ASSIGNMENT 1 GAUSS ELIMINATION

Transcription

1 ASSIGNMENT 1 GAUSS ELIMINATION SOURCE CODE: function []= gauss_elim(a,b) [m n]=size(a); [n p]=size(b); if(m==n) if(rank([a b])==rank(a)) f=1; fprintf('the given system of linear equations is consistent'); else fprintf('the given system of linear equations is inconsistent'); fprintf('thus no solution'); if(f==1) X = zeros(n,p); for i=1:n-1 mult=-a(i+1:n,i)/a(i,i); A(i+1:n,:)= A(i+1:n,:)+ mult*a(i,:); %Here mult is the multiplier b(i+1:n,:)= b(i+1:n,:)+ mult*b(i,:); % Back Substitution to find the unknowns X(n,:) = b(n,:)/a(n,n); for i = n-1:-1:1 X(i,:) = (b(i,:) - A(i,i+1:n)*X(i+1:n,:))/A(i,i); fprintf('the solution of the given system of equations is:\n'); disp(x); itr=n*(n+1)/2; fprintf('the number of iterations in which gauss elimination can be performed are: %d\n',itr); else fprintf('gauss Elimination cannot be done'); 1

2 OUTPUT 1: >> A = [ ; ; ; ] A = >> b = [5; 18; -4; 11] b = >> gauss_elim(a,b) The given system of linear equations is consistent. The solution of the given system of equations is: The number of iterations in which gauss elimination can be performed are: 10 2

3 ASSIGNMENT 2 POWER METHOD SOURCE CODE: function []=powereigen(y0,a,tolr) % Power Method % To find largest and smallest eigen values of given square matrix. % Inputs - % Y0 n * 1 matrix (Initial Guess) % A n * n matrix (Given Matrix) % tolr (Tolerance) %**************** Largest Eigen-Value *****************% dd = 1; x = Y0; n = 10; while dd > tolr y = A * x; dd = abs(norm(x) - n); n = norm(x); x = y./ n; lvalue = n %*************** Smallest Eigen-Value *****************% dd = 1; x = Y0; n = 10; while dd>tolr y = A \ x; dd = abs(norm(x) - n); n = norm(x); x = y./ n; if n == 0 sprintf('smallest eigen value not exist.'); else svalue = 1 / n 3

4 OUTPUT 2: >> A = [5-2 0; 1 2-3; 1-2 4] A = >> x = [1; 1; 1] x = >> tolr = tolr = e-03 >> powereigen(x,a,tolr) largest-value = smallest-value =

5 ASSIGNMENT 3 CURVE FITTING SOURCE CODE: % For the given set of points (x,y), the points to be fitted to a % polynomial linear / quadratic / cubic on user input and to plot graph of % the points with the polynomial clc; clear all; %Inputs: X = [0.02;0.03;0.04;0.05;0.06;0.07;0.08;0.09;0.10;0.11;0.13;0.15;0.17]; Y = [0.97;0.70;0.58;0.495;0.42;0.37;0.33;0.30;0.28;0.26;0.24;0.24;0.225]; a = zeros(13,1); f = [0,0.01,0.12,0.14,0.16]; s = sum(x); s2 = sum(x.*x); s3 = sum(x.*x.*x); s4 = sum(x.*x.*x.*x); s5 = sum(x.*x.*x.*x.*x); s6 = sum(x.*x.*x.*x.*x.*x); t = sum(y); t1 = sum(x.*y); t2 = sum(x.*x.*y); t3 = sum(x.*x.*x.*y); disp('welcome'); disp('1. Linear'); disp('2. Quadratic'); disp('3. Cubic'); disp('4. Exit'); ch = input('choose from above - '); disp(' '); while(ch ~= 4) if(ch == 1) A = [13,s;s,s2]; B = [t;t1]; sol = linsolve(a,b); plot(y,x,'--rs') 5

