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Golden Section Search Method

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

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  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

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

  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

  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 http://nm.mathforcollege.com

  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=> http://nm.mathforcollege.com

  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. http://nm.mathforcollege.com

  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 . http://nm.mathforcollege.com

  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=? http://nm.mathforcollege.com

  9. Solution Cont To check the stopping criteria the difference between and is calculated to be http://nm.mathforcollege.com

  10. X2 Xl Xu Solution Cont Iteration 2 X1 http://nm.mathforcollege.com

  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. http://nm.mathforcollege.com

  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

  13. THE END http://nm.mathforcollege.com

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