Simple relaxation method Matlab

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Rolland-Luigi Eva
评论： Rolland-Luigi Eva 2023-2-9
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.
  2 个评论
Torsten
Torsten 2023-2-9
Never seen this iteration scheme. Does it have a name ? Can you give a link where it is described ?
Rolland-Luigi Eva
Rolland-Luigi Eva 2023-2-9
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.

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采纳的回答

Alan Stevens
Alan Stevens 2023-2-9
编辑：Alan Stevens 2023-2-9
Ran in:
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
  3 个评论
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John D'Errico
John D'Errico 2023-2-9
Ran in:
@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
Rolland-Luigi Eva 2023-2-9
Yes, it couldn't been solved using the method I wanted to.
I apologise.

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