Home > Matrix & Vector calculators > Solving systems of linear equations using SOR (Successive over-relaxation) method example

12. SOR (Successive over-relaxation) method example ( Enter your problem )
  1. Example 3x-y+z=-1,-x+3y-z=7,x-y+3z=-7
  2. Example 4x+3y=24,3x+4y-z=30,-y+4z=-24
  3. Example 5x+y=10,2x+3y=4
  4. Example 10x+2y-z=7,x+8y+3z=-4,-2x-y+10z=9
Other related methods
  1. Inverse Matrix method
  2. Cramer's Rule method
  3. Gauss-Jordan Elimination method
  4. Gauss Elimination Back Substitution method
  5. Gauss Seidel method
  6. Gauss Jacobi method
  7. Elimination method
  8. LU decomposition using Gauss Elimination method
  9. LU decomposition using Doolittle's method
  10. LU decomposition using Crout's method
  11. Cholesky decomposition method
  12. SOR (Successive over-relaxation) method
  13. Relaxation method

1. Example 3x-y+z=-1,-x+3y-z=7,x-y+3z=-7
(Previous example)
3. Example 5x+y=10,2x+3y=4
(Next example)

2. Example 4x+3y=24,3x+4y-z=30,-y+4z=-24





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
Iterationxyz
17.52.34375-6.76758
23.427733.46069-4.72664
33.398673.8465-5.11631
43.044243.96056-4.98325
53.025923.9908-5.00706
63.002153.99808-4.99883
73.001263.99966-5.0004
833.99996-4.99991
93.000044-5.00002



This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then Submit Here



1. Example 3x-y+z=-1,-x+3y-z=7,x-y+3z=-7
(Previous example)
3. Example 5x+y=10,2x+3y=4
(Next example)





Share this solution or page with your friends.


 
Copyright © 2025. All rights reserved. Terms, Privacy
 
 

.