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Assignment No: 1

Statement: Write down the Matlab Program using Newton-Raphson method for any equation.



Solution:

Input :

clc; clear all; % Clears the workspace
format long;

% Take i/p from user
a=input('n Enter the function : ','s');
b=input('n Enter the derivative of function :         ','s');

% Converts the input string into symbolic function
ft=inline(a);
dft=inline(b);
n=input('enter no. of significant digits: ');
t0=0;
epsilon_s=(0.5*10^(2-n));
epsilon_a=100;
tr=fzero((ft),t0); % Solver
 disp (tr);
 varun=sprintf('BY NEWTON-RAPHSON METHOD:');
disp(varun);
  tx=input('Enter your initial guess for root: ');
 td=tx;
 head=sprintf('Time tttttepsilon_a tttttepsilon_t
tttttepsilon_s ');
disp(head);
while (epsilon_a>=epsilon_s)
    tnew=td-(ft(td)/dft(td));
    epsilon_a=abs((tnew-td)/tnew)*100;
    epsilon_t=abs((tr-tnew)/tr)*100;
    td=tnew;
    table=sprintf('%d ttt %ftttt %4.9f ttt %f ttt
%f',tnew,epsilon_a,epsilon_t,epsilon_s);
    disp(table);
end

% Prints the answer
fprintf('n n The root of the equation is :          %f n',tnew)




                                                     Third Year Mechanical Engineering
                                                  Computer Oriented Numerical Methods
                                                                             2011-12©
                                                        MITCOE Mechanical Engineering
Output :

Enter the function :   (exp(t))*cos(t)-1.4



 Enter the derivative of function :    (exp(t))*cos(t)-(exp(t))*sin(t)

enter no. of significant digits: 4

   0.433560875352657



BY NEWTON-RAPHSON METHOD:

Enter your initial guess for root: 0

Time              epsilon_a                  epsilon_t               epsilon_s

4.000000e-001     100.000000              7.740752743                 0.005000


4.327044e-001      7.558146               0.197537266                 0.005000


4.335602e-001     0.197392                0.000145353                 0.005000


4.335609e-001     0.000145                0.000000000                 0.005000




 The root of the equation is :   0.433561




                                                   Third Year Mechanical Engineering
                                                Computer Oriented Numerical Methods
                                                                           2011-12©
                                                      MITCOE Mechanical Engineering
Assignment No: 2

Statement: Write down the Matlab Program using Modified Newton-Raphson method for any
equation.



Solution:

Input :

clc; Clear all; % Clears the workspace

% Takes the   i/p from user
a=input('n   Enter the function : ','s');
b=input('n   Enter the derivative of function : ','s');
c=input('n   Enter second order derivative : ','s');
x0=0;

% Converts the input string into symbolic function
fx=inline(a);
dfx=inline(b);
d2fx=inline(c);
n=input('Enter number of significant digits: ');
epsilon_s=(0.5*10^(2-n));
tr=fzero((fx),0); % Using solver
 disp (tr);
 v=input('n Enter your initial guess for root :    ');
told=v;
varun=sprintf('BY MODIFIED NEWTON-RAPHSON METHOD:');
disp(varun);
head=sprintf('Time tttttepsilon_a tttttepsilon_t
tttttepsilon_s ');
disp(head);
while(1)
     tnew=told-((fx(told)*dfx(told)/((dfx(told)^2)-(fx(told)*d2fx(told)))));
     err=abs((tnew-told)/tnew)*100;
     epsilon_t=abs((tr-tnew)/tr)*100;
     told=tnew;
     table=sprintf('%d ttt %ftttt %4.9f ttt %f ttt
%f',tnew,err,epsilon_t,epsilon_s);
    disp(table);
     if (err<=epsilon_s)
         break;
     end
end
fprintf('n n The root of the equation is :    %f n',tnew)




                                                     Third Year Mechanical Engineering
                                                  Computer Oriented Numerical Methods
                                                                             2011-12©
                                                        MITCOE Mechanical Engineering
Output

 Enter the function :   x*sin(x)+cos(x)



 Enter the derivative of function :    x*cos(x)



 Enter second order derivative :    cos(x)-(x*sin(x))

Enter number of significant digits: 5

   -2.7984




 Enter your initial guess for root :      6

BY MODIFIED NEWTON-RAPHSON METHOD:

Time              epsilon_a            epsilon_t                epsilon_s

6.117645e+000      1.923040          318.613323399              0.000500


6.121248e+000      0.058870          318.742096720             0.000500


6.121250e+000      0.000035        318.742173765               0.000500




 The root of the equation is :     6.121250




                                                      Third Year Mechanical Engineering
                                                   Computer Oriented Numerical Methods
                                                                              2011-12©
                                                         MITCOE Mechanical Engineering
Assignment No: 3

Statement: Write down the Matlab Program using successive approximation method for any
equation.



Solution:

Input :

clc;
clear;
g=input('Enter the function:','s');
f=inline(g);              % Defining function
n=input('Enter number of significant digits: ');
es=(0.5*10^(2-n));        % Stopping criteria
ea=100;
t0=0;
t=input('Enter initial guess: ');
tr=fzero((f),t0);         % Calculating true roots
 disp (tr);
 head1=sprintf('BY SUCCESSIVE APPROXIMATION METHOD:');
disp(head1);
 head=sprintf('Time tttttepsilon_a ttttt epsilon_t    ttttt
epsilon_s ');
disp(head);
while (ea>=es)
     temp=t;
     t=f(t);
     ea=abs((t-temp)/t)*100; % Calc approximate error
     et=abs((tr-t)/tr)*100;   % Calc true error
     table=sprintf('%d ttt %ftttt %4.9f ttt %f ttt
%f',t,ea,et,es);
    disp(table);
end




                                                      Third Year Mechanical Engineering
                                                   Computer Oriented Numerical Methods
                                                                              2011-12©
                                                         MITCOE Mechanical Engineering
Output :

Enter the function:(exp(-x)-x)

Enter number of significant digits: 2

Enter initial guess: 0

BY SUCCESSIVE APPROXIMATION METHOD:

   0.567143290409784




                                           Third Year Mechanical Engineering
                                        Computer Oriented Numerical Methods
                                                                   2011-12©
                                              MITCOE Mechanical Engineering
Assignment No: 4

Statement: Write down the Matlab Program using Gauss Naïve elimination method.



Solution:

Input :

clc;
clear all;
a=input('enter matrix A[]: ')
b=input('enter column matrix B[]: ')
[m,n]=size(a); % determines size of matrix.
 if (m~=n) error('Matrix Must Be Square!'); end
%forward elimination
for k=1:n-1
    for i=k+1:n
        factor=a(i,k)/a(k,k);
        for j=k:n
            a(i,j)=a(i,j)-(factor*(a(k,j)));% calculates each element of matrix
A.
            end
            b(i)=b(i)-factor*(b(k)); % calculates each element of matrix B.
     end
     disp (a);
end
disp (a);
disp (b);

% backward substitution
for i=n:-1:1
    x(i)=b(i)/a(i,i); % calculates values of unknown matrix.
    for j=1:i-1
        b(j)=b(j)-x(i)*a(j,i);
    end
end
disp('VALUES ARE:')
disp(x)




                                                       Third Year Mechanical Engineering
                                                    Computer Oriented Numerical Methods
                                                                               2011-12©
                                                          MITCOE Mechanical Engineering
Output :


enter matrix A[]: [1 -1 1 ;3 4 2 ; 2 1 1 ]

a =

        1       -1       1
        3        4       2
        2        1       1

enter column matrix B[]: [6
9
7]

b =

        6
        9
        7

       1.0000        -1.0000    1.0000
            0         7.0000   -1.0000
            0              0   -0.5714

       6.0000
      -9.0000
      -1.1429

VALUES ARE:
    3.0000           -1.0000   2.0000




                                                Third Year Mechanical Engineering
                                             Computer Oriented Numerical Methods
                                                                        2011-12©
                                                   MITCOE Mechanical Engineering
Assignment No: 5

Statement: Write down the Matlab Program using Gauss with partial pivoting method.



Solution:

Input :

clc;
clear all;
a=input('enter matrix A[]: ');
b=input('enter column matrix B[]: ');
[m,n]=size(a); % calculates size of matrix A.
 if (m~=n) error('Matrix Must Be Square!'); end
%pivoting
for k=1:n-1
    [xyz,i]=max(abs(a(k:n,k))); % finds maximum element in matrix A.
    ipr=i+k-1;
    if ipr~=k
        a([k,ipr],:)=a([ipr,k],:); % interchanging of rows.
        b([k,ipr],:)=b([ipr,k],:); % interchanging of rows.
    end
    %forward elimination
    for i=k+1:n
        factor=a(i,k)/a(k,k);
        for j=k:n
            a(i,j)=a(i,j)-(factor*(a(k,j))); % calculates each element of matrix
A.
            end
            b(i)=b(i)-factor*(b(k)); % calculates each element of matrix B.
     end
     disp (a);
end
%disp (a);
disp (b);

% backward substitution
for i=n:-1:1
    x(i)=b(i)/a(i,i); % calculates values of unknown matrix.
    for j=1:i-1
        b(j)=b(j)-x(i)*a(j,i);
    end
end
disp('VALUES ARE:')
disp(x)




                                                       Third Year Mechanical Engineering
                                                    Computer Oriented Numerical Methods
                                                                               2011-12©
                                                          MITCOE Mechanical Engineering
Output :

enter matrix A[]: [2 -6 -1;-3 -1 7;-8 1 -2]

enter column matrix B[]: [-38;-34;-20]

  -8.00000000000000        1.00000000000000    -2.00000000000000

                       0   -1.37500000000000   7.75000000000000

                       0   -5.75000000000000   -1.50000000000000



  -8.00000000000000        1.00000000000000    -2.00000000000000

                       0   -5.75000000000000   -1.50000000000000

                       0                   0   8.10869565217391



 -20.00000000000000

 -43.00000000000000

 -16.21739130434783



VALUES ARE:

     4        8   -2




                                                       Third Year Mechanical Engineering
                                                    Computer Oriented Numerical Methods
                                                                               2011-12©
                                                          MITCOE Mechanical Engineering
Assignment No: 6

Statement: Write down the Matlab Program using Thomas Algorithm method.



Solution:

Input :

clc;
clear;
%format long;
e=input('Enter the value of e, ie. subdiagonal vector :');
f=input('Enter the value of f, ie. diagonal vector :');
g=input('Enter the value of g, ie. superdiagonal vector :');
r=input('Enter the value of r, ie. value vector :');
n=length(e);     % Size of matrix e
for k=1:n
     factor=e(k)/f(k);          % Multiplying factor
     f(k+1)=f(k+1)-factor*g(k); % Transforming diagonal vector
     r(k+1)=r(k+1)-factor*r(k); % Transforming value vector
end
x(n+1)=r(n+1)/f(n+1);      % Transforming unknown vector
for k=n:-1:1
     x(k)=(r(k)-g(k)*x(k+1))/f(k);   % Finding values of unknowns
end
disp('VALUES ARE:');
disp (x)




Output :
Enter the value of e, ie. subdiagonal vector :[-.4;-.4]

Enter the value of f, ie. diagonal vector :[0.8;0.8;0.8]

Enter the value of g, ie. superdiagonal vector :[-.4;-.4]

Enter the value of r, ie. value vector :[41;25;105]

VALUES ARE:

  173.7500    245.0000   253.7500




                                                      Third Year Mechanical Engineering
                                                   Computer Oriented Numerical Methods
                                                                              2011-12©
                                                         MITCOE Mechanical Engineering
Assignment No: 7

Statement: Write down the Matlab Program using Gauss Seidel without Relaxation method.



Solution:

Input :

clc;
clear all;
format long;
a = input('Enter Matrix A: ');
b = input('Enter Column Matrix B: ');
[m,n]= size(a); % calculates size of matrix A.
if (m~=n) error('Matrix Must Be Square!'); end
for i=1:n
     d(i)=b(i)/a(i,i);
end
d=d';
c=a;
for i=1:n
     for j=1:n
         c(i,j)=a(i,j)/a(i,i); % factor.
     end
     c(i,i)=0;
     x(i)=0;
end
x=x';
disp (a);
disp (b);
disp (d);
disp (c);
p = input('Enter No. of Iterations: ');
for k=1:p
     for i=1:n
         x(i)=d(i)-c(i,:)*x(:,1); % finds unknown value.
     end
     disp (x);
end




                                                       Third Year Mechanical Engineering
                                                    Computer Oriented Numerical Methods
                                                                               2011-12©
                                                          MITCOE Mechanical Engineering
Third Year Mechanical Engineering
Computer Oriented Numerical Methods
                           2011-12©
      MITCOE Mechanical Engineering
Output :

Enter Matrix A: [3 -0.1 -0.2;0.1 7 -.3;0.3 -0.2 10]

Enter Column Matrix B: [7.85;-19.3;71.4]

   3.00000000000000   -0.10000000000000    -0.20000000000000

   0.10000000000000   7.00000000000000     -0.30000000000000

   0.30000000000000   -0.20000000000000    10.00000000000000



   7.85000000000000

 -19.30000000000000

  71.40000000000001



                  0   -0.03333333333333    -0.06666666666667

   0.01428571428571                  0     -0.04285714285714

   0.03000000000000   -0.02000000000000                    0



Enter No. of Iterations: 3

   2.61666666666667

  -2.79452380952381

   7.00560952380952



   2.99055650793651

  -2.49962468480726

   7.00029081106576



   3.00003189791081

  -2.49998799235305

   6.99999928321562

                                                   Third Year Mechanical Engineering
                                                Computer Oriented Numerical Methods
                                                                           2011-12©
                                                      MITCOE Mechanical Engineering
Third Year Mechanical Engineering
Computer Oriented Numerical Methods
                           2011-12©
      MITCOE Mechanical Engineering
Assignment No: 8

Statement: Write down the Matlab Program using Gauss Seidel with relaxation method.



Solution:

Input :

clc;
clear all;
a=input('enter matrix A[]: ');
b=input('enter column matrix B[]: ');
[m,n]=size(a); % calculates size of matrix A.
 if (m~=n) error('Matrix Must Be Square!'); end
%pivoting
for k=1:n-1
     [xyz,i]=max(abs(a(k:n,k))); % finds maximum element in matrix A.
     ipr=i+k-1;
     if ipr~=k
         a([k,ipr],:)=a([ipr,k],:); % interchanging of rows.
         b([k,ipr],:)=b([ipr,k],:); % interchanging of rows.
     end
end
for i=1:n
     d(i)=b(i)/a(i,i);
end
d=d';
c=a;
for i=1:n
     for j=1:n
         c(i,j)=a(i,j)/a(i,i); % factor.
     end
     c(i,i)=0;
     x(i)=0;
end
x=x';
disp (a);
disp (b);
disp (d);
disp (c);
lambda = input('Enter the value of weighting factor: ');
es=0.05; % stopping criteria.
ea(i)=100;
head=sprintf('tttttttttValue of x ttttttttttttttValue
of ea ');
disp(head);
while (ea(i)>=es)
     for i=1:n
         y=x(i);
         x(i)=d(i)-c(i,:)*x(:,1);
           x(i)=lambda*x(i)+(1-lambda)*y; % calculates unknown value.

