Seron, M.(SLIDES)-2004-The Geometry of Quadratic Programming

8/2/2019 Seron, M.(SLIDES)-2004-The Geometry of Quadratic Programming

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The Geometry of Quadratic Programming

Mara M. Seron

September 2004

Centre for Complex DynamicSystems and Control


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Outline

1 Introduction

2 The Geometry of Quadratic Programming

QP Solution for N = 2

QP Solution for Arbitrary N

3 Solution Using the KKT Optimality Conditions

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Introduction

In the previous lecture we saw how dynamic programming can beused to derive a simple parameterisation of a finite horizon optimal

control problem and its RHC implementation.

Here we will analyse the geometric structure of the finite horizon

optimal control problem when seen as a quadratic

programme [QP].

Using this geometric structure, we will re-derive in the next lecture

the result obtained via dynamic programming.

We start with a simple case of horizon N = 2 and input constraintsand then show how the same ideas extend to more general cases.



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The Finite Horizon Optimal Control Problem

Let the system be given by

xk+1 = Axk + Buk, |uk| , (1)

where xk Rn and > 0 is the input constraint level. Consider the

following fixed horizon optimal control problem

P2(x) : V

2(x) min V2({xk}, {uk}), (2)

subject to:

xk+1 = Axk + Buk for k = 0, 1,

x0 = x,

uk U [, ] for k = 0, 1,

where the objective function in (2) is

V2({xk}, {uk}) 1

2x2

Px2 +1

2

1k=0

xkQxk + u

kRuk

. (3)



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Th A i d Q d i P


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The Associated Quadratic Program

For the above problem, the corresponding QP optimal solution is

QP: u(x) = arg minuR

12uHu + uFx, (6)

where

u =u0u1

, H = R

1

+(KB)

2

KBKB 1

, F = RK

+KBKAKA

, (7)

and R is the square

u

.

Note that the Hessian H is positive definite since R > 0 and

hence R = R + BPB > 0.


Th G t f Q d ti P i


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The Geometry of Quadratic Programming

The QP in (6) has a nice geometric interpretation in the u-space.

Consider the equation

1

2uHu + uFx = c, c constant . (8)

R

u0

u1

u(x)

u

(x)

This defines ellipsoids in R2

centred at u

(x) = H1Fx.

Then the QP in (6) amounts to

finding the smallest ellipsoid that

intersects the boundary of R,

and u(x) is the point of

intersection.



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G t f QP Mi i E lid Di t


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Geometry of QP as Minimum Euclidean Distance

Ru0

u1

u(x)

u

(x)

In the new coordinates, the level

sets are circlesin R2 centred at

u

(x) = H1/2Fx.

Then the QP in (6) amounts tofinding the point in R that is

closestto u

(x) in the Euclidean

distance.

This transformed geometric picture allows us to immediately write

down the solution to the QP for this special case.


Solution Using Geometry


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R

R1

R2

R3R4R5

R6

R7

R8

u0

u1

u(x)

u

(x)

The solution of the

QP is obtained bypartitioning R2 into

nine regions.

The first region is

the polytope R.

Regions R1 to R8are delimited by

lines that are

normal to the faces

of R and pass

through its vertices.




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The optimal constrainedsolution u(x) is determined by the

region in which the optimal unconstrainedsolution u

(x) lies, in

the following way:

R

R1

R3R5

R7

u0

u1

u(x)

u

(x)

First, the solution in R is

u(x) = u

(x);

Next, the solution in regionsR1, R3, R5 and R7 is simply

equal to the vertex that is

contained in the region.

Finally, the solution inregions R2, R4, R6 and R8 is

defined by the orthogonal

projection of u

(x) onto the

faces of R.


Geometric QP Characterisation


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Geometric QP Characterisation

As a result of above procedure, we obtain a characterisation of the

QP solution in an active region Ri as

u(x) = H1/2u(x) if u

(x) Ri,

where u(x) is an affine function of x and u

(x) = H1/2Fx.

