Efficiently solving SDPs originating from the KYP
lemma using standard software
Ragnar Wallin and Anders Hansson
Division of Automatic Control
Department of Electrical Engineering
Linköpings universitet
Lieven Vandenberghe
Electrical Engineering Department
UCLA
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Applications
■
Analysis and design of linear systems
■
Design of filters
■
Polynomial matrix problems in robust control
■
Robust control analysis using integral quadratic constraints
• uncertain dynamics
• parametric uncertainty
• time-varying parameters
• various nonlinearities
• structured uncertainty, for example, a combination of the above
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A standard semidefinite program
The optimization problem
min cT x
s. t. F = F0 +
q
X
xk F k ≤ 0
k=1
can be solved in polynomial time. The time required will be a polynomial in
1. The size of the matrices, Fk
2. The number of parameters, q
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The dual problem
− max hM0 , Zi
s. t. hMk , Zi = ck ,
k = 1, 2, . . . , q
Z≤0
The dual of the dual problem is the primal problem.
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Primal-dual methods
Features of primal-dual methods
■
Solves the primal and dual problems simultaneously
■
Non-heuristic stopping criteria
■
Better numerical properties
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SDP originating from the KYP lemma
The problem has a special structure
min cT x
T
p
X
A P + PA PB
+ M0 +
x k Mk ≤ 0
s. t. X :=
T
0
B P
k=1
and can be rewritten on a standard SDP form using q =
parameters. (P = P T ∈ Rn×n )
n(n+1)
2
+p
Assumptions: (A, B) controllable, λi (A) + λ∗j (A) 6= 0 ∀i, j, the Mk -matrices
linearly independent.
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The dual problem
− max hM0 , Zi
T
T
+ Z12
B=0
s. t. AZ11 + Z11 AT + BZ12
hMk , Zi = ck ,
Z=
Z11
T
Z12
k = 1, 2, . . . , p
Z12
≤0
Z22
This can be parameterized using q =
(Z ∈ R(n+m)×(n+m) )
(n+m)(n+m+1)
2
variables.
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The main idea
1. Reduce the number of variables in the dual
2. Solve an equivalent dual problem with fewer variables
3. The dual of the equivalent dual is equivalent to the original primal
problem
4. The primal-dual solver delivers Z and X, but NOT the variables of
interest, P and x
5. Recover the primal variables P and x from X
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Reducing the number of variables
A basis for Z that automatically fulfills the first dual constraint is
Fk =

Ek(11)



T

 Ek(12)


