Jacobi method in Matlab

조회 수: 10 (최근 30일)
gracias claude
gracias claude 2021년 2월 27일
댓글: Walter Roberson 2021년 2월 27일
My jacobi method not working. I feel like I'm missing something. Currently my numbers that are result back are m = 1 and be = 1.6. I wonder if I'm breaking out of one of the loops to fast
How to run :
n=100
[a,b]=sparsesetup(100);
[m,be] = Jacobi(a,b,0.00000025)
should result with -- getting wrong values
m = 36
be = 4.5785 * 10^-7
function [m,be] = Jacobi(a,b,tol)
n = length(b)
d= diag(a); % extract diagonal of a[i,j]
r=a-diag(d);
x=zeros(n,1);
p=zeros(n,1)
c=zeros([n,1]);
e=zeros([n,1]);
n1 =0;
err =0;
relerr = 1;
while(1)
x=b(b-r*x) ./d;
err =abs(norm(x-p));
relerr = err/(norm(x) + eps);
p=x;
n1= n1+1
if(err)
break
end
end
xc=x;
m=n1;
for i=1 : n
for j=1 : n
xa(j) = 1;
c(i) = c(i) + a(i,j) * xa(j);
e(i) = e(i) + a(i,j) * xc(j);
end
end
for i =1 : n
dif(i) = abs(xa(i) - xc(i));
dif2(i) = abs(e(i) - c(i));
end
fe = max (dif, [], 2);
be = max(dif2, [],2);
end
// sparesetup method
function [a,b] = sparsesetup (n)
e = ones(n,1);
a= spdiags ([-e 3*e -e], -1:1, n,n); %Entries of a{i,j}
b = zeros (n,1); % Entries of rhs b
b(1) = 2;
b(n) = 2;
b(2:n-1)=1;
end
  댓글 수: 3
gracias claude
gracias claude 2021년 2월 27일
I'm not sure what the threshold should be ?
Walter Roberson
Walter Roberson 2021년 2월 27일
x=b(b-r*x) ./d;
That is indexing. Is it possible that you want multiplication instead of indexing?

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