6 hold on for i = 1:13 a(i) = sol(1) + sol(2) * X(i); plot(a,x) hold off disp('linear - '); for i = 1:5 temp =sol(1) + sol(2) * f(i); disp(sprintf('y(%g) = %g',f(i),temp)); elseif(ch == 2) A = [13,s,s2;s,s2,s3;s2,s3,s4]; B = [t;t1;t2]; sol = linsolve(a,b); plot(y,x,'--rs') hold on for i = 1:13 a(i) = sol(1) + sol(2) * X(i) + sol(3) * X(i) * X(i); plot(a,x) hold off disp('quadratic - '); for i = 1:5 temp = sol(1) + sol(2) * f(i) + sol(3) * f(i) * f(i); disp(sprintf('y(%g) = %g',f(i),temp)); elseif(ch == 3) A = [13,s,s2,s3;s,s2,s3,s4;s2,s3,s4,s5;s3,s4,s5,s6]; B = [t;t1;t2;t3]; sol = linsolve(a,b); plot(y,x,'--rs') hold on for i = 1:13 a(i) = sol(1) + sol(2) * X(i) + sol(3) * X(i) * X(i) + sol(4) * X(i) * X(i) * X(i); plot(a,x) hold off disp('cubic - '); for i = 1:5 6

7 temp = sol(1) + sol(2) * f(i) + sol(3) * f(i) * f(i) + sol(4) * f(i) * f(i) * f(i); disp(sprintf('y(%g) = %g',f(i),temp)); else disp('incorrect input.'); disp(' '); disp(' '); disp('1. Linear'); disp('2. Quadratic'); disp('3. Cubic'); disp('4. Exit'); ch = input('choose from above - '); disp(' '); disp('thank You'); OUTPUT 3: Linear - Y(0) = Y(0.01) = Y(0.12) = Y(0.14) = Y(0.16) =

8 Quadratic - Y(0) = Y(0.01) = Y(0.12) = Y(0.14) = Y(0.16) = Cubic - Y(0) = Y(0.01) = Y(0.12) = Y(0.14) = Y(0.16) =

9 ASSIGNMENT 4 GRAM SCHMIDT SOURCE CODE: % the Gram Schmidt process is a method for orthonormalising % a set of vectors in an inner product space %input: %A: input matrix %m,n: size of matrix A %output: %R: matrix obtained by GramSchmidt Method clear all,close all,clc; %input: A=input('Enter matrix you want to apply Gram Schmidt: '); [m,n]=size(a); R=zeros(m,n); Q=A; %solution for j=1:n R(:,j)=Q(:,j); if(j>1) for i=1:j-1 R(:,j)=R(:,j)-sum(Q(:,j).*R(:,i))*R(:,i)/norm(R(:,i))^2;%Applying Formula %output disp('matrix obtained by Gram Schmidt method is :'); disp(' '); disp(r); OUTPUT 4: Enter matrix you want to apply Gram Schmidt: [1 2 3 ; ; 0-2 3] Matrix obtained by Gram Schmidt method is : 9

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11 ASSIGNMENT 5.1 GAUSS ELIMINATION SOURCE CODE: function []= gauss_elim(a,b) [m n]=size(a); [n p]=size(b); if(m==n) if(rank([a b])==rank(a)) f=1; fprintf('the given system of linear equations is consistent'); else fprintf('the given system of linear equations is inconsistent'); fprintf('thus no solution'); if(f==1) X = zeros(n,p); for i=1:n-1 mult=-a(i+1:n,i)/a(i,i); A(i+1:n,:)= A(i+1:n,:)+ mult*a(i,:); %Here mult is the multiplier b(i+1:n,:)= b(i+1:n,:)+ mult*b(i,:); % Back Substitution to find the unknowns X(n,:) = b(n,:)/a(n,n); for i = n-1:-1:1 X(i,:) = (b(i,:) - A(i,i+1:n)*X(i+1:n,:))/A(i,i); fprintf('the solution of the given system of equations is:\n'); disp(x); itr=n*(n+1)/2; fprintf('the number of iterations in which gauss elimination can be performed are: %d\n',itr); else fprintf('gauss Elimination cannot be done'); 11

12 OUTPUT 5.1: >> A = [ ; ; ; ] A = >> b = [5; 18; -4; 11] b = >> gauss_elim(a,b) The given system of linear equations is consistent. The solution of the given system of equations is: The number of iterations in which gauss elimination can be performed are: 10 12