                                                       Third Year Mechanical Engineering
                                                    Computer Oriented Numerical Methods
                                                                               2011-12©
                                                          MITCOE Mechanical Engineering
ea(i)=abs((x(i)-y)/x(i))*100;
end
               table1=sprintf('%d ttt %ftttt %4.9f ttt%f ttt
%ftttt %4.9f',x,ea);
disp(table1);
end




Output :


enter matrix A[]: [-3 1 12;6 -1 -1;6 9 1]

enter column matrix B[]: [50;3;40]

        6       -1      -1

        6        9       1

       -3        1      12



        3

       40

       50



       0.5000

       4.4444

       4.1667



            0        -0.1667     -0.1667

       0.6667                0   0.1111

      -0.2500         0.0833          0



Enter the value of weighting factor: 0.95

                                                   Third Year Mechanical Engineering
                                                Computer Oriented Numerical Methods
                                                                           2011-12©
                                                      MITCOE Mechanical Engineering
Value of x                     Value of ea
4.750000e-001 3.921389 3.760702546 100.000000 100.000000 100.000000000

1.715081e+000 2.935111   4.321337313   72.304517   33.602766   12.973640471

1.709692e+000 2.830032   4.356407772    0.315233   3.712986    0.805031598

1.698338e+000 2.828267   4.355604413    0.668543   0.062401    0.018444248




                                                  Third Year Mechanical Engineering
                                               Computer Oriented Numerical Methods
                                                                          2011-12©
                                                     MITCOE Mechanical Engineering
Assignment No: 42

Statement: Write down the Matlab Program to fit curve y = a0 + a1*x by using least square
techniques for given set of points.



Solution:

Input :

clc;
clear all;
x=input('Enter row matrix x : ');
y=input('Enter row matrix y : ');
[m,n]=size(x);
xy(1,1)=0;
i=1; X=0; Y=0; XY=0; Xsqr=0;
while i<=n;
     xy(1,i)=x(1,i)*y(1,i);
     xsqr(1,i)=x(1,i)^2;
     X=X+x(1,i);            % To calculate summation of x
     Y=Y+y(1,i);            % To calculate summation of y
     XY=XY+xy(1,i);         % To calculate summation of x*y
     Xsqr=Xsqr+xsqr(1,i);   % To calculate summation of x^2
     i=i+1;
end
disp(x);
disp(y);
a1=(n*XY-Y*X)/(n*Xsqr-X^2);
a0=(Y*Xsqr-X*XY)/(n*Xsqr-X^2);
ym=Y/n;
sr(1,1)=0;j=1;
while j<=n
     sr(1,j)=(y(1,j)-a0-a1*x(1,j))^2;     % To calculate sr for each x
     st(1,j)=(y(1,j)-ym)^2;               % To calculate st for each x
     j=j+1;
end
SR=sum(sr);
ST=sum(st);
r2=(ST-SR)/ST
s=sprintf('Best fit curve (straight line) for above data is given by : y = %f
* x + %f',a1,a0);
disp(s);


xp=linspace(min(x),max(x));
yp=a0+a1*xp;
plot(x,y,'o',xp,yp);
xlabel('values of x');
ylabel('values of y');
title('y=a0+a1*x');

                                                          Third Year Mechanical Engineering
                                                       Computer Oriented Numerical Methods
                                                                                  2011-12©
                                                             MITCOE Mechanical Engineering
grid on;




Output :

Enter row matrix x : [1.0 2.0 3.0 4.0 5.0 6.0 7.0]

Enter row matrix y : [0.5 2.5 2.0 4.0 3.5 6.0 5.5]

       1                      2      3     4         5      6          7



       0.5000                     2.5000       2.0000      4.0000          3.5000    6.0000     5.5000

r2 =

       0.8683

Best fit curve (straight line) for above data is given by :

 y = 0.839286 * x + 0.071429

                                                         y=a0+a1*x
                      6



                      5



                      4
        values of y




                      3



                      2



                      1



                      0
                          1          2           3            4            5         6         7
                                                         values of x




                                                                               Third Year Mechanical Engineering
                                                                            Computer Oriented Numerical Methods
                                                                                                       2011-12©
                                                                                  MITCOE Mechanical Engineering
Assignment No: 9

Statement: Write down the Matlab Program to fit curve y = a0 + a1*x+a2x2 by using least square
techniques for given set of points.



Solution:

Input :

clc;
clear all;
x = input('Enter values of x in row matrix form : ');
y = input('Enter values of y in row matrix form : ');
[m,n]=size(x);
sx = sum(x);
sy = sum(y);
sx2 = sum(x.*x);
sxy = sum(x.*y);
sx2y = sum(x.*x.*y);
sx3 = sum(x.*x.*x);
sx4 = sum(x.*x.*x.*x);
a = [sx2 sx n; sx3 sx2 sx; sx4 sx3 sx2];
b = [sy; sxy; sx2y];
z=inv(a)*b;
s=sprintf('Best fit curve (Quadratic) for above data is given by :y = %f + %f
* x + %f * x^2 ',z(1),z(2),z(3));
disp(s);
xp = linspace(min(x),max(x));
yp = z(3)*(xp.*xp)+z(2)*xp+z(1);
plot(x,y,'o',xp,yp);
grid on;
xlabel('Values of x');
ylabel('Values of function');
title('y=a0+ a1*x+ a2*(x^2)');




                                                          Third Year Mechanical Engineering
                                                       Computer Oriented Numerical Methods
                                                                                  2011-12©
                                                             MITCOE Mechanical Engineering
Output :

Enter values of x in row matrix form : [0.075 0.5 1 1.2 1.7 2.0 2.3]

Enter values of y in row matrix form : [600 800 1200 1400 2050 2650 3750]

Best fit curve (Quadratic) for above data is given by :y = 643.601494 + -
218.884701 * x + 685.248397 * x^2




                                           y=a0+ a1*x+ a2*(x 2)
                          4000


                          3500


                          3000
     Values of function




                          2500


                          2000


                          1500


                          1000


                          500
                                 0   0.5     1                 1.5            2          2.5
                                                 Values of x




                                                                        Third Year Mechanical Engineering
                                                                     Computer Oriented Numerical Methods
                                                                                                2011-12©
                                                                           MITCOE Mechanical Engineering
Assignment No: 10

Statement: Write down the Matlab Program to fit curve y = a1*(xb1) by using least square
techniques for given set of points.



Solution:

Input :

clc;
clear all;
xa=input('Enter row matrix x : ');
ya=input('Enter row matrix y : ');
[m,n]=size(xa);
xy(1,1)=0; y(1,1)=0;
i=1; X=0; Y=0; XY=0; Xsqr=0;
while (i<=n)
       y(1,i)=log10(ya(1,i));        % To calculate log of y
       x(1,i)=log10(xa(1,i));        % To calculate log of x
       xy(1,i)=x(1,i)*y(1,i);
       xsqr(1,i)=x(1,i)^2;
       X=X+x(1,i);                  % To calculate summation of x
       Y=Y+y(1,i);                  % To calculate summation of y
       XY=XY+xy(1,i);               % To calculate summation of x*y
       Xsqr=Xsqr+xsqr(1,i);         % To calculate summation of x^2
       i=i+1;
end


disp(xa);
disp(ya)
beta=(n*XY-Y*X)/(n*Xsqr-X^2);
a0=(Y*Xsqr-X*XY)/(n*Xsqr-X^2);
alpha=10^(a0);                      % To calculate co-eff of x^a0
ym=Y/n;
sr(1,1)=0;j=1;
while j<=n
       sr(1,j)=(y(1,j)-a0-beta*x(1,j))^2;         % To calculate sr for each x
                                                          Third Year Mechanical Engineering
                                                       Computer Oriented Numerical Methods
                                                                                  2011-12©
                                                             MITCOE Mechanical Engineering
st(1,j)=(y(1,j)-ym)^2;                   % To calculate st for each x
       j=j+1;
end
SR=sum(sr);
ST=sum(st);
r2=(ST-SR)/ST
s=sprintf('Best fit curve (polynomial) for above data is given by : y = %f *
x^(%f) ',alpha,beta);
disp(s);
xp = linspace(min(x),max(x));
yp = (xp.^beta)*alpha;
plot(xa,ya,'o')
hold on
plot(xp,yp)
grid on;
xlabel('values of x');
ylabel('values of y');
title('y=alpha*x^(beta)');




Output :
Enter row matrix x : [26.67 93.33 148.89 315.56]
Enter row matrix y : [1.35 0.085 0.012 0.00075]
      26.6700   93.3300   148.8900   315.5600


       1.3500    0.0850     0.0120     0.0008
r2 =


       0.9757
Best fit curve (polynomial) for above data is given by :
y = 38147.936083 * x^(-3.013376)




                                                      Third Year Mechanical Engineering
                                                   Computer Oriented Numerical Methods
                                                                              2011-12©
                                                         MITCOE Mechanical Engineering
y=alpha*(x)b
               2

              1.8

              1.6

              1.4

              1.2
values of y




               1

              0.8

              0.6

              0.4

              0.2

               0
                    0   50   100   150        200       250      300       350
                                    values of x




                                                       Third Year Mechanical Engineering
                                                    Computer Oriented Numerical Methods
                                                                               2011-12©
                                                          MITCOE Mechanical Engineering
Assignment No: 41

Statement: Write down the Matlab Program to fit curve y = a1 * e (b1*x) by using least square
techniques for given set of points.



Solution:

Input :
clc;
clear all;
x=input('Enter row matrix x : ');
ya=input('Enter row matrix y : ');
[m,n]=size(x);        % Defining size of matrix x
xy(1,1)=0; y(1,1)=0;        % Defining matrix xy & y
i=1; X=0; Y=0; XY=0; Xsqr=0; % Setting initial condition for loop
while i<=n;
       y(1,i)=log(ya(1,i));
       xy(1,i)=x(1,i)*y(1,i);
       xsqr(1,i)=x(1,i)^2;
       X=X+x(1,i);
       Y=Y+y(1,i);
       XY=XY+xy(1,i);
       Xsqr=Xsqr+xsqr(1,i);
       i=i+1;
end
disp(x);
disp(ya);
a1=(n*XY-Y*X)/(n*Xsqr-X^2);
a0=(Y*Xsqr-X*XY)/(n*Xsqr-X^2);
alpha=exp(a0);
ym=Y/n;         % Finding mean
sr(1,1)=0;j=1;
while j<=n;
       sr(1,j)=(y(1,j)-a0-a1*x(1,j))^2;
       st(1,j)=(y(1,j)-ym)^2;
       j=j+1;
                                                           Third Year Mechanical Engineering
                                                        Computer Oriented Numerical Methods
                                                                                   2011-12©
                                                              MITCOE Mechanical Engineering
end
xp = linspace(min(x),max(x));            % Condition for graph
yp= alpha*exp(a1*xp);                % Given function



SR=sum(sr);
ST=sum(st);
r2=(ST-SR)/ST         % Co-efficient of determination
s=sprintf('Best fit curve (exponential) for above data is given by : y = %f *
e^(%f * x) ',alpha,a1);
disp(s);


plot(x,ya,'o',xp,yp)         % Plots function & best fitted curve simultaneously
grid on;
xlabel('values of x'); % Defining specifications of graph
ylabel('values of y');
title('y=alpha*e^(beta*x)');
grid on;        % To display grid on graph




Output :

Enter row matrix x : [0.4 0.8 1.2 1.6 2.0 2.3]
Enter row matrix y : [800 975 1500 1950 2900 3600]
       0.4000       0.8000     1.2000        1.6000      2.0000      2.3000


       800            975         1500            1950            2900        3600




r2 =


       0.9933


Best fit curve (exponential) for above data is given by :
y = 546.590939 * e^(0.818651 * x)




                                                             Third Year Mechanical Engineering
                                                          Computer Oriented Numerical Methods
                                                                                     2011-12©
                                                                MITCOE Mechanical Engineering
y=alpha*e(beta*x)
              4000


              3500


              3000
values of y




              2500


              2000


              1500


              1000


              500
                0.4   0.6   0.8   1   1.2       1.4     1.6   1.8    2    2.2   2.4
                                            values of x




                                                                 Third Year Mechanical Engineering
                                                              Computer Oriented Numerical Methods
                                                                                         2011-12©
                                                                    MITCOE Mechanical Engineering
Assignment No: 11

Statement: Write down the Matlab Program for Lagrange Interpolation.

Solution:

Input :

clc;
clear all;
x = input('Enter the of Values of x: ');
y = input('Enter the of Values of y: ');
u = input('Value of x at which y is to be evaluated: ');
n = length(x); % Size of matrix x
p=1;
s=0;
for i=1:n
       p=y(i);
       for j=1:n
           if (i~=j)         % Condition for inequality
               p=p*(u-x(j))/(x(i)-x(j)); % Formula
           end
       end
       s=s+p;    % Summation
end
fprintf('n Value of y at required x is : %f ',s);



Output :
Enter the of Values of x: [1 4 5 7]

Enter the of Values of y: [21.746 438.171 1188.9147 8775.011]

Value of x at which y is to be evaluated: 4.2

Value of y at required x is : 490.360287




                                                       Third Year Mechanical Engineering
                                                    Computer Oriented Numerical Methods
                                                                               2011-12©
                                                          MITCOE Mechanical Engineering
Assignment No: 12

Statement: Write down the Matlab Program for Newton-Gregory Forward Difference
Interpolation.

Solution:

Input :
clc;
clear all;
x=input('Enter row matrix x : ');
y=input('Enter row matrix y : ');
X=input('Enter value of x at which value of function is to be calculated :
');
[m,n]=size(x);
dx=diff(x);             % Spatial diff.(for equally spaced data)
d(1,1)=y(1,1);
disp(x);
disp(y);
for j=1:(n-1)
       dy=diff(y);      % Delta matrix
       disp(dy);
       d(j+1)=dy(1);    % Stores 1st value of delta matrix.
       y=dy;
end
alpha=(X-x(1))/dx(1);     % Value of alpha is calculated.
a(1,1)=1; prod=1;
for k=1:(n-2)
       prod=prod*(alpha-k+1);
       a(k+1)=prod;
end
func=0;
for i=1:n-1
       fx=a(i)*d(i)/(factorial(i-1));
       func=func+fx;
end



                                                      Third Year Mechanical Engineering
                                                   Computer Oriented Numerical Methods
                                                                              2011-12©
                                                         MITCOE Mechanical Engineering
s=sprintf('Value of function calculated by N-G forward interpolation :
%f',func);
disp(s);


Output :

Enter row matrix x : [2 3 4 5 6 7 8 9]
Enter row matrix y : [19 48 99 178 291 444 643 894]
Enter value of x at which value of function is to be calculated : 3.5
     2        3    4     5     6     7     8    9


    19       48   99   178   291   444   643   894


    29       51   79   113   153   199   251


    22       28   34    40    46    52


     6        6    6     6     6


     0        0    0     0


     0        0    0


     0        0


     0


Value of function calculated by N-G forward interpolation : 70.375000




                                                  Third Year Mechanical Engineering
                                               Computer Oriented Numerical Methods
                                                                          2011-12©
                                                     MITCOE Mechanical Engineering
Assignment No: 13

Statement: Write down the Matlab Program for Newton-Gregory Backward Difference
Interpolation.

