mayank awasthi......iiit jabalpur ECE

1. EC 313: Numerical Methods Mayank Awasthi(2006033)
MATLAB Assignment – 2 B.Tech 3rd Year(ECE)
Pdpm IIITDM, JABALPUR
Ques1:
Gauss Siedel Method for Solving the Linear Equations:
% Implementing Gauss Siedel method
% a is matrix whose diagonal elements are non-zero else rearrange it
function x = gauss_siedel(a,b,x)
n=length(a);
l=zeros(n,n);
u=zeros(n,n);
d=zeros(n,n);
id=[1,0,0;0,1,0;0,0,1];
for i=1:n
for j=1:n
a(i,j)= a(i,j)/a(i,i); %making the diagonal elements 1.
% Breaking the matrix as a sum of ( L+U+I )
if i>j
l(i,j)=a(i,j);
end
if i<j
u(i,j)=a(i,j);
end
if i==j
d(i,j)=a(i,j);
end
end
end
%Implementing Norm of c where c = inverse(I+L)*U
norm2(inv2(id+l)*u);
if norm2(inv2(id+l)*u)>1
fprintf('Norm of c is greater than 1, Solution will diverge');
return;
end
for i=1:n
b(i,1)=b(i,1)/a(i,i);
end
% Calculating x using FORWARD DIFFERENCE CONCEPT
x=[0;0;0];
for i=1:100
x= forward_subs((l+d),(b-u*x)); m=a*x-b;

2. % Calculating(dis(A*x,b)<1e-6)to stop iteration at the given tolerance
temp=0;
for j=1:n
temp=temp+power(m(j,1),2);
end
temp=sqrt(temp);
if temp<0.0000001
break;
end
end
fprintf('nThe maximum no. of iteration required is %d',i);
end
MATLAB routine for Inverse of 3x3 matrix :
% Calculating Inverse of 3x3 matrix
function x = inv2(A)
I=[1,0,0;0,1,0;0,0,1];
n=length(A);
a21=A(2,1)/A(1,1);
for k = 1:n-1 % Elimination Phase
for i= k+1:n
if A(i,k) ~= 0
lambda = A(i,k)/A(k,k);
A(i,k:n) = A(i,k:n) - lambda*A(k,k:n);
I(i,k:n)= I(i,k:n) - lambda*I(k,k:n);
I(3,1)=I(2,1)*I(3,2)+I(3,1);
end
end
end
x1=back_subs(A,I(:,1)); % Backward Substitution
x2=back_subs(A,I(:,2));
x3=back_subs(A,I(:,3));
x=[x1,x2,x3];
end
MATLAB routine for Norm:
% Calculating Norm of matrix
function n0= norm2(c)
l=length(c);
n0=0;
for m=1:l
for n=1:l
n0 = n0 + c(m,n)*c(m,n);
end
end
n0=sqrt(n0);
end

3. MATLAB routine for Forward Substitution:
% Implementing Forward Substitution
function x=forward_subs(A,b)
n=length(A);
for i = 1:3
t=0;
for j = 1:(i-1)
t=t+A(i,j)*x(j);
end
x(i,1)=(b(i,1)-t)/A(i,i);
end
Part1: Part2:
>> a=[1 ,.2 .2;.2, 1, .2;.2, .2, 1] >> a=[1 ,.5 .5;.5, 1, .5;.5, .5, 1]
a= a=
1.0000 0.2000 0.2000 1.0000 0.5000 0.5000
0.2000 1.0000 0.2000 0.5000 1.0000 0.5000
0.2000 0.2000 1.0000 0.5000 0.5000 1.0000
>> b=[2;2;2] >> b=[2;2;2]
b= b=
2 2
2 2
2 2
>> x=[0;0;0] >> x=[0;0;0]
x= x=
0 0
0 0
0 0
>> gauss_siedel(a,b,x) >> gauss_siedel(a,b,x)
The maximum no. of iteration required is 8 The maximum no of iteration required is 17
ans = ans =
1.4286 1.0000
1.4286 1.0000
1.4286 1.0000

4. Part3:
>> a=[1 ,.9 .9;.9, 1, .9;.9, .9, 1]
a=
1.0000 0.9000 0.9000
0.9000 1.0000 0.9000
0.9000 0.9000 1.0000
>> b=[2;2;2]
b=
2
2
2
>> x=[0;0;0]
x=
0
0
0
>> gauss_siedel(a,b,x)
Norm of c is greater than 1, Solution will diverge
ans =
0
0
0
************************************************************************
Spectral radius of C:
Spectral radius of C is maximum eigenvalue of C*CT.
In this case C=(I+L)-1 * U

