linear matrix inequalities in control · state-feedback, estimation and output-feedback synthesis....

Linear Matrix Inequalities in Control

Carsten Scherer and Siep Weiland

7th Elgersburg School on Mathematical Systems Theory

Class 1

Carsten Scherer and Siep Weiland (Elgersburg) Linear Matrix Inequalities in Control Class 1 1 / 65

Part I

Course organization


Outline of Part I

1 Organization of the courseTopicsSoftware


course topics

• Facts from convex analysis.LMI’s: history, algorithms and software.

• The role of LMI’s in dissipativityStability and nominal performance. Analysis results.

• From analysis to synthesis.State-feedback, estimation and output-feedback synthesis.

• From nominal to robust stability, robust performance and robustsynthesis.

• IQC’s and multipliers.Relations to classical tests and to µ-theory.

• Mixed control problems and parameter-varying systems and controldesign.


software issues

Install Yalmip

• Modelling language for solving convex optimization problems

• Consistent with Matlab; free (!)

• Efficient to implement algorithms

• Developed by Johan Lofberg

Install SDP solver

• Small till medium size LP and QP problems: SeDuMi or SPDT3

• High level LP and QP problems: GUROBI

• Generic package for academic use: MOSEK

• Many more commercial and non-commercial solvers . . .

to Yalmip website


Part II

Let’s start . . .


Outline of Part II

2 What to expect?

3 Convex sets and convex functionsConvex setsConvex functions

4 Why is convexity important?ExamplesEllipsoidal algorithmDuality and convex programs

5 Linear Matrix InequalitiesDefinitionsLMI’s and convexityLMI’s in control

6 A design example


Outline

2 What to expect?




6 A design example


http://users.isy.liu.se/johanl/yalmip/

Merging control and optimization

Classical optimal control paradigm (LQG, H2, H∞, L1, MPC) isrestricted:

• Performance specs in terms of complete closed loop transfer matrix.

• One measure of performance only. Often multiple specs have beenimposed for controlled system

• Cannot incorporate structured time-varying/nonlinear uncertainties

• Can only design LTI controllers

• Can only synthesize one controller in one architecture.

Control vs. optimization

View control input and/or feedback controller as decision variable ofoptimization problem. Desired specifications are imposed as constraints oncontrolled system.


Major goals for control and optimization

• Distinguish easy from difficult problems. (Convexity is key!)

• What are consequences of convexity in optimization?

• What is robust optimization?

• How to check robust stability by convex optimization?

• Which performance measures can be dealt with?

• How can controller synthesis be convexified?

• What are limits for the synthesis of robust controllers?

• How can we perform systematic gain scheduling?


Outline

2 What to expect?




6 A design example


Optimization problems

Casting optimization problems in mathematics requires

• X : decision set

• S ⊆ X : feasible decisions

• f : S → R: cost function

f assigns to each decision x ∈ S a cost f(x) ∈ R.

Wish to select the decision x ∈ S that minimizes the cost f(x)


Optimization problems

1 What is least possible cost? Compute optimal value

fopt := infx∈S

f(x) = inf{f(x) | x ∈ S} ≥ −∞

Convention: S = ∅ then fopt = +∞Convention: If fopt = −∞ then problem is said to be unbounded

2 How to determine almost optimal solutions? For arbitrary ε > 0 find

xε ∈ S with fopt ≤ f(xε) ≤ fopt + ε.

3 Is there an optimal solution (or minimizer)? Does there exist

xopt ∈ S with fopt = f(xopt)

We write: f(xopt) = minx∈S f(x)

4 Can we calculate all optimal solutions? (Non)-uniqueness

arg min f(x) := {x ∈ S | fopt = f(x)}


Recap: infimum and minimum of functions

Infimum of a functionAny f : S → R has infimum L ∈ R ∪ {−∞} denoted as infx∈S f(x). It isdefined by the properties

• L ≤ f(x) for all x ∈ S• L finite: for all ε > 0 exists x ∈ S with f(x) < L+ εL infinite: for all ε > 0 there exist x ∈ S with f(x) < −1/ε

