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
