Background for Search Methods
An approach for finding the minimum of in
a given interval is to evaluate the function many times and search for a local
minimum. To reduce the number of function evaluations it is important to have a
good strategy for determining where is
to be evaluated. Two efficient bracketing methods are the
golden ratio
and Fibonacci
searches. To use either bracketing method for finding the minimum of , a
special condition must be met to ensure that there is a proper minimum in the
given interval.
Definition (Unimodal Function) The
function is
unimodal on , if
there exists a unique number
such
that
is
decreasing on ,
and
is
increasing on .
Minimization Using Derivatives
Suppose that is
unimodal over and
has a unique minimum at . Also, assume
that is
defined at all points in . Let
the starting value lie
in . If
, then
the minimum point p lies
to the right of . If
, then
the minimum point p lies
to the left of .
Background for Bracketing the
Minimum
Our first task is to obtain three test values,
(1) ,
so that
(2) .
Suppose that ; then and
the step size h should
be chosen positive. It is an easy task to find a value of h so
that the three points in (1) satisfy (2). Start with in
formula (1) (provided that ); if
not, take , and
so on.
Case (i) If
(2) is satisfied we are done.
Case (ii) If ,
then .
We need to check points that lie farther to the right. Double the step
size and repeat the process.
Case (iii) If ,
we have jumped over p and h is
too large.
We need to check values closer to . Reduce
the step size by a factor of and
repeat the process.
When , the
step size h should
be chosen negative and then cases similar to (i), (ii), and (iii) can be used.
Quadratic Approximation to Find p
Finally, we have three points (1) that satisfy (2). We will use quadratic
interpolation to find , which
is an approximation to p. The
Lagrange polynomial based on the nodes in (1) is
(3) ,
where .
The derivative of is
(4) .
Solving in
the form yields
(5) .
Multiply each term in (5) by and
collect terms involving :
This last quantity is easily solved for :
.
The value is
a better approximation to p than . Hence
we can replace with and
repeat the two processes outlined above to determine a new h and
a new . Continue
the iteration until the desired accuracy is achieved. In this algorithm the
derivative of the objective function was
used implicitly in (4) to locate the minimum of the interpolatory
quadratic. The reader should note that the subroutine makes no explicit use of
the derivative.
Cubic Approximation to Find p
We now consider an approach that utilizes functional evaluations of both and . An
alternative approach that uses both functional and derivative evaluations
explicitly is to find the minimum of a third-degree polynomial that interpolates
the objective function at
two points. Assume that is
unimodal and differentiable on , and
has a unique minimum at . Let . Any
good step size h can
be used to start the iteration. The
Mean Value Theorem
could be used to obtain and
if was
just to the right of the minimum, then the slope
might be twice
which would mean that
we
do not know how much further to the right
lies, so we can imagine that
is close to
and estimate h
with the formula:
.
Thus . The
cubic approximating polynomial
is expanded in a Taylor series about
(which is the abscissa of the minimum). At the minimum
we have ,
and we write
in the form:
(6) ,
and
(7) .
The introduction of in
the denominators of (6) and (7) will make further calculations less
tiresome. It is required that , , , and . To
find
we define:
(8) ,
and we must go through several intermediate calculations before we end up with
.
Use use (6) to obtain
Then use (8) to get
Then substitute
and we have
(9)
Use use (7) to obtain
Then use (8) to get
Then substitute
and we have
(10)
Finally, use (7) and write
Then use (8) to get
(11)
Now we will use the three nonlinear equations (9),
10), (11) listed below in (12). The order of determining the variables will
be (the
variable
will be eliminated).
(12)
First, we will find which
is accomplished by combining the equation in (12) as follows:
Straightforward simplification yields , therefore is
given by
(13) .
Second, we will eliminate by
combining the equation in (12) as follows, multiply the first equation by and
add it to the third equation
which can be rearranged in the form
Now the quadratic equation can be used to solve for
It will take a bit of effort to simplify this equation into its
computationally preferred form.
Hence,
(14)
Therefore, the value of is
found by substituting the calculated value of in
(14) into the formula . To
continue the iteration process, let and
replace
and
with
and
,
respectively, in formulas (12), (13), and (14). The algorithm outlined above is
not
a bracketing method. Thus determining stopping criteria becomes more
problematic. One technique would be to require that
, since .
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