Golden Section Search Method  Theory


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1 Numericl Methods Golden Section Serch Method  Theory
2 For more detils on this topic Go to Click on Keyword Click on Golden Section Serch Method
3 You re free to Shre to copy, distriute, disply nd perform the work to Remix to mke derivtive works
4 Under the following conditions Attriution You must ttriute the work in the mnner specified y the uthor or licensor (ut not in ny wy tht suggests tht they endorse you or your use of the work). Noncommercil You my not use this work for commercil purposes. Shre Alike If you lter, trnsform, or uild upon this work, you my distriute the resulting work only under the sme or similr license to this one.
5 5 Equl Intervl Serch Method Figure 1 Equl intervl serch method. x f(x) ε ε Choose n intervl [, ] over which the optim occurs Compute nd + + ε f If then the intervl in which the mximum occurs is otherwise it occurs in + ε f + > + + ε ε f f +, ε + +, ε Opt. + +
6 Golden Section Serch Method The Equl Intervl method is inefficient when ε is smll. Also, we need to compute interior points! The Golden Section Serch method divides the serch more efficiently closing in on the optim in fewer itertions. f f 1 f f u X X X 1 X u Figure. Golden Section Serch method 6
7 Golden Section Serch Method Selecting the Intermedite Points = 0.38( x u x ) f 1 f f 1 = 0.618( x u x ) f f u f  f u X X 1 X u Determining the first intermedite point X Determining the second intermedite point 1 = X l + = X u = = 0.618( why?); hence ( + = X X ) X = X = X et u = 0.618*( X + = = 1+ R = u,hence l X l ), nd = 0.38*( X X X X 1 X u Golden Rtio=> = u 1 ( 5 1) = 1+ R R + R 1 = 0 R = R R X l ) u l + =
8 Golden Section Serch Method f ( θ ) = f ( θ ) f ( θ ) 4sinθ (1 + cosθ ) = 4sinθ + sin(θ ) = 4cosθ + 4cos(θ ) = 0 4cosθ + 4[ cos θ 1] = 0 f (θ ) π 3 π θ = Hence, Opt. 3 fter solving qudrtic eqution, with initil guess = (0, rd) X X 1 π 3 π θ =Initil Itertion Second Itertion st 1 Only 1 new inserted loction need to e completed! 8
9 Golden Section Serch Determining the new serch region f f 1 f f 1 f f u f f u Cse1: If Cse: If X X X 1 X u Cse 1 f ( x) > f ( x1 ) f ( x) < f ( x1 ) then the new intervl is then the new intervl is X X X 1 X u Cse [ x, x, x 1] [ x, x1, x u ] 9
10 Golden Section Serch Determining the new serch region At ech new intervl,one needs to determine only 1(not ) new inserted loction (either compute the new,or new ) x1 Mx. f ( θ ) = 4sinθ (1 + cosθ ) Min. It is desirle to hve utomted procedure to compute x nd x initilly. u x f ( θ ) = 4sinθ (1 + cosθ ) 10
11 Min.g(α)=Mx.[g(α)] j th Golden Section Serch (1D) ine Serch Method j1 th j  α = Σδ(1.618) v V = 0 j th 0 δ.618δ 5.3δ 9.468δ Figure.4 Brcketing the minimum point. δ δ 1.618δ 1 st j _ α α = α α α U α α U = Σδ(1.618) v V = (α U  α l ) Figure.5 Golden section prtition. nd j  α = α (α U α l ) = Σδ(1.618) v δ(1.618) j1 ( ) V = 0 j  α = Σδ(1.618) v + 1δ(1.618) j1 = Σδ(1.618) v = lredy known! V = 0 j  1 V = 0 3 rd 4 th 11
12 Golden Section Serch (1D) ine Serch Method If g α ) = g( α ), Then the minimum will e etween α & α. If g( α ) g( α ) s shown in Figure.5, Then the minimum will e etween α & α α = α nd α = α. Notice tht: And α α ( = (0.36)( δ[1.618] α α α = j 1 = (0.38) ( δ[1.618] U j U α = αu α δ (1.618) = α α = (1 0.38)( α [ ]) = 0.618( δ[1.618] j U = 0.38( α U Thus α (wrt α U & α ) plys sme role s α (wrt α U & α )!! α ) U α ) = (0.36)( δ[1.618] j ) j 1 U + δ[1.618] j ) 1