13 ASSIGNMENT 5.2 LU DECOMPOSITION SOURCE CODE: function []=LU(A,b) [m n]=size(a); [n p]=size(b); X=zeros(n,p); Y=zeros(n,p); L=eye(n); if(m==n) if(rank([a b])==rank(a)) f=1; fprintf('the given system of linear equations is consistent \n'); else fprintf('the given system of linear equations is inconsistent \n'); fprintf('thus no solution'); if f==1 for i=1:n-1 mult=-a(i+1:n,i)/a(i,i); L(i+1:n,i)=-mult; A(i+1:n,:)=A(i+1:n,:)+ mult*a(i,:); U=A; fprintf('the unit lower triangulat matrix "L" is:\n'); disp(l); fprintf('the upper triangulat matrix "U" is:\n'); disp(u); %LUX=b, UX=Y,LY=b %to find Y from LY=b by forward substitution Y(1,:)=b(1,:); for i=2:n Y(i,:)=b(i,:)-L(i,1:i-1)*Y(1:i-1,:); %to find the the solution X of the given system of equations by backword substitution X(n,:)=Y(n,:)/U(n,n); for i=n-1:-1:1 X(i,:)=(Y(i,:)-U(i,i+1:n)*X(i+1:n,:))/U(i,i); 13

14 fprintf('hence the solution of the given system of equations is:\n'); disp(x); itr=(n*(n+1)/2)+n; fprintf('the number of iterations in which gauss elimination can be performed are: %d\n',itr); OUTPUT 5.2: >> A = [ ; ; ; ] A = >> b = [5; 18; -4; 11] b = >> LU(A,b) The given system of linear equations is consistent the unit lower triangulat matrix "L" is: the upper triangulat matrix "U" is:

15 Hence the solution of the given system of equations is: the number of iterations in which gauss elimination can be performed are: 14 15

16 ASSIGNMENT 5.3 SOR METHOD SOURCE CODE: function[]= sor(a, b, N) n = size(a,1); %splitting matrix A into the three matrices L, U and D D = diag(diag(a)); L = tril(-a,-1); U = triu(-a,1); Tj = inv(d)*(l+u); rho_tj = max(abs(eig(tj))); w = 2./(1+sqrt(1-rho_Tj^2)); disp('w ='); disp(w); disp('the rate of convergence is:'); disp(-log10(w-1)); %Jacobi iteration matrix %spectral radius of Tj %optimal overrelaxation parameter Tw = inv(d-w*l)*((1-w)*d+w*u); %SOR iteration matrix cw = w*inv(d-w*l)*b; %constant vector needed for iterations tol = 1e-05; k = 1; x = zeros(n,1); %starting vector while k <= N x(:,k+1) = Tw*x(:,k) + cw; if norm(x(:,k+1)-x(:,k)) <tol disp('the procedure was successful') disp('condition x^(k+1) - x^(k) <tol was met after k iterations') disp(k); disp('x = '); disp(x(:,k+1)); break k = k+1; k=1; if norm(x(:,k+1)- x(:,k)) >tol k > N disp('maximum number of iterations reached without satisfying condition:') disp(' x^(k+1) - x^(k) <tol'); disp(tol); disp('please, examine the sequence of iterates') 16

17 disp('in case you observe convergence, then increase the maximum number of iterations') disp('in case of divergence, the matrix may not be diagonally dominant') fprintf('\n\nthus, the solution of the given system of equations by SOR method is:\n'); disp(x); OUTPUT 5.3: >> A=[3 2 0;2 3-1;0-1 2] A = >> b=[4.5;5;-.5] b = >> N=3 N = 3 >>sor(a,b,n) w = the rate of convergence is: Maximum number of iterations reached without satisfying condition: x^(k+1) - x^(k) <tol e-05 17

18 Please, examine the sequence of iterates In case you observe convergence, then increase the maximum number of iterations In case of divergence, the matrix may not be diagonally dominant Thus, the solution of the given system of equations by SOR method is: