Duality Between Constrained Estimation and Control Jos e De Don a - - PowerPoint PPT Presentation

Duality Between Constrained Estimation and Control

Jos´ e De Don´ a September 2004

Centre of Complex Dynamic Systems and Control

Outline

1

Recap on Lagrangian Duality Primal and Dual Problems Strong Duality Theorem

2

Primal Estimation Problem System Definition Constrained Estimation Problem

3

Lagrangian Dual Problem Lagrangian Dual Function Formulation of the Lagrangian Dual Problem

4

Equivalent Formulation of the Primal Problem Preliminary Results Formulation of the Equivalent Primal Problem

5

Symmetry of Constrained Estimation and Control

6

More General Constraints

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Duality Between Constrained Estimation and Control

We have seen that the problem of constrained estimation can be formulated as a constrained optimisation problem. Indeed, this problem is remarkably similar to the constrained control problem—differing only with respect to the boundary

• conditions. (In control, the initial condition is fixed, whereas in

estimation, the initial condition can also be adjusted.) We will now derive the Lagrangian dual of a constrained estimation problem and show that it leads to a particular unconstrained nonlinear optimal control problem. We will then show that the original (primal) constrained estimation problem has an equivalent formulation as an unconstrained nonlinear optimisation problem, exposing a clear symmetry with its dual.

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Recap on Lagrangian Duality

Consider the following nonlinear programming problem: Primal Problem P minimise f(x), (1) subject to: gi(x) ≤ 0 for i = 1, . . . , m, hi(x) = 0 for i = 1, . . . , ℓ, x ∈ X.

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Recap on Lagrangian Duality

Then the Lagrangian dual problem is defined as the following nonlinear programming problem. Lagrangian Dual Problem D maximise θ(u, v), (2) subject to: u ≥ 0, where,

θ(u, v) = inf{f(x) +

m

• i=1

uigi(x) +

ℓ

• i=1

vihi(x) : x ∈ X}, (3) is the Lagrangian dual function.

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Recap on Lagrangian Duality

In the dual problem (2)–(3), the vectors u and v have as their components the Lagrange multipliers ui for i = 1, . . . , m, and vi for i = 1, . . . , ℓ. Note that the Lagrange multipliers ui, corresponding to the inequality constraints gi(x) ≤ 0, are restricted to be nonnegative, whereas the Lagrange multipliers vi, corresponding to the equality constraints hi(x) = 0, are unrestricted in sign. Given the primal problem P (1), several Lagrangian dual problems D of the form of (2)–(3) can be devised, depending

• n which constraints are handled as gi(x) ≤ 0 and hi(x) = 0,

and which constraints are handled by the set X. (An appropriate selection of the set X must be made, depending

• n the nature of the problem.)

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Recap on Lagrangian Duality

The following result, known as the strong duality theorem, shows that, under suitable convexity assumptions and under a constraint qualification, there is no duality gap between the primal and dual

• ptimal objective function values.

Theorem (Strong Duality Theorem) Let X be a nonempty convex set in Rn. Let f : Rn → R and g : Rn → Rm be convex, and h : Rn → Rℓ be affine. Suppose that the following constraint qualification is satisfied. There exists an

ˆ

x ∈ X such that g(ˆ x) < 0 and h(ˆ x) = 0, and 0 ∈ int h(X), where h(X) = {h(x) : x ∈ X}. Then, inf{f(x) : x ∈ X, g(x) ≤ 0, h(x) = 0} = sup{θ(u, v) : u ≥ 0}, (4) where θ(u, v) = inf{f(x) + ug(x) + vh(x) : x ∈ X}. Furthermore, if the inf is finite, then sup{θ(u, v) : u ≥ 0} is achieved at (¯ u, ¯ v) with

¯

u ≥ 0. If the inf is achieved at ¯ x, then ¯ ug(¯ x) = 0.

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System Definition

Consider the following system xk+1 = Axk + Bwk for k = 0, · · · , N − 1, yk = Cxk + vk for k = 1, · · · , N, (5) where xk ∈ Rn, wk ∈ Rm, yk ∈ Rp. For clarity of exposition, we will consider the case where only the process noise sequence {wk} is constrained. (The case of general constraints on wk, vk and x0 can be treated as well.)

