4. Example `10x+2y-z=7,x+8y+3z=-4,-2x-y+10z=9`
Solve Equations 10x+2y-z=7,x+8y+3z=-4,-2x-y+10z=9 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`
`10x+2y-z=7`
`x+8y+3z=-4`
`-2x-y+10z=9`
2. From the above equations, First write down the equations for Gauss Seidel method
`x_(k+1)=1/10(7-2y_(k)+z_(k))`
`y_(k+1)=1/8(-4-x_(k+1)-3z_(k))`
`z_(k+1)=1/10(9+2x_(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*1/10(7-2y_(k)+z_(k))`
`y_(k+1)=(1-w)*y_(k)+w*1/8(-4-x_(k+1)-3z_(k))`
`z_(k+1)=(1-w)*z_(k)+w*1/10(9+2x_(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*1/10[7-2(0)+(0)]=(-0.25)*0+1.25*1/10[7]=0+0.875=0.875`
`y_1=(1-1.25)*0+1.25*1/8[-4-(0.875)-3(0)]=(-0.25)*0+1.25*1/8[-4.875]=0+-0.7617=-0.7617`
`z_1=(1-1.25)*0+1.25*1/10[9+2(0.875)+(-0.7617)]=(-0.25)*0+1.25*1/10[9.9883]=0+1.2485=1.2485`
`2^(nd)` Approximation
`x_2=(1-1.25)*0.875+1.25*1/10[7-2(-0.7617)+(1.2485)]=(-0.25)*0.875+1.25*1/10[9.772]=-0.2188+1.2215=1.0027`
`y_2=(1-1.25)*-0.7617+1.25*1/8[-4-(1.0027)-3(1.2485)]=(-0.25)*-0.7617+1.25*1/8[-8.7484]=0.1904+-1.3669=-1.1765`
`z_2=(1-1.25)*1.2485+1.25*1/10[9+2(1.0027)+(-1.1765)]=(-0.25)*1.2485+1.25*1/10[9.829]=-0.3121+1.2286=0.9165`
`3^(rd)` Approximation
`x_3=(1-1.25)*1.0027+1.25*1/10[7-2(-1.1765)+(0.9165)]=(-0.25)*1.0027+1.25*1/10[10.2695]=-0.2507+1.2837=1.033`
`y_3=(1-1.25)*-1.1765+1.25*1/8[-4-(1.033)-3(0.9165)]=(-0.25)*-1.1765+1.25*1/8[-7.7825]=0.2941+-1.216=-0.9219`
`z_3=(1-1.25)*0.9165+1.25*1/10[9+2(1.033)+(-0.9219)]=(-0.25)*0.9165+1.25*1/10[10.1441]=-0.2291+1.268=1.0389`
`4^(th)` Approximation
`x_4=(1-1.25)*1.033+1.25*1/10[7-2(-0.9219)+(1.0389)]=(-0.25)*1.033+1.25*1/10[9.8827]=-0.2582+1.2353=0.9771`
`y_4=(1-1.25)*-0.9219+1.25*1/8[-4-(0.9771)-3(1.0389)]=(-0.25)*-0.9219+1.25*1/8[-8.0938]=0.2305+-1.2646=-1.0342`
`z_4=(1-1.25)*1.0389+1.25*1/10[9+2(0.9771)+(-1.0342)]=(-0.25)*1.0389+1.25*1/10[9.92]=-0.2597+1.24=0.9803`
`5^(th)` Approximation
`x_5=(1-1.25)*0.9771+1.25*1/10[7-2(-1.0342)+(0.9803)]=(-0.25)*0.9771+1.25*1/10[10.0486]=-0.2443+1.2561=1.0118`
`y_5=(1-1.25)*-1.0342+1.25*1/8[-4-(1.0118)-3(0.9803)]=(-0.25)*-1.0342+1.25*1/8[-7.9526]=0.2585+-1.2426=-0.9841`
`z_5=(1-1.25)*0.9803+1.25*1/10[9+2(1.0118)+(-0.9841)]=(-0.25)*0.9803+1.25*1/10[10.0396]=-0.2451+1.2549=1.0099`
`6^(th)` Approximation
`x_6=(1-1.25)*1.0118+1.25*1/10[7-2(-0.9841)+(1.0099)]=(-0.25)*1.0118+1.25*1/10[9.978]=-0.253+1.2472=0.9943`
`y_6=(1-1.25)*-0.9841+1.25*1/8[-4-(0.9943)-3(1.0099)]=(-0.25)*-0.9841+1.25*1/8[-8.0239]=0.246+-1.2537=-1.0077`
