Numerical Methods Golden Section Search Method - Theory nm.mathforcollege

1. Numerical MethodsGolden Section Search Method - Theoryhttp://nm.mathforcollege.com

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5. 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 Figure 1 Equal interval search method. http://nm.mathforcollege.com

6. f2 f1 fu fL XL Xu X1 X2 Golden Section Search Method • The Equal Interval method is inefficient when  is small. Also, we need to compute 2 interior points ! • 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

7. f2 f1 f1 fL fL fu fu a-b b X2 XL XL Xu Xu X1 X1 a b a Golden Section Search Method-Selecting the Intermediate Points b Determining the first intermediate point a Determining the second intermediate point Let ,hence Golden Ratio=> http://nm.mathforcollege.com

8. Golden Section Search Method Hence, after solving quadratic equation, with initial guess = (0, 1.5708 rad) =Initial Iteration Second Iteration Only 1 new inserted location need to be completed! http://nm.mathforcollege.com

9. f2 f1 f2 f1 fL fL fu fu X2 X2 XL XL Xu Xu X1 X1 Golden Section Search-Determining the new search region • Case1: If then the new interval is • Case2: If then the new interval is Case 2 Case 1 http://nm.mathforcollege.com

10. Golden Section Search-Determining the new search region • At each new interval ,one needs to determine only 1(not 2)new inserted location(either compute the new ,or new ) • Max. Min. • It is desirable to have automated procedure to compute and initially. http://nm.mathforcollege.com

11. Golden Section Search-(1-D) Line Search Method 0.382(αU -αl) • • Min.g(α)=Max.[-g(α)] 1st jth j j - 2 αL = Σδ(1.618)v αU = Σδ(1.618)v • • V = 0 j-2th V = 0 _ αb αU αL αa = αL j-1th Figure 2.5 Golden section partition. Figure 2.4 Bracketing the minimum point. α 2nd 0 δ 2.618δ 5.232δ 9.468δ 1.6182δ δ j - 2 j - 2 1.618δ αa = Σδ(1.618)v + 1δ(1.618)j-1 = Σδ(1.618)v = already known ! αa = αL + 0.382(αU – αl) = Σδ(1.618)v + 0.382δ(1.618)j-1(1+1.618) 3rd V = 0 V = 0 j - 1 4th V = 0 http://nm.mathforcollege.com

12. Golden Section Search-(1-D) Line Search Method • If, Then the minimum will be between αa & αb. • Ifas shown in Figure 2.5, Then the minimum will bebetween&and . Notice that: And Thusαb (wrtαU & αL ) plays same role as αa(wrtαU & αL ) !! _ _ http://nm.mathforcollege.com

13. Golden Section Search-(1-D) Line Search Method Step 1 : For a chosen small step size δ in α,say,let j be the smallest integer such that. The upper and lower bound on αi areand. Step 2: Compute g(αb) ,whereαa= αL+ 0.382(αU- αL) ,and αb = αL+ 0.618(αU- αL). Note that, so g(αa) is already known. Step 3: Compare g(αa) and g(αb) and go to Step 4,5, or 6. Step 4: If g(αa)<g(αb),thenαL ≤ αi ≤ αb. By the choice of αaandαb, the new pointsandhave . Compute , whereand go to Step 7. http://nm.mathforcollege.com

14. Golden Section Search-(1-D) Line Search Method Step 5: If g(αa) > g(αb),then αa≤ αi ≤ αU. Similar to the procedure in Step 4, put and . Compute ,where and go to Step 7. Step 6: If g(αa) = g(αb) put αL = αaand αu = αb and return to Step 2. Step 7: If is suitably small, put and stop. Otherwise, delete the bar symbols on ,and and return to Step 3. http://nm.mathforcollege.com

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16. Acknowledgement This instructional power point brought to you by Numerical Methods for STEM undergraduate http://nm.mathforcollege.com Committed to bringing numerical methods to the undergraduate

17. For instructional videos on other topics, go to http://nm.mathforcollege.com This material is based upon work supported by the National Science Foundation under Grant # 0717624. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.

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19. Numerical MethodsGolden Section Search Method - Examplehttp://nm.mathforcollege.com

20. For more details on this topic • Go to http://nm.mathforcollege.com • Click on Keyword • Click on Golden Section Search Method

21. You are free • to Share – to copy, distribute, display and perform the work • to Remix – to make derivative works

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23. Example . 2 2   2 The cross-sectional area A of a gutter with equal base and edge length of 2 is given by (trapezoidal area): Find the angle  which maximizes the cross-sectional area of the gutter. Using an initial interval of find the solution after 2 iterations. Convergence achieved if “ interval length ” is within http://nm.mathforcollege.com

24. 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

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

26. X2 XL Xu Solution Cont Iteration 2 X1 http://nm.mathforcollege.com

27. 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

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29. Acknowledgement This instructional power point brought to you by Numerical Methods for STEM undergraduate http://nm.mathforcollege.com Committed to bringing numerical methods to the undergraduate

30. For instructional videos on other topics, go to http://nm.mathforcollege.com This material is based upon work supported by the National Science Foundation under Grant # 0717624. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.

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