Solution:

Input :
clc;
clear all;
x=input('Enter row matrix x : ');
y=input('Enter row matrix y : ');
X=input('Enter value of x at which value of function is to be calculated :
');
[m,n]=size(x);
dx=diff(x);             % Spatial diff.(for equally spaced data)
d(1,1)=y(n);


newx(1,n:-1:1)=x(1,1:n);        % Reversing order of matrix x so that nth value is
brought 1st.
newy(1,n:-1:1)=y(1,1:n);        % Reversing order of matrix y so that nth value is
brought 1st.



disp(newx)
disp(newy)
for j=1:(n-1)
       dy=diff(newy);      % Delta matrix
       disp(dy);
       d(j+1)=dy(1);       % Stores 1st value of delta matrix.
       newy=dy;
end
alpha=(x(n)-X)/dx(1);        % Value of alpha is calculated.
a(1,1)=1; prod=1;
for k=1:(n-2)
       prod=prod*(alpha-k+1);
       a(k+1)=prod;
end

                                                         Third Year Mechanical Engineering
                                                      Computer Oriented Numerical Methods
                                                                                 2011-12©
                                                            MITCOE Mechanical Engineering
func=0;
for i=1:n-1
       fx=a(i)*d(i)/(factorial(i-1));
       func=func+fx;
end
s=sprintf('Value of function calculated by N-G backward interpolation :
%f',func);
disp(s);


Output :

Enter row matrix x : [0.1 0.2 0.3 0.4 0.5]
Enter row matrix y : [1.4 1.56 1.76 2 2.28]
Enter value of x at which value of function is to be calculated : 0.25
       0.5000    0.4000      0.3000    0.2000    0.1000


       2.2800    2.0000      1.7600     1.5600   1.4000


      -0.2800   -0.2400      -0.2000   -0.1600


       0.0400    0.0400      0.0400


  1.0e-015*0.2220         -0.2220


 -4.4409e-016


Value of function calculated by N-G backward interpolation : 1.655000




                                                     Third Year Mechanical Engineering
                                                  Computer Oriented Numerical Methods
                                                                             2011-12©
                                                        MITCOE Mechanical Engineering
Assignment No: 14

Statement: Write down the Matlab Program for Hermite interpolation method.



Solution:

Input :

clc;
clear all;
disp('HERMITE INTERPOLATION');
x=input('Enter the values of x: ');
xu=input('Enter the value unknown of x: ');
fx=input('Enter the values of fx: ');
dfx=input('Enter the values of dfx: ');
n=size(x); % Size of matrix
sum=0;suma=0;sumb=0;
for i=1:n
     pro=1;
     pro1=1;
     for j=1:n
     if i~=j
          pro=pro*(xu-x(j))/(x(i)-x(j));  % Lagrange formulation of unknown x.
          pro1=pro1*(x(i)-x(j)); % Derivative of Lagrange term
     end
         end
     L(i,1)=pro;    % Lagrange term
     dL(i,1)=pro1; % Derivative of Lagrange term
end
for k=1:n
     suma=suma+(1-2*(xu-x(k))*dL(k))*((L(k))^2)*fx(k); % Summation
     sumb=sumb+(xu-x(k))*((L(k))^2)*dfx(k);
end
sumf=suma+sumb;
disp('The value of fx at unknown x is: ');
disp(sumf);




Output:

HERMITE INTERPOLATION
Enter the values of x: [0;1]
Enter the value unknown of x: 0.4
Enter the values of fx: [0;1]
Enter the values of dfx: [0;2]
The value of fx at unknown x is:
    0.1600

                                                       Third Year Mechanical Engineering
                                                    Computer Oriented Numerical Methods
                                                                               2011-12©
                                                          MITCOE Mechanical Engineering
Third Year Mechanical Engineering
Computer Oriented Numerical Methods
                           2011-12©
      MITCOE Mechanical Engineering
Assignment No: 15

Statement: Write down the Matlab Program for interpolation by Cubic spline.



Solution:

Input :


% Clearing Workspace
clear all;
clc;
close;


% Defining Input points
x1=input('Enter matrix for values of x: ');
y1=input('Enter matrix for values of y: ');
xg=input('Enter value of x for which to find y: ');
m1=size(x1);
n=m1(1,2);
x=x1'; y=y1';
scatter(x,y);
hold on;

% MATLAB function plotting Cubic Interpolation
yy = spline(x,y,0:0.01:100);
plot(x,y,'o',0:0.01:100,yy);

% Defining end conditions f''(x)=0 @ 1st and last point
M(1:n+1)=0;


% First row of matrix to be solved
A(1,1:3)=[2*(x(3)-x(1)) (x(3)-x(2)) 0];
B(1,1)=6*(y(3)-y(2))/(x(3)-x(2))-6*(y(2)-y(1))/(x(2)-x(1));


% Subsequent rows till n-2
if n>3
    for l=2:n-2
        A(l,l-1:l+1)=[(x(l+1)-x(l)) 2*(x(l+2)-x(l)) (x(l+2)-x(l+1))];
        B(l,1)=6*(y(l+2)-y(l+1))/(x(l+2)-x(l+1))-6*(y(l+1)-y(l))/(x(l+1)-
x(l));
    end
end


% Last 1 row
A(n-1,n-2:n-1)=[(x(n)-x(n-1)) 2*(x(n)-x(n-1))];
B(n-1,1)=-6*(y(n)-y(n-1))/(x(n)-x(n-1));
                                                         Third Year Mechanical Engineering
                                                      Computer Oriented Numerical Methods
                                                                                 2011-12©
                                                            MITCOE Mechanical Engineering
% Finding other values of f''(x)
N=GaussSoln(A,B);

% Assigning Values to M
for i=1:n-1
    M(i+1)=N(i);
end


% Creating the interpolation function between intervals
f=inline('Ma/6/(xb-xa)*(xb-xx)^3-Mb/6/(xb-xa)*(xa-xx)^3+(ya/(xb-xa)-Ma*(xb-
xa)/6)*(xb-xx)-(yb/(xb-xa)-Mb*(xb-xa)/6)*(xa-
xx)','xx','Ma','Mb','xa','xb','ya','yb');


% Ploting the spline in intervals
xn(1:1000)=0;
yn(1:1000)=0;
for i=1:n-1
    j=1;
    dx=(x(i+1)-x(i))/1000;
    for k=x(i):dx:x(i+1)
         xn(j)=k;
         yn(j)=f(k,M(i),M(i+1),x(i),x(i+1),y(i),y(i+1));
         j=j+1;
    end
    if xg>=x(i) && xg<=x(i+1)
        yg=f(xg,M(i),M(i+1),x(i),x(i+1),y(i),y(i+1));
    end
    plot(xn,yn, 'LineWidth',2);
    xlim([min(x) max(x)]);
    ylim([min(y) max(y)]);
end
hold off;
fprintf('@x=%f, y=%fn',xg,yg);




GaussSoln:

function Soln=GaussSoln(x,y)
A1=x;
B=y;
n2=size(B);
n=n2(1,1);
clear x;
clear y;
if det(A1)==0
     disp('Either no solution or infinitely many solutions.');
else
     A=A1;
     A(:,n+1)=B(1:n);

                                                     Third Year Mechanical Engineering
                                                  Computer Oriented Numerical Methods
                                                                             2011-12©
                                                        MITCOE Mechanical Engineering
for i=1:n-1
         for j=i:n-1
              fac=A(j+1,i)/A(i,i);
              fac_mat=fac*A(i,:);
              A(j+1,:)=A(j+1,:)-fac_mat;
         end
    end
    i=0;j=0;
    if A(n,n)==0
         an(n)=0;
    else
         an(n)=A(n,n+1)/A(n,n);
    end
    for i=n-1:-1:1
         for j=n:-1:1
              x(j)=an(j)*A(i,j);
         end
         y=sum(x);
         if y==0
              an(i)=0;
         else
              an(i)=(A(i,n+1)-y)/A(i,i);
         end
    end
end
Soln=an;




Output:

Enter matrix for values of x: [1 2 3 4]

Enter matrix for values of y: [0 0.3 0.48 0.6]

Enter value of x for which to find y: 2.3

@x=2.300000, y=0.363014




                                                Third Year Mechanical Engineering
                                             Computer Oriented Numerical Methods
                                                                        2011-12©
                                                   MITCOE Mechanical Engineering
Third Year Mechanical Engineering
Computer Oriented Numerical Methods
                           2011-12©
      MITCOE Mechanical Engineering
Assignment No: 16

Statement: Write down the Matlab Program for Inverse Interpolation.

Solution:

Input :

clc;
clear all;
x = input('Enter the of Values of x: ');
y = input('Enter the of Values of y: ');
r = input('Value of y at which x is to be evaluated: ');
n = length(x); % determines size of matrix.
p=1;
s=0;
for j=1:n
     for i=1:n
         if i==j
             continue;
         end
         numerator=r-y(i);
         denominator=y(j)-y(i);
         v(j)=numerator/denominator;
         p=p*v(j);
     end
     s=s+p*x(j);
p=1;
end
fprintf('n Value is : %f ',s)




Output :

Enter the of Values of x: [0 1 2 3]
Enter the of Values of y: [0 1 7 25]
Value of y at which x is to be evaluated: 2


 Value is : 1.716138




                                                        Third Year Mechanical Engineering
                                                     Computer Oriented Numerical Methods
                                                                                2011-12©
                                                           MITCOE Mechanical Engineering
Assignment No: 17

Statement: Write down the Matlab Program for Newton Forward Differentiation.

Solution:

Input :

clc;
clear all;
x=input('Enter row matrix x : ');
y=input('Enter row matrix y : ');
r=input('Enter value of x at which value of function is to be calculated :
');
[m,n]=size(x);
p=1;
h=diff(x);         % Step size
disp(x);
disp(y);
for j=1:n
     if (r==x(j))
     p=j;
     end
end
d(1,1)=y(1,p);
for j=1:(n-p)
     dy=diff(y);   % Delta matrix
     disp(dy);
     y=dy;
     d(j+1)=y(1,p); % Stores p th value of delta matrix.
end
f=0;
for k=1:n-1
     fr=d(k+1)/k;
     f=f+((-1)^(k-1))*fr;
end
dx=(f/h(1));
s=sprintf('Value of dy/dx at %f is : % f',r,dx);
disp (s);




                                                       Third Year Mechanical Engineering
                                                    Computer Oriented Numerical Methods
                                                                               2011-12©
                                                          MITCOE Mechanical Engineering
Output :
Enter row matrix x : [1.5 2 2.5 3 3.5 4]

Enter row matrix y : [3.375 7 13.625 24 38.875 59]

Enter value of x at which value of function is to be calculated : 1.5

    1.5000       2.0000    2.5000    3.0000     3.5000    4.0000



    3.3750       7.0000   13.6250   24.0000    38.8750   59.0000



    3.6250       6.6250   10.3750   14.8750    20.1250



    3.0000       3.7500    4.5000    5.2500



    0.7500       0.7500    0.7500



     0       0



     0



Value of dy/dx at 1.500000 is :     4.750000




                                                    Third Year Mechanical Engineering
                                                 Computer Oriented Numerical Methods
                                                                            2011-12©
                                                       MITCOE Mechanical Engineering
Assignment No: 18

Statement: Write down the Matlab Program for Newton Backward Differentiation.

Solution:

Input :

clc;
clear all;
xin=input('Enter row matrix x : ');
yin=input('Enter row matrix y : ');
r=input('Enter value of x at which value of function is to be calculated :
');
[m,n]=size(xin);
p=1;
h=diff(xin);        % Step size
y(1,n:-1:1)=yin(1,1:n); % Reversing order of matrix y so that nth value is brought
1st.
x(1,n:-1:1)=xin(1,1:n);     % Reversing order of matrix x so that nth value is brought
1st.
disp(x)
disp(y)
for j=1:n
     if (r==x(j))
     p=j;
     end
end
d(1,1)=y(1,p);
for j=1:(n-p)
     dy=diff(y);   % Delta matrix
     y=(-1)*dy;
     d(j+1)=(y(1,p)); % Stores p th value of delta matrix.
     disp(y);
end
f=0;
for k=1:n-1
     fr=d(k+1)/k;
     f=f+fr;
end
dx=(f/h(1));
s=sprintf('Value of dy/dx at %f is : % f',r,dx);
disp (s);




                                                       Third Year Mechanical Engineering
                                                    Computer Oriented Numerical Methods
                                                                               2011-12©
                                                          MITCOE Mechanical Engineering
Output :

Enter row matrix x : [0 10 20 30 40]

Enter row matrix y : [1 0.984 0.939 0.866 0.766]

Enter value of x at which value of function is to be calculated : 40

    40       30      20     10        0



    0.7660         0.8660        0.9390   0.9840    1.0000



   -0.1000        -0.0730    -0.0450      -0.0160



   -0.0270        -0.0280    -0.0290



    0.0010         0.0010



 -2.2204e-016



Value of dy/dx at 40.000000 is : -0.011317




                                                        Third Year Mechanical Engineering
                                                     Computer Oriented Numerical Methods
                                                                                2011-12©
                                                           MITCOE Mechanical Engineering
Assignment No: 19

Statement: Write down the Matlab Program using Trapezoidal rule(single segment) for any
function.



Solution:

Input :

clear;
clc;
fprintf('NUMERICAL INTEGRATION BY TRAPEZOIDAL RULE nn');
x=input('Enter a function to integrate f(x)=','s');
a=input('Enter Lower Limit: ');
b=input('Enter Upper Limit: ');
n=2;               % No. of points
y=inline(x);       % Defining function
h=(b-a)/n;         % Step size
S=0;
for i=1:n-1;
     t=2*y(a+i*h);
     S=S+t;
end
A=h/2*(y(a)+y(b)+S);      % Calculation of area
fprintf('nAnswer= %fn',A);




Output :

NUMERICAL INTEGRATION BY TRAPEZOIDAL RULE



Enter a function to integrate f(x)=4*x+2

Enter Lower Limit: 1

Enter Upper Limit: 4



Answer= 36.000000




                                                       Third Year Mechanical Engineering
                                                    Computer Oriented Numerical Methods
                                                                               2011-12©
                                                          MITCOE Mechanical Engineering
Assignment No: 20

Statement: Write down the Matlab Program using Trapezoidal rule(multiple segment) for any
function.



