Numerical Methods Golden Section Search Method - Theory http://nm

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

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

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.

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 x f(x) a b Figure 1 Equal interval search method. http://nm.mathforcollege.com

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. X2 XL X1 Xu fu f2 f1 fL Figure 2. Golden Section Search method http://nm.mathforcollege.com

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

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

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

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

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

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

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 are and . Step 2: Compute g(αb) ,where αa= 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),then αL ≤ αi ≤ αb. By the choice of αa and αb, the new points and have . Compute , where and go to Step 7. http://nm.mathforcollege.com

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 = αa and 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

The End http://nm.mathforcollege.com

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

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.

The End - Really

Numerical Methods Golden Section Search Method - Example http://nm

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

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

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.

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

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: XL=X2 X2=X1 Xu X2 XL X1 Xu f2 f1 X1=? http://nm.mathforcollege.com

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

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

Theoretical Solution and Convergence Iteration xl xu x1 x2 f(x1) f(x2)  1 0.0000 1.5714 0.9712 0.6002 5.1657 4.1238 2 1.2005 5.0784 3 0.8295 4.9426 4 1.0588 5.1955 0.3710 5 1.1129 5.1740 0.2293 6 1.0253 5.1937 0.1417 7 1.0794 5.1908 0.0876 8 1.0460 5.1961 0.0541 9 1.0381 5.1957 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. http://nm.mathforcollege.com

The End http://nm.mathforcollege.com

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

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.

The End - Really