -2 0 60 0 -6 1 3 0 61 0 -1 0 1 0 60 0 3 0 -1 1 30
A Little More Complicated
After completing the pivot, we have
0 1 -1 2
3 0 0 100 0 -2 1
3 1 0 21 0 1 −1
3 0 0 40 0 1 1
3 0 1 32
Are we done?
0 1 -1 2
3 0 0 100 0 -2 1
3 1 0 21 0 1 −1
3 0 0 40 0 1 1
3 0 1 32
The maximum value of M = 32 is obtained at the point (4, 10).
A Little More Complicated
After completing the pivot, we have
0 1 -1 2
3 0 0 100 0 -2 1
3 1 0 21 0 1 −1
3 0 0 40 0 1 1
3 0 1 32
Are we done?
0 1 -1 2
3 0 0 100 0 -2 1
3 1 0 21 0 1 −1
3 0 0 40 0 1 1
3 0 1 32
The maximum value of M = 32 is obtained at the point (4, 10).
A Little More Complicated
After completing the pivot, we have
0 1 -1 2
3 0 0 100 0 -2 1
3 1 0 21 0 1 −1
3 0 0 40 0 1 1
3 0 1 32
Are we done?
0 1 -1 2
3 0 0 100 0 -2 1
3 1 0 21 0 1 −1
3 0 0 40 0 1 1
3 0 1 32
The maximum value of M = 32 is obtained at the point (4, 10).
A Little More Complicated
After completing the pivot, we have
0 1 -1 2
3 0 0 100 0 -2 1
3 1 0 21 0 1 −1
3 0 0 40 0 1 1
3 0 1 32
Are we done?
0 1 -1 2
3 0 0 100 0 -2 1
3 1 0 21 0 1 −1
3 0 0 40 0 1 1
3 0 1 32
The maximum value of M = 32 is obtained at the point (4, 10).
A Little More Complicated
After completing the pivot, we have
0 1 -1 2
3 0 0 100 0 -2 1
3 1 0 21 0 1 −1
3 0 0 40 0 1 1
3 0 1 32
Are we done?
0 1 -1 2
3 0 0 100 0 -2 1
3 1 0 21 0 1 −1
3 0 0 40 0 1 1
3 0 1 32
The maximum value of M = 32 is obtained at the point (4, 10).
An Extra Variable
ExampleMaximize 4x + 6y + 5z subject to the following constraints:
x + 2y + 3z ≤ 18x + 3y + z ≤ 12x ≥ 0, y ≥ 0, z ≥ 0
After introducing slack variables, we havex + 2y + 3z + u = 18x + 3y + z + v = 12−4x− 6y− 5z + M = 0x ≥ 0, y ≥ 0, z ≥ 0
An Extra Variable
ExampleMaximize 4x + 6y + 5z subject to the following constraints:
x + 2y + 3z ≤ 18x + 3y + z ≤ 12x ≥ 0, y ≥ 0, z ≥ 0
After introducing slack variables, we have
x + 2y + 3z + u = 18x + 3y + z + v = 12−4x− 6y− 5z + M = 0x ≥ 0, y ≥ 0, z ≥ 0
An Extra Variable
ExampleMaximize 4x + 6y + 5z subject to the following constraints:
x + 2y + 3z ≤ 18x + 3y + z ≤ 12x ≥ 0, y ≥ 0, z ≥ 0
After introducing slack variables, we havex + 2y + 3z + u = 18x + 3y + z + v = 12−4x− 6y− 5z + M = 0x ≥ 0, y ≥ 0, z ≥ 0
An Extra Variable
The initial tableau is ...
1 2 3 1 0 0 181 3 1 0 1 0 12-4 -6 -5 0 0 1 0
What is the first pivot column?
1 2 3 1 0 0 181 3 1 0 1 0 12-4 -6 -5 0 0 1 0
An Extra Variable
The initial tableau is ...
1 2 3 1 0 0 181 3 1 0 1 0 12-4 -6 -5 0 0 1 0
What is the first pivot column?
1 2 3 1 0 0 181 3 1 0 1 0 12-4 -6 -5 0 0 1 0
An Extra Variable
The initial tableau is ...
1 2 3 1 0 0 181 3 1 0 1 0 12-4 -6 -5 0 0 1 0
What is the first pivot column?