5. 1). >> a=[1 ,.2 .2;.2, 1, .2;.2, .2, 1]
a=
1.0000 0.2000 0.2000
0.2000 1.0000 0.2000
0.2000 0.2000 1.0000
>> L=[0,0,0;.2,0,0;.2,.2,0]
L=
0 0 0
0.2000 0 0
0.2000 0.2000 0
>> I=[1,0,0;0,1,0;0,0,1]
I=
1 0 0
0 1 0
0 0 1
>> U=[0,.2,.2;0,0,.2;0,0,0]
U=
0 0.2000 0.2000
0 0 0.2000
0 0 0
>> C=inv2(I+L)*U
C=
0 0.2000 0.2000
0 -0.0400 0.1600
0 -0.0320 -0.0720
>> lamda=eig(C*C')
lamda =
0.0000
0.0181
0.0953
>> SR=max(lamda)
SR = Spectral Radius
0.0953

6. 2).
>> a=[1 ,.5 .5;.5, 1, .5;.5, .5, 1]
a=
1.0000 0.5000 0.5000
0.5000 1.0000 0.5000
0.5000 0.5000 1.0000
>> L=[0,0,0;.5,0,0;.5,.5,0]
L=
0 0 0
0.5000 0 0
0.5000 0.5000 0
>> I=[1,0,0;0,1,0;0,0,1]
I=
1 0 0
0 1 0
0 0 1
>> U=[0,.5,.5;0,0,.5;0,0,0]
U=
0 0.5000 0.5000
0 0 0.5000
0 0 0
>> C=inv2(I+L)*U
C=
0 0.5000 0.5000
0 -0.2500 0.2500
0 -0.1250 -0.3750
>> lamda=eig(C*C')
lamda =
0.0000
0.1481
0.6332
>> SR=max(lamda)
SR = Spectral Radius
0.6332

7. 3).
>> a=[1 ,.9 .9;.9, 1, .9;.9, .9, 1]
a=
1.0000 0.9000 0.9000
0.9000 1.0000 0.9000
0.9000 0.9000 1.0000
>> L=[0,0,0;.9,0,0;.9,.9,0]
L=
0 0 0
0.9000 0 0
0.9000 0.9000 0
>> I=[1,0,0;0,1,0;0,0,1]
I=
1 0 0
0 1 0
0 0 1
>> U=[0,.9,.9;0,0,.9;0,0,0]
U=
0 0.9000 0.9000
0 0 0.9000
0 0 0
>> C=inv2(I+L)*U
C=
0 0.9000 0.9000
0 -0.8100 0.0900
0 -0.0810 -0.8910
>> lamda=eig(C*C')
lamda =
0.0000
0.7301
2.3546
>> SR=max(lamda)
SR = Spectral Radius
2.3546

8. Steps vs t :
Spectral radius vs t :

9. Ques2:
Jacobi Method for Solving the Linear Equations:
%Implementing Jacobi Method
function x = jacobi(A,b,x)
I=[1 0 0; 0 1 0; 0 0 1];
c=I-A;
%Checking Norm
if norm2(c) > 1
fprintf('nNorm is greater than one so it will diverge');
return;
end
for j=1:50
x=b+(I-A)*x;
n=A*x-b;
%checking the condition of tolerance to stop the iterations
temp=0;
for i = 1:3
temp= temp+ power(n(i,1),2);
end
temp=sqrt(temp);
if temp < 0.0001
break;
end
end
fprintf('nThe iteration required is %d',j);
end
MATLAB routine for Norm:
% Calculating Norm of matrix
function n0= norm2(c)
l=length(c);
n0=0;
for m=1:l
for n=1:l
n0 = n0 + c(m,n)*c(m,n);
end
end
n0=sqrt(n0);
end

10. Part1: Part2
>> a=[1 ,.2 .2;.2, 1, .2;.2, .2, 1] >> a=[1 ,.5 .5;.5, 1, .5;.5, .5, 1]
a= a=
1.0000 0.2000 0.2000 1.0000 0.5000 0.5000
0.2000 1.0000 0.2000 0.5000 1.0000 0.5000
0.2000 0.2000 1.0000 0.5000 0.5000 1.0000
>> b=[2;2;2] >> b=[2;2;2]
b= b=
2 2
2 2
2 2
>> x=[0;0;0] >> x=[0;0;0]
x= x=
0 0
0 0
0 0
>> jacobi(a,b,x) >> jacobi(a,b,x)
The iteration required is 12 Norm is greater than one so it will diverge
ans = ans =
1.4285 0
1.4285 0
1.4285 0

11. Part3 :
>> a=[1 ,.9 .9;.9, 1, .9;.9, .9, 1]
a=
1.0000 0.9000 0.9000
0.9000 1.0000 0.9000
0.9000 0.9000 1.0000
>> b=[2;2;2]
b=
2
2
2
>> x=[0;0;0]
x=
0
0
0
>> jacobi(a,b,x)
Norm is greater than one so it will diverge
ans =
0
0
0
From the Data we have calculated we find that Gauss_Siedel Method Diverge more
fastly than Jacobi Method.