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System Definition

We thus assume that

{vk} is an i.i.d. sequence having a Gaussian distribution

N(0, R) with covariance R > 0; x0 has a Gaussian distribution N(µ0, P0) with covariance P0 > 0; and,

{wk} is an i.i.d. sequence having a truncated Gaussian

distribution of the form: pw(wk) =

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ βw exp

• − 1

2w kQ−1wk

• βw
• Ω exp
• − 1

2νQ−1ν

• dν

for wk ∈ Ω,

• therwise.

(6) where Q > 0 and βw = (2π)− m

2 (det Q)− 1 2 . Centre of Complex Dynamic Systems and Control

Constrained Estimation Problem

Given the observations {yd

1 , . . . , yd N} and the knowledge of µ0 (the

mean value of x0), we consider the following optimisation problem, which can be shown to yield the joint a posteriori most probable state estimates:

Pe :

V

N (µ0, {yd k }) =

min

ˆ xk,ˆ vk, ˆ wk

VN({ˆ xk}, {ˆ vk}, { ˆ wk}), (7) subject to:

ˆ

xk+1 = A ˆ xk + B ˆ wk for k = 0, . . . , N − 1, (8)

ˆ

vk = yd

k − C ˆ

xk for k = 1, . . . , N, (9)

{ˆ

x0, . . . , ˆ xN, ˆ v1, . . . , ˆ vN, ˆ w0, . . . , ˆ wN−1} ∈ X, (10)

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Constrained Estimation Problem

In problem Pe, the constrain set X is given by: X = Rn × · · · × Rn

• N+1

× Rp × · · · × Rp

• N

× Ω × · · · × Ω

• N

,

(11) and the objective function is given by: VN({ˆ xk}, {ˆ vk}, { ˆ wk}) = 1 2(ˆ x0 − µ0)P−1

0 (ˆ

x0 − µ0)

+ 1

2

N−1

• k=0

ˆ

w

kQ−1 ˆ

wk + 1 2

N

• k=1

ˆ

v

kR−1 ˆ

vk. (12)

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Lagrangian Dual Problem

The Lagrangian dual function θ is given by:

θ{λk}, {uk} =

inf

ˆ wk∈Ω,ˆ xk,ˆ vk

L{ˆ xk}, {ˆ vk}, { ˆ wk}, {λk}, {uk}, (13) where the function L is defined as, L{ˆ xk}, {ˆ vk}, { ˆ wk}, {λk}, {uk} = VN({ˆ xk}, {ˆ vk}, { ˆ wk})

+

N−1

• k=0

λ

k

• ˆ

xk+1 − A ˆ xk − B ˆ wk

• +

N

• k=1

u

k

• yd

k − C ˆ

xk − ˆ vk

• .

(14) In (14), VN is the primal objective function and {λk} and {uk} are the Lagrange multipliers corresponding to the linear equalities.

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Lagrangian Dual Problem

Substituting the expression for the primal objective function VN given in (12) and rearranging terms we obtain: L{ˆ xk}, {ˆ vk}, { ˆ wk}, {λk}, {uk} = 1 2(ˆ x0 − µ0)P−1

0 (ˆ

x0 − µ0) − λ

0A ˆ

x0

+

N

• k=1

1

2 ˆ v

kR−1 ˆ

vk − u

k ˆ

vk + u

kyd k

• +

N−1

• k=0

1

2 ˆ w

kQ−1 ˆ

wk − λ

kB ˆ

wk

• +

N−1

• k=1

(λk−1 − Aλk − Cuk) ˆ

xk

• + (λN−1 − CuN) ˆ

xN. (15)

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Lagrangian Dual Problem

Recall that the Lagrangian dual function θ is given by:

θ{λk}, {uk} =

inf

ˆ wk∈Ω,ˆ xk,ˆ vk

L{ˆ xk}, {ˆ vk}, { ˆ wk}, {λk}, {uk}. We can see in L{ˆ xk}, {ˆ vk}, { ˆ wk}, {λk}, {uk} =

· · · +

N−1

• k=1

(λk−1 − Aλk − Cuk) ˆ

xk

• + (λN−1 − CuN) ˆ

xN, that the infimum is −∞ whenever {λk} and {uk} are such that

λk−1 − Aλk − Cuk 0

for k ∈ {1, · · · , N − 1},

• r,

λN−1 − CuN 0.

