15.093 Optimization Methods
Lecture 6: Duality Theory II
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B
c
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D
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Outline
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Slide 1
• Geometry of duality
• The dual simplex algorithm
• Farkas lemma
• Duality as a proof technique
2
The Geometry of Duality
Slide 2
min
s.t.
c′ x
a′i x ≥ bi ,
i = 1, . . . , m
max p′ b
m
�
pi ai = c
s.t.
i=1
p≥0
3
Dual Simplex Algorithm
3.1
Motivation
Slide 3
• In simplex method B −1 b ≥ 0
• Primal optimality condition
c′ − c′B B −1 A ≥ 0′
same as dual feasibility
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c
a1
a3
a3
a2
x *
a1
• Simplex is a primal algorithm: maintains primal feasibility and works
towards dual feasibility
• Dual algorithm: maintains dual feasibility and works towards primal
feasibility
Slide 4
−c′B xB
xB(1)
..
.
xB(m)
c̄1
|
...
c̄n
|
B −1 A1
|
...
B −1 An
|
• Do not require B −1 b ≥ 0
• Require c̄ ≥ 0 (dual feasibility)
• Dual cost is
p′ b = c′B B −1 b = c′B xB
• If B −1 b ≥ 0 then both dual feasibility and primal feasibility, and also
same cost ⇒ optimality
• Otherwise, change basis
3.2
An iteration
Slide 5
1. Start with basis matrix B and all reduced costs ≥ 0.
2. If B −1 b ≥ 0 optimal solution found; else, choose l s.t. xB(l) < 0.
2
x2
p2
1
.
D
c
1
.
.
. .
.
B
b
C
C
A
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A
E
D
1
x1
2
.
B
1/2
.
.
E
1
p1
(b)
(a)
3. Consider the lth row (pivot row) xB(l) , v1 , . . . , vn . If ∀i vi ≥ 0 then dual
optimal cost = +∞ and algorithm terminates.
Slide 6
4. Else, let j s.t.
c̄j
c̄i
= min
|vj | {i|vi <0} |vi |
5. Pivot element vj : Aj enters the basis and AB(l) exits.
3.3
An example
Slide 7
min x1 + x2
s.t. x1 + 2x2 ≥ 2
x1 ≥ 1
x1 , x2 ≥ 0
min
s.t.
x1 + x2
x1 + 2x2 − x3 = 2
x1 − x4 = 1
x1 , x2 , x3 , x4 ≥ 0
max
s.t.
2p1 + p2
p1 + p2 ≤ 1
2p1 ≤ 1
p1 , p2 ≥ 0
x1
x2
x3
x4
0
1
1
0
0
x3 =
−2
−1
−2*
1
0
x4 =
−1
−1
0
0
1
Slide 8
Slide 9
x1
x2
x3
x4
−1
1/2
0
1/2
0
x2 =
1
1/2
1
−1/2
0
x4 =
−1
−1*
0
0
1
3
A
1
A
A
3
2
p
b
.
4
x1
x2
x3
x4
−3/2
0
0
1/2
1/2
x2 =
1/2
0
1
−1/2
1/2
x1 =
1
1
0
0
−1
Duality as a proof method
4.1
Farkas lemma
Slide 10
Theorem:
Exactly one of the following two alternatives hold:
1. ∃x ≥ 0 s.t. Ax = b.
2. ∃p s.t. p′ A ≥ 0′ and p′ b < 0.
4.1.1
Proof
Slide 11
“ ⇒′′ If ∃x ≥ 0 s.t. Ax = b, and if p′ A ≥ 0′ , then p′ b = p′ Ax ≥ 0
“ ⇐′′ Assume there is no x ≥ 0 s.t. Ax = b
(P ) max 0′ x
s.t. Ax = b
x ≥ 0
(D) min
s.t.
(P) infeasible ⇒ (D) either unbounded or infeasible
Since p = 0 is feasible ⇒ (D) unbounded
⇒ ∃p : p′ A ≥ 0′ and p′ b < 0
4
p′ b
p′ A ≥ 0 ′
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15.093J / 6.255J Optimization Methods
Fall 2009
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