Successive over Relaxation (SOR) Method to Solve System of Equations

Problem: Develop a MATLAB code to solve the following system of algebraic equations using the Successive-over-Relaxation Method.

Solution:

Successive over Relaxation Method: This method is just the modification of the Gauss-Seidel method with an addition relaxation factor 𝛚. This method gives convergent solution as there is an option for under relaxation when 𝛚 is less than one. For different 𝛚, the following program can determine the solution. However, when 𝛚 is 0.05, the relatively better results are obtained.

MATLAB Program for Successive Over Relaxation (SOR) Method

function x = sor(a,b,x0,e,N,w)

% a - (nxn) matrix

% b - column vector of length n

% x0 - initial solution

% e - error definition

% N - no. of iterations

% w – relaxation factor

 

format long;

 

m=size(a,1); 

 

n=length(b);

 

p=length(x0);

 

q=length(e);

 

if (m ~= n)

    

    error('a and b do not have the same number of rows')

    

end

 

y0=x0;

 

for i=1:m

   

    [~,f]=max(abs(a(1:m,i:i))); 

                             

   % disp(f);  

 

    a([i,f],1:n)=a([f,i],1:n);                 

                                                                                       

end

 

for t=1:m

    

if a(t,t)==0

    

   [~,g]=max(abs(a(1:m,t:t)));            

        

   a([t,g],1:n)=a([g,t],1:n);

        

end

 

end

    

% disp(a);

 

c=1;

 

while(c<N)

 

for j=1:m  

    

    for k=1:p

        

        Z(1,k)=(a(j,1:n)*y0(1:p,1));

                                                                    

    end

    

    Y(j:j,1)=sum(Z(1,k)); 

       

   % disp(Y);

           

    R(j:j,1)=b(j:j,1)-Y(j:j,1);

    

   % disp (R);

                     

    x(j:j,1)=y0(j:j,1)+((R(j:j,1)*w)/a(j:j,j:j));

            

    y0(j:j,1)= x(j:j,1); 

    

   % disp (y0);    

   % disp (x);

               

end

 

if (abs(x(1:p,1)-x0(1:p,1)) < e(1:q,1))

    

    break;

    

end

 

disp (x);

disp(abs(x(1:p,1)-x0(1:p,1)));

    

x0=y0;

 

c=c+1;

 

end

 

x=x(1:p,1);

       

end

 

Program Outputs:

ans =

   2.322822494693882

   0.675693020694401

   2.160169880982512

  -0.350429027969758

  -0.428941223493063

  -0.394923172868219 

 

Error Percentage: 2.2%, 1.5%, 1.4%, 2.6%, 0.9% and 5.8%.