Numerical Methods Golden Section Search Method - Theory

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  • Numerical Methods Golden Section Search Method - Theory http://nm.mathforcollege.com http://nm.mathforcollege.com
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  • For more details on this topic Go to http://nm.mathforcollege.comhttp://nm.mathforcollege.com Click on Keyword Click on Golden Section Search Method
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  • You are free to Share to copy, distribute, display and perform the work to Remix to make derivative works
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  • Under the following conditions Attribution You must attribute the work in the manner specified by the author or licensor (but not in any way that suggests that they endorse you or your use of the work). Noncommercial You may not use this work for commercial purposes. Share Alike If you alter, transform, or build upon this work, you may distribute the resulting work only under the same or similar license to this one.
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  • http://nm.mathforcollege.com5 Equal Interval Search Method Figure 1 Equal interval search method. x f(x) a b 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
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  • 6 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. X2X2 XLXL X1X1 XuXu fufu f2f2 f1f1 fLfL Figure 2. Golden Section Search method http://nm.mathforcollege.com
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  • 7 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=> Let,hence b a http://nm.mathforcollege.com
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  • Golden Section Search Method 8 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
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  • Golden Section Search- Determining the new search region Case1: If then the new interval is Case2: If then the new interval is 9 X2X2 XLXL X1X1 XuXu fufu f2f2 f1f1 fLfL X2X2 XLXL X1X1 XuXu fufu f2f2 f1f1 fLfL Case 2 Case 1 http://nm.mathforcollege.com
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  • 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. 10 Golden Section Search- Determining the new search region http://nm.mathforcollege.com
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  • 11 0 2.6185.2329.468 1.618 1.618 2 Min.g()=Max.[-g()] j th j-1 th j-2 th L = (1.618) v V = 0 j - 2 U = (1.618) v V = 0 j LL aa bb UU 0.382( U - l ) 1 st 2 nd a = L + 0.382( U l ) = (1.618) v + 0.382(1.618) j-1 (1+1.618) V = 0 j - 2 3 rd a = (1.618) v + 1(1.618) j-1 = (1.618) v = already known ! V = 0 j - 2 V = 0 j - 1 4 th Figure 2.4 Bracketing the minimum point. Figure 2.5 Golden section partition. = L _ Golden Section Search- (1-D) Line Search Method http://nm.mathforcollege.com
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  • 12 If, Then the minimum will be between a & b. If as shown in Figure 2.5, Then the minimum will be between & and. Notice that : And Thus b ( wrt U & L ) plays same role as a ( wrt U & L ) !! _ _ Golden Section Search- (1-D) Line Search Method http://nm.mathforcollege.com
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  • 13 Step 1 : For a chosen small step size in ,say,let j be the smallest integer such that. The upper and lower bound on i are and. Step 2 : Compute g( b ),where a = L + 0.382( U - L ),and b = L + 0.618( U - L ). Note that, so g( a ) is already known. Step 3 : Compare g( a ) and g( b ) and go to Step 4,5, or 6. Step 4 : If g( a )
  • Step 5: If g( a ) > g( b ), then a i U. Similar to the procedure in Step 4, putand. Compute,where and go to Step 7. Step 6: If g( a ) = g( b ) put L = a and u = b 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. 14 Golden Section Search- (1-D) Line Search Method http://nm.mathforcollege.com
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  • THE END http://nm.mathforcollege.com
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  • This instructional power point brought to you by Numerical Methods for STEM undergraduate http://http://nm.mathforcollege.com Committed to bringing numerical methods to the undergraduate Acknowledgement
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  • For instructional videos on other topics, go to http://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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  • Numerical Methods Golden Section Search Method - Example http://nm.mathforcollege.com http://nm.mathforcollege.com
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  • For more details on this topic Go to http://nm.mathforcollege.comhttp://nm.mathforcollege.com Click on Keyword Click on Golden Section Search Method
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  • You are free to Share to copy, distribute, display and perform the work to Remix to make derivative works
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  • Under the following conditions Attribution You must attribute the work in the manner specified by the author or licensor (but not in any way that suggests that they endorse you or your use of the work). Noncommercial You may not use this work for commercial purposes. Share Alike If you alter, transform, or build upon this work, you may distribute the resulting work only under the same or similar license to this one.
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  • 23 Example 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 2 2 2 http://nm.mathforcollege.com
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  • 24 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 =? http://nm.mathforcollege.com
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  • 25 Solution Cont To check the stopping criteria the difference between and is calculated to be http://nm.mathforcollege.com
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  • 26 Solution Cont Iteration 2 XLXL X2X2 XuXu X1X1 http://nm.mathforcollege.com
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  • 27 Theoretical Solution and Convergence Iterationxlxl xuxu x1x1 x2x2 f(x 1 )f(x 2 ) 10.00001.57140.97120.60025.16574.12381.5714 20.60021.5714 1.2005 0.97125.07845.16570.9712 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. http://nm.mathforcollege.com
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  • THE END http://nm.mathforcollege.com
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  • This instructional power point brought to you by Numerical Methods for STEM undergraduate http://http://nm.mathforcollege.com Committed to bringing numerical methods to the undergraduate Acknowledgement
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  • For instructional videos on other topics, go to http://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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  • The End - Really