T-76.4115 Iteration Demo Apollo Crew PP Iteration 21.10.2008.
Math Models of OR: The Revised Simplex Method: An...
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Math Models of OR: The Revised Simplex Method: An Example John E. Mitchell Department of Mathematical Sciences RPI, Troy, NY 12180 USA September 2018 Mitchell The Revised Simplex Method: An Example 1 / 22
Transcript of Math Models of OR: The Revised Simplex Method: An...
Math Models of OR:The Revised Simplex Method: An Example
John E. Mitchell
Department of Mathematical SciencesRPI, Troy, NY 12180 USA
September 2018
Mitchell The Revised Simplex Method: An Example 1 / 22
Our example
Consider the following example with initial simplex tableau:
M0 =
#ratio x1 x2 x3 x4 x5 x6 x7
�4 0 1 �3 2 �5 1 02 2 1 �1 1� 1 2 �1 0� 1 0 4 �2 1 1 3 1
basic sequence S = (1, 7).
Mitchell The Revised Simplex Method: An Example 2 / 22
updated basic sequence : (3,7).
First iteration
Outline
1 First iteration
2 Second iteration
3 Third iteration
4 Checking the solution
Mitchell The Revised Simplex Method: An Example 3 / 22
First iteration
First iteration
In the first iteration, x3 replaces x1 in the basis.
M0 =
#ratio x1 x2 x3 x4 x5 x6 x7
�4 0 1 �3 2 �5 1 02 2 1 �1 1� 1 2 �1 0� 1 0 4 �2 1 1 3 1
S = (1, 7).
The initial pivot matrix is Q1, found by “doing unto the identity as youwould do unto M0”:
2
41 0 00 1 00 0 1
3
5 R0 + 3R1, R2 + 2R1�! Q1 =
2
41 3 00 1 00 2 1
3
5
Mitchell The Revised Simplex Method: An Example 4 / 22
First iteration
Find reduced costs
The updated tableau is M1 = Q1M0. We first calculate the reducedcosts:
Q1M0 =
2
41 3 00 1 00 2 1
3
5�4 0 1 �3 2 �5 1 0
2 1 �1 1 1 2 �1 01 0 4 �2 1 1 3 1
=2 3 �2 0 5 1 �2 0⇤ ⇤ ⇤ ⇤ ⇤ ⇤ ⇤ ⇤⇤ ⇤ ⇤ ⇤ ⇤ ⇤ ⇤ ⇤
=: M2 S = (3, 7).
Mitchell The Revised Simplex Method: An Example 5 / 22
First iteration
Find the pivot column
We will choose x2 as the next pivot column in the second iteration.
In order to finish the first iteration, we need to calculate the x2 columnof the tableau:
Q1M0 =
2
41 3 00 1 00 2 1
3
5�4 0 1 �3 2 �5 1 0
2 1 �1 1 1 2 �1 01 0 4 �2 1 1 3 1
=
#2 3 �2 0 5 1 �2 0⇤ ⇤ �1 ⇤ ⇤ ⇤ ⇤ ⇤⇤ ⇤ 2 ⇤ ⇤ ⇤ ⇤ ⇤
=: M2 S = (3, 7).
Mitchell The Revised Simplex Method: An Example 6 / 22
0
First iteration
Find b
To perform the minimum ratio test in the second iteration, we will needto know the entries in b corresponding to the positive entries in thepivot column:
Q1M0 =
2
41 3 00 1 00 2 1
3
5�4 0 1 �3 2 �5 1 0
2 1 �1 1 1 2 �1 01 0 4 �2 1 1 3 1
=
ratio #2 3 �2 0 5 1 �2 0
� ⇤ ⇤ �1 ⇤ ⇤ ⇤ ⇤ ⇤52 5 ⇤ 2� ⇤ ⇤ ⇤ ⇤ ⇤
=: M2 S = (3, 7).
Mitchell The Revised Simplex Method: An Example 7 / 22
Second iteration
Outline
1 First iteration
2 Second iteration
3 Third iteration
4 Checking the solution
Mitchell The Revised Simplex Method: An Example 8 / 22
Second iteration
Second iteration
In the second iteration, x2 replaces x7 in the basis.
