04/20/23 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
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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
22
baf
•If then the interval in which the maximum occurs is otherwise it occurs in
22
baf
2222
baf
baf
b
ba,
22
22
,ba
a
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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
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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
a
b
ba
a
b
ba
a
b
Golden Ratio=> ...618.0a
b
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.
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X2Xl X1 Xu
fu
f2f1
fl
],,[ 12 xxxl
],,[ 12 uxxx
)()( 12 xfxf
)()( 12 xfxf
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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
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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=?
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Solution Cont
2000.1)60000.05708.1(2
1560000.0)(
2
151
lul xxxx
To check the stopping criteria the difference between and is calculated to be
ux
lx
97080.060000.05708.1 lu xx
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Solution ContIteration 2
97080.0
2000.1
5708.1
60000.0
2
1
x
x
x
x
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
97080.0
2000.1
60000.0
1
x
x
x
u
l
9000.06000.02000.12
lu xx
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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 1.5714
2 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 0.2293
6 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
0588.10253.1
2
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
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