`z_6=(1-1.25)*1.0099+1.25*1/10[9+2(0.9943)+(-1.0077)]=(-0.25)*1.0099+1.25*1/10[9.9809]=-0.2525+1.2476=0.9951`
`7^(th)` Approximation
`x_7=(1-1.25)*0.9943+1.25*1/10[7-2(-1.0077)+(0.9951)]=(-0.25)*0.9943+1.25*1/10[10.0106]=-0.2486+1.2513=1.0027`
`y_7=(1-1.25)*-1.0077+1.25*1/8[-4-(1.0027)-3(0.9951)]=(-0.25)*-1.0077+1.25*1/8[-7.9882]=0.2519+-1.2482=-0.9962`
`z_7=(1-1.25)*0.9951+1.25*1/10[9+2(1.0027)+(-0.9962)]=(-0.25)*0.9951+1.25*1/10[10.0093]=-0.2488+1.2512=1.0024`
`8^(th)` Approximation
`x_8=(1-1.25)*1.0027+1.25*1/10[7-2(-0.9962)+(1.0024)]=(-0.25)*1.0027+1.25*1/10[9.9948]=-0.2507+1.2494=0.9987`
`y_8=(1-1.25)*-0.9962+1.25*1/8[-4-(0.9987)-3(1.0024)]=(-0.25)*-0.9962+1.25*1/8[-8.0058]=0.2491+-1.2509=-1.0018`
`z_8=(1-1.25)*1.0024+1.25*1/10[9+2(0.9987)+(-1.0018)]=(-0.25)*1.0024+1.25*1/10[9.9955]=-0.2506+1.2494=0.9988`
`9^(th)` Approximation
`x_9=(1-1.25)*0.9987+1.25*1/10[7-2(-1.0018)+(0.9988)]=(-0.25)*0.9987+1.25*1/10[10.0025]=-0.2497+1.2503=1.0007`
`y_9=(1-1.25)*-1.0018+1.25*1/8[-4-(1.0007)-3(0.9988)]=(-0.25)*-1.0018+1.25*1/8[-7.9972]=0.2505+-1.2496=-0.9991`
`z_9=(1-1.25)*0.9988+1.25*1/10[9+2(1.0007)+(-0.9991)]=(-0.25)*0.9988+1.25*1/10[10.0022]=-0.2497+1.2503=1.0006`
`10^(th)` Approximation
`x_10=(1-1.25)*1.0007+1.25*1/10[7-2(-0.9991)+(1.0006)]=(-0.25)*1.0007+1.25*1/10[9.9988]=-0.2502+1.2498=0.9997`
`y_10=(1-1.25)*-0.9991+1.25*1/8[-4-(0.9997)-3(1.0006)]=(-0.25)*-0.9991+1.25*1/8[-8.0014]=0.2498+-1.2502=-1.0004`
`z_10=(1-1.25)*1.0006+1.25*1/10[9+2(0.9997)+(-1.0004)]=(-0.25)*1.0006+1.25*1/10[9.9989]=-0.2501+1.2499=0.9997`
`11^(th)` Approximation
`x_11=(1-1.25)*0.9997+1.25*1/10[7-2(-1.0004)+(0.9997)]=(-0.25)*0.9997+1.25*1/10[10.0006]=-0.2499+1.2501=1.0002`
`y_11=(1-1.25)*-1.0004+1.25*1/8[-4-(1.0002)-3(0.9997)]=(-0.25)*-1.0004+1.25*1/8[-7.9993]=0.2501+-1.2499=-0.9998`
`z_11=(1-1.25)*0.9997+1.25*1/10[9+2(1.0002)+(-0.9998)]=(-0.25)*0.9997+1.25*1/10[10.0005]=-0.2499+1.2501=1.0001`
Solution By SOR (successive over-relaxation) method.
`x=1.0002~=1`
`y=-0.9998~=-1`
`z=1.0001~=1`
Iterations are tabulated as below
| Iteration | x | y | z | | 1 | 0.875 | -0.7617 | 1.2485 | | 2 | 1.0027 | -1.1765 | 0.9165 | | 3 | 1.033 | -0.9219 | 1.0389 | | 4 | 0.9771 | -1.0342 | 0.9803 | | 5 | 1.0118 | -0.9841 | 1.0099 | | 6 | 0.9943 | -1.0077 | 0.9951 | | 7 | 1.0027 | -0.9962 | 1.0024 | | 8 | 0.9987 | -1.0018 | 0.9988 | | 9 | 1.0007 | -0.9991 | 1.0006 | | 10 | 0.9997 | -1.0004 | 0.9997 | | 11 | 1.0002 | -0.9998 | 1.0001 |
This material is intended as a summary. Use your textbook for detail explanation.
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