Graphics Programs Reference
In-Depth Information
EXAMPLE 10.1
Use
goldSearch
to find
x
that minimizes
6
x
3
3
x
2
f
(
x
)
=
1
.
+
−
2
x
subject to the constraint
x
≥
0.Compare the result with the analytical solution.
Solution
This is a constrained minimizationproblem. Either the minimum of
f
(
x
) is
a stationary point in
x
≥
0, orit is locatedat the constraint boundary
x
=
0.Wehandle
x
)]
2
.
the constraint with the penalty functionmethod by minimizing
f
(
x
)
+
λ
[min(0
,
1 for the first step size in
goldBracket
(both
choices being rather arbitrary), we arrive at the following program:
Starting at
x
=
1 and choosing
h
=
0
.
% Example 10.1 (golden section minimization)
x=1.0;h=0.1;
[a,b] = goldBracket(@fex10
1,x,h);
[xMin,fMin] = goldSearch(@fex10
_
_
1,a,b)
The function to be minimizedis
functiony=fex10
1(x)
% Function used in Example 10.1.
lam = 1.0; % Penalty function multiplier
c = min(0.0,x); % Constraint penalty equation
y = 1.6*xˆ3 + 3.0*xˆ2 - 2.0*x + lam*cˆ2;
_
The outputfrom the program is
>> xMin =
0.2735
fMin =
-0.2899
Since the minimum wasfound to be a stationary point, the constraint was not
active. Therefore, the penalty functionwassuperfluous, but we did not know that at
the beginning.
The locationsofstationary points areobtained analyticallybysolving
f
(
x
)
8
x
2
=
4
.
+
6
x
−
2
=
0
The positive root of thisequationis
x
=
0
.
273 494.Asthis is the onlypositive root,
≥
there are no other stationary points in
x
0that we must check out. The only other
possible location of a minimum is the constraint boundary
x
=
0.Buthere
f
(0)
=
0
Search WWH ::
Custom Search