Solution:

Input :

clear;
clc;
fprintf('NUMERICAL INTEGRATION BY TRAPEZOIDAL RULE nn');
x=input('Enter a function to integrate f(x)=','s');
a=input('Enter Lower Limit: ');
b=input('Enter Upper Limit: ');
n=input('enter no. of segments');
y=inline(x);     % Defining function
h=(b-a)/n;       % Step size
S=0;
for i=1:n-1;
     t=2*y(a+i*h);
     S=S+t;
end
A=h/2*(y(a)+y(b)+S);     % Calculation of area
fprintf('nAnswer= %fn',A);




 Output :

NUMERICAL INTEGRATION BY TRAPEZOIDAL RULE



Enter a function to integrate f(x)=4*x+2

Enter Lower Limit: 1

Enter Upper Limit: 4

enter no. of segments6



Answer= 36.000000


                                                       Third Year Mechanical Engineering
                                                    Computer Oriented Numerical Methods
                                                                               2011-12©
                                                          MITCOE Mechanical Engineering
Third Year Mechanical Engineering
Computer Oriented Numerical Methods
                           2011-12©
      MITCOE Mechanical Engineering
Assignment No: 21

Statement: Write down the Matlab Program using Simpson’s 1/3rd (single segment) rule for any
function.



Solution:

Input :

clear;
clc;
fprintf('NUMERICAL INTEGRATION BY SIMPSONS 1/3 RULE nn');
x=input('Enter a function to integrate f(x)=','s');
a=input('Enter Lower Limit: ');
b=input('Enter Upper Limit: ');
n=2; % No. of segment
y=inline(x);
h=(b-a)/n;
S=0;
for i=1:n-1;
     if mod(i,2)==1 % Condition for even segments
          t=4*y(a+i*h);
     else
          t=2*y(a+i*h);
     end
     S=S+t;
end
A=h/3*(y(a)+y(b)+S);
fprintf('nAnswer= %fn',A);




Output :
NUMERICAL INTEGRATION BY SIMPSONS 1/3 RULE



Enter a function to integrate f(x)=exp(x)

Enter Lower Limit: 0

Enter Upper Limit: 4



Answer= 44.247402



                                                        Third Year Mechanical Engineering
                                                     Computer Oriented Numerical Methods
                                                                                2011-12©
                                                           MITCOE Mechanical Engineering
Assignment No: 22

Statement: Write down the Matlab Program using Simpson’s 1/3rd (multiple segment) rule for
any function.



Solution:

Input :

clear;
clc;
fprintf('NUMERICAL INTEGRATION BY SIMPSONS 1/3 RULE nn');
x=input('Enter a function to integrate f(x)=','s');
a=input('Enter Lower Limit: ');
b=input('Enter Upper Limit: ');
n=input(‘Enter no. of divisions: ’);
y=inline(x);
h=(b-a)/n;
S=0;
for i=1:n-1;
     if mod(i,2)==1
          t=4*y(a+i*h);
     else
          t=2*y(a+i*h);
     end
     S=S+t;
end
A=h/3*(y(a)+y(b)+S);
fprintf('nAnswer= %fn',A);




  Output :

NUMERICAL INTEGRATION BY SIMPSONS 1/3 RULE



Enter a function to integrate f(x)=exp(x)

Enter Lower Limit: 0

Enter Upper Limit: 4

enter no.of divisions:5



Answer= 44.683772
                                                        Third Year Mechanical Engineering
                                                     Computer Oriented Numerical Methods
                                                                                2011-12©
                                                           MITCOE Mechanical Engineering
Assignment No: 23

Statement: Write down the Matlab Program using Simpson’s 3/8th rule for any function.



Solution:

Input :

clear;
clc;
fprintf('NUMERICAL INTEGRATION BY SIMPSONS 3/8 RULE nn');
x=input('Enter a function to integrate f(x)=','s');
a=input('Enter Lower Limit: ');
b=input('Enter Upper Limit: ');
while mod(n,3)~=0            % Condition for no. of segments
     n=input('Enter No. of Divisions [Should be divisible by 3]: ');
end
y=inline(x);     % Defining function
h=(b-a)/n;       % Step size
S=0;
for i=1:n-1;
     if mod(i,3)==0      % Decision statement for usage of formula
          t=2*y(a+i*h);
     else
          t=3*y(a+i*h);
     end
     S=S+t;
end
A=3*h/8*(y(a)+y(b)+S); % Area calculation
fprintf('nAnswer= %fn',A);




Output :

Enter the function: 4*x-1

Initial Value of x :1

Final Value of x :4

Enter No. of Divisions [Should be divisible by 3]: 3

Answer: 27.000000>>




                                                        Third Year Mechanical Engineering
                                                     Computer Oriented Numerical Methods
                                                                                2011-12©
                                                           MITCOE Mechanical Engineering
Assignment No: 24

Statement: Write down the Matlab Program for Combined Simpson’s Rule.

Solution:

Input :

clear;
clc;
j=1;
fprintf('NUMERICAL INTEGRATION BY MULTIPLE SIMPSONS RULE nn');
x=input('Enter a function to integrate f(x)=','s');
a=input('Enter Lower Limit: ');
b=input('Enter Upper Limit: ');
while j==1
     n=input('Enter No. of Divisions [(n-3) divisible by 2]: ');         % Condition
for no. of segments
     if mod(n-3,2)==0
          j=0;
     end
end
y=inline(x);
h=(b-a)/n;
S=0;
if n>=3
     for i=1:2;
          t=3*y(a+i*h);
          S=S+t;
     end
     A=3*h/8*(y(a)+y(a+3*h)+S);
end
S=0;
for i=4:n-1;
     if mod(i,2)==0
          t=4*y(a+i*h);
     else
          t=2*y(a+i*h);
     end
     S=S+t;
end
A=A+h/3*(y(a+3*h)+y(b)+S);

fprintf('nAnswer= %fn',A);




                                                      Third Year Mechanical Engineering
                                                   Computer Oriented Numerical Methods
                                                                              2011-12©
                                                         MITCOE Mechanical Engineering
OUTPUT:


NUMERICAL INTEGRATION BY MULTIPLE SIMPSONS RULE

Enter   a function to integrate f(x)=x^0.1*(1.2-x)*(1-exp(20*(x-1)))
Enter   Lower Limit: 0
Enter   Upper Limit: 2
Enter   No. of Divisions [(n-3) divisible by 2]: 5

Answer= 55501691.391968
>>




                                                  Third Year Mechanical Engineering
                                               Computer Oriented Numerical Methods
                                                                          2011-12©
                                                     MITCOE Mechanical Engineering
Assignment No: 25

Statement: Write down the Matlab Program for Gauss-Legendre 2-pt method.

Solution:

Input :


clear;
clc;
fprintf('NUMERICAL INTEGRATION BY GAUSS-LEGENDRE 2-POINT FORMULA nn');
x=input('Enter a function to integrate f(x)=','s');
a=input('Enter Lower Limit: ');
b=input('Enter Upper Limit: ');
f=inline(x); % Defining function
c=(b-a)/2;   % Constants
d=(b+a)/2;   % Constants
x1=c/sqrt(3)+d;
x2=-c/sqrt(3)+d;
y1=f(x1);
y2=f(x2);
A=(y1+y2)*c;
fprintf('nAnswer= %fn',A);




OUTPUT:
NUMERICAL INTEGRATION BY GAUSS-LEGENDRE 2-POINT FORMULA

Enter a function to integrate f(x)=x^3+x-1
Enter Lower Limit: 1
Enter Upper Limit: 4

Answer= 68.250000




                                                      Third Year Mechanical Engineering
                                                   Computer Oriented Numerical Methods
                                                                              2011-12©
                                                         MITCOE Mechanical Engineering
Assignment No: 26

Statement: Write down the Matlab Program using Gauss Legendre 3-pt rule for any function.



Solution:

Input :

clear;
clc;
fprintf('NUMERICAL INTEGRATION BY GAUSS-LEGENDRE 3-POINT FORMULA nn');
x=input('Enter a function to integrate f(x)=','s');
a=input('Enter Lower Limit: ');
b=input('Enter Upper Limit: ');
f=inline(x);      % Defining function
c=(b-a)/2;
d=(b+a)/2;
x1=c*sqrt(3/5)+d;
x2=-c*sqrt(3/5)+d;
x3=d;
y1=f(x1);
y2=f(x2);
y3=f(x3);
A=(5/9*y1+5/9*y2+8/9*y3)*c;
fprintf('n Answer= %fn',A);




Output :

NUMERICAL INTEGRATION BY GAUSS-LEGENDRE 3-POINT FORMULA



Enter a function to integrate f(x)=x^2-5*x+2

Enter Lower Limit: 3

Enter Upper Limit: 5



Answer= -3.333333

                                                        Third Year Mechanical Engineering
                                                     Computer Oriented Numerical Methods
                                                                                2011-12©
                                                           MITCOE Mechanical Engineering
Third Year Mechanical Engineering
Computer Oriented Numerical Methods
                           2011-12©
      MITCOE Mechanical Engineering
Assignment No: 27

Statement: Write down the Matlab Program using Double integration by trapezoidal rule for
any function.



Solution:

Input :

clear;
clc;
% Taking Input
fprintf('DOUBLE INTEGRATION BY TRAPEZOIDAL RULE nn');
xy=input('Enter a function to integrate f(x,y)=','s');
ax=input('Enter Lower Limit of x: ');
bx=input('Enter Upper Limit of x: ');
ay=input('Enter Lower Limit of y: ');
by=input('Enter Upper Limit of y: ');
nx=input('No. of intervals for integration w.r.t. x: ');
ny=input('No. of intervals for integration w.r.t. y: ');
% Defining the function
f=inline(xy);
% Main Calculations
h=(bx-ax)/nx;
k=(by-ay)/ny;
an=0;
for i=0:nx-1
    for j=0:ny-1

tr=f(ax+i*h,ay+j*k)+f(ax+i*h,ay+(j+1)*k)+f(ax+(i+1)*h,ay+(j+1)*k)+f(ax+(i+1)*
h,ay+j*k);
        an=an+tr;
    end
end
A=h*k/4*an;
fprintf('nAnswer= %fn',A);




                                                        Third Year Mechanical Engineering
                                                     Computer Oriented Numerical Methods
                                                                                2011-12©
                                                           MITCOE Mechanical Engineering
Output :

DOUBLE INTEGRATION BY TRAPEZOIDAL RULE



Enter a function to integrate f(x,y)=x+y

Enter Lower Limit of x: 0

Enter Upper Limit of x: 2

Enter Lower Limit of y: 1

Enter Upper Limit of y: 3

No. of intervals for integration w.r.t. x: 2

No. of intervals for integration w.r.t. y: 2



Answer= 12.000000




                                                  Third Year Mechanical Engineering
                                               Computer Oriented Numerical Methods
                                                                          2011-12©
                                                     MITCOE Mechanical Engineering
Assignment No: 28

Statement: Write down the Matlab Program using double integration by Simpson’s 1/3rd rule
for any function.



Solution:

Input :

clear;
clc;
% Taking Input
fprintf('DOUBLE INTEGRATION BY SIMPSONS 1/3rd RULE nn');
xy=input('Enter a function to integrate f(x,y)=','s');
ax=input('Enter Lower Limit of x: ');
bx=input('Enter Upper Limit of x: ');
ay=input('Enter Lower Limit of y: ');
by=input('Enter Upper Limit of y: ');
nx=3; ny=3;
while mod(nx,2)~=0 || mod(ny,2)~=0
    nx=input('No. of intervals for integration w.r.t. x (Should be even): ');
    ny=input('No. of intervals for integration w.r.t. y (Should be even): ');
end
% Defining the function
f=inline(xy);
% Main Calculations
h=(bx-ax)/nx;
k=(by-ay)/ny;
an=0;
for i=0:2:nx-1
   for j=0:2:ny-1

tr1=f(ax+i*h,ay+j*k)+f(ax+i*h,ay+(j+2)*k)+f(ax+(i+2)*h,ay+(j+2)*k)+f(ax+(i+2)
*h,ay+j*k);

tr2=f(ax+i*h,ay+(j+1)*k)+f(ax+(i+1)*h,ay+(j+2)*k)+f(ax+(i+2)*h,ay+(j+1)*k)+f(
ax+(i+1)*h,ay+j*k);
        tr3=f(ax+(i+1)*h,ay+(j+1)*k);
        an=an+tr1+4*tr2+16*tr3;
    end
end
A=h*k/9*an;
fprintf('nAnswer= %fn',A);




                                                        Third Year Mechanical Engineering
                                                     Computer Oriented Numerical Methods
                                                                                2011-12©
                                                           MITCOE Mechanical Engineering
Output :

DOUBLE INTEGRATION BY SIMPSONS 1/3rd RULE



Enter a function to integrate f(x,y)=x-y+1

Enter Lower Limit of x: 6

Enter Upper Limit of x: 14

Enter Lower Limit of y: 1

Enter Upper Limit of y: 5

No. of intervals for integration w.r.t. x (Should be even): 4

No. of intervals for integration w.r.t. y (Should be even): 4



Answer= 256.000000




                                                Third Year Mechanical Engineering
                                             Computer Oriented Numerical Methods
                                                                        2011-12©
                                                   MITCOE Mechanical Engineering
Assignment No: 29

Statement: Write down the Matlab Program for Euler Method.



Solution:


Input :
clc; clear all;

dydx=input('Emter A Function dy/dx : ','s');
x0=input('Enter The Initial Value of x :');
y0=input('Enter The Initial Value of y :');
xf=input('Enter Value of "x" At Which Value of "y" Is To Be Found: ');
h=input('Enter Step Size :');

f=inline(dydx); % Defining function
n=(xf-x0)/h;

for i=1:n
    y(i) = y0 + h*(f(x0,y0)); % Evaluating function at given x & y
    y0 = y(i);
    x0 = x0 + h;
end
s=sprintf('n Value of y At x = %f Is : %f',xf,y(n));
disp(s);




Output :

Enter     A Function dy/dx : (x+y)/((y^2)-(sqrt(x*y)))
Enter     The Initial Value of x :1.3
Enter     The Initial Value of y :2
Enter     Value of "x" At Which Value of "y" Is To Be Found: 1.8
Enter     Step Size :.05

 Value of y At x = 1.800000 Is : 2.578164




                                                      Third Year Mechanical Engineering
                                                   Computer Oriented Numerical Methods
                                                                              2011-12©
                                                         MITCOE Mechanical Engineering
Assignment No: 30

Statement: Write down the Matlab Program for Heun’s method.



Solution:

Input :


clc;
clear all;
disp('HEUNS METHOD');

format long;
dydx=input('nEnter The Function dy/dx : ','s');
x0=input('Enter The Initial Value of x: ');
y0=input('Enter Initial Value of y: ');
h=input('Enter step size: ');
xf=input('Enter Value of x For Which y Is To Be Found: ');
fprintf('n');

f=inline(dydx);
n=(xf-x0)/h;

for i=1:n
    yf = y0 + h*f(x0,y0);
    yff = y0 + h*(f(x0,y0) + f(x0+h,yf))/2;
    y0 = yff;
    x0 = x0 + h;
    s = sprintf('Value y = %f At x%d',yff,i); disp(s);
end


Output :

HEUNS METHOD

Enter     The Function dy/dx : 4*exp(.8*x) - .5*y
Enter     The Initial Value of x: 0
Enter     Initial Value of y: 2
Enter     step size: 1
Enter     Value of x For Which y Is To Be Found: 4

Value     y   =   6.701082 At x1
Value     y   =   16.319782 At x2
Value     y   =   37.199249 At x3
Value     y   =   83.337767 At x4



                                                      Third Year Mechanical Engineering
                                                   Computer Oriented Numerical Methods
                                                                              2011-12©
                                                         MITCOE Mechanical Engineering
Third Year Mechanical Engineering
Computer Oriented Numerical Methods
                           2011-12©
      MITCOE Mechanical Engineering
Assignment No: 31

Statement: Write down the Matlab Program for Modified Euler method.

