《结构工程中的数学方法》课程教学课件（讲稿）example_GS_iteration

2AlgebraicproblemsinmatrixformIterative SolutionStrategiesGauss-Seidel iterativemethod-example(frompage33a)umstrukturiertesSystemausJacobi-Verfahren(Seite7Ob)[2.51.25025X,=-0.5.×,+10X0-40-1.00.6X2=0.6·X,+40X00.625-1.012.5X,=0.625.×,-12.5X.simultaneousupdatingofunknowns[k-1][k-1][k-1]X 2XX3[k][k-1[k-1]+x,+a13BXdil[k-]]LKXXa[k]XXXa32D70dMichaelBeer,EngineeringMathematics

umstrukturiertes System aus Jacobi-Verfahren (Seite 70b) Gauss-Seidel iterative method − example (from page 33a) 70d ● x 0.5 x 10 1 2 =− ⋅ + x 0.6 x 40 2 3 = ⋅+ x 0.625 x 12.5 3 2 = ⋅−           − ⋅ =−       −          1 2 3 2.5 1.25 0 x 25 0 1.0 0.6 x 40 0 0.625 1.0 x 12.5 − − −     =− + +              =− − +           =− + [k] [k 1] [k 1] 1 12 2 13 3 1 11 [k] [k] [k 1] 2 21 1 23 3 2 22 [k] [k] [k 3 31 1 32 2 33 1 x ax ax b a 1 x ax a x b a 1 x ax ax a       +       ] b3 ● simultaneous updating of unknowns [k 1] [k 1] −−− [k 1] x , x , x 123 Iterative Solution Strategies 2 Algebraic problems in matrix form Michael Beer, Engineering Mathematics

2AlgebraicproblemsinmatrixformIterative SolutionStrategiesGauss-Seidel iterative method - example (cont'd)initialeassumptionof solutionvector[0][0][O]X1=0X3=0》k=0:X2= 0iterativesolution》k=1:[1][1] [1][1][0][0]=-0.5=0.62512.5X2+10.0=0.6.X3+40.0X3X2X2X112.510.040.0》k= 2:[1][2] [1][2][2][2]=-0.5.= 0.6X3+40.012.5X2+10.0=0.625.X2X3X2X147.5-10.0= 17.188C.》 k = 8:[8][7][8][8][8][7]X1=-0.5.X2+10.0=0.6.X3+40.0=0.625.X2-12.5X2X3(20)=19.992=-15.983(-16)=51.987(52)70eMichael Beer,EngineeringMathematics

initiale assumption of solution vector 70e ● » k = 0: = ⋅+ [1] [0] x 0.6 x 40.0 2 3 ● iterative solution » k = 1: =− ⋅ + [2] [1] x 0.5 x 10.0 1 2 = ⋅− [2] [2] x 0.625 x 12.5 3 2 » k = 2: = = = [0] [0] [0] x 0, x 0, x 0 123 = ⋅+ [2] [1] x 0.6 x 40.0 2 3 = ⋅− [1] [1] =− ⋅ + x 0.625 x 12.5 3 2 [1] [0] x 0.5 x 10.0 1 2 = 10.0 = 40.0 = 12.5 = −10.0 = 47.5 = 17.188 ● ● ● ● ● ● ● ● ● » k = 8: =− ⋅ + [8] [7] x 0.5 x 10.0 1 2 = ⋅− [8] [8] = ⋅+ x 0.625 x 12.5 3 2 [8] [7] x 0.6 x 40.0 2 3 = −15.983 (−16) = 51.987 (52) = 19.992 (20) Iterative Solution Strategies 2 Algebraic problems in matrix form Michael Beer, Engineering Mathematics Gauss-Seidel iterative method − example (cont‘d)