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Lagrangian Dual Problem

However, since we will subsequently choose {λk} and {uk} so as to maximise θ({λk}, {uk}) (recall that the Lagrangian Dual Problem D consists of maximising θ({λk}, {uk})), we are only interested in those values of {λk} and {uk} satisfying:

λk−1 − Aλk − Cuk = 0

for k = 1, · · · , N − 1,

λN−1 − CuN = 0.

Introducing an extra variable, λN = 0, for ease of notation, we

• btain:

λN =

0, (16)

λk−1 =

Aλk + Cuk for k = 1, · · · , N. (17)

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Lagrangian Dual Problem

With the above choice of {λk} and {uk}, we are left with the problem:

θ{λk}, {uk} =

inf

ˆ wk∈Ω,ˆ xk,ˆ vk

L{ˆ xk}, {ˆ vk}, { ˆ wk}, {λk}, {uk}. where, L{ˆ xk}, {ˆ vk}, { ˆ wk}, {λk}, {uk} = 1 2(ˆ x0 − µ0)P−1

0 (ˆ

x0 − µ0) − λ

0A ˆ

x0

+

N

• k=1

1

2 ˆ v

kR−1 ˆ

vk − u

k ˆ

vk + u

kyd k

• +

N−1

• k=0

1

2 ˆ w

kQ−1 ˆ

wk − λ

kB ˆ

wk

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Lagrangian Dual Problem

L{ˆ xk}, {ˆ vk}, { ˆ wk}, {λk}, {uk} = 1 2(ˆ x0 − µ0)P−1

0 (ˆ

x0 − µ0) − λ

0A ˆ

x0

+

N

• k=1

1

2 ˆ v

kR−1 ˆ

vk − u

k ˆ

vk + u

kyd k

• +

N−1

• k=0

1

2 ˆ w

kQ−1 ˆ

wk − λ

kB ˆ

wk

• Notice that the first two sets of terms on the right hand side are

convex and involve the unconstrained variables ˆ x0 and {ˆ vk}. Hence, the minimisers can be computed from:

∂L(·) ∂ˆ

x0

=

P−1

• ˆ

x∗

0 − µ0

• − Aλ0 = 0,

∂L(·) ∂ˆ

vk

=

R−1 ˆ v∗

k − uk = 0

for k = 1, · · · , N.

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Lagrangian Dual Problem

From the previous equations, we easily obtain:

ˆ

x∗

=

P0Aλ0 + µ0, (18)

ˆ

v∗

k

=

Ruk for k = 1, · · · , N. (19) Finally, we have to minimise the remaining terms in L: L{ˆ xk}, {ˆ vk}, { ˆ wk}, {λk}, {uk} =

· · · +

N−1

• k=0

1

2 ˆ w

kQ−1 ˆ

wk − λ

kB ˆ

wk

• and the minimisation has to be carried out subject to the

constraints ˆ wk ∈ Ω for k = 0, · · · , N − 1.

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Lagrangian Dual Problem

We next define the variables ζk and s, and the set ˜

Ω: ζk = Bλk,

s = Q−1/2 ˆ wk,

˜ Ω = {z : Q1/2z ∈ Ω},

Note that ˆ wk ∈ Ω ⇔ s = Q−1/2 ˆ wk ∈ ˜

Ω.