Minimum of a functionIf exists x0 ∈ S with f(x0) = infx∈S f(x) we say that f attains itsminimum on S and write L = minx∈S f(x).If exists, the minimum is uniquely defined by the properties

• L ≤ f(x) for all x ∈ S• There exists some x0 ∈ S with f(x0) = L


A classical result

Theorem (Weierstrass)

If f : S → R is continuous and S is a compact subset of the normed linearspace X , then there exists xmin, xmax ∈ S such that for all x ∈ S

infx∈S

f(x) = f(xmin) ≤ f(x) ≤ f(xmax) = supx∈S

f(x)

Comments:

• Answers problem 3 for “special” S and f

• No clue on how to find xmin, xmax

• No answer to uniqueness issue

• S compact if for every sequence xn ∈ S a subsequence xnm existswhich converges to a point x ∈ S

• Continuity and compactness overly restrictive!


Convex sets

Definition

A set S in a linear vector space X is convex if

x1, x2 ∈ S =⇒ αx1 + (1− α)x2 ∈ S for all α ∈ (0, 1)

The point αx1 + (1− α)x2 with α ∈ (0, 1) is a convex combination of x1and x2.

Definition

The point x ∈ X is a convex combination of x1, . . . , xn ∈ X if

x :=

n∑i=1

αixi, αi ≥ 0,

n∑i=1

αi = 1

• Note: set of all convex combinations of x1, . . . , xn is convex.


Examples of convex sets

Convex sets Non-convex sets


Basic properties of convex sets

Theorem

Let S and T be convex. Then

• αS := {x | x = αs, s ∈ S} is convex

• S + T := {x | x = s+ t, s ∈ S, t ∈ T } is convex

• closure of S and interior of S are convex

• S ∩ T := {x | x ∈ S and x ∈ T } is convex.

• x ∈ S is interior point of S if there exist ε > 0 such that all y with‖x− y‖ ≤ ε belong to S

• x ∈ X is closure point of S ⊆ X if for all ε > 0 there exist y ∈ S with‖x− y‖ ≤ ε


Examples of convex sets

• With a ∈ Rn\{0} and b ∈ R, the hyperplane

H = {x ∈ Rn | a>x = b}

and the half-space

H− = {x ∈ Rn | a>x ≤ b}

are convex.

• The intersection of finitely many hyperplanes and half-spaces is apolyhedron. Any polyhedron is convex and can be described as

{x ∈ Rn | Ax ≤ b,Dx = e}

for suitable matrices A and D and vectors b, e.

• A compact polyhedron is a polytope.


The convex hull

Definition

The convex hull of a set S ⊂ X is

conv(S) := ∩{T | T is convex and S ⊆ T }

• conv(S) is convex for any set S• conv(S) is set of all convex combinations of points of S• The convex hull of finitely many points conv(x1, . . . , xn) is a

polytope. Moreover, any polytope can be represented in this way!!

Latter property allows explicit representation of polytopes. For example{x ∈ Rn | a ≤ x ≤ b} consists of 2n inequalities and requires 2n

generators for its representation as convex hull!


Convex functions

Definition

A function f : S → R is convex if

• S is convex and

• for all x1, x2 ∈ S, α ∈ (0, 1) there holds

f(αx1 + (1− α)x2) ≤ αf(x1) + (1− α)f(x2)

• We have

f : S → R convex =⇒ {x ∈ S | f(x) ≤ γ}︸︷︷︸Sublevel sets

convex for all γ ∈ R

Derives convex sets from convex functions

• Converse ⇐= is not true!

