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

Golden Section Search Method

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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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