We next proceed to find an algebraic description of a particular

region and an expression for u

(x) in it.


Region Calculation


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Region Calculation

R

R8

n1

n4

f4

u0

u1

v1v4 u(x)

u

(x)

Consider the active region R8associated with face f4.

It is delimited by face f4 and its normals

n1 and n4 passing through the vertices

v1 and v4.

The line that contains face f4corresponds to the second control u1equal to the saturation level , that is

u1 = [0 1]u = in the original

u-coordinates.

In the new u-coordinates (9), it is then

defined by the equation

f4 : [0 1]H1/2

u = .


Region Calculation


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Region Calculation

R

R8

n1n4

f4

u0

u1

v1

v4 u

(x)

u

(x)

Hence, face f4 is:

f4 : [0 1]H1/2

u = .

The normal lines n1 and n2 are:

n1 : [1 0]H1/2u = H1[1 1].

n4 : [1 0]H1/2u = H1[1 1]

.

Combining the above three equations, region R8 is

R8 :

[0 1]H1/2u

H1[1 1] [1 0]H1/2u H1[1 1]


Solution Calculation


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R

R8

n1

n4

f4

u0

u1

v1v4 u(x)

u

(x)

The optimal constrained solution in R8is given by the normal projection of the

unconstrained solution u

(x) onto

face f4.

That is, u(x) satisfies the equation offace f4 and that of the normal to it

passing through u

(x):

[0 1]H1/2u(x) = ,

[1 0]H1/2u(x) = [1 0]H1/2u

(x).




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Hence, u(x) satisfies:

[0 1]H1/2u(x) = ,

[1 0]H1/2u(x) = [1 0]H1/2u

(x).

Using u = H1/2u = H1/2[u0

(x) u1

(x)] and

u

(x) = H1

/2

Fx in the above equations, and solving for the firstcomponent u

0(x) yields

u0

(x) = Gx h. (10)

where

G = K + KBKA1 + (KB)2

h = KB1 + (KB)2

.

Note that the above matrices coincide with the previously derived

using Dynamic Programming.


Recapitulation


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Recapitulation

The solution in active region R8 is

u(x) =

Gx h

if

R8 :

[0 1]u

(x)

H1[1 1] [1 0]Hu

(x) H1[1 1]

.

where u

(x) = H1Fx.

A similar characterisation can be derived in the remaining regions.


Geometric QP Characterisation More General Cases


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Geometric QP Characterisation. More General Cases

We will now consider arbitrary horizon N, m 1 inputs, and more

general constraints, leading to a QP of the form

u(x) = arg min

ux

1

2uHu + uFx, (11)

where u RNm, and where the constraint set

u x, RqNm (12)

is a polyhedron inRNm

.




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Again, it is convenient to work in the u-coordinates u = H1/2u to

turn the QP into a minimum Euclidean distance problem.

Hence the constraint set, R, is the polyhedron given by:

u x, H1/2.

R

Figure: Example of constraint set u x in R3



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Region Calculation


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Region Calculation

Recall the constraint set

u x, RqNm.

A face is determined by a subset of the above q inequality

constraints that are active, that is, they hold as equality constraints:

u = x, (13)

{1, 2, . . . , q} is the set of indices of active constraints and the

notation M indicates selection of rows of M with indices in .

The active region associated with face (13) is

[

]1u [

]1[ x]

s[I

[

]1]u s sx s

[

]1[x ],

(14)

where s is the set of indices of inactive constraints.Centre for Complex DynamicSystems and Control



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Ca a

The optimal constrained solution u(x) in the above active region

is given by the point of the corresponding face that is closest to the

unconstrained solution u (x) = H1/2Fx.

This point is given by

u(x) =

[

]1( x) [I

[

]1]H

1/2Fx.

In the original coordinates u = H1/2u we have

u(x) = H1/2

[

]1( x)

H1/2[I [

]1]H

1/2Fx,

Gx + h.