M2 =
ratio #2 3 �2 0 5 1 �2 0
� ⇤ ⇤ �1 ⇤ ⇤ ⇤ ⇤ ⇤52 5 ⇤ 2� ⇤ ⇤ ⇤ ⇤ ⇤
S = (3, 7).
We can now calculate the new pivot matrix Q2, by “doing unto theidentity as you would do unto M1”:2
41 0 00 1 00 0 1
3
512R2 then R0 + 2R2, R1 + R2
�! Q2 =
2
41 0 10 1 1
20 0 1
2
3
5
Mitchell The Revised Simplex Method: An Example 9 / 22
Second iteration
Use products of pivot matrices
The updated tableau is M2 = Q2M1.
However, we don’t know M1 explicitly, so we need to exploit the factthat M1 = Q1M0, so M2 = (Q2Q1)M0, by the associativity property ofmatrix multiplication.
We first calculate the matrix product P2 = Q2Q1:
P2 = Q2Q1 =
2
41 0 10 1 1
20 0 1
2
3
5
2
41 3 00 1 00 2 1
3
5 =
2
41 5 10 2 1
20 1 1
2
3
5
Mitchell The Revised Simplex Method: An Example 10 / 22
M r sQin, = Qe(Q,Mu)= (QQ,) Mo
M - = R Mo
Second iteration
Find the reduced costs
We can now calculate the reduced costs:
Q2M1 = P2M0 =
2
41 5 10 2 1
20 1 1
2
3
5�4 0 1 �3 2 �5 1 0
2 1 �1 1 1 2 �1 01 0 4 �2 1 1 3 1
=7 5 0 0 8 6 �1 1⇤ ⇤ ⇤ ⇤ ⇤ ⇤ ⇤ ⇤⇤ ⇤ ⇤ ⇤ ⇤ ⇤ ⇤ ⇤
=: M2
S = (3, 2).
Mitchell The Revised Simplex Method: An Example 11 / 22
O
X , replace, o n e of these.
Second iteration
Calculate the pivot column
We choose x6 as the pivot column in the third iteration, so we need tocalculate the x6 column of the tableau as part of the second iteration:
P2M0 =
2
41 3 00 1 00 2 1
3
5�4 0 1 �3 2 �5 1 0
2 1 �1 1 1 2 �1 01 0 4 �2 1 1 3 1
=
#7 5 0 0 8 6 �1 1⇤ ⇤ ⇤ ⇤ ⇤ ⇤ �1
2 ⇤⇤ ⇤ ⇤ ⇤ ⇤ ⇤ 1
2� ⇤
=: M2
S = (3, 2).
Mitchell The Revised Simplex Method: An Example 12 / 22
K Oshould be P ,
New basic sequence wi l l be (3,6)
Second iteration
Calculate b
To perform the minimum ratio test in the third iteration, we will need toknow the entries in b corresponding to the positive entries in the pivotcolumn:
P2M0 =
2
41 3 00 1 00 2 1
3
5�4 0 1 �3 2 �5 1 0
2 1 �1 1 1 2 �1 01 0 4 �2 1 1 3 1
=
ratio #7 5 0 0 8 6 �1 1
� ⇤ ⇤ ⇤ ⇤ ⇤ ⇤ �12 ⇤
5 52 ⇤ ⇤ ⇤ ⇤ ⇤ 1
2� ⇤
=: M2
S = (3, 2).
Mitchell The Revised Simplex Method: An Example 13 / 22
Third iteration
Outline
1 First iteration
2 Second iteration
3 Third iteration
4 Checking the solution
Mitchell The Revised Simplex Method: An Example 14 / 22
Third iteration
Third iteration
In the third iteration, x6 replaces x2 in the basis.
ratio #7 5 0 0 8 6 �1 1
� ⇤ ⇤ ⇤ ⇤ ⇤ ⇤ �12 ⇤
5 52 ⇤ ⇤ ⇤ ⇤ ⇤ 1
2� ⇤
=: M2 S = (3, 2).
We can now calculate the new pivot matrix Q3, by “doing unto theidentity as you would do unto M2”:2
41 0 00 1 00 0 1
3
5 2R2 then R0 + R2, R1 +12R2
�! Q3 =
2
41 0 20 1 10 0 2
3
5
Mitchell The Revised Simplex Method: An Example 15 / 22
Third iteration
Calculate product of pivot matrices
The updated tableau is M3 = Q3M2.
