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8/8/2019 Lecture 4 Maximum Principle
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LECTURE 4
Maximum Principle
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Pontryagin's maximum principle is used inoptimal control theory to find the best possible
control for taking a dynamic system from one
state to another, especially in the presence of
constraints for the state or input controls. One
of these methods is the max principle which
was initially proposed by Pontryagin. It was
formulated by the Russian mathematician LevSemenovich Pontryagin and his students. It has
a general case as the Euler-Lagrangian
equation of the calculus of variations.
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Example: Planners problem Maximization of consumption over time
Notation:
K(t) = capital (only factor of production)
Q(K) =well-behaved output function i.e. Q K)>0, and Q(K)0
C(t) = consumption
I(t) = investments = Q(K) C(t)
H = constant rate ofdepreciation of capital.r = constant rate ofinterest of capital
We are not looking for a single optimal value C*, but for values C(t) that produce an
optimal value for the integral (or aggregate discounted consumption over time).
T
rt dttCe
0
)(
0x
x
K
Q0
2
2
x
x
K
Q
Max
subject to Q = Q(K) with
change in the capital stock
and
HKCQK !y
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The principle states informally that the Hamiltonian
(1-23)
where(t) is a vector of co-state variables of the same dimension as the state
variablesx(t).
must be minimized over the set of all permissiblecontrols U, as seen in the second Euler-Lagrange
equation.
Which is the necessary condition for optimum value.
),,(),,(),,,( tuxftuxLtuxH TPP !
.
),,,(
x
tuxH
x
x
!
y PP
u
tuxH
x
x!
),,,(0
P
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First Pontryagin proved that a necessary conditionfor solving the optimal control problem is that the
control should be chosen so as to optimize theHamiltonian; that is:
(1-24)
In case of closed and bounded control region, the optimalcontrol is found by optimizing w.r.t uin the given control region U, while treating other variables asif they are constants.
Therefore, is the admissible control vector for which
is maximum or (minimum).
]t,[tt,),,,(),,,( f0e UutuxHtuxH PP
),,( txu P
),,,( tux P
),,( txu P
),,,( tux P
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Then he generalized Pontryagin's maximumprinciple which states that the optimal statetrajectoryx*, optimal control u *, andcorresponding Lagrange multiplier vector *must optimize the Hamiltonian Hso that:
(1-25)
Then after he introduced the general theorem: ifu*(.) is an optimal control, then there exist afunction H*(.), is called the co-state, that
satisfies a certain maximization principle.
),),,,(,(),,( ttxuxHtxH PPP !
),,,(min tuxUu
P
!
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Using Pontryagin Principle in solving Optimal ControlProblems can be summarized in the following steps:
Given the plant equation as in equation (1-15)
Given the performance index as in equation (1-14)
Given the control variable constraints
Step 1: Form the Pontryagin function as in equation (1-17)
Step 2: Minimize H w.r.t all admissible control vectors tofind u*
Step 3: Find the Pontryagin function H* by substitute u*in step 1.
Step 4: Solve the state and co-state equations with theboundary conditions given.
Step 5: Substitution of results 4 into step 2 results in the
Optimum Control u*.
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Example: Given the plant equation as
The performance index to be minimized
The control inequality constraints are given by.
Solution:
Step 1. pontryagin function
dtuxJ
tf
t
)(
2
1 2
0
2
!
)()()()()( 2221 tutxtxtxtx !!yy
uxxuxtH 2222122
2
1
2
1),( PPP !u,x,
],[1)(0 ftttfortu e
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Step 2: minimize H subject to inequality constraint which results in
the following equation (whenu is unsaturated)
Thus for the control u is given by
and when we find that the control minimizes H is:
The optimal control strategy is given in the figure below:
If the boundary conditions are given then optimal control u(t) can be found
solving the state and co-state equations.
)()( 2* ttu P!
)()( 2* ttu P!
1)(2 "tP
1)(2 etP
1)(1
1)(1)(
2
2*
"!
tfor
tfortu
P
P
1
1
-1
-1
)(*2 tP
)(* tu
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The Maximum Principle in Nonlinear
Programming Problems
We begin by starting a general form of a
nonlinear programming problem.
y: be an n-component column vector,
a: be an r-component column vector,
b: be an s-component column vector.h, g,w : be given functions.
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We assume functions gand w to be column
vectors with components rand s ,respectively.
We consider the nonlinear programmingproblem:
subject to
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With equality constraints, we can use the
Lagrangian as:
( 4)
where P is an r-component row vector.
The necessary condition fory* to be a
(maximum) solution to be (1) and (2) is that
there exists an r- component row vectorP such
that:
and
Finding out and
0),( !x
xyyL P 0
),(!
xx
P
P
yL
*P
*y
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For Inequality Constraints
(5)
(6)
(7)
(8)
However, the conditions in (8) are new and are
particular to the inequality-constrained
problem.
0!x
x
x
x!
x
x
yy
H
y
L [Q
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Example
Solution. We form the Lagrangian
The necessary conditions (6)-(8) become
(9)
(10)
(11)
028!
!
x
x
Qxx
L
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Case 1:
From (9) we getx= 4, which also satisfies (10).
Hence, this solution, which makes h(4)=16, is a
possible candidate for the maximum solution.
Case 2:
Here from (9) we get Q = - 4, which does not
satisfy the inequality Q u 0 in (11).
From these two cases we conclude that the
optimum solution isx*= 4 and
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Example : solve the problem:
Solution. The Lagrangian is
The necessary conditions are
(12)
(13)
(14)
028 !
!x
x
xx
L
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Case 1: Q = 0
From (12) we obtainx= 4 , which does not
satisfy (13), thus, infeasible solution.
Case 2:x=6
(13) holds. From (12) we get Q = 4, so that (14)
holds. The optimal solution is then
since it is the only solution satisfying thenecessary conditions.
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Example: Find the shortest distance between
the point (2.2) and the point solving the
following :
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The Lagrangian function for this problem is
(15)
The necessary conditions are(16)
(17)
(18)
(19)
(20)
(21)
From (20) we see that eitherQ =0 orx2+y2=1,i.e., we
are on the boundary of the semicircle. IfQ =0, we see
from (16) thatx=2. Butx=2 does not satisfy (18) for
any y , and hence we conclude Q>0andx2
+y2
=1.
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From (21) we conclude that either ory=0. If
, then from (16), (17) and Q >0, we get
x = y. Solving the latter withx2
+y2
=1 , gives; and
;
Ify=0, then (17) is not satisfied the two points areshown in figure. Of the two points found that satisfythe necessary conditions, clearly the point
is the nearest point and solves the closest-
point problem. The other point is in factthe farthest point.
)2
1,
2
1(),( !yx )
2
1,
2
1(),( !yx
249 !h 249
)2
1,
2
1(),( !yx
)2
1,
2
1(),( !yx
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