0


 0
Ek(12)
0
0
Ek(22)
k = 1, 2, . . . , mn
k = mn + 1, mn + 2, . . . , mn +
m(m+1)
2
where Ek(12) is the standard basis for unstructured n × m matrices, Ek(22) is
the standard basis for symmetric m × m matrices and Ek(11) is related to
Ek(12) through
T
+ Ek(12) B T = 0
AEk(11) + Ek(11) AT + BEk(12)
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Further reducing the number of variables
As we now have
mn+
m(m+1)
2
X
Z=
zj F j
j=1
the second dual constraint can be written as
mn+
hMk , Zi =
m(m+1)
2
X
hMk , Fj izj = ck
j=1
This is a linear system of equations
Gz = c
with
Gj,k = hMk , Fj i
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Further reducing the number of variables
Doing a QR-factorization
GT = Q 1
R
Q2
0
where Q1 and Q2 are orthogonal, we can express z and Z as
mn+
z = Q1 R
−T
c + Q2 z̃,
Z = H0 +
m(m+1)
−p
2
X
z̃k Hk
k=1
where the Hk s are linear combinations of the Fj s. H0 depends on Q1 , R and
c and the other Hk depend on Q2 ,
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Equivalent dual
We are back at a standard SDP
min dT z̃
s. t. Z = H0 +
q
X
z̃k Hk ≤ 0
k=1
with
dk = hM0 , Hk i
k = 1, 2, . . . , q
m(m + 1)
−p
q = mn +
2
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General purpose SDP solvers
The equivalent dual can be solved using any primal-dual SDP solver like
■
■
■
■
■
SeDuMi
SDPT3
CSDP
SDPA
SDPSOL
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Recovering x
Define the linear operators
F(P ) =
T
A P + PA PB
0
BT P
T
+ Z12T B
F ∗ (Z) = AZ11 + Z11 AT + BZ12
As each and every one of the first basis matrices for Z, Fk , fulfills F ∗ (Fk ) = 0
the following is true
hFk , F(P )i = hF ∗ (Fk ), P i = 0
| {z }
=0
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Recovering x, continued
Form the inner products between the Fk s and the constraint of the original
primal
hFk , Xi = hFk , F(P ) + M0 +
p
X
xj Mj i = hFk , M0 i +
j1
p
X
hFk , Mj ixj
j=1
This is an overdetermined consistent linear system of equations
GT x = g
where Gk,j = hMj , Fk i and g = hFk , X − M0 i.
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Recovering P
Having x we can find P by solving the Lyapunov equation corresponding to
the (1,1)-block of
X=
T
A P + PA PB
+ M0 +
0
BT P
p
X
x k Mk
k=1
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Example
Every point is an average over ten randomly generated systems with one inut
and one Mk -matrix.
3
10
Primal
Reduced dual
2
time
10
1
10
0
10
1
10
2
n
10
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Generalizations
■
Several constraints
■
λi (A) + λ∗j (A) = 0
■
(A, B) only stabilizable but not controllable
■
Linearly dependent Mk -matrices
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Several constraints
min cT x
s. t. X1 :=
X2 :=
AT1 P1
+ P 1 A1
B1T P1
AT2 P2
+ P 2 A2
B2T P2
P1 B 1
+ M10 +
0
P2 B 2
+ M20 +
0
p
X
k=1
p
X
xk M1k ≤ 0
xk M2k ≤ 0
k=1
WARNING! The method doesn’t work for several constraints with the same P .
(Quadratic stability for polytopic systems)
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λi (A) + λ∗j (A) = 0
Can be handled by
1. Finding a state-feedback such that A − BL Hurwitz
2. Transform the constraints
X̃ =
I
−L
0
I
T
X
I
−L
0
≤0
I
This results in the same P and x
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(A, B) only stabilizable
Can be handled by
1. Find a transformation, T , such that
à = T T AT =
A1
0
A12
A2
B̃ = T T A =
B1
0
2. Solve the optimization problem with the constraint
AT1 P1
+ P 1 A1
B1T P1
P1 B 1
+ M̃0 +
0
p
X
xk M̃k ≤ 0
k=1
Results in the same x. It is possible to construct a full P by solving a Lyapunov
equation involving A2 .
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Linearly dependent Mk -matrices
If the Mk -matrices are linearly dependent there does not exist a strictly
feasible dual point. There are two cases
1. The problem can be reduced to one with fewer variables
2. The problem is unbounded from below
Results in the same P but an x̃ of smaller dimension. The original x can be
found by multiplying x̃ by one of the matrices used in the reduction of variables.
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Standard SDP is a special case of KYP-SDP
min cT x
s. t.
T
A P + PA PB
+ M0 +
0
BT P
p
X
x k Mk ≤ 0
k=1
If the B-matrix has zero rows ⇒ M (x) ≤ 0, standard LMIs can be constraints.
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Lyapunov equations is a special case of KYP-SDP
min cT x
s. t.
T
A P + PA PB
+ M0 +
0
BT P
p
X
x k Mk ≤ 0
k=1
If the B-matrix has zero columns ⇒ AT P + P A + M (x) ≤ 0
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Package for KYP-SDP
A package for solving KYP-SDPs is downloadable from
http://www.control.isy.liu.se/publications/doc?id=1470
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