11/30/2015 1 Secant Method Industrial Engineering Majors Authors: Autar Kaw, Jai Paul .
2/24/2016 1 Golden Section Search Method Major: All Engineering Majors Authors: Autar Kaw, Ali...
-
Upload
lisa-quinn -
Category
Documents
-
view
229 -
download
0
description
Transcript of 2/24/2016 1 Golden Section Search Method Major: All Engineering Majors Authors: Autar Kaw, Ali...
05/06/23 http://nm.mathforcollege.com 1
Golden Section Search Method
Major: All Engineering Majors
Authors: Autar Kaw, Ali Yalcin
http://nm.mathforcollege.comTransforming Numerical Methods Education for STEM
Undergraduates
http://nm.mathforcollege.com3
Equal Interval Search Method
Figure 1 Equal interval search method.
x
f(x)
a b
2
2
(a+b)/2
•Choose an interval [a, b] over which the optima occurs
•Compute and
22baf
•If then the interval in which the maximum occurs is otherwise it occurs in
22baf
2222 bafbaf
bba ,22
22
, baa
http://nm.mathforcollege.com4
Golden Section Search Method The Equal Interval method is inefficient when is small.
The Golden Section Search method divides the search more efficiently closing in on the optima in fewer iterations.
X2Xl X1 Xu
fu
f2 f1
fl
Figure 2. Golden Section Search method
http://nm.mathforcollege.com5
Golden Section Search Method-Selecting the Intermediate Points
a bXl X1 Xu
fu
f1
fl
Determining the first intermediate point
a-b
b
X2
aXl X1 Xu
fu
f2f1
fl
Determining the second intermediate point
ab
baa
b
baab
Golden Ratio=> ...618.0ab
Golden Section Search-Determining the new search
region
If then the new interval is If then the new interval is All that is left to do is to determine the
location of the second intermediate point.
http://nm.mathforcollege.com6
X2Xl X1 Xu
fu
f2f1
fl
],,[ 12 xxxl
],,[ 12 uxxx
)()( 12 xfxf
)()( 12 xfxf
http://nm.mathforcollege.com7
Example
The cross-sectional area A of a gutter with equal base and edge length of 2 is given by
)cos1(sin4 A
05.0
.
Find the angle which maximizes the cross-sectional area of the gutter. Using an initial interval of find the solution after 2 iterations. Use an initial .
]2/,0[
2
2
2
http://nm.mathforcollege.com8
Solution)cos1(sin4)( f
60000.0)5708.1(2
155708.1)(2
15
97080.0)5708.1(2
150)(2
15
2
1
luu
lul
xxxx
xxxx
The function to be maximized is
Iteration 1: Given the values for the boundaries of we can calculate the initial intermediate points as follows:
2/ 0 ul xandx
1654.5)97080.0( f
1227.4)60000.0( f
X2XlX1 Xu
f2f1
Xl=X2X2=X1 Xu
X1=?
http://nm.mathforcollege.com9
Solution Cont2000.1)60000.05708.1(
21560000.0)(
215
1
lul xxxx
To check the stopping criteria the difference between and is calculated to be
uxlx
97080.060000.05708.1 lu xx
http://nm.mathforcollege.com10
Solution ContIteration 2
97080.02000.15708.160000.0
2
1
xxxx
u
l
0791.5)2000.1( f
1654.5)97080.0( f
)()( 21 xfxf
82918.0)6000.02000.1(2
152000.1)(2
152
luu xxxx
XlX2 XuX1
97080.02000.160000.0
1
xxx
u
l
9000.06000.02000.12
lu xx
http://nm.mathforcollege.com11
Theoretical Solution and Convergence
Iteration xl xu x1 x2 f(x1) f(x2) 1 0.0000 1.5714 0.9712 0.6002 5.1657 4.1238 1.57142 0.6002 1.5714 1.2005 0.9712 5.0784 5.1657 0.97123 0.6002 1.2005 0.9712 0.8295 5.1657 4.9426 0.60024 0.8295 1.2005 1.0588 0.9712 5.1955 5.1657 0.37105 0.9712 1.2005 1.1129 1.0588 5.1740 5.1955 0.22936 0.9712 1.1129 1.0588 1.0253 5.1955 5.1937 0.14177 1.0253 1.1129 1.0794 1.0588 5.1908 5.1955 0.08768 1.0253 1.0794 1.0588 1.0460 5.1955 5.1961 0.05419 1.0253 1.0588 1.0460 1.0381 5.1961 5.1957 0.0334
0420.12
0588.10253.12
lu xx 1960.5)0420.1( f
The theoretically optimal solution to the problem happens at exactly 60 degrees which is 1.0472 radians and gives a maximum cross-
sectional area of 5.1962.
Additional ResourcesFor all resources on this topic such as digital audiovisual lectures, primers, textbook chapters, multiple-choice tests, worksheets in MATLAB, MATHEMATICA, MathCad and MAPLE, blogs, related physical problems, please visit
http://nm.mathforcollege.com/topics/opt_golden_section_search.html