1. 2/24/2016 http://nm.mathforcollege.com 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

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

3. 3 Equal Interval Search Method Figure 1 Equal interval search method. x f(x) a b (a+b)/2 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

4. http://nm.mathforcollege.com4 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. X2X2 XlXl X1X1 XuXu fufu f2f2 f1f1 flfl Figure 2. Golden Section Search method

5. http://nm.mathforcollege.com5 Golden Section Search Method- Selecting the Intermediate Points ab XlXl X1X1 XuXu fufu f1f1 flfl Determining the first intermediate point a-b b X2X2 a XlXl X1X1 XuXu fufu f2f2 f1f1 flfl Determining the second intermediate point Golden Ratio=>

6. Golden Section Search- Determining the new search region 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.com6 X2X2 XlXl X1X1 XuXu fufu f2f2 f1f1 flfl

7. 7 Example 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. 2 2 2  

8. http://nm.mathforcollege.com8 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: X2X2 XlXl X1X1 XuXu f2f2 f1f1 X l =X 2 X 2 =X 1 XuXu X 1 =?

9. http://nm.mathforcollege.com9 Solution Cont To check the stopping criteria the difference between and is calculated to be

10. http://nm.mathforcollege.com10 Solution Cont Iteration 2 XlXl X2X2 XuXu X1X1

11. http://nm.mathforcollege.com11 Theoretical Solution and Convergence Iterationxlxl xuxu x1x1 x2x2 f(x 1 )f(x 2 )  10.00001.57140.97120.60025.16574.12381.5714 20.60021.57141.20050.97125.07845.16570.9712 30.60021.20050.97120.82955.16574.94260.6002 40.82951.20051.05880.97125.19555.16570.3710 50.97121.20051.11291.05885.17405.19550.2293 60.97121.11291.05881.02535.19555.19370.1417 71.02531.11291.07941.05885.19085.19550.0876 81.02531.07941.05881.04605.19555.19610.0541 91.02531.05881.04601.03815.19615.19570.0334 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.

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

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