Golden Section Search Method Major All Engineering Majors

Golden Section Search Method Major: All Engineering Majors Authors: Autar Kaw, Ali Yalcin http: //nm. mathforcollege. com Transforming Numerical Methods Education for STEM Undergraduates 11/30/2020 http: //nm. mathforcollege. com 1

Golden Section Search Method http: //nm. mathforcollege. com

Equal Interval Search Method • Choose an interval [a, b] over which the optima occurs • Compute and f(x) • If then the interval in which the maximum occurs is otherwise it occurs in a b (a+b)/2 x Figure 1 Equal interval search method. 3 forcollege. com http: //nm. math

Golden Section Search Method n n 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. f f 2 1 fu fl Xl X 2 X 1 Xu Figure 2. Golden Section Search method 4 forcollege. com http: //nm. math

Golden Section Search Method. Selecting the Intermediate Points f 2 f 1 fl fu fl f 1 fu a-b b Xl a X 1 b Xu Determining the first intermediate point Xl X 2 a X 1 Xu Determining the second intermediate point Golden Ratio=> 5 forcollege. com http: //nm. math

Golden Section Search. Determining the new search region f 2 f 1 fl Xl n n n 6 fu X 2 X 1 Xu If then the new interval is All that is left to do is to determine the location of the second intermediate point. forcollege. com http: //nm. math

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. 7 forcollege. com http: //nm. math

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: f 2 f 1 X 1=? Xl 8 X 2 X 1 Xu Xl=X 2 X 2=X 1 forcollege. com Xu http: //nm. math

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

Solution Cont Iteration 2 Xl 10 X 2 X 1 Xu forcollege. com http: //nm. math

Theoretical Solution and Convergence Iteration 1 2 3 4 5 6 7 8 9 xl 0. 0000 0. 6002 0. 8295 0. 9712 1. 0253 xu 1. 5714 1. 2005 1. 1129 1. 0794 1. 0588 x 1 0. 9712 1. 2005 0. 9712 1. 0588 1. 1129 1. 0588 1. 0794 1. 0588 1. 0460 x 2 0. 6002 0. 9712 0. 8295 0. 9712 1. 0588 1. 0253 1. 0588 1. 0460 1. 0381 f(x 1) 5. 1657 5. 0784 5. 1657 5. 1955 5. 1740 5. 1955 5. 1908 5. 1955 5. 1961 f(x 2) 4. 1238 5. 1657 4. 9426 5. 1657 5. 1955 5. 1937 5. 1955 5. 1961 5. 1957 1. 5714 0. 9712 0. 6002 0. 3710 0. 2293 0. 1417 0. 0876 0. 0541 0. 0334 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. 11 forcollege. com http: //nm. math

Additional Resources For all resources on this topic such as digital audiovisual lectures, primers, textbook chapters, multiple-choice tests, worksheets in MATLAB, MATHEMATICA, Math. Cad and MAPLE, blogs, related physical problems, please visit http: //nm. mathforcollege. com/topics/opt_golden_section_search. html

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