Solve Equations 5x+y=10,2x+3y=4 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 `2`

`5x+y=10`

`2x+3y=4`


2. From the above equations, First write down the equations for Gauss Seidel method
`x_(k+1)=1/5(10-y_(k))`

`y_(k+1)=1/3(4-2x_(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*1/5(10-y_(k))`

`y_(k+1)=(1-w)*y_(k)+w*1/3(4-2x_(k+1))`

4. Initial gauss `(x,y) = (0,0)` and `w=1.25`

Solution steps are
`1^(st)` Approximation

`x_1=(1-1.25)*0+1.25*1/5[10-(0)]=(-0.25)*0+1.25*1/5[10]=0+2.5=2.5`

`y_1=(1-1.25)*0+1.25*1/3[4-2(2.5)]=(-0.25)*0+1.25*1/3[-1]=0+-0.4167=-0.4167`

`2^(nd)` Approximation

`x_2=(1-1.25)*2.5+1.25*1/5[10-(-0.4167)]=(-0.25)*2.5+1.25*1/5[10.4167]=-0.625+2.6042=1.9792`

`y_2=(1-1.25)*-0.4167+1.25*1/3[4-2(1.9792)]=(-0.25)*-0.4167+1.25*1/3[0.0417]=0.1042+0.0174=0.1215`

`3^(rd)` Approximation

`x_3=(1-1.25)*1.9792+1.25*1/5[10-(0.1215)]=(-0.25)*1.9792+1.25*1/5[9.8785]=-0.4948+2.4696=1.9748`

`y_3=(1-1.25)*0.1215+1.25*1/3[4-2(1.9748)]=(-0.25)*0.1215+1.25*1/3[0.0503]=-0.0304+0.021=-0.0094`

`4^(th)` Approximation

`x_4=(1-1.25)*1.9748+1.25*1/5[10-(-0.0094)]=(-0.25)*1.9748+1.25*1/5[10.0094]=-0.4937+2.5024=2.0086`

`y_4=(1-1.25)*-0.0094+1.25*1/3[4-2(2.0086)]=(-0.25)*-0.0094+1.25*1/3[-0.0173]=0.0024+-0.0072=-0.0049`

`5^(th)` Approximation

`x_5=(1-1.25)*2.0086+1.25*1/5[10-(-0.0049)]=(-0.25)*2.0086+1.25*1/5[10.0049]=-0.5022+2.5012=1.9991`

`y_5=(1-1.25)*-0.0049+1.25*1/3[4-2(1.9991)]=(-0.25)*-0.0049+1.25*1/3[0.0019]=0.0012+0.0008=0.002`

`6^(th)` Approximation

`x_6=(1-1.25)*1.9991+1.25*1/5[10-(0.002)]=(-0.25)*1.9991+1.25*1/5[9.998]=-0.4998+2.4995=1.9997`

`y_6=(1-1.25)*0.002+1.25*1/3[4-2(1.9997)]=(-0.25)*0.002+1.25*1/3[0.0005]=-0.0005+0.0002=-0.0003`

`7^(th)` Approximation

`x_7=(1-1.25)*1.9997+1.25*1/5[10-(-0.0003)]=(-0.25)*1.9997+1.25*1/5[10.0003]=-0.4999+2.5001=2.0001`

`y_7=(1-1.25)*-0.0003+1.25*1/3[4-2(2.0001)]=(-0.25)*-0.0003+1.25*1/3[-0.0003]=0.0001+-0.0001=0`


Solution By SOR (successive over-relaxation) method.
`x=2.0001~=2`

`y=0~=0`

Iterations are tabulated as below

Iteration	x	y
1	2.5	-0.4167
2	1.9792	0.1215
3	1.9748	-0.0094
4	2.0086	-0.0049
5	1.9991	0.002
6	1.9997	-0.0003
7	2.0001	0

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