However, we don’t know M2 explicitly, so we need to exploit the factthat M2 = Q2Q1M0 = P2M0, so M3 = (Q3P2)M0, by the associativityproperty of matrix multiplication.
We first calculate the matrix product P3 = Q3P2:
P3 = Q3P2 =
2
41 0 20 1 10 0 2
3
5
2
41 5 10 2 1
20 1 1
2
3
5 =
2
41 7 20 3 10 2 1
3
5
Mitchell The Revised Simplex Method: An Example 16 / 22
= Qs(P,Mu)=@R)Mo
Third iteration
Calculate reduced costs
We can now calculate the reduced costs:
Q3M2 = P3M0 =
2
41 7 20 3 10 2 1
3
5�4 0 1 �3 2 �5 1 0
2 1 �1 1 1 2 �1 01 0 4 �2 1 1 3 1
=12 7 2 0 11 11 0 2⇤ ⇤ ⇤ ⇤ ⇤ ⇤ ⇤ ⇤⇤ ⇤ ⇤ ⇤ ⇤ ⇤ ⇤ ⇤
=: M3
S = (3, 6).
Mitchell The Revised Simplex Method: An Example 17 / 22
0I 'o9×3=7,
x , = 5
Third iteration
Get optimal solutionThis tableau is in optimal form, so we have an optimal solution withbasic variables x3 and x6. We need to calculate the values of the basicvariables:
P3M0 =
2
41 3 00 1 00 2 1
3
5�4 0 1 �3 2 �5 1 0
2 1 �1 1 1 2 �1 01 0 4 �2 1 1 3 1
=12 7 2 0 11 11 0 2
7 ⇤ ⇤ ⇤ ⇤ ⇤ ⇤ ⇤5 ⇤ ⇤ ⇤ ⇤ ⇤ ⇤ ⇤
=: M3
S = (3, 6).
Thus, the optimal solution is x3 = 7, x6 = 5, with nonbasicsx1 = x2 = x4 = x5 = x7 = 0 and value �12.
Mitchell The Revised Simplex Method: An Example 18 / 22
Checking the solution
Outline
1 First iteration
2 Second iteration
3 Third iteration
4 Checking the solution
Mitchell The Revised Simplex Method: An Example 19 / 22
Checking the solution
Check the optimal solution
The initial tableau is
M0 =
x1 x2 x3 x4 x5 x6 x7�4 0 1 �3 2 �5 1 0
2 1 �1 1 1 2 -1� 01 0 4 -2� 1 1 3 1
S = (1, 7).
We found an optimal solution with basic variables x3 = 7 and x6 = 5.
We can first check that x3 = 7, x6 = 5 does indeed satisfy theconstraints, with value �12. X
Mitchell The Revised Simplex Method: An Example 20 / 22
Checking the solution
Pivoting to create canonical form
By pivoting on the two indicated entries, we can construct a tableauwhere x3 and x6 are the basic variables.
Provided our answer above was correct, the resulting tableau shouldhave b � 0 and c � 0. Pivot first on entry a16:
M0 =
x1 x2 x3 x4 x5 x6 x7�4 0 1 �3 2 �5 1 0
2 1 �1 1 1 2 -1� 01 0 4 -2� 1 1 3 1
S = (1, 7).
�R1 then R0 � R1, R2 � 3R1�!
x1 x2 x3 x4 x5 x6 x7�2 1 0 �2 3 �3 0 0�2 �1 1 �1 �1 �2 1� 0
7 3 1 1� 4 7 0 1
basic sequence S = (6, 7).
Mitchell The Revised Simplex Method: An Example 21 / 22
Checking the solution
The second pivot to create canonical form
Now pivot on entry a23:
x1 x2 x3 x4 x5 x6 x7�2 1 0 �2 3 �3 0 0�2 �1 1 �1 �1 �2 1� 0
7 3 1 1� 4 7 0 1
basic sequence S = (6, 7).
R0 + 2R2, R1 + R2�!
x1 x2 x3 x4 x5 x6 x712 7 2 0 11 11 0 25 2 2 0 3 5 1� 17 3 1 1� 4 7 0 1
S = (6, 3).
Thus, we do indeed get a tableau in optimal form. X
Mitchell The Revised Simplex Method: An Example 22 / 22
