Lecture 6
Simplex Method:
Artifical Starting Solution
and Some Special Cases
September 4, 2009
Lecture 6
Outline:
• Initial table vs initial simplex table
• Artificial start: Two-phase method
• Special cases:
• Degeneracy
• Multiple optimal solutions
• Infinite optimal value
• Infeasibility of the problem
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Lecture 6
Simplex method: started at a feasible basic solution
Illustrated on the Reddy Mikks problem
Original LP formulation
Standard LP form
maximize z = 5x1 + 4x2
subject to 6x1 + 4x2 ≤ 24
x1 + 2x2 ≤ 6
x1 , x 2 ≥ 0
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maximize z = 5x1 + 4x2
subject to 6x1 + 4x2 + x3
= 24
x1 + 2x2
+ x4 = 6
x1 , x 2 , x 3 , x 4 ≥ 0
2
Lecture 6
Initial table vs Initial simplex table
Basis
z
??
??
x1
−5
6
1
x2
−4
4
2
x3
0
1
0
x4
0
0
1
RHS Values
= 0
= 24
= 6
x1
−5
6
1
x2
−4
4
2
x3
0
1
0
x4
0
0
1
RHS Values
= 0
= 24
= 6
Basis: x3 and x4
Basis
z
x3
x4
Initial table can be used as initial simplex table
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Lecture 6
Basis: x1 and x4
Basis
z
x1
x4
x1
−5
6
1
x2
−4
4
2
x3
0
1
0
x4
0
0
1
RHS Values
0
24
6
This table cannot be used as the initial simplex table! We have to transform
the table (Gauss-Jordan elimination using x1-column elements)
Basis
z
x1
0
x2
− 23
x3
x4
0
RHS Values
20
x1
1
2
3
1
6
0
4
x4
0
4
3
− 61
1
2
5
6
This table is an initial simplex table, i.e., the simplex method can start.
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Lecture 6
Artificial Start: Two-phase method
• Sometimes, it is not easy to find an initial feasible solution (i.e., to
choose initial bases yielding a feasible point)
• Two-phase method is used in such situations
• In first phase, a feasibility problem associated with the LP is solved
by a simplex method
• In the second phase, the solution from the first phase is used to start
running the simplex method
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Lecture 6
Two-phase Method: Example
Original LP formulation
Standard LP form
minimize z = 4x1 + x2
minimize z = 4x1 + x2
subject to 3x1 + x2
=3
subject to 3x1 + x2 = 3
4x1 + 3x2 − x3
=6
4x1 + 3x2 ≥ 6
x1 + 2x2
+ x4 = 4
x1 + 2x2 ≤ 4
x1 , x 2 , x 3 , x 4 ≥ 0
x1 , x 2 ≥ 0
It is not easy to find a basis yielding a basic feasible solution!
Phase I: we first solve a feasibility problem associated with the LP:
introduce new variables R1 and R2
minimize r = R1 + R2
subject to 3x1 + x2
+ R1
=3
4x1 + 3x2 − x3
+ R2 = 6
x1 + 2x2
+ x4
=4
x1 , x 2 , x 3 , x 4 , R 1 , R 2 ≥ 0
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Lecture 6
Phase I: Initial table
Basis
r
R1
R2
x4
x1
0
3
4
1
x2
0
1
3
2
x3
0
0
−1
0
x4
0
0
0
1
R1
−1
1
0
0
R2
−1
0
1
0
RHS Values
0
3
6
4
R1
0
1
0
0
R2
0
0
1
0
RHS Values
9
3
6
4
Initial simplex table
Basis
r
R1
R2
x4
x1
7
3
4
1
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x2
4
1
3
2
x3
−1
0
−1
0
x4
0
0
0
1
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Lecture 6
Applying the simplex method, we will obtain an optimal solution
Basis
r
x1
x2
x4
x1
0
1
0
0
x2
0
0
1
0
x3
0
1
5
− 53
1
x4
0
0
0
1
R1
−1
3
5
4
−5
1
R2
−1
− 15
RHS Values
0
3
5
3
5
6
5
−1
1
Q: How do we know this is optimal solution to the feasibility problem?
What is the solution? What is the optimal value for r?
Use the solution of Phase I to start Phase II (solving the original LP)
Basis
z
x1
x2
x4
x1
−4
1
0
0
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x2
−1
0
1
0
x3
0
1
5
3
−5
1
x4
0
0
0
1
RHS Values
0
3
5
6
5
1
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Lecture 6
The initial table of Phase II
Basis
z
x1
x2
x4
x1
0
1
0
0
x2
0
0
1
0
x3
1
5
1
5
3
−5
1
x4
0
0
0
1
RHS Values
18
5
3
5
6
5
1
We continue with simplex iterations until we find an optimal solution.