Thus, for k = 0, · · · , N − 1, we obtain

ˆ

w∗

k

= arg min

ˆ wk∈Ω

1

2 ˆ w

kQ−1 ˆ

wk − λ

kB ˆ

wk

• =

arg min

ˆ wk∈Ω

1

2 ˆ w

kQ−1/2

• s

Q−1/2 ˆ wk

• s

− λ

kB

• ζ

k

Q1/2 Q−1/2 ˆ wk

• s
• =

Q1/2 arg min

s∈ ˜ Ω,

1

2ss − (ζ

kQ1/2)s

• (20)

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Lagrangian Dual Problem

We can then express the solution of problem (20), for k = 0, · · · , N − 1, as

ˆ

w∗

k

=

Q1/2 arg min

s∈ ˜ Ω,

1

2ss − (ζ

kQ1/2)s

• (21)

=

Q1/2Π ˜

ΩQ1/2ζk

(22) where Π ˜

Ω is the minimum Euclidean distance projection:

Π ˜

Ω : Rm −→ ˜

Ω

s −→ ¯ s = Π ˜

Ωs = arg min z∈ ˜ Ω

z − s.

(23)

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Lagrangian Dual Problem

Finally, we define

¯ ζk = Q−1 ˆ

w∗

k = Q−1/2Π ˜ ΩQ1/2ζk.

Thus, we have:

ˆ

w∗

k

=

Q ¯

ζk

for k = 0, · · · , N − 1, (24)

¯ ζk =

Q−1/2Π ˜

ΩQ1/2ζk

for k = 0, · · · , N − 1, (25)

ζk =

Bλk for k = 0, · · · , N − 1. (26)

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Lagrangian Dual Problem

Summarising, we have: To avoid having the infimum equal to −∞

λN =

(27)

λk−1 =

Aλk + Cuk, k = 1, · · · , N (28)

∂L(·) ∂ˆ

x0

= 0 ∂L(·) ∂ˆ

vk

= 0 ˆ

x∗

=

P0Aλ0 + µ0 (29)

ˆ

v∗

k

=

Ruk, k = 1, · · · , N (30)

arg min

ˆ wk∈Ω

• ·
• ˆ

w∗

k

=

Q ¯

ζk,

k = 0, · · · , N − 1 (31)

¯ ζk =

Q−1/2Π ˜

ΩQ1/2ζk,

k = 0, · · · , N − 1 (32)

ζk =

Bλk, k = 0, · · · , N − 1 (33)

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Lagrangian Dual Problem

The next step is the substitution of (27)–(33) into the function L, which was given by: L{ˆ xk}, {ˆ vk}, { ˆ wk}, {λk}, {uk} = 1 2(ˆ x0 − µ0)P−1

0 (ˆ

x0 − µ0) − λ

0A ˆ

x0

+

N

• k=1

1

2 ˆ v

kR−1 ˆ

vk − u

k ˆ

vk + u

kyd k

• +

N−1

• k=0

1

2 ˆ w

kQ−1 ˆ

wk − λ

kB ˆ

wk

• +

N−1

• k=1

(λk−1 − Aλk − Cuk) ˆ

xk

• + (λN−1 − CuN) ˆ

xN.

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Lagrangian Dual Problem

Performing the substitution, completing the squares and after some algebraic manipulations, we obtain that the Lagrangian dual function is:

θ{λk}, {uk} = −1

2(Aλ0 + P−1

0 µ0)P0(Aλ0 + P−1 0 µ0)

− 1

2

N

• k=1

(uk − R−1yd

k )R(uk − R−1yd k )

−

N−1

• k=0

1

2

¯ ζ

kQ ¯

ζk + (ζk − ¯ ζk)Q ¯ ζk

• − γ,

where γ is the constant term given by

γ = −1

2µ

0P−1 0 µ0 − 1

2

N

• k=1

(yd

k )R−1yd k .

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Lagrangian Dual Problem

Recall that the dual problem is defined as: Lagrangian Dual Problem D maximise θ(u, v), subject to: u ≥ 0, Defining φ = −θ, and using the fact that max θ = − min(−θ)

= − min φ and the optimisers are the same, we can convert the

above maximisation problem into a minimisation problem. Also, it follows from the Strong Duality Theorem that there is no duality gap, that is, the optimal primal objective value is equal to minus the minimum achieved by the dual problem. The above results are summarised in the following theorem.

