1. Solve Equations 3x-y+z=-1,-x+3y-z=7,x-y+3z=-7 using SOR (Successive over-relaxation) method
Solution:
We know that, for symmetric positive definite matrix the SOR method converges for values of the relaxation parameter `w` from the interval `0 < w < 2`
The iterations of the SOR method
1. Total Equations are `3`
`3x-y+z=-1`
`-x+3y-z=7`
`x-y+3z=-7`
2. From the above equations, First write down the equations for Gauss Seidel method
`x_(k+1)=1/3(-1+y_(k)-z_(k))`
`y_(k+1)=1/3(7+x_(k+1)+z_(k))`
`z_(k+1)=1/3(-7-x_(k+1)+y_(k+1))`
3. Now multiply the right hand side by the parameter `w` and add to it the vector `x_k` from the previous iteration multiplied by the factor of `(1-w)`
`x_(k+1)=(1-w)*x_(k)+w*1/3(-1+y_(k)-z_(k))`
`y_(k+1)=(1-w)*y_(k)+w*1/3(7+x_(k+1)+z_(k))`
`z_(k+1)=(1-w)*z_(k)+w*1/3(-7-x_(k+1)+y_(k+1))`
4. Initial gauss `(x,y,z) = (0,0,0)` and `w=1.25`
Solution steps are
`1^(st)` Approximation
`x_1=(1-1.25)*0+1.25*1/3[-1+(0)-(0)]=(-0.25)*0+1.25*1/3[-1]=0+-0.41667=-0.41667`
`y_1=(1-1.25)*0+1.25*1/3[7+(-0.41667)+(0)]=(-0.25)*0+1.25*1/3[6.58333]=0+2.74306=2.74306`
`z_1=(1-1.25)*0+1.25*1/3[-7-(-0.41667)+(2.74306)]=(-0.25)*0+1.25*1/3[-3.84028]=0+-1.60012=-1.60012`
`2^(nd)` Approximation
`x_2=(1-1.25)*-0.41667+1.25*1/3[-1+(2.74306)-(-1.60012)]=(-0.25)*-0.41667+1.25*1/3[3.34317]=0.10417+1.39299=1.49715`
`y_2=(1-1.25)*2.74306+1.25*1/3[7+(1.49715)+(-1.60012)]=(-0.25)*2.74306+1.25*1/3[6.89704]=-0.68576+2.87377=2.188`
`z_2=(1-1.25)*-1.60012+1.25*1/3[-7-(1.49715)+(2.188)]=(-0.25)*-1.60012+1.25*1/3[-6.30915]=0.40003+-2.62881=-2.22878`
`3^(rd)` Approximation
`x_3=(1-1.25)*1.49715+1.25*1/3[-1+(2.188)-(-2.22878)]=(-0.25)*1.49715+1.25*1/3[3.41679]=-0.37429+1.42366=1.04937`
`y_3=(1-1.25)*2.188+1.25*1/3[7+(1.04937)+(-2.22878)]=(-0.25)*2.188+1.25*1/3[5.82059]=-0.547+2.42524=1.87824`
`z_3=(1-1.25)*-2.22878+1.25*1/3[-7-(1.04937)+(1.87824)]=(-0.25)*-2.22878+1.25*1/3[-6.17113]=0.5572+-2.5713=-2.01411`
`4^(th)` Approximation
`x_4=(1-1.25)*1.04937+1.25*1/3[-1+(1.87824)-(-2.01411)]=(-0.25)*1.04937+1.25*1/3[2.89235]=-0.26234+1.20515=0.9428`
`y_4=(1-1.25)*1.87824+1.25*1/3[7+(0.9428)+(-2.01411)]=(-0.25)*1.87824+1.25*1/3[5.9287]=-0.46956+2.47029=2.00073`
`z_4=(1-1.25)*-2.01411+1.25*1/3[-7-(0.9428)+(2.00073)]=(-0.25)*-2.01411+1.25*1/3[-5.94207]=0.50353+-2.47586=-1.97234`
`5^(th)` Approximation
`x_5=(1-1.25)*0.9428+1.25*1/3[-1+(2.00073)-(-1.97234)]=(-0.25)*0.9428+1.25*1/3[2.97307]=-0.2357+1.23878=1.00308`