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Lecture 6
Special Case: Degeneracy
Degeneracy is a term used for a basic feasible solution having one or
more basic variables at value 0
Original LP formulation
Standard LP form
maximize z = 3x1 + 9x2
maximize z = 3x1 + 9x2
subject to x1 + 4x2 ≤ 8
subject to x1 + 4x2 + x3
=8
x1 + 2x2 ≤ 4
x1 + 2x2
+ x4 = 4
x1 , x 2 ≥ 0
x1 , x 2 , x 3 , x 4 ≥ 0
At the end of the simplex method, (started with basis x3 and x4), we have
Basis
z
x2
x1
x1
0
0
1
x2
0
1
0
x3
x4
3
2
1
2
3
2
1
−2
−1
2
RHS Values
18
2
0
There is another basis corresponding to the same solution!
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Lecture 6
Special Case: Multiple Solutions
Original LP formulation
Standard LP form
maximize z = 2x1 + 4x2
maximize z = 2x1 + 4x2
subject to x1 + 2x2 ≤ 5
subject to x1 + 2x2 + x3
=5
x1 + 2x2 ≤ 4
x1 + x2
+ x4 = 4
x1 , x 2 ≥ 0
x1 , x 2 , x 3 , x 4 ≥ 0
At the end of the simplex method, (started with basis x3 and x4), we have
Basis
z
x2
x4
x1
0
1
2
1
2
x2
0
1
0
x3
2
1
2
− 12
x4
0
0
1
RHS Values
10
5
2
3
2
There is a nonbasic variable with 0 coefficient in the optimal table.
This means we can bring that variable in the basis without changing the
z -value.
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Lecture 6
Choosing x1 to enter the basis, we perform another simplex iteration and
find an alternative optimal solution
Basis
z
x2
x4
x1
0
0
1
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x2
0
1
0
x3
2
1
−1
x4
0
−1
2
RHS Values
10
1
3
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Lecture 6
Special Case: Unbounded Optimal Value
Original LP formulation
Standard LP form
maximize z = 2x1 + x2
maximize z = 2x1 + x2
subject to x1 − x2 ≤ 10
subject to x1 − x2 + x3
= 10
2x1
≤ 40
2x1
+ x4 = 40
x1 , x 2 ≥ 0
x1 , x 2 , x 3 , x 4 ≥ 0
At some iteration of the simplex method, (in this example, it happened
to be the initial iteration) a nonbasic variable with negative coefficient
can enter the basis without a bound on its value (maximization)
Basis x1
x2
x3
x4
RHS Values
z −2 −1
0
0
0
1
0
10
x3
1 −1
x4
2
0
0
1
40
This means we can bring that variable in the basis and increase the z -value
to +∞ (since the variable can be increased to +∞).
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Lecture 6
Special Case: Infeasibility
Original LP formulation
Standard LP form
maximize z = 3x1 + 2x2
maximize
subject to 2x1 + x2 ≤ 2
subject to
3x1 + 4x2 ≥ 12
x1 , x 2 ≥ 0
None of the basic solutions is feasible!
z = 3x1 + 2x2
2x1 + x2 + x3
=2
3x1 + 4x2
− x4 = 12
x1 , x 2 , x 3 , x 4 ≥ 0
Here, it is not as easy to read off a feasible basis -apply Two phase method
Phase I Solve the feasibility problem
minimize r = R
subject to 2x1 + x2 + x3
=2
3x1 + 4x2
− x4 + R = 12
x1 , x 2 , x 3 , x 4 , R ≥ 0
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Lecture 6
Initial table for feasibility problem
Basis
r
x3
R
x1
0
2
3
x2
0
1
4
x3
0
1
0
x4
R
0 −1
0
0
−1
1
RHS Values
0
2
12
Initial simplex table for feasibility problem
Basis
r
x3
R
x1
3
2
3
x2
4
1
4
x3
0
1
0
x4
−1
0
−1
R
0
0
1
RHS Values
12
2
12
The table is not optimal.
Choose a nonbasic variable with positive r-coefficient (minimization), say
choose x2
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Lecture 6
In the next simplex iteration, we have
Basis
r
x2
R
x1
−5
2
3
x2
0
1
4
x3
−4
1
0
x4
−1
0
−1
R
0
0
1
RHS Values
4
2
12
This table is optimal (the feasibility problem involves minimization)
Indication of infeasibility:
In the optimal (phase I) table,
• The optimal value is positive
• The basis solution contains an artificial variable with a positive value
The material is in Sections 3.4.2 and 3.5
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