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Lagrangian Dual Problem

Theorem (Dual Problem) Assume Ω is a nonempty closed convex set. Given the primal constrained fixed horizon estimation problem Pe defined by equations (7)–(12), the Lagrangian dual problem is

De : φ(µ0, {yd

k }) = min λk,uk φ({λk}, {uk}),

(34) subject to:

λk−1 = Aλk + Cuk

for k = 1, · · · , N, (35)

λN = 0,

(36)

ζk = Bλk

for k = 0, · · · , N − 1, (37)

¯ ζk = Q−1/2Π ˜

ΩQ1/2ζk

for k = 0, · · · , N − 1. (38)

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Lagrangian Dual Problem

Theorem (Dual Problem Ctd.) In (34), the objective function is

φ({λk}, {uk}) = 1

2(Aλ0 + P−1

0 µ0)P0(Aλ0 + P−1 0 µ0)

+ 1

2

N

• k=1

(uk − R−1yd

k )R(uk − R−1yd k )

+

N−1

• k=0

1

2

¯ ζ

kQ ¯

ζk + (ζk − ¯ ζk)Q ¯ ζk

• + γ

(39) where γ is the constant term given by

γ = −1

2µ

0P−1 0 µ0 − 1

2

N

• k=1

(yd

k )R−1yd k .

(40)

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Lagrangian Dual Problem

Theorem (Dual Problem Ctd.) In (38), Π ˜

Ω denotes the minimum Euclidean distance projection

• nto ˜

Ω = {z : Q1/2z ∈ Ω}, that is, Π ˜

Ω : Rm −→ ˜

Ω

s −→ ¯ s = Π ˜

Ωs = arg min z∈ ˜ Ω

z − s.

(41) Moreover, there is no duality gap, that is, the minimum achieved in (7) is equal to minus the minimum achieved in (34).

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Lagrangian Dual Problem

We can think of (35)–(37) as the state equations of a system (running in reverse time) with input uk and output ζk. The theorem then shows that the dual of the primal estimation problem of minimisation with constraints on the system inputs (the process noise wk) is an unconstrained optimisation problem using projected outputs ¯

ζk in the objective function.

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Equivalent Primal Problem

We will present now some preliminary results that will facilitate the formulation of the primal estimation problem Pe in an equivalent form P′

e.

First, recall the following result. Theorem (Closest Point Theorem) Let S be a nonempty, closed convex set in Rn and y S. Then, there exists a unique point ¯ x ∈ S with minimum distance from y. Furthermore, ¯ x is the minimising point, or closest point to y, if and

• nly if (y − ¯

x)(x − ¯ x) ≤ 0 for all x ∈ S.

• xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

x ¯ x S y

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Equivalent Primal Problem

A related result can be stated as follows. Lemma (I) Let ˜

Ω ⊂ Rm be a closed convex set with a

nonempty interior. Let s ∈ Rm such that s ˜

Ω.

Then there exists a unique point ¯ s ∈ ˜

Ω with

minimum Euclidean distance from s. Furthermore, s and ¯ s satisfy the inequality

(s − ¯

s)(¯ s − ξ) > 0 (42) for any point ξ in the interior of ˜

Ω.

• s

¯ s ˜ Ω ξ

> π/2

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Equivalent Primal Problem

Lemma (II) Let f : Rm → R be any function and let Ω ⊂ Rm be a closed convex set that contains zero in its interior. Consider the optimisation problem

P′

1 :

min

w V(w),

(43) with V(w) = f( ¯ w) + (w − ¯ w)Q−1 ¯ w, (44)

¯

w = Q1/2Π ˜

ΩQ−1/2w,

(45) where Π ˜

Ω is the minimum Euclidean distance projection onto ˜

Ω

and the set ˜

Ω is defined as ˜ Ω = {z : Q1/2z ∈ Ω}.

Then any solution w∗ to (43)–(45) satisfies w∗ ∈ Ω.

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Proof of Lemma (II)

Suppose, by contradiction, that w∗ Ω. Let,

¯

w∗ = Q1/2Π ˜

ΩQ−1/2w∗.

