the convergence rate of the penalty function method - Science Direct

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Feb 17, 1970 - without estimating the rate of convergence. One of the few exceptions is [3] in which the closeness of the solution of the problem with a penalty ...
THE CONVERGENCE

RATE OF THE PENALTY FUNCTION METHOD* B.T.POLYAK Moscow

(Receiwd

17 February

1970)

THE closeness of the solution of the problem with a penalty the original problem is estimated at a conditional minimum. match the accuracy penalty

function

of the discovery

with the value

The idea of the penalty conditional

extremum

of the minimum in the problem uith

of the coefficient

function

to problems

to the solution of It is shown how to

of the penalty

method - the reduction

without

constraints

a

function.

of problems

by the introduction

at a of a

penalty on the infringement of constraints - has been known for a long time and has been discussed by many authors. 4t the present time there are numerous papers devoted to various modifications of the method, their applications to particular

problems

etc, (references

found in the book Ill,

specially

or in [21, section

the method is far from complete,

12).

since

devoted However,

the investigation

the proof of its convergence, without estimating the few exceptions is [3] in which the closeness with a penalty

to the solution

\31 only the convex to non-linear

case

constraints

was there obtained

of the original

is considered

with respect

problem

type),

to a functional

with respect to the actual variables. following. In the practical realization

the theoretical is usually

study of confined

to

the rate of convergence. One of of the solution of the problem

(that is, these

of the equality

to this method can be

is estimated. results

However,

in

are not applicable

and also the estimate and to the constraints,

of closeness but not

Another important question left open is the of the penalty function method the discovery

of the exact solution of each problem at an unconditional minimum is impossible. How must the accuracy of the solution of this auxiliary problem be chosen, in order that, on the one hand, the computing

time is not increased

too greatly,

and

on the other, the convergence of the method is not destroyed? Finally, it is known that in the penalty function method we obtain approximations for the dual

* Zh.

uychisl.

Mat.

mat.

fiz.,

11, 1, 3-11,

1971.

B. T. Polyak

2

variables (Lagrange multipliers) simultaneously. approximations has not been investigated.

But the accuracy of these

In the present paper an answer to these questions is given for the case of constraints given by equations and the simplest form of penalty. 1. Convergence

of the penalty

function

method

We consider the problem of the minimum of the functional f(x) in the Hilbert space E, subject to the constraints g(x) = 0, where g(x) is a non-linear operator from E, into some Hilbert space E,. Instead of this the minimum of the functional

over the whole of E, is sought for some K, > 0. Let the following assumptions be satisfied. A. There exists a local point of minimum x* in the original problem, that is, g(x*)

= 0,

f(Y)

< f(x)

if

2(X) = 0,

lb - 5’11
0,

for all

that is, CE, = E,. In

y EI Ez.

When these assumptions have been made the Lagrange multiplier rule (141, p. 490), is applicable, that is, there exists a y* E E, such that

_&(x’, y’) = f/(x’)

D. The linear

L(x7Y) = /!x) + (!I?g(x)).

+ C’y* = 0,

self-adjoint

operator

A .= L,.

(x*, y*) is positive

definite,

that is.

(AS, cig=(f”(X*)Z, for all

It must be mentioned conditions for a minimum gence of methods Theorem

2) + (f,

g”(F)

(5,5))

< nllsll”,

m > 0,

f E Et.

that these assumptions are close to the sufficient at the point x* and are usual in the study of the conver-

of solving

problems

at a conditional

extremum

[51.

1

Subject

to conditions

.4 - D for sufficiently

(a) there exists an xi which is the unique an c-neighbourhood of x*,

great Ki > 0: point of local minimum of pi

in

(b) IIs, - X*II < CillY*Il/ 2Ki, (c) llK,g(si) - f/‘ll < CiIIY’II/ Xi(Here and below the constants Proof.

