Simple relaxation method Matlab

Hi.

I want to solve an ecuation system using the simple relaxation method.

I wrote the code using an exemple but the result is wrong. ( Or null? )

A=[ 67 -8 -3 78; -8 65 12 12 ; -3 12 64 -17; 78 12 -17 61];

b=[4; 6 ; 8 ;10];

n=length (b);

x=zeros (n,1) ;

r=A*x-b;

while norm (r, inf) > (10^ (-10) )

p=zeros (n,1);

[c, j]=max(abs (r) );

p ( j) =1;

t= ( -r (j) ) / A (j,j) ;

x=x+t*p;

r=A*x-b;

end

x

i

This is what I have managed to write and the result is NaN, NaN, NaN, Inf.

Is there anything I'm doing wrong? I apologise if this seems like a silly question. I'm just new into it and trying to discover how to solve it.

Thank you! I have tried both r2009a and the online version.

[Torsten]

Never seen this iteration scheme. Does it have a name ? Can you give a link where it is described ?

[Rolland-Luigi Eva]

Unfortunately it is not described, probably that's why I could't solve it, I didn't understand the solving steps properly.

It is from an university course.

Answer 1

[Alan Stevens]

For the simple relaxation technique to work the A matrix needs to be diagonally dominant. Your A is not diagonally dominant - see first and last rows.

For example, if we make A diagonally dominant (I've just arbitrarily modified your A so that it is), your program generates the following:

A=[ 78 -8 -3 67; -8 65 12 12 ; -3 12 64 -17; 61 12 -17 78];
b=[4; 6 ; 8  ;10];
n=length (b);
x=zeros (n,1) ;
r=A*x-b;
while norm (r, inf) > (10^ (-10) )
             p=zeros (n,1);
[c, j]=max(abs (r) );
p ( j) =1;
t= ( -r (j) ) / A (j,j) ;
x=x+t*p;
r=A*x-b;
end
disp(x)

   -0.3524
   -0.0841
    0.2494
    0.4711

Compare this with

disp(A\b)

   -0.3524
   -0.0841
    0.2494
    0.4711

[Rolland-Luigi Eva]

You are right. Thank you.

I have also asked a teacher from my former university who said my exemple has no solution.

Good to learn new things everyday.

[John D'Errico]

@Rolland-Luigi Eva - What you said is not technically correct. The system of equations DOES have a solution.

A=[ 78 -8 -3 67; -8 65 12 12 ; -3 12 64 -17; 61 12 -17 78];
b=[4; 6 ; 8  ;10];
x = A\b

x = 4×1

-0.3524 -0.0841 0.2494 0.4711

That is the solution of that system of equations. It is just not achievable using a simple relaxation scheme, as that appears to be divergent for this problem.

norm(A*x-b)

ans = 3.9721e-15

So zero to within floating point trash for the problem.

The problem is that some linear systems of equations will not converge under a relaxation scheme.

https://en.wikipedia.org/wiki/Successive_over-relaxation

[Rolland-Luigi Eva]

Yes, it couldn't been solved using the method I wanted to.

I apologise.