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Golden Section Search Method






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Golden Section Search Method. Major: All Engineering Majors Authors: Autar Kaw, Ali Yalcin http://nm.mathforcollege.com Transforming Numerical Methods Education for STEM Undergraduates. Golden Section Search Method http://nm.mathforcollege.com. f(x). x. a. b. Equal Interval Search Method.
Golden Section Search Method

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Slide 1

Golden Section Search Method

Major: All Engineering Majors

Authors: Autar Kaw, Ali Yalcin

http://nm.mathforcollege.com

Transforming Numerical Methods Education for STEM Undergraduates

http://nm.mathforcollege.com

Slide 2

Golden Section Search Methodhttp://nm.mathforcollege.com

Slide 3

f(x)

x

a

b

Equal Interval Search Method

  • Choose an interval [a, b] over which the optima occurs

  • Compute and

  • If

  • then the interval in which the maximum occurs is otherwise it occurs in

(a+b)/2

Figure 1 Equal interval search method.

http://nm.mathforcollege.com

Slide 4

f2

f1

fu

fl

Xl

Xu

X1

X2

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.

Figure 2. Golden Section Search method

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Slide 5

f2

f1

f1

fl

fl

fu

fu

a-b

b

X2

Xl

Xl

Xu

Xu

X1

X1

a

a

b

Golden Section Search Method-Selecting the Intermediate Points

Determining the first intermediate point

Determining the second intermediate point

Golden Ratio=>

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Slide 6

f2

f1

fl

fu

X2

Xl

Xu

X1

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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Slide 7

Example

.

2

2

2

The cross-sectional area A of a gutter with equal base and edge length of 2 is given by

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 .

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Slide 8

f1

f2

X2

X2=X1

X1

Xl=X2

Xl

Xu

Xu

Solution

The function to be maximized is

Iteration 1: Given the values for the boundaries of

we can calculate the initial intermediate points as follows:

X1=?

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Slide 9

Solution Cont

To check the stopping criteria the difference between and is calculated to be

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Slide 10

X2

Xl

Xu

Solution Cont

Iteration 2

X1

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Slide 11

Theoretical Solution and Convergence

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.

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Slide 12

Additional Resources

For 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

Slide 13

THE END

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