`y_5=(1-1.25)*2.00073+1.25*1/3[7+(1.00308)+(-1.97234)]=(-0.25)*2.00073+1.25*1/3[6.03074]=-0.50018+2.51281=2.01263`
`z_5=(1-1.25)*-1.97234+1.25*1/3[-7-(1.00308)+(2.01263)]=(-0.25)*-1.97234+1.25*1/3[-5.99045]=0.49308+-2.49602=-2.00294`
`6^(th)` Approximation
`x_6=(1-1.25)*1.00308+1.25*1/3[-1+(2.01263)-(-2.00294)]=(-0.25)*1.00308+1.25*1/3[3.01556]=-0.25077+1.25648=1.00572`
`y_6=(1-1.25)*2.01263+1.25*1/3[7+(1.00572)+(-2.00294)]=(-0.25)*2.01263+1.25*1/3[6.00278]=-0.50316+2.50116=1.998`
`z_6=(1-1.25)*-2.00294+1.25*1/3[-7-(1.00572)+(1.998)]=(-0.25)*-2.00294+1.25*1/3[-6.00771]=0.50073+-2.50321=-2.00248`
`7^(th)` Approximation
`x_7=(1-1.25)*1.00572+1.25*1/3[-1+(1.998)-(-2.00248)]=(-0.25)*1.00572+1.25*1/3[3.00048]=-0.25143+1.2502=0.99877`
`y_7=(1-1.25)*1.998+1.25*1/3[7+(0.99877)+(-2.00248)]=(-0.25)*1.998+1.25*1/3[5.99629]=-0.4995+2.49845=1.99895`
`z_7=(1-1.25)*-2.00248+1.25*1/3[-7-(0.99877)+(1.99895)]=(-0.25)*-2.00248+1.25*1/3[-5.99982]=0.50062+-2.49992=-1.9993`
`8^(th)` Approximation
`x_8=(1-1.25)*0.99877+1.25*1/3[-1+(1.99895)-(-1.9993)]=(-0.25)*0.99877+1.25*1/3[2.99826]=-0.24969+1.24927=0.99958`
`y_8=(1-1.25)*1.99895+1.25*1/3[7+(0.99958)+(-1.9993)]=(-0.25)*1.99895+1.25*1/3[6.00028]=-0.49974+2.50012=2.00038`
`z_8=(1-1.25)*-1.9993+1.25*1/3[-7-(0.99958)+(2.00038)]=(-0.25)*-1.9993+1.25*1/3[-5.9992]=0.49983+-2.49967=-1.99984`
`9^(th)` Approximation
`x_9=(1-1.25)*0.99958+1.25*1/3[-1+(2.00038)-(-1.99984)]=(-0.25)*0.99958+1.25*1/3[3.00022]=-0.2499+1.25009=1.0002`
`y_9=(1-1.25)*2.00038+1.25*1/3[7+(1.0002)+(-1.99984)]=(-0.25)*2.00038+1.25*1/3[6.00035]=-0.50009+2.50015=2.00005`
`z_9=(1-1.25)*-1.99984+1.25*1/3[-7-(1.0002)+(2.00005)]=(-0.25)*-1.99984+1.25*1/3[-6.00014]=0.49996+-2.50006=-2.0001`
Solution By SOR (successive over-relaxation) method.
`x=1.0002~=1`
`y=2.00005~=2`
`z=-2.0001~=-2`
Intertions are tabulated as below
| Iteration | x | y | z |
| 1 | -0.41667 | 2.74306 | -1.60012 |
| 2 | 1.49715 | 2.188 | -2.22878 |
| 3 | 1.04937 | 1.87824 | -2.01411 |
| 4 | 0.9428 | 2.00073 | -1.97234 |
| 5 | 1.00308 | 2.01263 | -2.00294 |
| 6 | 1.00572 | 1.998 | -2.00248 |
| 7 | 0.99877 | 1.99895 | -1.9993 |
| 8 | 0.99958 | 2.00038 | -1.99984 |
| 9 | 1.0002 | 2.00005 | -2.0001 |
This material is intended as a summary. Use your textbook for detail explanation.
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