1 2 3 1 0 0 181 3 1 0 1 0 12-4 -6 -5 0 0 1 0
An Extra Variable
The initial tableau is ...
1 2 3 1 0 0 181 3 1 0 1 0 12-4 -6 -5 0 0 1 0
What is the first pivot column?
1 2 3 1 0 0 181 3 1 0 1 0 12-4 -6 -5 0 0 1 0
An Extra Variable
What is the pivot element? 1 2 3 1 0 0 181 3 1 0 1 0 12-4 -6 -5 0 0 1 0
182 = 9
123 = 4
1 2 3 1 0 0 181 3 1 0 1 0 12-4 -6 -5 0 0 1 0
An Extra Variable
What is the pivot element? 1 2 3 1 0 0 181 3 1 0 1 0 12-4 -6 -5 0 0 1 0
182 = 9
123 = 4
1 2 3 1 0 0 181 3 1 0 1 0 12-4 -6 -5 0 0 1 0
An Extra Variable
What is the pivot element? 1 2 3 1 0 0 181 3 1 0 1 0 12-4 -6 -5 0 0 1 0
182 = 9
123 = 4
1 2 3 1 0 0 181 3 1 0 1 0 12-4 -6 -5 0 0 1 0
An Extra Variable
After we complete the first pivot, we have ...
13 0 7
3 1 −23 0 10
13 1 1
3 0 13 0 4
-2 0 -3 0 2 1 24
Are we done? What is the next pivot column?
13 0 7
3 1 −23 0 10
13 1 1
3 0 13 0 4
-2 0 -3 0 2 1 24
An Extra Variable
After we complete the first pivot, we have ...
13 0 7
3 1 −23 0 10
13 1 1
3 0 13 0 4
-2 0 -3 0 2 1 24
Are we done? What is the next pivot column?
13 0 7
3 1 −23 0 10
13 1 1
3 0 13 0 4
-2 0 -3 0 2 1 24
An Extra Variable
After we complete the first pivot, we have ...
13 0 7
3 1 −23 0 10
13 1 1
3 0 13 0 4
-2 0 -3 0 2 1 24
Are we done?
What is the next pivot column?
13 0 7
3 1 −23 0 10
13 1 1
3 0 13 0 4
-2 0 -3 0 2 1 24
An Extra Variable
After we complete the first pivot, we have ...
13 0 7
3 1 −23 0 10
13 1 1
3 0 13 0 4
-2 0 -3 0 2 1 24
Are we done? What is the next pivot column?
13 0 7
3 1 −23 0 10
13 1 1
3 0 13 0 4
-2 0 -3 0 2 1 24
An Extra Variable
After we complete the first pivot, we have ...
13 0 7
3 1 −23 0 10
13 1 1
3 0 13 0 4
-2 0 -3 0 2 1 24
Are we done? What is the next pivot column?
13 0 7
3 1 −23 0 10
13 1 1
3 0 13 0 4
-2 0 -3 0 2 1 24
An Extra Variable
Now we find the pivot element. 13 0 7
3 1 −23 0 10
13 1 1
3 0 13 0 4
-2 0 -3 0 2 1 24
1073= 30
7413= 12
So, the pivot element is ...
13 0 7
3 1 − 23 0 10
13 1 1
3 0 13 0 4
-2 0 -3 0 2 1 24
An Extra Variable
Now we find the pivot element. 13 0 7
3 1 −23 0 10
13 1 1
3 0 13 0 4
-2 0 -3 0 2 1 24
1073= 30
7413= 12
So, the pivot element is ...
13 0 7
3 1 − 23 0 10
13 1 1
3 0 13 0 4
-2 0 -3 0 2 1 24
An Extra Variable
Now we find the pivot element. 13 0 7
3 1 −23 0 10
13 1 1
3 0 13 0 4
-2 0 -3 0 2 1 24
1073= 30
7413= 12
So, the pivot element is ...
13 0 7
3 1 − 23 0 10
13 1 1
3 0 13 0 4
-2 0 -3 0 2 1 24
An Extra Variable
Now we find the pivot element. 13 0 7
3 1 −23 0 10
13 1 1
3 0 13 0 4
-2 0 -3 0 2 1 24
1073= 30
7413= 12
So, the pivot element is ...
