LECTURE 5: CONJUGATE DUALITY...0. (ii) strong duality. 0 = f(x*) + h(y*) Where is the duality...
Transcript of LECTURE 5: CONJUGATE DUALITY...0. (ii) strong duality. 0 = f(x*) + h(y*) Where is the duality...
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LECTURE 5: CONJUGATE DUALITY 1. Primal problem and its conjugate dual 2. Duality theory and optimality conditions 3. Relation to other type of dual problems 4. Linear conic programming problems
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Motivation of conjugate duality Min f(x) (over R) = Max g(y) (over R) and h(y) = - g(y) • f(x) + h(y) = f(x) – g(y) can be viewed as “duality gap” • Would like to have (i) weak duality 0 (ii) strong duality 0 = f(x*) + h(y*)
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Where is the duality information • Recall the fundamental result of Fenchel’s conjugate inequality
• Need a structure such that in general
and at optimal solutions 0 = <x*, y*> = f(x*) + h(y*)
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Concept of dual cone • Let X be a cone in
• Define its dual cone • Properties: (i) Y is a cone. (ii) Y is a convex set. (iii) Y is a closed set.
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Observations
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Conjugate (Geometric) duality
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Dual side information • Conjugate dual function
• Dual cone
Properties: 1. Y is a cone in 2. Y is closed and convex 3. both are closed and convex.
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Conjugate (Geometric) dual problem
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Observations
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Conjugate duality theory
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Conjugate duality theory
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Proof
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Conjugate duality theory
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Example – Standard form LP
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Conjugate dual problem
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Dual LP
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Example – Karmarkar form LP
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Example – Karmarkar form LP
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Example – Karmarkar form LP • Conjugate dual problem becomes
which is an unconstrained convex programming problem.
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Illustration
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Example – Posynomial programming • Nonconvex programming problem
• Transformation
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Posynomial programming • Primal problem: Conjugate dual problem:
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Dual Posynomial Programming
h(y) , supx∈En[< x , y > −f (x)] < +∞
Let g(x) =∑n
j=1 xjyj − log∑n
j=1 cjexj
∂g∂xj
= 0 ⇒ yj =cje
x∗j∑nj=1 cje
x∗j
⇒ yj > 0 and∑n
j=1 yj = 1.⇒ Ω is closed.⇒ Ω = y ∈ En|yj > 0 and
∑nj=1 yj = 1.
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log(yj) = log(cjex∗
j )− log∑n
j=1 cjex∗
j
= log(cj) + x∗j − log∑n
j=1 cjex∗
j
⇒ log(yjcj
) + log∑n
j=1 cjex∗
j = x∗j
h(y) = supx∈En [< x , y > −f (x)] = g(x∗)=
∑nj=1 x∗j yj − log
∑nj=1 cje
x∗j
=∑n
j=1 yj [log(yjcj
) + log∑n
j=1 cjex∗
j ]− log∑n
j=1 cjex∗
j
=∑n
j=1 yj log(yjcj
) +∑n
j=1 yj log∑n
j=1 cjex∗
j − log∑n
j=1 cjex∗
j
=∑n
j=1 yj log(yjcj
) + (∑n
j=1 yj)log∑n
j=1 cjex∗
j − log∑n
j=1 cjex∗
j
=∑n
j=1 yj log(yjcj
)
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Conjugate dual problem
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Degree of difficulties • When degree of difficulty = 0, we have a system of linear
equations:
• When degree of difficulty = k, we have
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Duality gap • Definition:
• Observation:
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Extremality conditions • Definition:
• Corollary:
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Proof of Corollary
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Necessary and sufficient conditions • Corollary
• Observation
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When will the duality gap vanish?
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Nonlinear complementarity problem
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Lagrangian Function and Saddle Point
I Let f (x) : S⋂
X and h(y) : Ω⋂
Y be a known conjugatepair with X being closed and convex and f ∈ C′(S).
I The lagrangian function is defined as
L(x , y) , f (x)− < x , y > .
I A point pair (x , y) ∈ S × Y is called a saddle point ofL(x , y) if
L(x , y) > L(x , y) > L(x , y), ∀x ∈ S, y ∈ Y ,
or
infx∈S
L(x , y) = L(x , y) = supy∈Y
L(x , y).
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Saddle Point and Optimality
Theorem: (saddle point⇒ Optimality)If (x , y) ∈ S × Y is a saddle point of L(x , y), then x ∈ S ∗ andy ∈ T ∗.Proof: By definition,
infx∈S
L(x , y) = L(x , y) = supy∈Y
L(x , y)
1L(x , y) = infx∈S L(x , y)
= infx∈S[f (x)− < x , y >]= − supx∈S[< x , y > −f (x)]
Hence y ∈ Ω and L(x , y) = −h(y), (now, y ∈ Y⋂
Ω = T )
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2L(x , y) = supy∈Y L(x , y)
= supy∈Y [f (x)− < x , y >]
= f (x) + supy∈Y [− < x , y >]
= f (x)− infy∈Y [< x , y >]
Since Y is a cone, x ∈ dual(Y ) = X and infy∈Y [< x , y >] = 0.Hence L(x , y) = f (x). (Now, x ∈ S
⋂X = S )
3 Putting 1 and 2 together, we have
f (x) = L(x , y) = −h(y).
Hence f (x) + h(y) = 0, and x ∈ S ∗, y ∈ T ∗.
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Saddle Point and Optimality
Theorem (Convex Optimality with No Gap⇒ Saddle Point)Let f : S
⋂X and h : Ω
⋂Y be a closed convex conjugate dual
pair with no duality gap. Then (x , y) is a saddle point of L(x , y)if and only if x ∈ S ∗ and y ∈ T ∗.Proof: We need only to prove that “If x ∈ S ∗ and y ∈ T ∗, then(x , y) is a saddle point of L(x , y).” When x ∈ S ∗ and y ∈ T ∗,we have < x , y >= 0. Since there is no duality gap,L(x , y) = f (x)− < x , y >= −h(y). Hence,
−h(y)︸ ︷︷ ︸infx∈S L(x ,y)
= L(x , y) = f (x)︸︷︷︸supy∈Y L(x ,y)
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Observations
1. If x ∈ S ∗ is known, then y ∈ T ∗ can be found by solving
max L(x , y) = f (x)− < x , y >s. t. y ∈ Y
When Y is linearly structures, then it is a linearly programmingproblem.2. If y ∈ T ∗ is known, then x ∈ S ∗ can be found by solving
min L(x , y) = f (x)− < x , y >s. t. x ∈ S
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Linear conic programming problems • Linear Conic Programming (LCoP) A general conic optimization problem is as follows:
( )
where is a closed and convex co
minimize subject to
" " is a linear operator line an
ke "inner product."d
P c xA x b
x∈=
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Dual linear conic dual problem
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Duality theorems for Linear CP
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Duality theorems for linear CP • Theorem 3 (Conic Duality Theorem): (i) If problems (CP) and (CD) are both feasible, then they have optimal solutions. (ii) If one of the two problems has an interior feasible solution with a finite optimal objective value, then the other one is feasible and has the same optimal objective value. (iii) If one of the two problems is unbounded, then the other has no feasible solution. (iv) If (CP) and (CD) both have interior feasible solutions, then they have optimal solutions with zero duality gap.
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Examples of conic programs
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Example of conic programs
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Example of conic programs
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Semi-definite Programming (SDP)
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Duality theorems for SDP
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Quadratic programming problem
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Conjugate dual QP
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Conjugate dual QP
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Conjugate dual QP