12. Ques3:
Cholesky Method for Solving the Linear Equations:
% Implementing Cholesky Method
function x = cholesky(A,b,x)
n=length(A)
l(1,1)=sqrt(A(1,1));
for i=2:n
l(i,1)=A(i,1)/l(1,1);
end
for j=2:(n-1)
temp=0;
for k=1:j-1
temp=temp+l(j,k)*l(j,k);
end
l(j,j)= sqrt(A(j,j)-temp);
end
for j=2:(n-1)
for i=j+1:n
temp=0;
for k=1:j-1
temp=temp+l(i,k)*l(j,k);
end
l(i,j)=(A(i,j)-temp)/l(j,j);
end
end
temp1=l(n,1)*l(n,1)+l(n,2)*l(n,2);
l(n,n)=sqrt(A(n,n)-temp1);
y=forward_subs(l,b); %Using Forward Substitution Concept
x=back_subs(l',y);
end

13. Part1: Part2
>> a=[1 ,.2 .2;.2, 1, .2;.2, .2, 1] >> a=[1 ,.5 .5;.5, 1, .5;.5, .5, 1]
a= a=
1.0000 0.2000 0.2000 1.0000 0.5000 0.5000
0.2000 1.0000 0.2000 0.5000 1.0000 0.5000
0.2000 0.2000 1.0000 0.5000 0.5000 1.0000
>> b=[2;2;2] >> b=[2;2;2]
b= b=
2 2
2 2
2 2
>> x=[0;0;0] >> x=[0;0;0]
x= x=
0 0
0 0
0 0
>> cholesky(a,b,x) >> cholesky(a,b,x)
ans = ans =
1.4286 1.0000
1.4286 1.0000
1.4286 1.0000

14. Part3:
>> a=[1 ,.9 .9;.9, 1, .9;.9, .9, 1]
a=
1.0000 0.9000 0.9000
0.9000 1.0000 0.9000
0.9000 0.9000 1.0000
>> b=[2;2;2]
b=
2
2
2
>> x=[0;0;0]
x=
0
0
0
>> cholesky(a,b,x)
ans =
0.7143
0.7143
0.7143

15. Ques4:
Matlab Subroutine to calculate Condition Number:
% Calculating Condition Number
function [] = cond(a,b,b1);
x0=[0;0;0];
val1=max(abs(eig(a)));
val2=min(abs(eig(a)));
val=val1/val2 %definition of Condition number
x=gauss_siedel(a,b,x0);
x1=gauss_siedel(a,b1,x0);
errip=norm(x-x1)/norm(x);
errop=norm(b-b1)/norm(b);
fprintf('nerror in i/p is %d nand error in o/p is
%dn',errip,errop);
end
1).
>> a=[1 ,.2 .2;.2, 1, .2;.2, .2, 1]
a=
1.0000 0.2000 0.2000
0.2000 1.0000 0.2000
0.2000 0.2000 1.0000
>> b=[2;2;2]
b=
2
2
2
>> b1=[1.1;2.1;3.1]
b1 =
1.1000
2.1000
3.1000

16. >> cond(a,b,b1)
a=
1.0800 0.4400 0.4400
0.4400 1.0800 0.4400
0.4400 0.4400 1.0800
val =
3.0625 Condition Number.
The maximum no. of iteration required is 14
error in i/p is 1.309812e+000
and error in o/p is 4.112988e-001
2).
>> a=[1 ,.5 .5;.5, 1, .5;.5, .5, 1]
a=
1.0000 0.5000 0.5000
0.5000 1.0000 0.5000
0.5000 0.5000 1.0000
>> b=[2;2;2]
b=
2
2
2
>> b1=[1.1;2.1;3.1]
b1 =
1.1000
2.1000
3.1000

17. >> cond(a,b,b1)
a=
1.5000 1.2500 1.2500
1.2500 1.5000 1.2500
1.2500 1.2500 1.5000
val =
16.0000 Condition Number
Norm of c is greater than 1, Solution will diverge.
3).
>> a=[1 ,.9 .9;.9, 1, .9;.9, .9, 1]
a=
1.0000 0.9000 0.9000
0.9000 1.0000 0.9000
0.9000 0.9000 1.0000
>> b1=[1.1;2.1;3.1]
b1 =
1.1000
2.1000
3.1000
>> b1=[1.1;2.1;3.1]
b1 =
1.1000
2.1000
3.1000
>> b=[2;2;2]
b=
2
2
2

18. >> cond(a,b,b1)
a=
2.6200 2.6100 2.6100
2.6100 2.6200 2.6100
2.6100 2.6100 2.6200
val =
28.0000 Condition Number.
Norm of c is greater than 1, Solution will diverge.