Optimal Control Lecture Prof. Daniela Iacoviello Department of Computer, Control, and Management Engineering Antonio Ruberti Sapienza University of Rome
Optimal Control
Lecture
Prof. Daniela Iacoviello
Department of Computer, Control, and Management Engineering Antonio Ruberti
Sapienza University of Rome
21/11/2016 Controllo nei sistemi biologici
Lecture 1
Pagina 2
Prof. Daniela Iacoviello
Department of computer, control and management
Engineering Antonio Ruberti
Office: A219 Via Ariosto 25
http://www.dis.uniroma1.it/
Prof.Daniela Iacoviello- Optimal Control
Examples taken from Bruni et al. 2005
Prof.Daniela Iacoviello- Optimal Control
Example 1 (exercise 3.3)
Prof.Daniela Iacoviello- Optimal Control
Consider a vehicle of mass m=1 moving along a straight street.
Assume null initial position and velocity.
Indicate with u the force acting on the vehicle having the same
direction of the street.
The aim is to transfer the vehicle to the maximum
distance with final null velocity in a fixed time interval [0, tf],
with a fixed
energy:
ft
Kdttu0
2 0)(
Prof.D.Iacoviello
Optimal Control
Indicate with:
x1(t) the position of the vehicle
x2(t) the velocity of the vehicle
Cost Index
ft
f
Kdttu
txxx
tutx
txtx
0
2
221
2
21
0)(
0)(0)0()0(
)()(
)()(
ft
f dttxtxxJ0
21 )()()(
Prof.D.Iacoviello
Optimal Control
The problem is not convex since the isoperimetric constraint is not
linear
only necessary conditions
Define the Hamiltonian in the normal case:
)()()()()()(),,1,,( 22212 tututtxttxuxH
0)(
)()(20
)(1)(
0)(
)()(
)()(
1
2
12
1
2
21
ft
ttu
tt
t
tutx
txtx
Prof.D.Iacoviello
Optimal Control
)()(20
0)(
)(
)(1)(
0)(
)()(
)()(
2
*
1
12
1
2
21
ttu
txt
tt
t
tutx
txtx
T
ff
u doesn’t have discontinuity points
Ctt
t
)(
0)(
2
1
can’t be zero
22
)()( 2 Ctt
tu
Substitute in the state
equation
Prof.D.Iacoviello
Optimal Control
)()(
)()(
2
21
tutx
txtx
22
)()( 2 Ctt
tu
tCt
tx
tCt
tx
24)(
412)(
3
2
23
1
0)(2 ftx
2
ftC
2
3
0
2
2
0
2
2
4844
1)(
f
t
f
f
tt
dtt
tttdttuK
ff
Prof.D.Iacoviello
Optimal Control
2
3
0
2
2
0
2
2
4844
1)(
f
t
f
f
tt
dtt
tttdttuK
ff
KK
tt ff
34
tt
tu
ttt
tx
ttt
tx
f
f
f
22
1)(
44)(
812)(
3
2
23
1
Kttt
txfff
f3224
)(
3
1
Prof.D.Iacoviello
Optimal Control
No abnormal extremum exists.
In fact: define the Hamiltonian in the abnormal case:
)()()()()(),,1,,( 2221 tututtxtuxH
0)(
)()(20
)()(
0)(
)()(
)()(
1
2
12
1
2
21
ft
ttu
tt
t
tutx
txtx
Ct
t
)(
0)(
2
1
0C
2)(
Ctu
Prof.D.Iacoviello
Optimal Control
2)(
Ctu
tC
tx2
)(2
That is not compatible with 0)(2 ftx
Prof.Daniela Iacoviello- Optimal Control
Consider a vehicle of mass m=1 moving along a straight street.
Assume null initial position and velocity.
Indicate with u the force acting on the vehicle having the same
direction of the street.
Assume NOT fixed the final instant.
