Golden Section Search Method

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# Golden Section Search Method - PowerPoint PPT Presentation

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.

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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

http://nm.mathforcollege.com

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.

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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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f2

f1

f1

fl

fl

fu

fu

a-b

b

X2

Xl

Xl

Xu

Xu

X1

X1

a

a

b

Determining the first intermediate point

Determining the second intermediate point

Golden Ratio=>

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f2

f1

fl

fu

X2

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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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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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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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Solution Cont

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

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X2

Xl

Xu

Solution Cont

Iteration 2

X1

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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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