Solution:

Input:
clc; clear all; % Clears the workspace
disp('MODIFIED EULER METHOD');
format long

% Take the input from user
eq=input('nEnter the diff. eqn in x and y: ','s');
s=inline(eq);
y0=input('Enter y: ');
x0=input('Enter x: ');
xu=input('Enter unknown x: ');
acc=input('Enter accuracy required: ');

% Calculatoins
h=(xu-x0)/2;n=2;
for i=1:n
    x1=x0+h;
    y1=y0+h*s(x0,y0);
    y1n=y0+(h/2)*(s(x0,y0)+s(x1,y1));
    dy=abs(y1-y1n);
    while dy>acc
     y1=y1n;
     y1n=y0+(h/2)*(s(x0,y0)+s(x1,y1));
     dy=abs(y1-y1n);
     end
    x0=x1;
    y0=y1n;
end

% Prints the answer
disp('The value of the diff eqn at unkown x is: ');
disp(y1n);




                                                      Third Year Mechanical Engineering
                                                   Computer Oriented Numerical Methods
                                                                              2011-12©
                                                         MITCOE Mechanical Engineering
Output:

MODIFIED EULER METHOD

Enter   the diff. eqn in x and y: sqrt(x+y)
Enter   y: 2.2
Enter   x: 1
Enter   unknown x: 1.2
Enter   accuracy required: 0.0001

          The value of the diff eqn at unkown x is:   2.573186212370175




                                                  Third Year Mechanical Engineering
                                               Computer Oriented Numerical Methods
                                                                          2011-12©
                                                     MITCOE Mechanical Engineering
Assignment No: 32

Statement: Write down the Matlab Program for Runge-Kutta 2nd order method.



Solution:

Input:
clc; clear all; % Clears the workspace
disp('RUNGE KUTTA METHOD 2ND ORDER');
format long

% Takes the input from user
eq=input('Enter the diff. eqn in x and y: ','s');
s=inline(eq); % Converts the i/p string into symbolic function
y0=input('Enter y: ');
x0=input('Enter x: ');
xu=input('Enter unknown x: ');
h=input('Enter step size: ');
n=(xu-x0)/h;
for i=1:n+1
    x1=x0+h;
    y1=y0+h*s(x0,y0);
    c1=h*s(x0,y0);
    c2=h*s(x1,y1);
    c=(c1+c2)/2;
    yans=y0+c;
    y0=yans;
    x0=x1;
end

% Prints the answer
disp('The value of the diff eqn at unkown x is: ');
disp(yans);




                                                       Third Year Mechanical Engineering
                                                    Computer Oriented Numerical Methods
                                                                               2011-12©
                                                          MITCOE Mechanical Engineering
Third Year Mechanical Engineering
Computer Oriented Numerical Methods
                           2011-12©
      MITCOE Mechanical Engineering
Output:

RUNGE KUTTA METHOD 2ND ORDER
Enter the diff. eqn in x and y: -(y+x*y^2)
Enter y: 1
Enter x: 0
Enter unknown x: 0.3
Enter step size: 0.1
The value of the diff eqn at unkown x is:
   0.715327926979073




                                                Third Year Mechanical Engineering
                                             Computer Oriented Numerical Methods
                                                                        2011-12©
                                                   MITCOE Mechanical Engineering
Assignment No: 33

Statement: Write down the Matlab Program for Runge-Kutta 4th order method.

Solution:

Input:
clc; clear all; % Clears the workspace
disp('RUNGE KUTTA METHOD 4TH ORDER');
format long

% Takes the input from user
eq=input('Enter the diff. eqn in x and y: ','s');
s=inline(eq); % Converts the i/p string into symbolic function
y0=input('Enter y: ');
x0=input('Enter x: ');
xu=input('Enter unknown x: ');
h=input('Enter step size: ');

% Calculation
n=(xu-x0)/h;
for i=1:n
    x1=x0+h;
    y1=y0+h*s(x0,y0);
    c1=h*s(x0,y0);
    c2=h*s((x0+(h/2)),(y0+(c1/2)));
    c3=h*s((x0+(h/2)),(y0+(c2/2)));
    c4=h*s(x1,(y0+c3));
    c=(c1+2*c2+2*c3+c4)/6;
    yans=y0+c;
    y0=yans;
    x0=x1;
end

% Prints the answer
disp('The value of the diff eqn at unkown x is: ');
disp(yans);




                                                      Third Year Mechanical Engineering
                                                   Computer Oriented Numerical Methods
                                                                              2011-12©
                                                         MITCOE Mechanical Engineering
Output:
RUNGE KUTTA METHOD 4TH ORDER
Enter the diff. eqn in x and y: 0*x+y
Enter y: 2
Enter x: 0
Enter unknown x: 0.2
Enter step size: 0.1
The value of the diff eqn at unkown x is:
   2.442805141701389




                                               Third Year Mechanical Engineering
                                            Computer Oriented Numerical Methods
                                                                       2011-12©
                                                  MITCOE Mechanical Engineering
Assignment No: 34

Statement: Write down the Matlab Program for Milne’s correct prediction method .

Solution:

Input:
clc; clear all; % Clears the workspace
disp('MILNE PREDICTION');
format long
% Take the input from user
eq=input('Enter the 1st diff. eqn in x, y: ','s');
s=inline(eq);
y=input('Enter y: ');
x=input('Enter x: ');
xu=input('Enter unknown x: ');
h=input('Enter step size: ');

%calculation
n=(xu-x(4))/h;
f1=s(x(2),y(2));
f2=s(x(3),y(3));
f3=s(x(4),y(4));
for i=1:n+1
    y4pr=y(1)+(4*h/3)*(2*f1-f2+2*f3);
    f4pr=s(xu-h*(n-i),y4pr);
    y4cr=y(3)+(h/3)*(f2+4*f3+f4pr);
    if y4pr~=y4cr
        y4pr=y4cr;
        y4=y4cr;
    end
        f4=s(xu-h*(n-i),y4);
    f1=f2;f2=f3;f3=f4;
    y(1)=y(2); y(3)=y(4);
    yans=y4cr;
end
disp('The value of the diff eqn at unkown x is: '); disp(yans);



Output:
MILNE    PREDICTION
Enter    the 1st diff. eqn in x, y: x-y+1
Enter    y: [0;0.1951;0.3812;0.5591]
Enter    x: [1;1.1;1.2;1.3]
Enter    unknown x: 1.5
Enter    step size: 0.1

                                                        Third Year Mechanical Engineering
                                                     Computer Oriented Numerical Methods
                                                                                2011-12©
                                                           MITCOE Mechanical Engineering
The value of the diff eqn at unkown x is:
   0.893399346172840
                                   Assignment No: 35

Statement: Write down the Matlab Program for Runge-Kutta simultaneous method.

Solution:

Input:
clc; clear all; % Clears the workspace
disp('RUNGE KUTTA METHOD 4TH ORDER FOR SIMLTANEOUS EQUATONS');
format long

% Takes the input from user
eq=input('Enter the 1st diff. eqn in x, y, z: ','s');
eq1=input('Enter the 2nd diff. eqn in x, y, z: ','s');
s=inline(eq,'x','y','z'); % Converts the i/p string into symbolic function
s1=inline(eq1,'x','y','z'); % Converts the i/p string into symbolic function
y0=input('Enter y: ');
x0=input('Enter x: '); z0=input('Enter z: ');
xu=input('Enter unknown x: ');
h=input('Enter step size: ');

% Calculation
n=(xu-x0)/h;
for i=1:n
    x1=x0+h;
    c1=h*s(x0,y0,z0);
    d1=h*s1(x0,y0,z0);
    c2=h*s((x0+(h/2)),(y0+(c1/2)),(z0+(d1/2)));
    d2=h*s1((x0+(h/2)),(y0+(c1/2)),(z0+(d1/2)));
    c3=h*s((x0+(h/2)),(y0+(c2/2)),(z0+(d2/2)));
    d3=h*s1((x0+(h/2)),(y0+(c2/2)),(z0+(d2/2)));
    c4=h*s(x1,(y0+c3),(z0+d3));
    d4=h*s1(x1,(y0+c3),(z0+d3));
    c=(c1+2*c2+2*c3+c4)/6;
    d=(d1+2*d2+2*d3+d4)/6;
    yans=y0+c;
    zans=z0+d;
    y0=yans;
    z0=zans;
    x0=x1;
end

% Prints the answer
disp('The value of the diff eqn at unknown x is: ');
disp(yans);

                                                      Third Year Mechanical Engineering
                                                   Computer Oriented Numerical Methods
                                                                              2011-12©
                                                         MITCOE Mechanical Engineering
disp('The value of the differential at unknown x is: ');
disp(zans);




                                                Third Year Mechanical Engineering
                                             Computer Oriented Numerical Methods
                                                                        2011-12©
                                                   MITCOE Mechanical Engineering
Output:
RUNGE KUTTA METHOD 4TH ORDER FOR SIMLTANEOUS EQUATONS
Enter the 1st diff. eqn in x, y, z: x+y*z
Enter the 2nd diff. eqn in x, y, z: x^2-y^2
Enter y: 1
Enter x: 0
Enter z: 0.5
Enter unknown x: 1.2
Enter step size: 1.2
The value of the diff eqn at unknown x is:
   1.352724056760832

The value of the differential at unknown x is:
  -0.775714711925248




                                                Third Year Mechanical Engineering
                                             Computer Oriented Numerical Methods
                                                                        2011-12©
                                                   MITCOE Mechanical Engineering
Assignment No:

Statement: Write down the Matlab Program for Adams Bashforth.


Solution:

Input:
clear; clc; % Clears the work space

% Get the input from user
g=input('Enter the function dy/dx: ','s');
x=input('Enter values of x: ');
y=input('Enter values of y: ');
xg=input('Enter x at which value is to be found: ');
h=input('Enter step size: ');

f=inline(g); % Convert the input string into a symbolic function

m=size(x); % Calculate the size of matrix x

% Main calculation
n=(xg-x(4))/h;
for i=1:n
    ya=y(4)+(h/24)*(-9*(f(x(1),y(1)))+(37*(f(x(2),y(2))))-
(59*(f(x(3),y(3))))+(55*(f(x(4),y(4)))));
    ya1=y(4)+(h/24)*((f(x(2),y(2)))-
(5*(f(x(3),y(3))))+(19*(f(x(4),y(4))))+(9*f(x(4)+h,ya)));
    while(ya1~=ya)
        ya=ya1;
        ya1=y(4)+(h/24)*((f(x(2),y(2)))-
(5*(f(x(3),y(3))))+(19*(f(x(4),y(4))))+(9*f(x(4)+h,ya)));
    end
    for j=1:m-1
    x(j)=x(j+1);
    y(j)=y(j+1);
    end
    x(m)=x(m)+h;
    y(4)=ya1;
end

fprintf('The value at given x is :      %f n',ya1); % Prints the answer




                                                      Third Year Mechanical Engineering
                                                   Computer Oriented Numerical Methods
                                                                              2011-12©
                                                         MITCOE Mechanical Engineering
OUTPUT:

Enter the function dy/dx: 1+x*y^2
Enter values of x: [0 0.1 0.2 0.3]
Enter values of y: [0.2 0.3003 0.4022 0.5075]
Enter x at which value is to be found: 0.5
Enter step size: 0.1
The value at given x is : 0.740490




                                                   Third Year Mechanical Engineering
                                                Computer Oriented Numerical Methods
                                                                           2011-12©
                                                      MITCOE Mechanical Engineering
Assignment No: 36

Statement: Write down the Matlab Program for Parabolic method.



Solution:

Input :

% Program for Parabollic Equation (Schmidt Method)
clear all;
clc;
a=1; b=1;
% input
xi=input('Enter initial value of x: ');
xf=input('Enter final value of x: ');
while a==1
     h=input('Enter step size for x: ');
     co=(xf-xi)*10000/(h*10000);
     if mod(co,1)==0
         a=0;
     end
end
ti=input('Enter initial value of t: ');
tf=input('Enter final value of t: ');
while b==1
     k=input('Enter step size for t: ');
     ro=(tf-ti)*10000/(k*10000);
     if mod(ro,1)==0
         b=0;
     end
end
s=input('For all values of x at t=0, u(x)=','s');
f=inline(s);
C=input('Enter value of C: ');
r=k/h^2*C^2;
% Assign side values in matrix
u(1,2:co+2)=xi:h:xf;
u(2:ro+2,1)=ti:k:tf;
u(2:ro+2,2)=input('Enter constant value of u for x=xi: ');
u(2:ro+2,co+2)=input('Enter constant value of u for x=xf: ');
% Assign central values in matrix by finding them
for i=3:co+1
     u(2,i)=f(u(1,i));
end

for i=3:ro+2
    for j=3:co+1
        u(i,j)=r*u(i-1,j-1)+(1-2*r)*u(i-1,j)+r*u(i-1,j+1);
    end
end
% display output
                                                      Third Year Mechanical Engineering
                                                   Computer Oriented Numerical Methods
                                                                              2011-12©
                                                         MITCOE Mechanical Engineering
disp(u);



Output:

Enter initial value of x: 0

Enter final value of x: 1

Enter step size for x: 0.2

Enter initial value of t: 0

Enter final value of t: 0.006

Enter step size for t: 0.002

For all values of x at t=0, u(x)=sin(pi*x)

Enter value of C: 1

Enter constant value of u for x=xi: 0

Enter constant value of u for x=xf: 0

           0          0   0.2000   0.4000    0.6000    0.8000     1.0000

           0          0   0.5878   0.9511    0.9511    0.5878          0

    0.0020            0   0.5766   0.9329    0.9329    0.5766          0

    0.0040            0   0.5655   0.9151    0.9151    0.5655          0

    0.0060            0   0.5547   0.8976    0.8976    0.5547          0




                                                 Third Year Mechanical Engineering
                                              Computer Oriented Numerical Methods
                                                                         2011-12©
                                                    MITCOE Mechanical Engineering
Assignment No: 37

Statement: Write down the Matlab Program for Crank-Nicholeson method.



