presentazione di powerpoint - uniroma1.itiacoviel/materiale/lecture_esempio_2017... · = quantity...

Optimal Control

Prof. Daniela Iacoviello

Examples

Example

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

28/11/2017 Pagina 2

Minimize:

* * *

Solution of the first state equation:

Prof.Daniela Iacoviello- Optimal Control Pagina 3

Obviously:

x2=x’2 +x’’2

intravenous

Transferred from

the gastrointestinal compartment

Since:

Define:

It can be defined the following tracking problem:

0,0,

0,00,

''2

''2

QTTtQtx

QTtQtx

It could be verified that:

28/11/2017 Prof.Daniela Iacoviello- Optimal Control

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:

28/11/2017 Prof.Daniela Iacoviello- Optimal Control Pagina 6

where:

28/11/2017 Pagina 7

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:

28/11/2017 Pagina 9

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:

nd the transverality conditions will be particularized:

1,0)(1)()(1)( **** ttttu

Ctsigntu 112

1)(*


1)(* t

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

28/11/2017 Pagina 13

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 t The transversality condition

yields:

1C

01)1(* C

0)(* tu 0)(* tx Not possible

3) . The transversality condition yields:

0

**

*

T

ft t

Hf

1)(,1 ff txt

From the stationarity of the problem:

4C 312

1)(* tsigntu

01)1(

)1(

**

C

x

T

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

28/11/2017

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:


Prof.Daniela Iacoviello- Optimal Control

As far as the abnormal solution is concerned, for the case

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,)( CCt28/11/2017 Pagina 19

1)( * ftx

1)(* tu ttx )(*1ft

The unique possible solution is the extremum

That is both a normal and abnormal solution.


presentazione di powerpoint - uniroma1.itiacoviel/materiale/lecture_esempio_2017... · = quantity...

Documents