13 0 7
3 1 − 23 0 10
13 1 1
3 0 13 0 4
-2 0 -3 0 2 1 24
An Extra Variable
After we take care of this pivot, we have
17 0 1 3
7 −27 0 30
727 1 0 −1
737 0 18
7−11
7 0 0 97
87 1 258
7
Are we done? Which is the new pivot column?
17 0 1 3
7 −27 0 30
727 1 0 −1
737 0 18
7−11
7 0 0 97
87 1 258
7
An Extra Variable
After we take care of this pivot, we have
17 0 1 3
7 −27 0 30
727 1 0 −1
737 0 18
7−11
7 0 0 97
87 1 258
7
Are we done? Which is the new pivot column?
17 0 1 3
7 −27 0 30
727 1 0 −1
737 0 18
7−11
7 0 0 97
87 1 258
7
An Extra Variable
After we take care of this pivot, we have
17 0 1 3
7 −27 0 30
727 1 0 −1
737 0 18
7−11
7 0 0 97
87 1 258
7
Are we done?
Which is the new pivot column?
17 0 1 3
7 −27 0 30
727 1 0 −1
737 0 18
7−11
7 0 0 97
87 1 258
7
An Extra Variable
After we take care of this pivot, we have
17 0 1 3
7 −27 0 30
727 1 0 −1
737 0 18
7−11
7 0 0 97
87 1 258
7
Are we done? Which is the new pivot column?
17 0 1 3
7 −27 0 30
727 1 0 −1
737 0 18
7−11
7 0 0 97
87 1 258
7
An Extra Variable
After we take care of this pivot, we have
17 0 1 3
7 −27 0 30
727 1 0 −1
737 0 18
7−11
7 0 0 97
87 1 258
7
Are we done? Which is the new pivot column?
17 0 1 3
7 −27 0 30
727 1 0 −1
737 0 18
7−11
7 0 0 97
87 1 258
7
An Extra Variable
Now we find the pivot element. 17 0 1 3
7 −27 0 30
727 1 0 −1
737 0 18
7−11
7 0 0 97
87 1 258
7
30717= 30
18727= 9
So, the pivot element is17 0 1 3
7 − 27 0 30
727 1 0 − 1
737 0 18
7
−117 0 0 9
787 1 258
7
An Extra Variable
Now we find the pivot element. 17 0 1 3
7 −27 0 30
727 1 0 −1
737 0 18
7−11
7 0 0 97
87 1 258
7
30717= 30
18727= 9
So, the pivot element is17 0 1 3
7 − 27 0 30
727 1 0 − 1
737 0 18
7
−117 0 0 9
787 1 258
7
An Extra Variable
Now we find the pivot element. 17 0 1 3
7 −27 0 30
727 1 0 −1
737 0 18
7−11
7 0 0 97
87 1 258
7
30717= 30
18727= 9
So, the pivot element is17 0 1 3
7 − 27 0 30
727 1 0 − 1
737 0 18
7
−117 0 0 9
787 1 258
7
An Extra Variable
And, after we complete this pivot, we have
0 −12 1 1
2 −12 0 3
1 72 0 −1
232 0 9
0 112 0 1
272 1 51
Are we done?
0 −12 1 1
2 −12 0 3
1 72 0 −1
232 0 9
0 112 0 1
272 1 51
The maximum value of M = 51 is obtained at the point (9, 0, 3).
An Extra Variable
And, after we complete this pivot, we have
0 −12 1 1
2 −12 0 3
1 72 0 −1
232 0 9
0 112 0 1
272 1 51
Are we done?
0 −12 1 1
2 −12 0 3
1 72 0 −1
232 0 9
0 112 0 1
272 1 51
The maximum value of M = 51 is obtained at the point (9, 0, 3).
An Extra Variable
And, after we complete this pivot, we have
0 −12 1 1
2 −12 0 3
1 72 0 −1
232 0 9
0 112 0 1
272 1 51
Are we done?
0 −12 1 1
2 −12 0 3
1 72 0 −1
232 0 9
0 112 0 1
272 1 51
The maximum value of M = 51 is obtained at the point (9, 0, 3).
An Extra Variable
And, after we complete this pivot, we have
0 −12 1 1
2 −12 0 3
1 72 0 −1
232 0 9
0 112 0 1
272 1 51
Are we done?