13 Golden Section Serch (1D) ine Serch Method 1 Step 1 : For chosen smll step size δ in α,sy δ = ,let j e the j V j 1 V smllest integer such tht g( δ (1.618) ) g( δ (1.618) ) The upper nd lower ound on α i V re αu = δ (1.618) nd V α = δ (1.618). = 0 Step : Compute g(α ),where α = α (α U  α ),nd α = α (α U  α ). Note tht j 1 = α δ (1.618) = 0 V, so g(α ) is lredy known. Step 3: Compre g(α ) nd g(α ) nd go to Step 4,5, or 6. V V = 0 Step 4: If g(α )<g(α ),then α α i α. By the choice of α nd α, the new points α = α nd α u = α hve α = α. Compute g α ), where α = α +.38( α α ) nd go to Step 7. ( J V = 0 V = 0 0 u j V 13
14 Golden Section Serch (1D) ine Serch Method Step 5: If g(α ) > g(α ), then α α i α U. Similr to the procedure in Step 4, put α nd. = α α u = α u Compute g α ),where α = α +.618( α α ) nd go to Step 7. ( 0 u Step 6: If g(α ) = g(α ) put α = α nd α u = α nd return to Step. i 1 Step 7: If α u α is suitly smll, put α = ( α u + α ) nd stop. Otherwise, delete the r symols on α, α, α,nd α u nd return to Step
15 THE END
16 Acknowledgement This instructionl power point rought to you y Numericl Methods for STEM undergrdute Committed to ringing numericl methods to the undergrdute
17 For instructionl videos on other topics, go to This mteril is sed upon work supported y the Ntionl Science Foundtion under Grnt # Any opinions, findings, nd conclusions or recommendtions expressed in this mteril re those of the uthor(s) nd do not necessrily reflect the views of the Ntionl Science Foundtion.
18 The End  Relly
19 Numericl Methods Golden Section Serch Method  Exmple
20 For more detils on this topic Go to Click on Keyword Click on Golden Section Serch Method
21 You re free to Shre to copy, distriute, disply nd perform the work to Remix to mke derivtive works
22 Under the following conditions Attriution You must ttriute the work in the mnner specified y the uthor or licensor (ut not in ny wy tht suggests tht they endorse you or your use of the work). Noncommercil You my not use this work for commercil purposes. Shre Alike If you lter, trnsform, or uild upon this work, you my distriute the resulting work only under the sme or similr license to this one.
23 Exmple. θ θ The crosssectionl re A of gutter with equl se nd edge length of is given y (trpezoidl re): Mx. f ( θ ) = A = 4sinθ (1 + cosθ ) = 4sinθ + sin(θ ) Find the ngle θ which mximizes the crosssectionl re of the gutter. Using n initil intervl of π [0, ] find the solution fter itertions. ε = 0.05 Convergence chieved if intervl length is within 3
24 Solution The function to e mximized is f ( θ ) = 4sinθ (1 + cosθ ) Itertion 1: Given the vlues for the oundries of x = 0 nd xu = π / we cn clculte the initil intermedite points s follows: x x 1 = x = x u ( x 5 1 ( x u u x x ) = 0 + ) = (1.5708) = (1.5708) = f ( ) = f ( ) = 4.17 f 1 f X 1 =? X X X 1 X u X =X X =X 1 X u 4
25 Solution Cont x1 = x + ( xu x ) = ( ) = To check the stopping criteri the difference etween nd is clculted to e x x u x u x = =
26 Solution Cont Itertion x x x x u 1 = = = = f ( 1.000) = f ( ) = f ( x1 ) < f ( x) X X X 1 X u x xu = x = x = ( x ) = ( ) = xu u x = x u + x = =
27 Theoreticl Solution nd Convergence Itertion x l x u x 1 x f(x 1 ) f(x ) ε x u + x = = f ( 1.040) = The theoreticlly optiml solution to the prolem hppens t exctly 60 degrees which is rdins nd gives mximum crosssectionl re of
28 THE END
29 Acknowledgement This instructionl power point rought to you y Numericl Methods for STEM undergrdute Committed to ringing numericl methods to the undergrdute
30 For instructionl videos on other topics, go to This mteril is sed upon work supported y the Ntionl Science Foundtion under Grnt # Any opinions, findings, nd conclusions or recommendtions expressed in this mteril re those of the uthor(s) nd do not necessrily reflect the views of the Ntionl Science Foundtion.
31 The End  Relly
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