We introduce

cj are independent

the new variables %= x -x1,

It then follows

of the Ki).

rl = t&g(z)

-

y’.

from B that

P(s) = f’W) + P’(x*)%+ ri(%)* where rj (0) = 0, the rj (t) are differentiable

g’(s) = g’(4 + s”W% + Q(E)* for 1I$\ 6 c, j = 1, 2, and

\I?-,‘(%) 11 = Ilf”(Z) - p’(s*) II < JLIIEII, jirz’(%)II = Ilg”(s) - g”(x*) II G Ll1511.

B. T. Polyak

4

Hence,

using Q’(S)

the definitions =f’(Z)

+

of T], y* and A, we obtain Kig’(z)*g(2;)

=

f’(Y)

fyF)g +

+

+ k’@*) + g”(z*) g + rz (5)) * (11 + if)= r3Grl)

On

=h(E)

+r2(E)*rl

+h(E)*y*

+

ri(E) +

& + C*r + r3( t, q)* (g”(z*)t)*q.

the other hand, rl+y -

g(x) = g(c*)+g’(4E+r4(EJ

=

Kt

= Ct + rb(E),

where r,(O)= 0, h’(E) Ii = llg’(z) - g’(q II d CillEll, c2= llg”@*) I/ + Therefore, the solution of the equation 4; (x) = 0 is equivalent to the solution

LiS.

(for c, 7) of the system

We write this system

in the “matrix”

BZ

form:

=$d+r(z),

(2)

1

where z, d, r(z) E E = E, X E,, B: E-+E,

d=(;*)

z= (i)’

B=

From the properties

(

- rk(E)

of rj((,

(L

We now prove two lemmas

)

*

7) cited above,

r(0) =O, =



-r3(E9d

r(z) =

c3”

lIzlIE = ll~llE,2+ !lqllEt2,

lb-‘(z)II < +

Lne

$-

we obtain

that

c3ii41,

-&?lly’ll+ llg”(X*) II)” +

on the properties

of equation

(2).

Lemma 1 The operator

B introduced

above has an inverse

and

2c22.

IIB-‘II < Cl.

Proof

For an arbitrary

a E

E we consider

the equation

Bz = a or, in greater

(3)

detail,

1

cE-Fq=az.

Ag+C'q=a,,

1

Since A-’ exists, we obtain

choosing

: from the first equation

CA-V

We consider the operator We have, using

Hence D” exists

= ++Z)q t D = CA-Y*

CA-'ai

and substituting

in the second,

- a2.

4)

+ (I/ Kg) Z 0 is the unit operator).

C and D,

in 1lD-‘jl sg (p” / 11 AlI + I/

Ki) -’ < 11 AlI / Q”. From (4) we

then obtain

Hence

llgll= II- A-‘C’q -I- A%ll IICII c$=-c&+;. m

1

< y

llqll + 1

Ilaill Q dall,

6

B. T. Polyak

Therefore, the solution of (3) exists and

+ llyll” < Ci211~l12, llzllz= 11511”

ci2 = c‘* + cs2.

But this also means that B” exists and 1lB-‘j 14 c,. Lemma 2

Let B be a linear operator from E into E, possessing an inverse, and IIR”I I\( c,; let r k) be a linear operator from E into E such that r(0) = 0, I/r’(z) II < c,llzll. Then for every a, IjaIl < 1/ 4ci8cJ, the equation

BZ = a + r(z) has in the sphere llzll (

4c,Ilall

a unique solution z*, where

ll2*ll < $ Ibll. Proof. Beginning with z” = 0 we will construct a sequence of approximations by the formula pt1 = B-‘a + B-9@“).

Let it have already been proved that ~1~“~~ < 2c,((a/ since r (0) = 0, we have

(this is true for n = 0). Then

llz”+‘lI Q ll~-‘all+ll~-‘~(z”) II d c,Ilall+ c,csllznl12 < < c; llnlj+ 4c,“cJnll* < ClMl + (4c,3c3/4c,2c3) llall = 2c,Ilull. Therefore,

for all n we have ilz;ll < 2ciIlall.

Moreoever,

IlZ“+’ - znll= IjB-‘(r(z”) - ?‘(z’i-‘)) II < 4 ci sup Ilr’(02” + (1 - O)P’) IIllzn-

z”-‘II G

O