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Chapter
19
BARRIER FUNCTION METHODS
1
Introduction
Penalty function methods generate a sequen ce of infeasible points
{xk
which
com e closer to the constraints as the iterations proceed. By contrast, barrier
function methods
-
which ca n only be applied to problems with inequalities but
no equality constraints
-
generate points which lie inside the feasible region.
Hence, for the remainder of the chapter, we consider the problem
Minimize F x )
19.1.1)
subject to c i x ) i=
1 ...,
m )
19.1.2)
2
Barrier functions
Definition On e form of barrier function for the problem 19.1. ) , 19.1.2) is
Because the barrier term inc ludes reciprocals of the constraints w e see that
B
is
very much greater than
F
when any c i x ) s near zero .e. wh en
x
is near the
boundary of the feasible region. Similarly,
B =
F when all the q x ) are much
greater than zero and x is in the interior of the feasible region.
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Definition A second (more popular) barrier function for (19.1.1), (19.1.2) is
m
B(x, r)
=
F (x)-
log(cj(x))
(19.2.2)
i
1
When 1 > ci(x)
>
0 then log(ci(x)
< 0.
Hence the second term on the right
of (19.2.2) implies B >>
F
when any of the constraint functions approaches
zero. N ote, however, that (19.2.2) is undefined when any c j(x ) is negative.
There is a relationship, similar to that for penalty functions, between uncon-
strained m inima of B(x , r ) and the solution if (19.1. I) , (19.1.2).
Proposition
Suppose that (19.1. l), (19.1.2) has a unique solu tion x,*
,
A*. Sup-
pose also that
p
is a positive constant and, for all rk
0. Hence (19.2.12) gives one positive and on e negative value for XI But
(19.2.11) means that x2 must have the same sign as xl. However, a solution
with
both
xl and x2 negative cannot satisfy (19.2.8). Therefore the uncon-
strained minimum of B(x ,
r
is at
Hence, as
r
0 we have xl
-+
and x2
.
These values satisfy the
optimality conditions for problem (19.2.7), (19.2.8).
Exercises
1. Deduce the Lag range multiplier fo r the worked exam ple above.
2. U se a log-barrier function approach to solve the problem
Minimize
xl +x2 sub jec t to
x; +x;
3. A log-barrier approach is used to solve the problem
Minimize
Ty
subject to yTQy
5 V,.
Suppose that the barrier parameter r is chosen s o the minimum of B(y,r ) oc-
curs where yTQy= kV,, where k
0 , x2 >
1
using the barrier function
(1
9.2.1).
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Barrier function methods
3 Numerical results with B-SUMT
We use B-SUMT to denote the SAMPO implementation of the barrier SUMT
algorithm using the log-barrier function
(1
9.2.2).
In
B-SUMT
he unconstrained
minim izations are done by QNw or QNp
A safeguard is needed in the line search for the unconstrained minimization
technique used in
B-SUMT.
The log-barrier function is undefined if any of the
constraints ci x)are non-positive and therefore the line search must reject trial
poin ts where this occurs. This can be done within the framework of the Arm ijo
line-search by re-setting
B ( x ,
r ) to a very large value at any point x which vio-
lates one or more of the constraints.
In this section w e consider the minimum-risk and maximum-return problems
in the forms Minrisk4 and Maxret4. Program
sample11
allows us to solve
such problems using
B-SUMT
and we begin with Problem T l
lb.
We note first
that B-SUMT must be started with a feasible point. Hence the automatic initial
guess
y = l l n , i =
1 .
..,n
(suitable for the other two SU M T algorithms) w ill
probably not be appropriate for B-SUMT. n the present case, it is fairly easy to
find a feasib le starting guess. Since we have the expected return fo r each asset
4.4.4),we can set
y3
= 0.95 and
yi
= 0.01, i = 1,2,4,5. This ensures
C y i
< 1
and also that the expected portfolio return exceeds
Rp l
).
The progress m ade
by
B-SUMT
s show n in Table
19.1.
For com parison, this table also summ arises
the behaviour of P-SUMT from the same starting point. In both cases the un-
constrained minimizer was QNw and the initial penalty parameter and rate of
reduction were given by
r = 0.1, P = 0.25.
B-SUMT P-SUMT
Table
19.1.
B-SUMT
and
P-SUMT solutions to problem
Tl
l b
Th e fact that P-SUMT converges much more quickly than B-SUMT s largely due
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to the fact that the minimum of B(x,O. 1 ) is very much further from the opti-
mu m than the m inimum of P (x , 0. 1) . For this problem, therefore, the initial
choice
rl
=0.1 is a bad o ne for the barrier approach.
The solution to problem T I l b obtained with
B-SUMT
s approximately
with
V w
1.083 and R
w
1.044 . Th e solution from
P-SUMT
s
giving V w 1.083 and
R
w 1.026 . These both lie between the two solutions
(17 .5.5) and (17.5.6). Both solutions are both equally valid because they are
feasible points giving the same value of the objective function V. Th e fact that
B-SUMT and P-SUMT erminate at different points seems to be due to the fact
that one w orks inside the feasible region while the other operates outside.
We now apply
B-SUMT
to the maximum-return problem T13b. We need to
find a feasible starting values for the
yi,
which is not so straightforward as
it was for
Minrisk4
However, if we have already solved
Minrisk4
then its
solution will be appropriate as an initial guess for
Maxret4
provided it gives
V