Solution:

Input :

 % Crank Nicoleson
clear all;
clc;
a=1; b=1; c=1;
% input
xi=input('Enter initial value of x: ');
xf=input('Enter final value of x: ');
h=input('Enter step size for x: ');
co=(xf-xi)*10000/(h*10000);
ti=input('Enter initial value of t: ');
tf=input('Enter final value of t: ');
k=input('Enter step size for t: ');
ro=(tf-ti)*10000/(k*10000);
s=input('For all values of x at t=0, u(x)=','s');
f=inline(s);
C=input('Enter value of C: ');
r=k*C^2/h^2;
% define side values of matrix
u(1,2:co+2)=xi:h:xf;
u(2:ro+2,1)=ti:k:tf;
u(2:ro+2,2)=input('Enter constant value of u for x=xi: ');
u(2:ro+2,co+2)=input('Enter constant value of u for x=xf: ');
for i=3:co+1
     u(2,i)=f(u(1,i));
end
ui=u;
k=1;
% find central values of matrix
while c==1 && k<=1000
     ui=u;
     for i=2:ro+1
         for j=3:co+1
              %u(i+1,j)=r/(2*(1+r))*(u(i+1,j-1)+u(i+1,j+1)+u(i,j-1)-2*u(i,j)-
u(i,j+1))+u(i,j)/(1+r);
              u(i+1,j)=1/4*(u(i+1,j-1)+u(i+1,j+1)+u(i,j-1)+u(i,j+1));
         end
     end
     k=k+1;
     uf=(u-ui)./u;
     if max(max(uf))<=0.001
         c=0;
     end
end

                                                      Third Year Mechanical Engineering
                                                   Computer Oriented Numerical Methods
                                                                              2011-12©
                                                         MITCOE Mechanical Engineering
disp(u);




Output:



Enter initial value of x: 0

Enter final value of x: 3

Enter step size for x: 1

Enter initial value of t: 0

Enter final value of t: .3

Enter step size for t: .1

For all values of x at t=0, u(x)=x^2

Enter value of C: 1

Enter constant value of u for x=xi: 0

Enter constant value of u for x=xf: 0

           0          0    1.0000   2.0000   3.0000

           0          0    1.0000   4.0000        0

    0.1000            0    1.1333   0.5333        0

    0.2000            0    0.2178   0.3378        0

    0.3000            0    0.1046   0.0806        0




                                                 Third Year Mechanical Engineering
                                              Computer Oriented Numerical Methods
                                                                         2011-12©
                                                    MITCOE Mechanical Engineering
Assignment No: 38

Statement: Write down the Matlab Program for Hyperbolic method.



Solution:

Input :

% Program to solve Hyperbolic Partial Differential Equation
clear all;
clc;
a=1; b=1;
% input
xi=input('Enter initial value of x: ');
xf=input('Enter final value of x: ');
h=input('Enter step size for x: ');
co=(xf-xi)*10000/(h*10000);
ti=input('Enter initial value of t: ');
tf=input('Enter final value of t: ');
k=input('Enter step size for t: ');
ro=(tf-ti)*10000/(k*10000);
s=input('For all values of x at t=0, u(x)=','s');
f=inline(s);
C=input('Enter value of C: ');
r=h/k;
if r~=C
     error('r is not equal to C');
end
% Assign side values in matrix
u(1,2:co+2)=xi:h:xf;
u(2:ro+2,1)=ti:k:tf;
u(2:ro+2,2)=input('Enter constant value of u for x=xi: ');
u(2:ro+2,co+2)=input('Enter constant value of u for x=xf: ');
% Assign unknown values in matrix
for i=3:co+1
     u(2,i)=f(u(1,i));
end
for i=3:co+1
     u(3,i)=(u(2,i-1)+u(2,i+1))/2;
end
for i=4:ro+2
     for j=3:co+1
         u(i,j)=u(i-1,j-1)+u(i-1,j+1)-u(i-2,j);
     end
end
% display output
disp(u);




                                                      Third Year Mechanical Engineering
                                                   Computer Oriented Numerical Methods
                                                                              2011-12©
                                                         MITCOE Mechanical Engineering
Output:



Enter initial value of x: 0

Enter final value of x: 4

Enter step size for x: 1

Enter initial value of t: 0

Enter final value of t: 2.5

Enter step size for t: 0.5

For all values of x at t=0, u(x)=(x^2)*(2-x)

Enter value of C: 2

Enter constant value of u for x=xi: 0

Enter constant value of u for x=xf: 0

          0           0    1.0000   2.0000    3.0000     4.0000

          0           0    1.0000        0    -9.0000         0

    0.5000            0         0   -4.0000         0         0

    1.0000            0   -5.0000         0    5.0000         0

    1.5000            0         0   4.0000         0          0

    2.0000            0    9.0000        0    -1.0000         0

    2.5000            0         0   4.0000         0          0




                                                   Third Year Mechanical Engineering
                                                Computer Oriented Numerical Methods
                                                                           2011-12©
                                                      MITCOE Mechanical Engineering
Assignment No: 39

Statement: Write down the Matlab Program for Elliptical method.



Solution:

Input :

clear all;
clc;
% take user input
u=input('Temperature of upper surface: ');
l=input('Temperature of left surface: ');
r=input('Temperature of right surface: ');
b=input('Temperature of lower surface: ');
cs=input('No. of elements in a row: ');
n=cs-1;
% Create a equation matrix
an(n,n)=0;
for i=1:n^2
     for j=1:n^2
         if i==j
             an(i,j)=4;
         elseif mod(i,n)==1 && j==i+1
             an(i,j)=-1;
         elseif j==i-n && j>0
             an(i,j)=-1;
         elseif j==i+n && j<=n^2
             an(i,j)=-1;
         elseif mod(i,n)==0 && j==i-1
             an(i,j)=-1;
         elseif mod(i,n)>1 && ( j==i+1 || j==i-1 )
             an(i,j)=-1;
         end
     end
end
so(n)=0;
for i=1:n^2
     if i==1
         so(i)=u+l;
     elseif i>1 && i<n
         so(i)=u;
     elseif i==n
         so(i)=u+r;
     elseif mod(i,n)==1 && i>n && i<=n^2-n
         so(i)=l;
     elseif mod(i,n)>1 && mod(i,n)<n && i>n && i<=n^2-n
         so(i)=0;
     elseif mod(i,n)==0 && i>n && i<=n^2-n
         so(i)=r;
     elseif i==n^2-n+1

                                                        Third Year Mechanical Engineering
                                                     Computer Oriented Numerical Methods
                                                                                2011-12©
                                                           MITCOE Mechanical Engineering
so(i)=l+b;
    elseif i>n^2-n+1 && i<n^2
        so(i)=b;
    elseif i==n^2
        so(i)=b+r;
    end
end
an
so
an1=an;
clear an;
% solve the matrix
t=GaussSoln(an1,so,n^2);
k=1;
% interpret the answers
for i=1:n
     for j=1:n
         t1(i,j)=t(k);
         k=k+1;
     end
end
t1
hold off;
% plot the answers
for i=1:n
     for j=1:n
         scatter(i,j,80,[0.5 0 0],'filled');
         s=sprintf('n %1.2f',(t1(i,j)));
         text(j,i,s);
         hold on;
     end
end
axis ij;
axis ([ 0 n+1 0 n+1]);
hold off;


GaussSoln:

function Soln=GaussSoln(x,y,n1)
A1=x;
B=y;
n=n1;
clear x;
clear y;
% Check the conditions
if det(A1)==0
     disp('Either no solution or infinitely many solutions.');
else
% forward elimination
     A=A1;
     A(:,n+1)=B(1:n);
     for i=1:n-1
         for j=i:n-1
             fac=A(j+1,i)/A(i,i);
                                                  Third Year Mechanical Engineering
                                               Computer Oriented Numerical Methods
                                                                          2011-12©
                                                     MITCOE Mechanical Engineering
fac_mat=fac*A(i,:);
                 A(j+1,:)=A(j+1,:)-fac_mat;
         end
    end
    i=0;j=0;
% Back substitution
    if A(n,n)==0
         an(n)=0;
    else
         an(n)=A(n,n+1)/A(n,n);
    end
    for i=n-1:-1:1
         for j=n:-1:1
              x(j)=an(j)*A(i,j);
         end
         y=sum(x);
         if y==0
              an(i)=0;
         else
              an(i)=(A(i,n+1)-y)/A(i,i);
         end
    end
end
% answer
Soln=an;




Output:

Temperature of upper surface: 100

Temperature of left surface: 100

Temperature of right surface: 0

Temperature of lower surface: 0

No. of elements in a row: 3



an =



        4   -1      -1      0

       -1    4       0     -1

       -1    0       4     -1

        0   -1      -1      4

                                                 Third Year Mechanical Engineering
                                              Computer Oriented Numerical Methods
                                                                         2011-12©
                                                    MITCOE Mechanical Engineering
so =



   200   100     100     0




t1 =



   75.0000     50.0000

   50.0000     25.0000




                                Third Year Mechanical Engineering
                             Computer Oriented Numerical Methods
                                                        2011-12©
                                   MITCOE Mechanical Engineering
Third Year Mechanical Engineering
Computer Oriented Numerical Methods
                           2011-12©
      MITCOE Mechanical Engineering
FLOWCHART: Newton Raphson method



                                     START


                                   Set v=1500
                                   vr=2500
                                   g=9.81
                                   m=2,00,000
                                   uf=300



                                            `
                               Read function (a)
                               Derivative of function(b)




                                    ft=inline(a)

                                    dft=inline(b)



                                Read significant
                                digits ‘n’




                           epsilon_s= (0.5*10^(2-n))

                           epsilon_a=100



                                Input initial
                                guess



                                    td=tx


                                                       Third Year Mechanical Engineering
                                                    Computer Oriented Numerical Methods
                                                                               2011-12©
                                                          MITCOE Mechanical Engineering
A
                      A



        while epsilon_a >=epsilon_s




tnew= td-(ft(td)/dft(td))

epsilon_a= abs((tnew-td)/tnew)*100

td=tnew




            print error,tnew




             print tnew




                 END




                                   Third Year Mechanical Engineering
                                Computer Oriented Numerical Methods
                                                           2011-12©
                                      MITCOE Mechanical Engineering
FLOWCHART: Modified Newton Raphson method



                                      START


                              Read function (a)
                              Derivative of function(b)
                              Second derivative (c)




                                    t0=0
                                    f=inline(a)
                                    df=inline(b)
                                    ddf=inline(c)



                                Read significant
                                digits ‘n’




                           epsilon_s= (0.5*10^(2-n))
                           epsilon_a=100
                           tr=fzero(inline ft)



                                    disp tr



                              Input initial guess




                                  print head




                                 disp head

                                                        Third Year Mechanical Engineering
                                                     Computer Oriented Numerical Methods
                                                                                2011-12©
                                                           MITCOE Mechanical Engineering
A
                     A



                  while (1)




tnew=told-((fx(told)*dfx(told)/((dfx(told)^2)-(fx(told)*d2fx(told)))))
err=abs((tnew-told)/tnew)*100
epsilon_t=abs((tr-tnew)/tr)*100
 told=tnew




              disp table



                                                                         NO

           if err<=epsilon_s

                         YES

              print tnew




                  END




                                     Third Year Mechanical Engineering
                                  Computer Oriented Numerical Methods
                                                             2011-12©
                                        MITCOE Mechanical Engineering
FLOWCHART: Successive Approximation


                                   START



                              Read function (g)




                                f=inline(g)




                             Read significant
                             digits ‘n’




                            Input initial guess




                         epsilon_s= (0.5*10^(2-n))
                         epsilon_a=100
                         tr=fzero(inline (g))



                                 disp tr



                               set abcd




                              disp abcd




                              set head
                                                     Third Year Mechanical Engineering
                                                  Computer Oriented Numerical Methods
                                                                             2011-12©
                                                        MITCOE Mechanical Engineering
A



                     A
                 disp head


               while ea>=es




temp=t
t=f(t)
ea=abs((t-temp)/t)*100
et=abs((tr-t)/tr)*100




               disp table




                  END




                                 Third Year Mechanical Engineering
                              Computer Oriented Numerical Methods
                                                         2011-12©
                                    MITCOE Mechanical Engineering
FLOWCHART: Gauss-Naïve Elimination method
                                             Start


                                          Input
                                          matrices
                                          A&B



                                   [m,n]=size [A]




                                                             NO          Print “Matrix
                                        If m~=n
                                                                         must be
                                                                         square!”
                                                 YES

                                      For k=1:n-1
                                                                              A




                                      For i=k+1:n




                               Factor a(i,k)/a(k,k)

                               9i,k

                                        For j=k:n




                              a(I,j)=a(I,j)-factor a(k,j)



                               b(i)=b(i)- factor*b(k)
                                                               Third Year Mechanical Engineering
                                                            Computer Oriented Numerical Methods
                                                                                       2011-12©
                                                                  MITCOE Mechanical Engineering
                                           M
M



      Display A & B




         For i=n:-1:1




      x(i) = b(i) / a(I,i)




          for j=1:i-1




  b(j) = b(j) - x(i)*a(j,i)




Display values of x




                                      A


    End
                                 Third Year Mechanical Engineering
                              Computer Oriented Numerical Methods
                                                         2011-12©
                                    MITCOE Mechanical Engineering
FLOWCHART: Gauss with Partial Pivoting method


                                             Start




                                  Input matrices A & B




                                     [m,n] = size (a)




                                                            NO
                                       If m~=n                         Print “Matrix
                                                                       must be
                                                                       square!”

                                                 YES

                                                                         E
                A                     For k=1:n-1




                               [xyz,i]=max(abs(a(k:n,k)))
                                 ipr=i+k-1;




                                     if ipr~=k




                                                           Third Year Mechanical Engineering
                                                        Computer Oriented Numerical Methods
                                                                                   2011-12©
                                                              MITCOE Mechanical Engineering
B

            a([k,ipr],:)=a([ipr,k],:)
            b([k,ipr],:)=b([ipr,k],:)

                  For i=k+1:n
                           B




              factor=a(i,k)/a(k,k)




                   For j=k:n




        a(i,j)=a(i,j)-(factor*(a(k,j)))




           b(i)=b(i)-factor*(b(k))



A

                 Display A & B




    D
                 for i=n:-1:1




                                             Third Year Mechanical Engineering
                       C                  Computer Oriented Numerical Methods
                                                                     2011-12©
                                                MITCOE Mechanical Engineering
C




       x(i)=b(i)/a(i,i)




          For j=1:i-1




    b(j)=b(j)-x(i)*a(j,i)


D

      Display x


                                     E


         End




                               Third Year Mechanical Engineering
                            Computer Oriented Numerical Methods
                                                       2011-12©
                                  MITCOE Mechanical Engineering
FLOWCHART: Thomas Algorithm


                                 START


                          Input matix e,f,g,r




                              n=length (e)




                                 for k=1:n




                       factor =e(k)/f(k)
                       f(k+1)=f(k+1)-xg(k)
                       r(k+1)= r(k+1)-factor*r(k)




                         x(n+1)=r(n+1)/f(n+1)




                               for k=n:1




                        x(k)=r(k)-g(k)*x(k+1)/k




                                display




                                   end
                                                     Third Year Mechanical Engineering
                                                  Computer Oriented Numerical Methods
                                                                             2011-12©
                                                        MITCOE Mechanical Engineering
Third Year Mechanical Engineering
Computer Oriented Numerical Methods
                           2011-12©
      MITCOE Mechanical Engineering
FLOWCHART: Gauss-Seidel without Relaxation method

                                           Start



                                Input matrices
                                A&B


                                [m,n]=size(a)




                                                     NO             Print “Matrix Must
                                   if (m~=n)
                                                                    Be Square!”