The aim is to perform an automatic coupling of a
second vehicle moving with a uniform move
(with velocity v) along the same
street,
assuming as cost index:
ft
f dttutuJ
0
2 )(2
1),(
Example 2- same as example 1 with a change
Prof.D.Iacoviello
Optimal Control
The final instant is not fixed, therefore the problem is not convex.
)()()()()(2
1),,1,,( 221
2 tuttxttuuxH
)()(0
)()(
0)(
2
12
1
ttu
tt
t
0)(
)(),(
2
1
vtx
rvttxttx
f
ff
ff
21
212
11
)(
)(
)(
CtCtu
CtCt
Ct
Prof.D.Iacoviello
Optimal Control
Integrate the state equations
with null initial conditions
vt
H
txt
T
ft
T
ff
f1
*
2
1
*
,,)(
)(
tCt
Ctx
tC
tCtx
2
2
12
2
2
3
11
2)(
26)(
From the transversality conditions and taking into account the
stationarity of the problem:
01 )( Hvt f 2212 CvC
Prof.D.Iacoviello
Optimal Control
0)(
)(),(
2
1
vtx
rvttxttx
f
ff
ff
2212 CvC
vtCv
tC
rvtt
Cv
tC
f
f
f
ff
2
2
22
2
2
3
22
4
212
2
3
1
2
29
2
3
23
r
vC
r
vC
v
rt f
Prof.Daniela Iacoviello- Optimal Control
Consider a vehicle of mass m=1 moving along a straight street.
Assume null initial position and velocity.
Indicate with u the force acting on the vehicle having the same
direction of the street.
Assume NOT fixed the final instant.
The aim is to perform an automatic coupling of a second
vehicle moving with a uniform move (with velocity v) along the
same street,
assuming fixed the energy for the control
and minimizing the cost index: ff ttJ )(
Example 3- same as example 2 with a change
0)(
0
2 Kdttu
ft
Prof.D.Iacoviello
Optimal Control
Define the Hamiltonian in the normal case:
)()()()()(1),,1,,( 2221 tututtxtuxH
)()(20
)()(
0)(
)()(
)()(
2
12
1
2
21
ttu
tt
t
tutx
txtx
212
11
)(
)(
CtCt
Ct
)0()0(1 221 uuCvC 22
)( 21 CtCtu
Prof.D.Iacoviello
Optimal Control
tCtC
tx
tCtC
tx
24)(
412)(
22
12
223
11
vtCt
v
C
v
rvttCt
v
C
v
ff
f
ff
244
1
4124
1
2
2
2222
2
22
2
3222
)0()0(1 221 uuCvC
v
C
vC
4
122
1
Prof.D.Iacoviello
Optimal Control
2
2
2
22
222
2
3222
0
2
444
1
124
1
)(
fff
t
tCtC
v
C
v
t
v
C
v
dttuK
f
vtCt
v
C
v
rvttCt
v
C
v
ff
f
ff
244
1
4124
1
2
2
2222
2
22
2
3222
vtCt
v
C
v
ff
244
1 2
2
2222
f
ff
t
rvttC 1282
2
2
2
22
222
2
3222
0
2
444
1
124
1
)(
fff
t
tCtC
v
C
v
t
v
C
v
dttuK
f
32232212324212 fffff Ktrvtrvtrvtrvt
If 1r 85.3ft
26.0
83.1
74.0
2
1
C
C
If 1r 1ft
vtCt
v
C
v
ff
244
1 2
2
2222
f
ff
t
rvttC 1282
The two equations
don’t have solution
1231 KvrAssume
Prof.D.Iacoviello
Optimal Control
Define the Hamiltonian in the abnormal case:
Substituting into the admissibility equations and the isoperimetric
Constraint:
)()()()()(),,1,,( 2221 tututtxtuxH
)0()0( 221 uuCvC v
CC
4
22
1
Ktv
Ct
Ct
v
C
vtC
tv
C
rvttC
tv
C
fff
ff
fff
2
3
32
2
223
42
42
22
2
22
223
2
22
164192
216
448
ft
vC 42
K
vt f
3
4 2
Ktv
Ct
Ct
v
C
vtC
tv
C
rvttC
tv
C
fff
ff
fff
2
3
32
2
223
42
42
22
2
22
223
2
22
164192
216
448
r
vK
9
4 3
Abnormal extremal exist only if
the parameters satisfy this relation.
