Solve Equations 4x+3y=24,3x+4y-z=30,-y+4z=-24 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
4x+3y-0z=24
3x+4y-z=30
0x-y+4z=-24
2. From the above equations, First write down the equations for Gauss Seidel method
x_(k+1)=1/4(24-3y_(k)-0z_(k))
y_(k+1)=1/4(30-3x_(k+1)+z_(k))
z_(k+1)=1/4(-24-0x_(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*1/4(24-3y_(k)-0z_(k))
y_(k+1)=(1-w)*y_(k)+w*1/4(30-3x_(k+1)+z_(k))
z_(k+1)=(1-w)*z_(k)+w*1/4(-24-0x_(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*1/4[24-3(0)-0(0)]=(-0.25)*0+1.25*1/4[24]=0+7.5=7.5
y_1=(1-1.25)*0+1.25*1/4[30-3(7.5)+(0)]=(-0.25)*0+1.25*1/4[7.5]=0+2.34375=2.34375
z_1=(1-1.25)*0+1.25*1/4[-24-0(7.5)+(2.34375)]=(-0.25)*0+1.25*1/4[-21.65625]=0+-6.76758=-6.76758
2^(nd) Approximation
x_2=(1-1.25)*7.5+1.25*1/4[24-3(2.34375)-0(-6.76758)]=(-0.25)*7.5+1.25*1/4[16.96875]=-1.875+5.30273=3.42773
y_2=(1-1.25)*2.34375+1.25*1/4[30-3(3.42773)+(-6.76758)]=(-0.25)*2.34375+1.25*1/4[12.94922]=-0.58594+4.04663=3.46069
z_2=(1-1.25)*-6.76758+1.25*1/4[-24-0(3.42773)+(3.46069)]=(-0.25)*-6.76758+1.25*1/4[-20.53931]=1.69189+-6.41853=-4.72664
3^(rd) Approximation
x_3=(1-1.25)*3.42773+1.25*1/4[24-3(3.46069)-0(-4.72664)]=(-0.25)*3.42773+1.25*1/4[13.61792]=-0.85693+4.2556=3.39867
y_3=(1-1.25)*3.46069+1.25*1/4[30-3(3.39867)+(-4.72664)]=(-0.25)*3.46069+1.25*1/4[15.07736]=-0.86517+4.71168=3.8465
z_3=(1-1.25)*-4.72664+1.25*1/4[-24-0(3.39867)+(3.8465)]=(-0.25)*-4.72664+1.25*1/4[-20.1535]=1.18166+-6.29797=-5.11631
4^(th) Approximation
x_4=(1-1.25)*3.39867+1.25*1/4[24-3(3.8465)-0(-5.11631)]=(-0.25)*3.39867+1.25*1/4[12.46049]=-0.84967+3.8939=3.04424
y_4=(1-1.25)*3.8465+1.25*1/4[30-3(3.04424)+(-5.11631)]=(-0.25)*3.8465+1.25*1/4[15.75098]=-0.96163+4.92218=3.96056
z_4=(1-1.25)*-5.11631+1.25*1/4[-24-0(3.04424)+(3.96056)]=(-0.25)*-5.11631+1.25*1/4[-20.03944]=1.27908+-6.26233=-4.98325
5^(th) Approximation
x_5=(1-1.25)*3.04424+1.25*1/4[24-3(3.96056)-0(-4.98325)]=(-0.25)*3.04424+1.25*1/4[12.11833]=-0.76106+3.78698=3.02592
y_5=(1-1.25)*3.96056+1.25*1/4[30-3(3.02592)+(-4.98325)]=(-0.25)*3.96056+1.25*1/4[15.93899]=-0.99014+4.98093=3.9908
z_5=(1-1.25)*-4.98325+1.25*1/4[-24-0(3.02592)+(3.9908)]=(-0.25)*-4.98325+1.25*1/4[-20.0092]=1.24581+-6.25288=-5.00706
6^(th) Approximation
x_6=(1-1.25)*3.02592+1.25*1/4[24-3(3.9908)-0(-5.00706)]=(-0.25)*3.02592+1.25*1/4[12.02761]=-0.75648+3.75863=3.00215
y_6=(1-1.25)*3.9908+1.25*1/4[30-3(3.00215)+(-5.00706)]=(-0.25)*3.9908+1.25*1/4[15.98649]=-0.9977+4.99578=3.99808
z_6=(1-1.25)*-5.00706+1.25*1/4[-24-0(3.00215)+(3.99808)]=(-0.25)*-5.00706+1.25*1/4[-20.00192]=1.25177+-6.2506=-4.99883
7^(th) Approximation
x_7=(1-1.25)*3.00215+1.25*1/4[24-3(3.99808)-0(-4.99883)]=(-0.25)*3.00215+1.25*1/4[12.00576]=-0.75054+3.7518=3.00126
y_7=(1-1.25)*3.99808+1.25*1/4[30-3(3.00126)+(-4.99883)]=(-0.25)*3.99808+1.25*1/4[15.99737]=-0.99952+4.99918=3.99966
z_7=(1-1.25)*-4.99883+1.25*1/4[-24-0(3.00126)+(3.99966)]=(-0.25)*-4.99883+1.25*1/4[-20.00034]=1.24971+-6.25011=-5.0004
8^(th) Approximation
x_8=(1-1.25)*3.00126+1.25*1/4[24-3(3.99966)-0(-5.0004)]=(-0.25)*3.00126+1.25*1/4[12.00102]=-0.75032+3.75032=3
y_8=(1-1.25)*3.99966+1.25*1/4[30-3(3)+(-5.0004)]=(-0.25)*3.99966+1.25*1/4[15.99959]=-0.99991+4.99987=3.99996
z_8=(1-1.25)*-5.0004+1.25*1/4[-24-0(3)+(3.99996)]=(-0.25)*-5.0004+1.25*1/4[-20.00004]=1.2501+-6.25001=-4.99991
9^(th) Approximation
x_9=(1-1.25)*3+1.25*1/4[24-3(3.99996)-0(-4.99991)]=(-0.25)*3+1.25*1/4[12.00013]=-0.75+3.75004=3.00004
y_9=(1-1.25)*3.99996+1.25*1/4[30-3(3.00004)+(-4.99991)]=(-0.25)*3.99996+1.25*1/4[15.99997]=-0.99999+4.99999=4
z_9=(1-1.25)*-4.99991+1.25*1/4[-24-0(3.00004)+(4)]=(-0.25)*-4.99991+1.25*1/4[-20]=1.24998+-6.25=-5.00002
Solution By SOR (successive over-relaxation) method.
x=3.00004~=3
y=4~=4
z=-5.00002~=-5
Intertions are tabulated as below
Iteration | x | y | z |
1 | 7.5 | 2.34375 | -6.76758 |
2 | 3.42773 | 3.46069 | -4.72664 |
3 | 3.39867 | 3.8465 | -5.11631 |
4 | 3.04424 | 3.96056 | -4.98325 |
5 | 3.02592 | 3.9908 | -5.00706 |
6 | 3.00215 | 3.99808 | -4.99883 |
7 | 3.00126 | 3.99966 | -5.0004 |
8 | 3 | 3.99996 | -4.99991 |
9 | 3.00004 | 4 | -5.00002 |
This material is intended as a summary. Use your textbook for detail explanation.
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