                                               YES


                                 For i= m:n




                            d(i)=b(i)/a(i,i)




                                    d=d’
                                    c=a




                                 For i=1:n
          B




                               For j=1:n



                           c(i,j)=a(i,j)/a(i,i)          Third Year Mechanical Engineering
                                                      Computer Oriented Numerical Methods
                                                                                 2011-12©
                                                            MITCOE Mechanical Engineering
                                    A
MITCOE 2011-12 conm-submission
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MITCOE 2011-12 conm-submission

  • 1. Assignment No: 1 Statement: Write down the Matlab Program using Newton-Raphson method for any equation. Solution: Input : clc; clear all; % Clears the workspace format long; % Take i/p from user a=input('n Enter the function : ','s'); b=input('n Enter the derivative of function : ','s'); % Converts the input string into symbolic function ft=inline(a); dft=inline(b); n=input('enter no. of significant digits: '); t0=0; epsilon_s=(0.5*10^(2-n)); epsilon_a=100; tr=fzero((ft),t0); % Solver disp (tr); varun=sprintf('BY NEWTON-RAPHSON METHOD:'); disp(varun); tx=input('Enter your initial guess for root: '); td=tx; head=sprintf('Time tttttepsilon_a tttttepsilon_t tttttepsilon_s '); disp(head); while (epsilon_a>=epsilon_s) tnew=td-(ft(td)/dft(td)); epsilon_a=abs((tnew-td)/tnew)*100; epsilon_t=abs((tr-tnew)/tr)*100; td=tnew; table=sprintf('%d ttt %ftttt %4.9f ttt %f ttt %f',tnew,epsilon_a,epsilon_t,epsilon_s); disp(table); end % Prints the answer fprintf('n n The root of the equation is : %f n',tnew) Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 2. Output : Enter the function : (exp(t))*cos(t)-1.4 Enter the derivative of function : (exp(t))*cos(t)-(exp(t))*sin(t) enter no. of significant digits: 4 0.433560875352657 BY NEWTON-RAPHSON METHOD: Enter your initial guess for root: 0 Time epsilon_a epsilon_t epsilon_s 4.000000e-001 100.000000 7.740752743 0.005000 4.327044e-001 7.558146 0.197537266 0.005000 4.335602e-001 0.197392 0.000145353 0.005000 4.335609e-001 0.000145 0.000000000 0.005000 The root of the equation is : 0.433561 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 3. Assignment No: 2 Statement: Write down the Matlab Program using Modified Newton-Raphson method for any equation. Solution: Input : clc; Clear all; % Clears the workspace % Takes the i/p from user a=input('n Enter the function : ','s'); b=input('n Enter the derivative of function : ','s'); c=input('n Enter second order derivative : ','s'); x0=0; % Converts the input string into symbolic function fx=inline(a); dfx=inline(b); d2fx=inline(c); n=input('Enter number of significant digits: '); epsilon_s=(0.5*10^(2-n)); tr=fzero((fx),0); % Using solver disp (tr); v=input('n Enter your initial guess for root : '); told=v; varun=sprintf('BY MODIFIED NEWTON-RAPHSON METHOD:'); disp(varun); head=sprintf('Time tttttepsilon_a tttttepsilon_t tttttepsilon_s '); disp(head); while(1) tnew=told-((fx(told)*dfx(told)/((dfx(told)^2)-(fx(told)*d2fx(told))))); err=abs((tnew-told)/tnew)*100; epsilon_t=abs((tr-tnew)/tr)*100; told=tnew; table=sprintf('%d ttt %ftttt %4.9f ttt %f ttt %f',tnew,err,epsilon_t,epsilon_s); disp(table); if (err<=epsilon_s) break; end end fprintf('n n The root of the equation is : %f n',tnew) Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 4. Output Enter the function : x*sin(x)+cos(x) Enter the derivative of function : x*cos(x) Enter second order derivative : cos(x)-(x*sin(x)) Enter number of significant digits: 5 -2.7984 Enter your initial guess for root : 6 BY MODIFIED NEWTON-RAPHSON METHOD: Time epsilon_a epsilon_t epsilon_s 6.117645e+000 1.923040 318.613323399 0.000500 6.121248e+000 0.058870 318.742096720 0.000500 6.121250e+000 0.000035 318.742173765 0.000500 The root of the equation is : 6.121250 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 5. Assignment No: 3 Statement: Write down the Matlab Program using successive approximation method for any equation. Solution: Input : clc; clear; g=input('Enter the function:','s'); f=inline(g); % Defining function n=input('Enter number of significant digits: '); es=(0.5*10^(2-n)); % Stopping criteria ea=100; t0=0; t=input('Enter initial guess: '); tr=fzero((f),t0); % Calculating true roots disp (tr); head1=sprintf('BY SUCCESSIVE APPROXIMATION METHOD:'); disp(head1); head=sprintf('Time tttttepsilon_a ttttt epsilon_t ttttt epsilon_s '); disp(head); while (ea>=es) temp=t; t=f(t); ea=abs((t-temp)/t)*100; % Calc approximate error et=abs((tr-t)/tr)*100; % Calc true error table=sprintf('%d ttt %ftttt %4.9f ttt %f ttt %f',t,ea,et,es); disp(table); end Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 6. Output : Enter the function:(exp(-x)-x) Enter number of significant digits: 2 Enter initial guess: 0 BY SUCCESSIVE APPROXIMATION METHOD: 0.567143290409784 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 7. Assignment No: 4 Statement: Write down the Matlab Program using Gauss Naïve elimination method. Solution: Input : clc; clear all; a=input('enter matrix A[]: ') b=input('enter column matrix B[]: ') [m,n]=size(a); % determines size of matrix. if (m~=n) error('Matrix Must Be Square!'); end %forward elimination for k=1:n-1 for i=k+1:n factor=a(i,k)/a(k,k); for j=k:n a(i,j)=a(i,j)-(factor*(a(k,j)));% calculates each element of matrix A. end b(i)=b(i)-factor*(b(k)); % calculates each element of matrix B. end disp (a); end disp (a); disp (b); % backward substitution for i=n:-1:1 x(i)=b(i)/a(i,i); % calculates values of unknown matrix. for j=1:i-1 b(j)=b(j)-x(i)*a(j,i); end end disp('VALUES ARE:') disp(x) Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 8. Output : enter matrix A[]: [1 -1 1 ;3 4 2 ; 2 1 1 ] a = 1 -1 1 3 4 2 2 1 1 enter column matrix B[]: [6 9 7] b = 6 9 7 1.0000 -1.0000 1.0000 0 7.0000 -1.0000 0 0 -0.5714 6.0000 -9.0000 -1.1429 VALUES ARE: 3.0000 -1.0000 2.0000 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 9. Assignment No: 5 Statement: Write down the Matlab Program using Gauss with partial pivoting method. Solution: Input : clc; clear all; a=input('enter matrix A[]: '); b=input('enter column matrix B[]: '); [m,n]=size(a); % calculates size of matrix A. if (m~=n) error('Matrix Must Be Square!'); end %pivoting for k=1:n-1 [xyz,i]=max(abs(a(k:n,k))); % finds maximum element in matrix A. ipr=i+k-1; if ipr~=k a([k,ipr],:)=a([ipr,k],:); % interchanging of rows. b([k,ipr],:)=b([ipr,k],:); % interchanging of rows. end %forward elimination for i=k+1:n factor=a(i,k)/a(k,k); for j=k:n a(i,j)=a(i,j)-(factor*(a(k,j))); % calculates each element of matrix A. end b(i)=b(i)-factor*(b(k)); % calculates each element of matrix B. end disp (a); end %disp (a); disp (b); % backward substitution for i=n:-1:1 x(i)=b(i)/a(i,i); % calculates values of unknown matrix. for j=1:i-1 b(j)=b(j)-x(i)*a(j,i); end end disp('VALUES ARE:') disp(x) Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 10. Output : enter matrix A[]: [2 -6 -1;-3 -1 7;-8 1 -2] enter column matrix B[]: [-38;-34;-20] -8.00000000000000 1.00000000000000 -2.00000000000000 0 -1.37500000000000 7.75000000000000 0 -5.75000000000000 -1.50000000000000 -8.00000000000000 1.00000000000000 -2.00000000000000 0 -5.75000000000000 -1.50000000000000 0 0 8.10869565217391 -20.00000000000000 -43.00000000000000 -16.21739130434783 VALUES ARE: 4 8 -2 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 11. Assignment No: 6 Statement: Write down the Matlab Program using Thomas Algorithm method. Solution: Input : clc; clear; %format long; e=input('Enter the value of e, ie. subdiagonal vector :'); f=input('Enter the value of f, ie. diagonal vector :'); g=input('Enter the value of g, ie. superdiagonal vector :'); r=input('Enter the value of r, ie. value vector :'); n=length(e); % Size of matrix e for k=1:n factor=e(k)/f(k); % Multiplying factor f(k+1)=f(k+1)-factor*g(k); % Transforming diagonal vector r(k+1)=r(k+1)-factor*r(k); % Transforming value vector end x(n+1)=r(n+1)/f(n+1); % Transforming unknown vector for k=n:-1:1 x(k)=(r(k)-g(k)*x(k+1))/f(k); % Finding values of unknowns end disp('VALUES ARE:'); disp (x) Output : Enter the value of e, ie. subdiagonal vector :[-.4;-.4] Enter the value of f, ie. diagonal vector :[0.8;0.8;0.8] Enter the value of g, ie. superdiagonal vector :[-.4;-.4] Enter the value of r, ie. value vector :[41;25;105] VALUES ARE: 173.7500 245.0000 253.7500 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 12. Assignment No: 7 Statement: Write down the Matlab Program using Gauss Seidel without Relaxation method. Solution: Input : clc; clear all; format long; a = input('Enter Matrix A: '); b = input('Enter Column Matrix B: '); [m,n]= size(a); % calculates size of matrix A. if (m~=n) error('Matrix Must Be Square!'); end for i=1:n d(i)=b(i)/a(i,i); end d=d'; c=a; for i=1:n for j=1:n c(i,j)=a(i,j)/a(i,i); % factor. end c(i,i)=0; x(i)=0; end x=x'; disp (a); disp (b); disp (d); disp (c); p = input('Enter No. of Iterations: '); for k=1:p for i=1:n x(i)=d(i)-c(i,:)*x(:,1); % finds unknown value. end disp (x); end Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 13. Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 14. Output : Enter Matrix A: [3 -0.1 -0.2;0.1 7 -.3;0.3 -0.2 10] Enter Column Matrix B: [7.85;-19.3;71.4] 3.00000000000000 -0.10000000000000 -0.20000000000000 0.10000000000000 7.00000000000000 -0.30000000000000 0.30000000000000 -0.20000000000000 10.00000000000000 7.85000000000000 -19.30000000000000 71.40000000000001 0 -0.03333333333333 -0.06666666666667 0.01428571428571 0 -0.04285714285714 0.03000000000000 -0.02000000000000 0 Enter No. of Iterations: 3 2.61666666666667 -2.79452380952381 7.00560952380952 2.99055650793651 -2.49962468480726 7.00029081106576 3.00003189791081 -2.49998799235305 6.99999928321562 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 15. Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 16. Assignment No: 8 Statement: Write down the Matlab Program using Gauss Seidel with relaxation method. Solution: Input : clc; clear all; a=input('enter matrix A[]: '); b=input('enter column matrix B[]: '); [m,n]=size(a); % calculates size of matrix A. if (m~=n) error('Matrix Must Be Square!'); end %pivoting for k=1:n-1 [xyz,i]=max(abs(a(k:n,k))); % finds maximum element in matrix A. ipr=i+k-1; if ipr~=k a([k,ipr],:)=a([ipr,k],:); % interchanging of rows. b([k,ipr],:)=b([ipr,k],:); % interchanging of rows. end end for i=1:n d(i)=b(i)/a(i,i); end d=d'; c=a; for i=1:n for j=1:n c(i,j)=a(i,j)/a(i,i); % factor. end c(i,i)=0; x(i)=0; end x=x'; disp (a); disp (b); disp (d); disp (c); lambda = input('Enter the value of weighting factor: '); es=0.05; % stopping criteria. ea(i)=100; head=sprintf('tttttttttValue of x ttttttttttttttValue of ea '); disp(head); while (ea(i)>=es) for i=1:n y=x(i); x(i)=d(i)-c(i,:)*x(:,1); x(i)=lambda*x(i)+(1-lambda)*y; % calculates unknown value. Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 17. ea(i)=abs((x(i)-y)/x(i))*100; end table1=sprintf('%d ttt %ftttt %4.9f ttt%f ttt %ftttt %4.9f',x,ea); disp(table1); end Output : enter matrix A[]: [-3 1 12;6 -1 -1;6 9 1] enter column matrix B[]: [50;3;40] 6 -1 -1 6 9 1 -3 1 12 3 40 50 0.5000 4.4444 4.1667 0 -0.1667 -0.1667 0.6667 0 0.1111 -0.2500 0.0833 0 Enter the value of weighting factor: 0.95 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 18. Value of x Value of ea 4.750000e-001 3.921389 3.760702546 100.000000 100.000000 100.000000000 1.715081e+000 2.935111 4.321337313 72.304517 33.602766 12.973640471 1.709692e+000 2.830032 4.356407772 0.315233 3.712986 0.805031598 1.698338e+000 2.828267 4.355604413 0.668543 0.062401 0.018444248 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 19. Assignment No: 42 Statement: Write down the Matlab Program to fit curve y = a0 + a1*x by using least square techniques for given set of points. Solution: Input : clc; clear all; x=input('Enter row matrix x : '); y=input('Enter row matrix y : '); [m,n]=size(x); xy(1,1)=0; i=1; X=0; Y=0; XY=0; Xsqr=0; while i<=n; xy(1,i)=x(1,i)*y(1,i); xsqr(1,i)=x(1,i)^2; X=X+x(1,i); % To calculate summation of x Y=Y+y(1,i); % To calculate summation of y XY=XY+xy(1,i); % To calculate summation of x*y Xsqr=Xsqr+xsqr(1,i); % To calculate summation of x^2 i=i+1; end disp(x); disp(y); a1=(n*XY-Y*X)/(n*Xsqr-X^2); a0=(Y*Xsqr-X*XY)/(n*Xsqr-X^2); ym=Y/n; sr(1,1)=0;j=1; while j<=n sr(1,j)=(y(1,j)-a0-a1*x(1,j))^2; % To calculate sr for each x st(1,j)=(y(1,j)-ym)^2; % To calculate st for each x j=j+1; end SR=sum(sr); ST=sum(st); r2=(ST-SR)/ST s=sprintf('Best fit curve (straight line) for above data is given by : y = %f * x + %f',a1,a0); disp(s); xp=linspace(min(x),max(x)); yp=a0+a1*xp; plot(x,y,'o',xp,yp); xlabel('values of x'); ylabel('values of y'); title('y=a0+a1*x'); Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 20. grid on; Output : Enter row matrix x : [1.0 2.0 3.0 4.0 5.0 6.0 7.0] Enter row matrix y : [0.5 2.5 2.0 4.0 3.5 6.0 5.5] 1 2 3 4 5 6 7 0.5000 2.5000 2.0000 4.0000 3.5000 6.0000 5.5000 r2 = 0.8683 Best fit curve (straight line) for above data is given by : y = 0.839286 * x + 0.071429 y=a0+a1*x 6 5 4 values of y 3 2 1 0 1 2 3 4 5 6 7 values of x Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 21. Assignment No: 9 Statement: Write down the Matlab Program to fit curve y = a0 + a1*x+a2x2 by using least square techniques for given set of points. Solution: Input : clc; clear all; x = input('Enter values of x in row matrix form : '); y = input('Enter values of y in row matrix form : '); [m,n]=size(x); sx = sum(x); sy = sum(y); sx2 = sum(x.