Recapitulation


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p

The solution in an active region is

u(x) =

[

]1( x) [I

[

]1]H

1/2Fx

(15)

H1/2(Gx + h).

if

[

]1 u

(x) [

]1[ x]

s[I

[

]1] u

(x) [s + s

[

]1]x

+s + s[]1,(16)

where u

(x) = H1/2Fx and and s are the indices of active and

inactive constraints, respectively.


Solution Using the KKT Optimality Conditions


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g p y

The geometric structure of QP has allowed us to derive its solution

in an active region.

The same solution can be derived using the KKT optimality

conditions, as we show next.


The KKT Optimality Conditions


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p y

Consider the following nonlinear programming problem:

minimise f(z),

subject to:g(z) 0,

where f : Rr R, g = [g1, . . . , gq] : Rr Rq are differentiable.

Let z be a feasible solution, and let = {i : gi(z) = 0}.

Suppose that gi(z) for i are linearly independent and let

g [g1

, . . . , gq].

If z is a local optimal solution, then there exist Lagrange multipliers

= [1, . . . , q] such that the following KKT conditionshold:

f(z) + g(z) = 0,

g(z) = 0,

0.

(KKT)


KKT Conditions for QP


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For the QP problem

minimise1

2 zHz + zc,

subject to:

Lz W,

the KKT conditions (KKT) are:

Primal Feasibility (PF): L z W,

Dual Feasibility (DF): Hz + c + L = 0,

0,

Complementary Slackness (CS):

(L z W) = 0.(17)

If H is positive definite, the KKT conditions are necessary and

sufficient for z to be the unique global minimum.


QP Characterisation Using KKT Conditions


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g

Recall the QP (11) used in MPC:

In the original coordinates

minimise1

2uHu + uFx,

subject to:

u x.

In the u-coordinates u = H1/2u

minimise1

2uu + u

H1/2Fx,

subject to:

u x.

When solved for different values of x, the above problem is

referred to as a multiparametric quadratic program[mp-QP].

An mp-QP is a QP in which the linear term in the objective function

and the right hand side of the constraints depend linearly on a

vector of parameters (the state vector x in this case).




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Let u(x) be the optimal solution of the above problem. The KKT

conditions (17) then are

u

(x) + x , (PF)u(x) + H1/2Fx + = 0, (DF1)

0, (DF2)

[u(x) + x] = 0. (CS)

Consider now an active set of indices for u(x), and let s be thecorresponding inactive set. We then have that u(x) satisfies the

equality constraints (13), that is,

u(x) = x. (18)

From (18), (CS) and (DF2), we have that

s = 0, 0.




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Using s = 0, 0 in the (first) dual feasibility condition (DF1)

and solving foru

(x) yields

u(x) = H1/2Fx . (19)

Using (19) in the active constraint equality (18) and solving for gives

= [

]1[H

1/2Fx + x ]. (20)

Substituting the above equation into (19) we obtain

u(x) = []1( x) [I []1]H1/2Fx, (21)

which is identical to (15).




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The inequalities (16) that define the region in the state space

where the solution (21) is valid can be recovered as follows:

Combining the expression for the Lagrange multipliers (20)

= [

]1[H

1/2Fx + x ].

and the sign condition 0 yields the first inequality in (16) .

Finally, the second inequality in (16) follows from primal feasibility

(see (PF)) of the inactive constraints, that is,

su

(x) + sx s,

upon substitution of the expression (21) for the optimal solution.


Connection Between Geometry and KKT


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In summary, we can see that the geometric characterisation

given before is a particular arrangement of the KKT optimalityconditions.

R

n1n4

R8

f4

u0

u1In particular:

The halfspace above a face (for

example, f4) is given by the

nonnegativity of the Lagrange

multipliers corresponding to active

constraints on the face.

The band between normals to theface (for example, between n1 and

n4) is given by feasibility of the

inactive constraints on the face.


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Seron, M.(SLIDES)-2004-The Geometry of Quadratic Programming