(46) Notice that ¯ w∗ ∈ Ω since (45), together with the definitions of Π ˜

Ω

and ˜

Ω, defines a projection of Rm onto Ω.

Define, s∗ = Q−1/2w∗,

¯

s∗ = Q−1/2 ¯ w∗. (47) Then, by construction, s∗ and ¯ s∗ satisfy,

¯

s∗ = Π ˜

Ωs∗,

(48) and, in particular, ¯ s∗ ∈ ˜

Ω.

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Proof of Lemma (II) (Ctd.)

Using (46) and (47) in (44)–(45), we obtain, V(w∗) = f( ¯ w∗) + (w∗ − ¯ w∗)Q−1 ¯ w∗ = f(Q1/2¯ s∗) + (s∗ − ¯ s∗)¯ s∗. (49) Also, since ¯ w∗ ∈ Ω, we have ( ¯ w∗) = Q1/2Π ˜

ΩQ−1/2 ¯

w∗ = ¯ w∗. Thus, V( ¯ w∗) = f( ¯ w∗) + ( ¯ w∗ − ¯ w∗)Q−1 ¯ w∗ = f(Q1/2¯ s∗). (50) It is easy to see, from the assumptions on Ω, that ˜

Ω is a closed

convex set and 0 ∈ int ˜

Ω since Q1/2 > 0. From Lemma (I), from

equation (48), the definition of Π ˜

Ω, and noticing that

s∗ = Q−1/2w∗ ˜

Ω, we conclude that (s∗ − ¯

s∗)¯ s∗ > 0. Hence, from (49) and (50), we have V(w∗) − V( ¯ w∗) = (s∗ − ¯ s∗)¯ s∗ > 0. We have thus found a point ¯ w∗ ∈ Ω that yields a strictly lower value for the objective function, which contradicts the fact that w∗ is a solution of (43)–(45). It follows that w∗ must be in Ω.

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Equivalent Problems

We consider two optimisation problems to be equivalent if they both achieve the same optimum and if the optimisers are the same. Corollary Under the conditions of Lemma (II), problem P′

1 defined by

(43)–(45) is equivalent to the following problem

P1 :

min

w∈Ω f(w).

(51) Proof. It was shown in Lemma (II) that any solution to (43)–(45) belongs to Ω, and hence we can perform the minimisation of (44) in Ω without losing global optimal solutions. Since the mapping Q1/2Π ˜

ΩQ−1/2 used in (45) reduces to the identity mapping in Ω,

we conclude that (44) is equal to the objective function in (51) for all w ∈ Ω, and thus the problems are equivalent.

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Restatement of the Primal Estimation Problem

Recall that the primal estimation problem is defined as:

Pe :

V

N (µ0, {yd k }) =

min

ˆ xk,ˆ vk, ˆ wk

VN({ˆ xk}, {ˆ vk}, { ˆ wk}), subject to:

ˆ

xk+1 = A ˆ xk + B ˆ wk for k = 0, . . . , N − 1,

ˆ

vk = yd

k − C ˆ

xk for k = 1, . . . , N,

ˆ

wk ∈ Ω for k = 0, · · · , N − 1, where and the objective function is given by: VN({ˆ xk}, {ˆ vk}, { ˆ wk}) = 1 2(ˆ x0 − µ0)P−1

0 (ˆ

x0 − µ0)

+ 1

2

N−1

• k=0

ˆ

w

kQ−1 ˆ

wk + 1 2

N

• k=1

ˆ

v

kR−1 ˆ

vk.