If no solution exists 1r
If
there exist infinite
number of abnormal
solutions
1r 1ft 1221 CC
1231 KvrFor the choice
Example 4
Consider a simple model of phamacodynamic:
x1= quantity of drug in the gastrintestinal section
x2= quantity of drug in the blood
r=reference value of the quantity of drug in the blood
u1 flow rate of the drug by oral administration
u2 flow rate of the drug by administered intravenously
Assume
Minimize:
* * *
Solution of the state equation:
Obviously:
x2=x’2 +x’’2
intravenous Gastrintestinal
Since:
Define:
It can be defined the following tracking problem:
0,0,
0,00,
''2
''2
QTTtQtx
QTtQtx
It could be verified that:
The problem is convex and the Lagrangian is strictly convex
necessary and sufficient conditions and,
if the solution exists, it is unique
By using the theory of optimal tracking:
where:
where
If the admissibility condition is satisfied then the couple
Is the unique optimal solution of the subproblem
0)(2 tuo
To determine Q>0 it is possible to solve another
quadratic problem:
where are suitable affine functions depending
on the Problem; the cost index is strictly convex:
If a solution exists
it is unique and is given by
the condition:
Exercise
Consider the system:
The aim is to have:
with a bounded control
minimizing
)()( tutx 0)0( x
1,1)( ff ttx 1,0)( tu
ft
f dttxtutuJ0
)()(3,
The final instant is not fixed the problem is not convex
The Hamiltonian is:
)()()()(3)(,1),(),( tuttxtuttutxH
Ctt )(*The costate equation yields
The minimum condition is:
The problem may be analyzed for four different cases:
and the transverality conditions will be particularized:
1,0)(1)()(1)( **** ttttu
Ctsigntu 112
1)(*
1) . Three alternatives are possible:
1Ca) . From Ctsigntu 11
2
1)(*
01 Ct
0)(* tu 0)(* tx Not possible
b) . There exists a discontinuity instant
for the control with
1,2 C 1,0t
Ct 1
2C
Not possible
c) . 2C 1,0,01 tforCt
1)(* tu ttx )(*acceptable
In case 1) we have obtained one normal extremum:
1)(* tu ttx )(*1ft
2,)( CCtt
2) . The transversality condition yields:
1C
01)1(* C
0)(* tu 0)(* tx Not possible
3) . The transversality condition yields:
0** ft
H
1)(,1 ff txt
From the stationarity of the problem:
4C 312
1)(* tsigntu
AS far as the value of is concerned two alternatives are possible:
i) . From
*ft
3,1* ft 312
1)(* tsigntu
ttx )(*
Not possible
1)(* tu
1)( *** ff ttx
ii) 3* ft
with 3)( * ftx Not possible
4) . . The transversality condition yields:
0)( *** Ctt ff
1)(,1 ff txt
0)(3 **** ft
txHf
From 312
1)(* tsigntu
There exists an instant of discontinuity
for the control such that:
** ,01 ff ttt
From the transversality condition: 13)( ** ff ttx
With: 1)( tt
In case 4) we have obtained one normal extremum:
The value of the cost index evaluated at point of case 4) is bigger than the
value of the cost index evaluated at the solution of point found in case 1).
1)(* tu ttx )(*1ft
2,)( CCttThe normal extremum is:
As far as the abnormal solution is concerned, all the solutions with
must be rejected .
For
0)( ** ft
1)( * ftx 0)(* Ct
0)( * ftxNot acceptable:
1ftNot possible with the
condition
1ftIf one obtains
with:
1)(* tu ttx )(*1ft
0,)( CCt
1)(* tu ttx )(*1ft
The unique possible solution is the extremum
That is both a normal and abnormal solution.