0 −12 1 1
2 −12 0 3
1 72 0 −1
232 0 9
0 112 0 1
272 1 51
The maximum value of M = 51 is obtained at the point (9, 0, 3).
An Extra Variable
And, after we complete this pivot, we have
0 −12 1 1
2 −12 0 3
1 72 0 −1
232 0 9
0 112 0 1
272 1 51
Are we done?
0 −12 1 1
2 −12 0 3
1 72 0 −1
232 0 9
0 112 0 1
272 1 51
The maximum value of M = 51 is obtained at the point (9, 0, 3).
As Big As We Will Do
ExampleMaximize x + 2y− z subject to the following constraints:
2x + y + z ≤ 144x + 2y + 3z ≤ 282x + 5y + 5z ≤ 30x ≥ 0, y ≥ 0, z ≥ 0
After introducing slack variables, we have2x + y + z + u = 144x + 2y + 3z + v = 282x + 5y + 5z + w = 30−x− 2y + z + M = 0x ≥ 0, y ≥ 0, z ≥ 0
As Big As We Will Do
ExampleMaximize x + 2y− z subject to the following constraints:
2x + y + z ≤ 144x + 2y + 3z ≤ 282x + 5y + 5z ≤ 30x ≥ 0, y ≥ 0, z ≥ 0
After introducing slack variables, we have
2x + y + z + u = 144x + 2y + 3z + v = 282x + 5y + 5z + w = 30−x− 2y + z + M = 0x ≥ 0, y ≥ 0, z ≥ 0
As Big As We Will Do
ExampleMaximize x + 2y− z subject to the following constraints:
2x + y + z ≤ 144x + 2y + 3z ≤ 282x + 5y + 5z ≤ 30x ≥ 0, y ≥ 0, z ≥ 0
After introducing slack variables, we have2x + y + z + u = 144x + 2y + 3z + v = 282x + 5y + 5z + w = 30−x− 2y + z + M = 0x ≥ 0, y ≥ 0, z ≥ 0
As Big As We Will Do
The initial tableau is ...
2 1 1 1 0 0 0 144 2 3 0 1 0 0 282 5 5 0 0 1 0 30-1 -2 1 0 0 0 1 0
What is the first pivot column?
2 1 1 1 0 0 0 144 2 3 0 1 0 0 282 5 5 0 0 1 0 30-1 -2 1 0 0 0 1 0
As Big As We Will Do
The initial tableau is ...
2 1 1 1 0 0 0 144 2 3 0 1 0 0 282 5 5 0 0 1 0 30-1 -2 1 0 0 0 1 0
What is the first pivot column?
2 1 1 1 0 0 0 144 2 3 0 1 0 0 282 5 5 0 0 1 0 30-1 -2 1 0 0 0 1 0
As Big As We Will Do
The initial tableau is ...
2 1 1 1 0 0 0 144 2 3 0 1 0 0 282 5 5 0 0 1 0 30-1 -2 1 0 0 0 1 0
What is the first pivot column?
2 1 1 1 0 0 0 144 2 3 0 1 0 0 282 5 5 0 0 1 0 30-1 -2 1 0 0 0 1 0
As Big As We Will Do
The initial tableau is ...
2 1 1 1 0 0 0 144 2 3 0 1 0 0 282 5 5 0 0 1 0 30-1 -2 1 0 0 0 1 0
What is the first pivot column?
2 1 1 1 0 0 0 144 2 3 0 1 0 0 282 5 5 0 0 1 0 30-1 -2 1 0 0 0 1 0
As Big As We Will Do
Now, we determine the pivot element.
2 1 1 1 0 0 0 144 2 3 0 1 0 0 282 5 5 0 0 1 0 30-1 -2 1 0 0 0 1 0
141 = 14
282 = 14
305 = 6
So, the pivot element is ...2 1 1 1 0 0 0 144 2 3 0 1 0 0 282 5 5 0 0 1 0 30-1 -2 1 0 0 0 1 0
As Big As We Will Do
Now, we determine the pivot element.
2 1 1 1 0 0 0 144 2 3 0 1 0 0 282 5 5 0 0 1 0 30-1 -2 1 0 0 0 1 0
141 = 14
282 = 14
305 = 6
So, the pivot element is ...2 1 1 1 0 0 0 144 2 3 0 1 0 0 282 5 5 0 0 1 0 30-1 -2 1 0 0 0 1 0
As Big As We Will Do
Now, we determine the pivot element.
2 1 1 1 0 0 0 144 2 3 0 1 0 0 282 5 5 0 0 1 0 30-1 -2 1 0 0 0 1 0
141 = 14
282 = 14
305 = 6
So, the pivot element is ...2 1 1 1 0 0 0 144 2 3 0 1 0 0 282 5 5 0 0 1 0 30-1 -2 1 0 0 0 1 0
As Big As We Will Do
Now, we determine the pivot element.
2 1 1 1 0 0 0 144 2 3 0 1 0 0 282 5 5 0 0 1 0 30-1 -2 1 0 0 0 1 0
141 = 14
282 = 14
305 = 6
So, the pivot element is ...
2 1 1 1 0 0 0 144 2 3 0 1 0 0 282 5 5 0 0 1 0 30-1 -2 1 0 0 0 1 0
As Big As We Will Do
Now, we determine the pivot element.
2 1 1 1 0 0 0 144 2 3 0 1 0 0 282 5 5 0 0 1 0 30-1 -2 1 0 0 0 1 0
141 = 14
282 = 14
305 = 6
So, the pivot element is ...2 1 1 1 0 0 0 144 2 3 0 1 0 0 282 5 5 0 0 1 0 30-1 -2 1 0 0 0 1 0
As Big As We Will Do
After we complete the first pivot, we have
85 0 0 1 0 − 1
5 0 8165 0 1 0 1 − 2
5 0 1625 1 1 0 0 1
5 0 6− 1
5 0 3 0 0 25 1 12
Are we done? What is the next pivot column?
85 0 0 1 0 − 1
5 0 8165 0 1 0 1 − 2
5 0 1625 1 1 0 0 1
5 0 6− 1
5 0 3 0 0 25 1 12
As Big As We Will Do
After we complete the first pivot, we have
85 0 0 1 0 − 1
5 0 8165 0 1 0 1 − 2
5 0 1625 1 1 0 0 1
5 0 6− 1
5 0 3 0 0 25 1 12
Are we done? What is the next pivot column?
85 0 0 1 0 − 1
5 0 8165 0 1 0 1 − 2
5 0 1625 1 1 0 0 1
5 0 6− 1
5 0 3 0 0 25 1 12
As Big As We Will Do
After we complete the first pivot, we have
85 0 0 1 0 − 1
5 0 8165 0 1 0 1 − 2
5 0 1625 1 1 0 0 1
5 0 6− 1
5 0 3 0 0 25 1 12
Are we done?
What is the next pivot column?
85 0 0 1 0 − 1
5 0 8165 0 1 0 1 − 2
5 0 1625 1 1 0 0 1
5 0 6− 1
5 0 3 0 0 25 1 12
As Big As We Will Do
After we complete the first pivot, we have
85 0 0 1 0 − 1
5 0 8165 0 1 0 1 − 2
5 0 1625 1 1 0 0 1
5 0 6− 1
5 0 3 0 0 25 1 12
Are we done? What is the next pivot column?
85 0 0 1 0 − 1
5 0 8165 0 1 0 1 − 2
5 0 1625 1 1 0 0 1
5 0 6− 1
5 0 3 0 0 25 1 12
As Big As We Will Do
After we complete the first pivot, we have
85 0 0 1 0 − 1
5 0 8165 0 1 0 1 − 2
5 0 1625 1 1 0 0 1
5 0 6− 1
5 0 3 0 0 25 1 12
Are we done? What is the next pivot column?
85 0 0 1 0 − 1
5 0 8165 0 1 0 1 − 2
5 0 1625 1 1 0 0 1
5 0 6− 1
5 0 3 0 0 25 1 12
As Big As We Will Do
Now we find the pivot element.85 0 0 1 0 − 1
5 0 8165 0 1 0 1 − 2
5 0 1625 1 1 0 0 1
5 0 6−1
5 0 3 0 0 25 1 12
885= 5
16165= 5
625= 15
Since we get the same positive ratio for two elements, we can chooseeither.
85 0 0 1 0 −1
5 0 8165 0 1 0 1 −2
5 0 1625 1 1 0 0 1
5 0 6− 1
5 0 3 0 0 25 1 12
As Big As We Will Do
Now we find the pivot element.85 0 0 1 0 − 1
5 0 8165 0 1 0 1 − 2
5 0 1625 1 1 0 0 1
5 0 6−1
5 0 3 0 0 25 1 12
885= 5
16165= 5
625= 15
Since we get the same positive ratio for two elements, we can chooseeither.
85 0 0 1 0 −1
5 0 8165 0 1 0 1 −2
5 0 1625 1 1 0 0 1
5 0 6− 1
5 0 3 0 0 25 1 12
As Big As We Will Do
Now we find the pivot element.85 0 0 1 0 − 1
5 0 8165 0 1 0 1 − 2
5 0 1625 1 1 0 0 1
5 0 6−1
5 0 3 0 0 25 1 12
885= 5
16165= 5
625= 15
Since we get the same positive ratio for two elements, we can chooseeither.
85 0 0 1 0 −1
5 0 8165 0 1 0 1 −2
5 0 1625 1 1 0 0 1
5 0 6− 1
5 0 3 0 0 25 1 12
As Big As We Will Do
Now we find the pivot element.85 0 0 1 0 − 1
5 0 8165 0 1 0 1 − 2
5 0 1625 1 1 0 0 1
5 0 6−1
5 0 3 0 0 25 1 12
885= 5
16165= 5
625= 15
Since we get the same positive ratio for two elements, we can chooseeither.
85 0 0 1 0 −1
5 0 8165 0 1 0 1 −2
5 0 1625 1 1 0 0 1
5 0 6− 1
5 0 3 0 0 25 1 12
As Big As We Will Do
Once we complete this pivot, we will have
1 0 0 5
8 0 − 18 0 5
0 0 1 -2 1 0 0 00 1 1 −1
4 0 14 0 4
0 0 3 18 0 3
8 1 13
Are we done?
1 0 0 5
8 0 − 18 0 5
0 0 1 -2 1 0 0 00 1 1 −1
4 0 14 0 4
0 0 3 18 0 3
8 1 13
The maximum value of M = 13 is obtained at the point (5, 4, 0).
As Big As We Will Do
Once we complete this pivot, we will have
1 0 0 5
8 0 − 18 0 5
0 0 1 -2 1 0 0 00 1 1 −1
4 0 14 0 4
0 0 3 18 0 3
8 1 13
Are we done?
1 0 0 5
8 0 − 18 0 5
0 0 1 -2 1 0 0 00 1 1 −1
4 0 14 0 4
0 0 3 18 0 3
8 1 13
The maximum value of M = 13 is obtained at the point (5, 4, 0).
As Big As We Will Do
Once we complete this pivot, we will have
1 0 0 5
8 0 − 18 0 5
0 0 1 -2 1 0 0 00 1 1 −1
4 0 14 0 4
0 0 3 18 0 3
8 1 13
Are we done?
1 0 0 5
8 0 − 18 0 5
0 0 1 -2 1 0 0 00 1 1 −1
4 0 14 0 4
0 0 3 18 0 3
8 1 13
The maximum value of M = 13 is obtained at the point (5, 4, 0).
As Big As We Will Do
Once we complete this pivot, we will have
1 0 0 5
8 0 − 18 0 5
0 0 1 -2 1 0 0 00 1 1 −1
4 0 14 0 4
0 0 3 18 0 3
8 1 13
Are we done?
1 0 0 5
8 0 − 18 0 5
0 0 1 -2 1 0 0 00 1 1 −1
4 0 14 0 4
0 0 3 18 0 3
8 1 13
The maximum value of M = 13 is obtained at the point (5, 4, 0).
As Big As We Will Do
Once we complete this pivot, we will have
1 0 0 5
8 0 − 18 0 5
0 0 1 -2 1 0 0 00 1 1 −1
4 0 14 0 4
0 0 3 18 0 3
8 1 13
Are we done?
1 0 0 5
8 0 − 18 0 5
0 0 1 -2 1 0 0 00 1 1 −1
4 0 14 0 4
0 0 3 18 0 3
8 1 13
The maximum value of M = 13 is obtained at the point (5, 4, 0).