• f is strictly convex if < instead of ≤Carsten Scherer and Siep Weiland (Elgersburg) Linear Matrix Inequalities in Control Class 1 21 / 65

Examples of convex functions

Convex functions

• f(x) = ax2 + bx+ c convex ifa > 0

• f(x) = |x|• f(x) = ‖x‖• f(x) = sinx on [π, 2π]

Non-convex functions

• f(x) = x3 on R• f(x) = −|x|• f(x) =

√x on R+

• f(x) = sinx on [0, π]


Convex matrix-valued functions

An interesting generalization:

• Define H and S to be the sets of Hermitian and Symmetric matrices,i.e., sets of square matrices A for which

A = A∗ or A = A>

• Define partial ordering on H and S

A1 ≺ A2, A1 4 A2, A1 � A2, A1 < A2

by requiring x∗(A1 −A2)x to be negative, non-positive, positive ornon-positive for all x 6= 0.

Definition

A matrix-valued function F : S → H is convex if S is a convex set and

F (αx1 + (1− α)x2) 4 αF (x1) + (1− α)F (x2)

for any x1, x2 ∈ S and 0 < α < 1.


Recap: Hermitian and symmetric matrices

Definition

For a real or complex matrix A the inequality A ≺ 0 means that A isHermitian and negative definite.

• A is Hermitian if A = A∗ = A>. If A is real this amounts to A = A>

and we call A symmetric.

Set of n× n Hermitian or symmetric matrices: Hn and Sn.

• All eigenvalues of Hermitian and symmetric matrices are real.

• By definition a Hermitian matrix A is negative definite if

u∗Au < 0 for all complex vectors u 6= 0

A is negative definite if and only if all its eigenvalues are negative.

• A 4 B, A � B and A < B defined and characterized analogously.


Affine sets

Definition

A subset S of a linear vector space is affine if x = αx1 + (1− α)x2belongs to S for every x1, x2 ∈ S and α ∈ R

• Geometric idea: line through any two points belongs to set

• Every affine set is convex

• S affine if and only if there exists x0 such that

S = {x | x = x0 +m, m ∈M}

with M a linear subspace


Affine functions

Definition

A function f : S → T is affine if

f(αx1 + (1− α)x2) = αf(x1) + (1− α)f(x2)

for all x1, x2 ∈ S and for all α ∈ R

Theorem

If S and T are finite dimensional, then f : S → T is affine if and only if

f(x) = f0 + T (x)

where f0 ∈ T and T : S → T a linear map (a matrix).


Cones and convexity

Definition

A convex cone is a set K ⊂ Rn with the property that

x1, x2 ∈ K =⇒ α1x1 + α2x2 ∈ K for all α1, α2 ≥ 0.

• Since x1, x2 ∈ K implies αx1 + (1− α)x2 ∈ K for all α ∈ (0, 1) aconvex cone is convex.

• If S ⊂ X is an arbitrary set, then

K := {y ∈ Rn | 〈x, y〉 ≥ 0 for all x ∈ S}

is a cone. Also denoted K = S∗ and called dual cone of S.


Cones and convexity

• Cones are unboundedsets.

• If K1 and K2 are convexcones, then so are

αK1,K1 ∩ K2,K1 +K2

for any α ∈ R.

• The intersection of a conewith a hyperplane is aconic section.


Outline

2 What to expect?




6 A design example


Why is convexity interesting ???

Reason 1: absence of local minima

Definition

f : S → H. Then x0 ∈ S is a

• local minimum if ∃ε > 0 such that

f(x0) 4 f(x) for all x ∈ S with ‖x− x0‖ ≤ ε

• global minimum if f(x0) 4 f(x) for all x ∈ S

Theorem

If f : S → H is convex then every local minimum x0 is a global minimumof f . If f is strictly convex, then the global minimum x0 is unique.



Reason 2: uniform bounds

Theorem

Suppose S = conv(S0) and f : S → H is convex. Then equivalent are:

• f(x) 4 γ for all x ∈ S• f(x) 4 γ for all x ∈ S0

Very interesting if S0 consists of finite number of points, i.e,S0 = {x1, . . . , xn}. A finite test!!



Reason 3: subgradients

Definition

A vector g = g(x0) ∈ Rn is a subgradient of f at x0 if

f(x) ≥ f(x0) + 〈g, x− x0〉

for all x ∈ S

Geometric idea: graph of affine function x 7→ f(x0) + 〈g, x− x0〉 tangentto graph of f at (x0, f(x0)).