Centre of Complex Dynamic Systems and Control

Formulation of the Equivalent Primal Problem

Applying the result of the previous corollary to the primal estimation problem we obtain the following theorem. Theorem (Equivalent Primal Formulation) Assume that Ω is a closed convex set that contains zero in its

• interior. Then the primal estimation problem Pe defined by

equations (7)–(12) is equivalent to the following unconstrained

• ptimisation problem:

P′

e :

V

N (µ0, yd k ) =

min

ˆ xk,ˆ vk, ˆ wk

V′

N({ˆ

xk}, {ˆ vk}, { ˆ wk}), (52) subject to:

ˆ

xk+1 = A ˆ xk + B ¯ wk for k = 0, · · · , N − 1, (53)

ˆ

vk = yd

k − C ˆ

xk for k = 1, · · · , N, (54)

¯

wk = Q1/2Π ˜

ΩQ−1/2 ˆ

wk for k = 0, . . . , N − 1. (55)

Centre of Complex Dynamic Systems and Control

Formulation of the Equivalent Primal Problem

Theorem (Equivalent Primal Formulation Ctd.) The objective function in (52) is V′

N({ˆ

xk}, {ˆ vk}, { ˆ wk}) = 1 2(ˆ x0 − µ0)P−1

0 (ˆ

x0 − µ0) + 1 2

N

• k=1

ˆ

v

kR−1 ˆ

vk

+

N−1

• k=0

1

2 ¯ w

kQ−1 ¯

wk + ( ˆ wk − ¯ wk)Q−1 ¯ wk

• .

(56)

• The theorem shows that the primal estimation problem of

minimisation with constraints on the system inputs (the process noise wk) can be transformed into an equivalent unconstrained minimisation problem using projected inputs ¯ wk both in the objective function and in the state equations (53).

Centre of Complex Dynamic Systems and Control

Symmetry of Constrained Estimation and Control

Configuration for the equivalent primal problem P′

e:

− +

yd

k

ˆ

vk

ˆ

wk Q−1/2

¯

wk

Π ˜

Ω

Q1/2

(A, B, C)

C ˆ xk

Configuration for the dual problem De:

ζk =Bλk

Q1/2

¯ ζk Π ˜

Ω

Q−1/2

(A, C, B)

uk R−1yd

k

+ − ˆ

uk

Centre of Complex Dynamic Systems and Control

Symmetry of Constrained Estimation and Control

The previous figures illustrate the primal equivalent problem P′

e

and the dual problem De, respectively. Note from the figures and corresponding equations the symmetry between both problems. Namely, input variables take the role of output variables in the objective function; system matrices are swapped: A −→ AT, B −→ CT, C −→ BT; time is reversed; and, input projections become output projections.

Centre of Complex Dynamic Systems and Control

More General Constraints

The connections for the case of more general constraints wk ∈ Ω1, vk ∈ Ω2 and x0 ∈ Ω3 (not all of them developed in these notes) are summarised in the table below.

Primal Equivalent Primal Dual State equations ˆ xk+1 = A ˆ xk + B ˆ wk ¯ xk+1 = A ¯ xk + B ¯ wk λk−1 = Aλk + Cuk, λ−1 = Aλ0 Output equation ˆ vk = yd

k − C ˆ

xk ¯ vk = yd

k − C ¯

xk ζk = Bλk Input/output connection Input constraints ˆ wk ∈ Ω1 Unconstrained minimisation using the projected input ¯ wk in the objective function. Projected input used in the state equations: ¯ xk+1 = A ¯ xk + B ¯ wk . Unconstrained minimisa- tion using projected output ¯ ζk in the objective function.

Centre of Complex Dynamic Systems and Control

More General Constraints

Connections for the case of more general constraints wk ∈ Ω1, vk ∈ Ω2 and x0 ∈ Ω3 (Ctd.).

Primal Equivalent Primal Dual Output/input connection Output constraints ˆ vk ∈ Ω2 Unconstrained minimisation using the projected output ¯ vk in the objective function. Projected output required to satisfy the

• utput

equation: ¯ vk = yd

k − C ¯

xk . Unconstrained minimisa- tion using the projected input ¯ uk in the objective function. Initial/final state connection Initial state constraints ˆ x0 ∈ Ω3 Unconstrained minimisation using the projected initial state ¯ x0 in the objective function. Projected initial state used as initial state for the state equa- tions. Unconstrained minimisa- tion using the projected terminal state ¯ λ−1 in the

• bjective function.

Centre of Complex Dynamic Systems and Control