Numericl Methods Golden Section Serch Method - Theory http://nm.mthforcollege.com
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http://nm.mthforcollege.com 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. + +
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 http://nm.mthforcollege.com
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=> = 0.618... u 1 ( 5 1) = 1+ R R + R 1 = 0 R = R R X l ) u l + = 0.61803 http://nm.mthforcollege.com 7
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, 1.5708 rd) X X 1 π 3 π θ =Initil Itertion Second Itertion st 1 Only 1 new inserted loction need to e completed! http://nm.mthforcollege.com 8
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 http://nm.mthforcollege.com
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 http://nm.mthforcollege.com
Min.g(α)=Mx.[-g(α)] j- th Golden Section Serch- (1-D) ine Serch Method j-1 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 1.618 δ 1.618δ 1 st j _ α α = α α α U α α U = Σδ(1.618) v V = 0 0.38(α U - α l ) Figure.5 Golden section prtition. nd j - α = α + 0.38(α U α l ) = Σδ(1.618) v + 0.38δ(1.618) j-1 (1+1.618) V = 0 j - α = Σδ(1.618) v + 1δ(1.618) j-1 = Σδ(1.618) v = lredy known! V = 0 j - 1 V = 0 3 rd 4 th 11 http://nm.mthforcollege.com
Golden Section Serch- (1-D) 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 [1 + 1.618]) = 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 1 1.618 ) 1.618 j 1 U + δ[1.618] j ) 1 http://nm.mthforcollege.com
Golden Section Serch- (1-D) ine Serch Method 1 Step 1 : For chosen smll step size δ in α,sy δ = + 10 10,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 + 0.38(α U - α ),nd α = α + 0.618(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 http://nm.mthforcollege.com
Golden Section Serch- (1-D) 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 3. 14 http://nm.mthforcollege.com
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Acknowledgement This instructionl power point rought to you y Numericl Methods for STEM undergrdute http://nm.mthforcollege.com Committed to ringing numericl methods to the undergrdute
For instructionl videos on other topics, go to http://nm.mthforcollege.com This mteril is sed upon work supported y the Ntionl Science Foundtion under Grnt # 071764. 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.
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Numericl Methods Golden Section Serch Method - Exmple http://nm.mthforcollege.com
For more detils on this topic Go to http://nm.mthforcollege.com Click on Keyword Click on Golden Section Serch Method
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Exmple. θ θ The cross-sectionl 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 cross-sectionl 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 http://nm.mthforcollege.com
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 + 5 1 ( x 5 1 ( x u u x x ) = 0 + ) = 1.5708 5 1 (1.5708) = 0.97080 5 1 (1.5708) = 0.60000 f ( 0.97080) = 5.1654 f ( 0.60000) = 4.17 f 1 f X 1 =? X X X 1 X u X =X X =X 1 X u 4 http://nm.mthforcollege.com
Solution Cont 5 1 5 1 x1 = x + ( xu x ) = 0.60000 + (1.5708 0.60000) = 1.000 To check the stopping criteri the difference etween nd is clculted to e x x u x u x = 1.5708 0.60000 = 0.97080 5 http://nm.mthforcollege.com
Solution Cont Itertion x x x x u 1 = 0.60000 = 1.5708 = 1.000 = 0.97080 f ( 1.000) = f ( 0.97080) = 5.0791 5.1654 f ( x1 ) < f ( x) X X X 1 X u x xu = 1.000 x = 1 0.97080 x = 0.60000 5 1 ( x ) = 1.000 5 1 (1.000 0.6000) = xu u x = x u + x = 1.000 + 0.6000 = 0.9000 0.8918 6 http://nm.mthforcollege.com
Theoreticl Solution nd Convergence Itertion x l x u x 1 x f(x 1 ) f(x ) ε 1 0.0000 1.5714 0.971 0.600 5.1657 4.138 1.5714 0.600 1.5714 1.005 0.971 5.0784 5.1657 0.971 3 0.600 1.005 0.971 0.895 5.1657 4.946 0.600 4 0.895 1.005 1.0588 0.971 5.1955 5.1657 0.3710 5 0.971 1.005 1.119 1.0588 5.1740 5.1955 0.93 6 0.971 1.119 1.0588 1.053 5.1955 5.1937 0.1417 7 1.053 1.119 1.0794 1.0588 5.1908 5.1955 0.0876 8 1.053 1.0794 1.0588 1.0460 5.1955 5.1961 0.0541 9 1.053 1.0588 1.0460 1.0381 5.1961 5.1957 0.0334 x u + x 1.053 + 1.0588 = = 1.040 f ( 1.040) = 5.1960 The theoreticlly optiml solution to the prolem hppens t exctly 60 degrees which is 1.047 rdins nd gives mximum cross-sectionl re of 5.196. 7 http://nm.mthforcollege.com
THE END http://nm.mthforcollege.com
Acknowledgement This instructionl power point rought to you y Numericl Methods for STEM undergrdute http://nm.mthforcollege.com Committed to ringing numericl methods to the undergrdute
For instructionl videos on other topics, go to http://nm.mthforcollege.com This mteril is sed upon work supported y the Ntionl Science Foundtion under Grnt # 071764. 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.
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