Theorem

A convex function f : S → R has a subgradient at every interior point x0of S.


Examples and properties of subgradients

• f differentiable, then g = g(x0) = ∇f(x0) is subgradientSo, for differentiable functions every gradient is a subgradient

• the non-differentiable function f(x) = |x| has any real numberg ∈ [−1, 1] as its subgradient at x0 = 0.

• f(x0) is global minimum of f if and only if 0 is subgradient of f at x0.

• Since〈g, x− x0〉 > 0 =⇒ f(x) > f(x0),

all points in half space H := {x | 〈g, x− x0〉 > 0} can be discarded insearching for minimum of f .

Used explicitly in ellipsoidal algorithm


Ellipsoidal algorithm

Aim: Minimize convex function f : Rn → RStep 0 Let x0 ∈ Rn and P0 � 0 such that all minimizers of f are

located in the ellipsoid

E0 := {x ∈ Rn | (x− x0)>P−10 (x− x0) ≤ 1}.

Set k = 0.Step 1 Compute a subgradient gk of f at xk. If gk = 0 then stop,

otherwise proceed to Step 2.Step 2 All minimizers are contained in

Hk := Ek ∩ {x | 〈gk, x− xk〉 ≤ 0}.

Step 3 Compute xk+1 ∈ Rn and Pk+1 � 0 with minimaldeterminant detPk+1 such that

Ek+1 := {x ∈ Rn | (x− xk+1)>P−1k+1(x− xk+1) ≤ 1}

contains Hk.Step 4 Set k to k + 1 and return to Step 1.


Ellipsoidal algorithm

Remarks on ellipsoidal algorithm:

• Convergence f(xk)→ infx f(x).

• Exist explicit equations for

xk, Pk, Ek

such that volume of Ek decreases with factor e−1/2n at each step.(See lecture notes).

• Simple, robust, reliable, easy to implement, but slow convergence.



Reason 4: Duality and convex programsSet of feasible decisions often described by equality and inequalityconstraints:

S = {x ∈ X | gk(x) ≤ 0, k = 1, . . . ,K, h`(x) = 0, ` = 1, . . . , L}

Primal optimization:

Popt = infx∈S

f(x)

• One of index sets K or L infinite: semi-infinite optimization

• Both index sets K and L finite: nonlinear program



• Examples: saturation constraints, safety margins, constitutive andbalance equations all assume form S = {x | g(x) 4 0, h(x) = 0}.

• semi-definite program:

minimize f(x) subject to g(x) 4 0, h(x) = 0

f , g and h affine.

• quadratic program:

minimize x>Qx+ 2s>x+ r subject to g(x) 4 0, h(x) = 0

with g and h affine.

• quadratically constraint quadratic program:

f(x) = x>Qx+2s>x+r, gj(x) = x>Qjx+2s>j x+rj , h(x) = h0+Hx


Upper and lower bounds for convex programs

Primal optimization problem

Popt = infx∈X

f(x)

subject to g(x) 4 0, h(x) = 0

Can we obtain bounds on optimal value Popt ?

• Upper bound on optimal valueFor any x0 ∈ S we have

Popt ≤ f(x0)

which defines an upper bound on Popt.


Upper and lower bounds for convex programs

• Lower bound on optimal valueLet x ∈ S. Then for arbitrary y < 0 and z we have

L(x, y, z) := f(x) + 〈y, g(x)〉+ 〈z, h(x)〉 ≤ f(x)

and, in particular,

`(y, z) := infx∈X

L(x, y, z) ≤ infx∈S

L(x, y, z) ≤ infx∈S

f(x) = Popt.

so that

Dopt := supy<0, z

`(y, z) = supy<0, z

infx∈X

L(x, y, z) ≤ Popt

defines a lower bound for Popt.


Duality and convex programs

Some terminology:

• Lagrange function: L(x, y, z)

• Lagrange dual cost: `(y, z)

• Lagrange dual optimization problem:

Dopt := supy<0, z

`(y, z)

Remarks:

• `(y, z) computed by solving an unconstrained optimization problem.Is concave function.

• Dual problem is concave maximization problem. Constraints aresimpler than in primal problem

• Main question: when is Dopt = Popt?


Example of duality

• Primal Linear Program

Popt = infx

c>x

subject to x ≥ 0, b−Ax = 0

Lagrange dual cost

`(y, z) = infxc>x− y>x+ z>(b−Ax)

=

{b>z if c−A>z − y = 0

−∞ otherwise

• Dual Linear Program

Dopt = supz

b>z

subject to y = c−A>z ≥ 0


Karush-Kuhn-Tucker and duality

We need the following property:

Definition

Suppose f , g convex and h affine.(g, h) satisfy the constraint qualification if ∃x0 in the interior of X withg(x0) 4 0, h(x0) = 0 such that gj(x0) < 0 for all component functions gjthat are not affine.

Example: (g, h) satisfies constraint qualification if g and h are affine.



Theorem (Karush-Kuhn-Tucker)

If (g, h) satisfies the constraint qualification, then we have strong duality:

Dopt = Popt.

There exist yopt < 0 and zopt, such that Dopt = `(yopt, zopt).Moreover, xopt is an optimal solution of the primal optimization problemand (yopt, zopt) is an optimal solution of the dual optimization problem, ifand only if

1 g(xopt) 4 0, h(xopt) = 0,

2 yopt < 0 and xopt minimizes L(x, yopt, zopt) over all x ∈ X and

3 〈yopt, g(xopt)〉 = 0.



Remarks:

• Very general result, strong tool in convex optimization

• Dual problem simpler to solve, (yopt, zopt) called Kuhn Tucker point.

• The triple (xopt, yopt, zopt) exist if and only if it defines a saddle pointof the Lagrangian L in that

L(xopt, y, z) ≤ L(xopt, yopt, zopt)︸︷︷︸=Popt=Dopt

≤ L(x, yopt, zopt)

for all x, y < 0 and z.


Outline

2 What to expect?




6 A design example


Linear Matrix Inequalities

Definition

A linear matrix inequality (LMI) is an expression

F (x) = F0 + x1F1 + . . .+ xnFn ≺ 0

where

• x = col(x1, . . . , xn) is a vector of real decision variables,

• Fi = F>i are real symmetric matrices and

• ≺ 0 means negative definite, i.e.,

F (x) ≺ 0 ⇔ u>F (x)u < 0 for all u 6= 0

⇔ all eigenvalues of F (x) are negative

⇔ λmax (F (x)) < 0

• F is affine function of decision variables


Simple examples of LMI’s

• 1 + x < 0

• 1 + x1 + 2x2 < 0

•(

1 00 1

)+ x1

(2 −1−1 2

)+ x2

(1 00 0

)≺ 0.

All the same with � 0, 4 0 and < 0.

Only very simple cases can be treated analytically.

Need to resort to numerical techniques!


Main LMI problems

1 LMI feasibility problem: Test whether there exists x1, . . . , xn suchthat F (x) ≺ 0.

2 LMI optimization problem: Minimize f(x) over all x for which theLMI F (x) ≺ 0 is satisfied.

How is this solved?F (x) ≺ 0 is feasible iff minx λmax(F (x)) < 0 and therefore involvesminimizing the function

x 7→ λmax (F (x))

• Possible because this function is convex!

• There exist efficient algorithms (Interior point, ellipsoid).


Linear Matrix Inequalities

Definition

(More general:) A linear matrix inequality is an inequality

F (X) ≺ 0

where F is an affine function mapping a finite dimensional vector spaceX to the set H of Hermitian matrices.

• Allows defining matrix valued LMI’s.

• F affine means F (X) = F0 + T (X) with T a linear map (a matrix).

• With Ej basis of X , any X ∈ X can be expanded as X =∑n

j=1 xjEj

so that

F (X) = F0 + T (X) = F0 +

n∑j=1

xjFj with Fj = T (Ej)

which is the standard form.


Why are LMI’s interesting?

• Reason 1: LMI’s define convex constraints on x, i.e.,S := {x | F (x) ≺ 0} is convex.

Indeed, F (αx1 + (1− α)x2) = αF (x1) + (1− α)F (x2) ≺ 0.• Reason 2: Solution set of multiple LMI’s

F1(x) ≺ 0, . . . , Fk(x) ≺ 0

is convex and representable as one single LMI

F (x) =

F1(x) 0 . . . 0

0. . . 0

0 . . . 0 Fk(x)

≺ 0

Allows to combine LMI’s!• Reason 3: Incorporate affine constraints such asF (x) ≺ 0 and Ax = bF (x) ≺ 0 and x = Ay + b for some yF (x) ≺ 0 and x ∈ S with S an affine set.



Reason 3: Affinely constrained LMI is LMI

Combine LMI constraint F (x) ≺ 0 and x ∈ S with S affine:

• Write S = x0 +M with M a linear subspace of dimension k

• Let {ej}kj=1 be a basis for M• Write F (x) = F0 + T (x) with T linear

• Then for any x ∈ S:

F (x) = F0 + T

x0 +k∑

j=1

xjej

= F0 + T (x0)︸︷︷︸constant

+k∑

j=1

xjT (ej)︸︷︷︸linear

= G0 + x1G1 + . . .+ xkGk = G(x)

Result: x unconstrained and x has lower dimension than x !!



Reason 4: conversion nonlinear constraints to linear ones

Theorem (Schur complement)

Let F be an affine function with

F (x) =

(F11(x) F12(x)F21(x) F22(x)

), F11(x) is square.

Then

F (x) ≺ 0 ⇐⇒

{F11(x) ≺ 0

F22(x)− F21(x) [F11(x)]−1 F12(x) ≺ 0.

⇐⇒

{F22(x) ≺ 0

F11(x)− F12(x) [F22(x)]−1 F21(x) ≺ 0.


First examples in control

Example 1: Stability

Consider autonomous system

x = Ax.

Verify stability through feasibility. We have:

x = Ax asymptotically stable ⇐⇒(−X 0

0 A>X +XA

)≺ 0 feasible

Here X = X> defines a Lyapuov function

V (x) := x>Xx

for the flow x = Ax.



Example 2: Joint stabilization

Given (A1, B1), . . . , (Ak, Bk), find F such that(A1 +B1F ), . . . , (Ak +BkF ) are asymptotically stable.

Equivalent to finding F ,X1, . . . , Xk such that for j = 1, . . . , k:(−Xj 0

0 (Aj +BjF )Xj +Xj(Aj +BjF )>

)≺ 0 not an LMI!!

Sufficient condition: X = X1 = . . . = Xk, K = FX, yields(−X 0

0 AjX +XA>j +BjK +K>B>j

)≺ 0 an LMI!!

Set feedback F = KX−1.



Example 3: Eigenvalue problemGiven F : V→ S affine, minimize over all x

f(x) = λmax (F (x)) .

Observe that, with γ > 0, and using Schur complement:

λmax(F>(x)F (x)) < γ2 ⇔ 1

γF>(x)F (x)−γI ≺ 0⇔

(−γI F>(x)F (x) −γI

)≺ 0

We can define

y :=

(xγ

); G(y) :=

(−γI F>(x)F (x) −γI

); g(y) := γ

then G is affine in y and minx f(x) = minG(y)≺0 g(y). This is a LP!


Outline

2 What to expect?




6 A design example


Truss topology design

0 50 100 150 200 250 300 350 400

0

50

100

150

200

250


Trusses

• Trusses consist of straight members (‘bars’) connected at joints.

• One distinguishes free and fixed joints.

• Connections at the joints can rotate.

• The loads (or the weights) are assumed to be applied at the freejoints.

• This implies that all internal forces are directed along the members,(so no bending forces occur).

• Construction reacts based on principle of statics: the sum of theforces in any direction, or the moments of the forces about any joint,are zero.

• This results in a displacement of the joints and a new tensiondistribution in the truss.

Many applications (roofs, cranes, bridges, space structures, . . . ) !!

Design your own bridge



Problem features:

• Connect nodes by N bars of length ` = col(`1, . . . , `N ) (fixed) andcross sections s = col(s1, . . . , sN ) (to be designed)

• Impose bounds on cross sections ak ≤ sk ≤ bk and total volume`>s ≤ v (and hence an upperbound on total weight of the truss).Let a = col(a1, . . . , aN ) and b = col(b1, . . . , bN ).

• Distinguish fixed and free nodes.

• Apply external forces f = col(f1, . . . fM ) to some free nodes. Theseresult in a node displacements d = col(d1, . . . , dM ).

Mechanical model defines relation A(s)d = f where A(s) < 0 is thestiffness matrix which depends linearly on s.

Goal:

Maximize stiffness or, equivalently, minimize elastic energy f>d



Problem

Find s ∈ RN which minimizes elastic energy f>d subject to the constraints

A(s) � 0, A(s)d = f, a ≤ s ≤ b, `>s ≤ v

• Data: Total volume v > 0, node forces f , bounds a, b, lengths ` andsymmetric matrices A1, . . . , AN that define the linear stiffness matrixA(s) = s1A1 + . . .+ sNAN .

• Decision variables: Cross sections s and displacements d (bothvectors).

• Cost function: stored elastic energy d 7→ f>d.

• Constraints:• Semi-definite constraint: A(s) � 0• Non-linear equality constraint: A(s)d = f• Linear inequality constraints: a ≤ s ≤ b and `>s ≤ v.


http://www.jhu.edu/~virtlab/bridge/truss.htm

From truss topology design to LMI’s

• First eliminate affine equality constraint A(s)d = f :

minimize f> (A(s))−1 fsubject to A(s) � 0, `>s ≤ v, a ≤ s ≤ b

• Push objective to constraints with auxiliary variable γ:

minimize γ

subject to γ > f> (A(s))−1 f , A(s) � 0, `>s ≤ v, a ≤ s ≤ b• Apply Schur lemma to linearize

minimize γ

subject to

(γ f>

f A(s)

)� 0, `>s ≤ v, a ≤ s ≤ b

Note that the latter is an LMI optimization problem as all constraints on sare formulated as LMI’s!!


Yalmip coding for LMI optimization problem

Equivalent LMI optimization problem:

minimize γ

subject to

(γ f>

f A(s)

)� 0, `>s ≤ v, a ≤ s ≤ b

The following YALMIP code solves this problem:

gamma=sdpvar(1,1); x=sdpvar(N,1,’full’);

lmi=set([gamma f’; f A*diag(x)*A’]);

lmi=lmi+set(l’*x<=v);

lmi=lmi+set(a<=x<=b);

options=sdpsettings(’solver’,’csdp’);

solvesdp(lmi,gamma,options);

s=double(x);


Result: optimal truss

0 50 100 150 200 250 300 350 400

0

50

100

150

200

250


Useful software:

General purpose MATLAB interface Yalmip

• Free code developed by J. Lofberg accessible here

Get Yalmip now

Run yalmipdemo.m for a comprehensive introduction.Run yalmiptest.m to test settings.

• Yalmip uses the usual Matlab syntax to define optimization problems.Basic commands sdpvar, set, sdpsettings and solvesdp.Truely easy to use!!!

• Yalmip needs to be connected to solver for semi-definite programming.There exist many solvers:

SeDuMi PENOPT OOQPDSDP CSDP MOSEK

• Alternative Matlab’s LMI toolbox for dedicated control applications.


http://control.ee.ethz.ch/~joloef/wiki/pmwiki.php

http://sedumi.ie.lehigh.edu/

http://www.penopt.com

http://www.cs.wisc.edu/~swright/ooqp/

http://www-unix.mcs.anl.gov/DSDP/

http://www.nmt.edu/~borchers/csdp.html

http://www.mosek.com

Gallery

Joseph-Louis Lagrange (1736) Aleksandr Mikhailovich Lyapunov (1857)

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