*x); sxy = sum(x.*y); sx2y = sum(x.*x.*y); sx3 = sum(x.*x.*x); sx4 = sum(x.*x.*x.*x); a = [sx2 sx n; sx3 sx2 sx; sx4 sx3 sx2]; b = [sy; sxy; sx2y]; z=inv(a)*b; s=sprintf('Best fit curve (Quadratic) for above data is given by :y = %f + %f * x + %f * x^2 ',z(1),z(2),z(3)); disp(s); xp = linspace(min(x),max(x)); yp = z(3)*(xp.*xp)+z(2)*xp+z(1); plot(x,y,'o',xp,yp); grid on; xlabel('Values of x'); ylabel('Values of function'); title('y=a0+ a1*x+ a2*(x^2)'); Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 22. Output : Enter values of x in row matrix form : [0.075 0.5 1 1.2 1.7 2.0 2.3] Enter values of y in row matrix form : [600 800 1200 1400 2050 2650 3750] Best fit curve (Quadratic) for above data is given by :y = 643.601494 + - 218.884701 * x + 685.248397 * x^2 y=a0+ a1*x+ a2*(x 2) 4000 3500 3000 Values of function 2500 2000 1500 1000 500 0 0.5 1 1.5 2 2.5 Values of x Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 23. Assignment No: 10 Statement: Write down the Matlab Program to fit curve y = a1*(xb1) by using least square techniques for given set of points. Solution: Input : clc; clear all; xa=input('Enter row matrix x : '); ya=input('Enter row matrix y : '); [m,n]=size(xa); xy(1,1)=0; y(1,1)=0; i=1; X=0; Y=0; XY=0; Xsqr=0; while (i<=n) y(1,i)=log10(ya(1,i)); % To calculate log of y x(1,i)=log10(xa(1,i)); % To calculate log of x xy(1,i)=x(1,i)*y(1,i); xsqr(1,i)=x(1,i)^2; X=X+x(1,i); % To calculate summation of x Y=Y+y(1,i); % To calculate summation of y XY=XY+xy(1,i); % To calculate summation of x*y Xsqr=Xsqr+xsqr(1,i); % To calculate summation of x^2 i=i+1; end disp(xa); disp(ya) beta=(n*XY-Y*X)/(n*Xsqr-X^2); a0=(Y*Xsqr-X*XY)/(n*Xsqr-X^2); alpha=10^(a0); % To calculate co-eff of x^a0 ym=Y/n; sr(1,1)=0;j=1; while j<=n sr(1,j)=(y(1,j)-a0-beta*x(1,j))^2; % To calculate sr for each x Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 24. st(1,j)=(y(1,j)-ym)^2; % To calculate st for each x j=j+1; end SR=sum(sr); ST=sum(st); r2=(ST-SR)/ST s=sprintf('Best fit curve (polynomial) for above data is given by : y = %f * x^(%f) ',alpha,beta); disp(s); xp = linspace(min(x),max(x)); yp = (xp.^beta)*alpha; plot(xa,ya,'o') hold on plot(xp,yp) grid on; xlabel('values of x'); ylabel('values of y'); title('y=alpha*x^(beta)'); Output : Enter row matrix x : [26.67 93.33 148.89 315.56] Enter row matrix y : [1.35 0.085 0.012 0.00075] 26.6700 93.3300 148.8900 315.5600 1.3500 0.0850 0.0120 0.0008 r2 = 0.9757 Best fit curve (polynomial) for above data is given by : y = 38147.936083 * x^(-3.013376) Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 25. y=alpha*(x)b 2 1.8 1.6 1.4 1.2 values of y 1 0.8 0.6 0.4 0.2 0 0 50 100 150 200 250 300 350 values of x Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 26. Assignment No: 41 Statement: Write down the Matlab Program to fit curve y = a1 * e (b1*x) by using least square techniques for given set of points. Solution: Input : clc; clear all; x=input('Enter row matrix x : '); ya=input('Enter row matrix y : '); [m,n]=size(x); % Defining size of matrix x xy(1,1)=0; y(1,1)=0; % Defining matrix xy & y i=1; X=0; Y=0; XY=0; Xsqr=0; % Setting initial condition for loop while i<=n; y(1,i)=log(ya(1,i)); xy(1,i)=x(1,i)*y(1,i); xsqr(1,i)=x(1,i)^2; X=X+x(1,i); Y=Y+y(1,i); XY=XY+xy(1,i); Xsqr=Xsqr+xsqr(1,i); i=i+1; end disp(x); disp(ya); a1=(n*XY-Y*X)/(n*Xsqr-X^2); a0=(Y*Xsqr-X*XY)/(n*Xsqr-X^2); alpha=exp(a0); ym=Y/n; % Finding mean sr(1,1)=0;j=1; while j<=n; sr(1,j)=(y(1,j)-a0-a1*x(1,j))^2; st(1,j)=(y(1,j)-ym)^2; j=j+1; Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 27. end xp = linspace(min(x),max(x)); % Condition for graph yp= alpha*exp(a1*xp); % Given function SR=sum(sr); ST=sum(st); r2=(ST-SR)/ST % Co-efficient of determination s=sprintf('Best fit curve (exponential) for above data is given by : y = %f * e^(%f * x) ',alpha,a1); disp(s); plot(x,ya,'o',xp,yp) % Plots function & best fitted curve simultaneously grid on; xlabel('values of x'); % Defining specifications of graph ylabel('values of y'); title('y=alpha*e^(beta*x)'); grid on; % To display grid on graph Output : Enter row matrix x : [0.4 0.8 1.2 1.6 2.0 2.3] Enter row matrix y : [800 975 1500 1950 2900 3600] 0.4000 0.8000 1.2000 1.6000 2.0000 2.3000 800 975 1500 1950 2900 3600 r2 = 0.9933 Best fit curve (exponential) for above data is given by : y = 546.590939 * e^(0.818651 * x) Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 28. y=alpha*e(beta*x) 4000 3500 3000 values of y 2500 2000 1500 1000 500 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 values of x Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 29. Assignment No: 11 Statement: Write down the Matlab Program for Lagrange Interpolation. Solution: Input : clc; clear all; x = input('Enter the of Values of x: '); y = input('Enter the of Values of y: '); u = input('Value of x at which y is to be evaluated: '); n = length(x); % Size of matrix x p=1; s=0; for i=1:n p=y(i); for j=1:n if (i~=j) % Condition for inequality p=p*(u-x(j))/(x(i)-x(j)); % Formula end end s=s+p; % Summation end fprintf('n Value of y at required x is : %f ',s); Output : Enter the of Values of x: [1 4 5 7] Enter the of Values of y: [21.746 438.171 1188.9147 8775.011] Value of x at which y is to be evaluated: 4.2 Value of y at required x is : 490.360287 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 30. Assignment No: 12 Statement: Write down the Matlab Program for Newton-Gregory Forward Difference Interpolation. Solution: Input : clc; clear all; x=input('Enter row matrix x : '); y=input('Enter row matrix y : '); X=input('Enter value of x at which value of function is to be calculated : '); [m,n]=size(x); dx=diff(x); % Spatial diff.(for equally spaced data) d(1,1)=y(1,1); disp(x); disp(y); for j=1:(n-1) dy=diff(y); % Delta matrix disp(dy); d(j+1)=dy(1); % Stores 1st value of delta matrix. y=dy; end alpha=(X-x(1))/dx(1); % Value of alpha is calculated. a(1,1)=1; prod=1; for k=1:(n-2) prod=prod*(alpha-k+1); a(k+1)=prod; end func=0; for i=1:n-1 fx=a(i)*d(i)/(factorial(i-1)); func=func+fx; end Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 31. s=sprintf('Value of function calculated by N-G forward interpolation : %f',func); disp(s); Output : Enter row matrix x : [2 3 4 5 6 7 8 9] Enter row matrix y : [19 48 99 178 291 444 643 894] Enter value of x at which value of function is to be calculated : 3.5 2 3 4 5 6 7 8 9 19 48 99 178 291 444 643 894 29 51 79 113 153 199 251 22 28 34 40 46 52 6 6 6 6 6 0 0 0 0 0 0 0 0 0 0 Value of function calculated by N-G forward interpolation : 70.375000 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 32. Assignment No: 13 Statement: Write down the Matlab Program for Newton-Gregory Backward Difference Interpolation. Solution: Input : clc; clear all; x=input('Enter row matrix x : '); y=input('Enter row matrix y : '); X=input('Enter value of x at which value of function is to be calculated : '); [m,n]=size(x); dx=diff(x); % Spatial diff.(for equally spaced data) d(1,1)=y(n); newx(1,n:-1:1)=x(1,1:n); % Reversing order of matrix x so that nth value is brought 1st. newy(1,n:-1:1)=y(1,1:n); % Reversing order of matrix y so that nth value is brought 1st. disp(newx) disp(newy) for j=1:(n-1) dy=diff(newy); % Delta matrix disp(dy); d(j+1)=dy(1); % Stores 1st value of delta matrix. newy=dy; end alpha=(x(n)-X)/dx(1); % Value of alpha is calculated. a(1,1)=1; prod=1; for k=1:(n-2) prod=prod*(alpha-k+1); a(k+1)=prod; end Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 33. func=0; for i=1:n-1 fx=a(i)*d(i)/(factorial(i-1)); func=func+fx; end s=sprintf('Value of function calculated by N-G backward interpolation : %f',func); disp(s); Output : Enter row matrix x : [0.1 0.2 0.3 0.4 0.5] Enter row matrix y : [1.4 1.56 1.76 2 2.28] Enter value of x at which value of function is to be calculated : 0.25 0.5000 0.4000 0.3000 0.2000 0.1000 2.2800 2.0000 1.7600 1.5600 1.4000 -0.2800 -0.2400 -0.2000 -0.1600 0.0400 0.0400 0.0400 1.0e-015*0.2220 -0.2220 -4.4409e-016 Value of function calculated by N-G backward interpolation : 1.655000 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 34. Assignment No: 14 Statement: Write down the Matlab Program for Hermite interpolation method. Solution: Input : clc; clear all; disp('HERMITE INTERPOLATION'); x=input('Enter the values of x: '); xu=input('Enter the value unknown of x: '); fx=input('Enter the values of fx: '); dfx=input('Enter the values of dfx: '); n=size(x); % Size of matrix sum=0;suma=0;sumb=0; for i=1:n pro=1; pro1=1; for j=1:n if i~=j pro=pro*(xu-x(j))/(x(i)-x(j)); % Lagrange formulation of unknown x. pro1=pro1*(x(i)-x(j)); % Derivative of Lagrange term end end L(i,1)=pro; % Lagrange term dL(i,1)=pro1; % Derivative of Lagrange term end for k=1:n suma=suma+(1-2*(xu-x(k))*dL(k))*((L(k))^2)*fx(k); % Summation sumb=sumb+(xu-x(k))*((L(k))^2)*dfx(k); end sumf=suma+sumb; disp('The value of fx at unknown x is: '); disp(sumf); Output: HERMITE INTERPOLATION Enter the values of x: [0;1] Enter the value unknown of x: 0.4 Enter the values of fx: [0;1] Enter the values of dfx: [0;2] The value of fx at unknown x is: 0.1600 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 35. Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 36. Assignment No: 15 Statement: Write down the Matlab Program for interpolation by Cubic spline. Solution: Input : % Clearing Workspace clear all; clc; close; % Defining Input points x1=input('Enter matrix for values of x: '); y1=input('Enter matrix for values of y: '); xg=input('Enter value of x for which to find y: '); m1=size(x1); n=m1(1,2); x=x1'; y=y1'; scatter(x,y); hold on; % MATLAB function plotting Cubic Interpolation yy = spline(x,y,0:0.01:100); plot(x,y,'o',0:0.01:100,yy); % Defining end conditions f''(x)=0 @ 1st and last point M(1:n+1)=0; % First row of matrix to be solved A(1,1:3)=[2*(x(3)-x(1)) (x(3)-x(2)) 0]; B(1,1)=6*(y(3)-y(2))/(x(3)-x(2))-6*(y(2)-y(1))/(x(2)-x(1)); % Subsequent rows till n-2 if n>3 for l=2:n-2 A(l,l-1:l+1)=[(x(l+1)-x(l)) 2*(x(l+2)-x(l)) (x(l+2)-x(l+1))]; B(l,1)=6*(y(l+2)-y(l+1))/(x(l+2)-x(l+1))-6*(y(l+1)-y(l))/(x(l+1)- x(l)); end end % Last 1 row A(n-1,n-2:n-1)=[(x(n)-x(n-1)) 2*(x(n)-x(n-1))]; B(n-1,1)=-6*(y(n)-y(n-1))/(x(n)-x(n-1)); Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 37. % Finding other values of f''(x) N=GaussSoln(A,B); % Assigning Values to M for i=1:n-1 M(i+1)=N(i); end % Creating the interpolation function between intervals f=inline('Ma/6/(xb-xa)*(xb-xx)^3-Mb/6/(xb-xa)*(xa-xx)^3+(ya/(xb-xa)-Ma*(xb- xa)/6)*(xb-xx)-(yb/(xb-xa)-Mb*(xb-xa)/6)*(xa- xx)','xx','Ma','Mb','xa','xb','ya','yb'); % Ploting the spline in intervals xn(1:1000)=0; yn(1:1000)=0; for i=1:n-1 j=1; dx=(x(i+1)-x(i))/1000; for k=x(i):dx:x(i+1) xn(j)=k; yn(j)=f(k,M(i),M(i+1),x(i),x(i+1),y(i),y(i+1)); j=j+1; end if xg>=x(i) && xg<=x(i+1) yg=f(xg,M(i),M(i+1),x(i),x(i+1),y(i),y(i+1)); end plot(xn,yn, 'LineWidth',2); xlim([min(x) max(x)]); ylim([min(y) max(y)]); end hold off; fprintf('@x=%f, y=%fn',xg,yg); GaussSoln: function Soln=GaussSoln(x,y) A1=x; B=y; n2=size(B); n=n2(1,1); clear x; clear y; if det(A1)==0 disp('Either no solution or infinitely many solutions.'); else A=A1; A(:,n+1)=B(1:n); Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 38. for i=1:n-1 for j=i:n-1 fac=A(j+1,i)/A(i,i); fac_mat=fac*A(i,:); A(j+1,:)=A(j+1,:)-fac_mat; end end i=0;j=0; if A(n,n)==0 an(n)=0; else an(n)=A(n,n+1)/A(n,n); end for i=n-1:-1:1 for j=n:-1:1 x(j)=an(j)*A(i,j); end y=sum(x); if y==0 an(i)=0; else an(i)=(A(i,n+1)-y)/A(i,i); end end end Soln=an; Output: Enter matrix for values of x: [1 2 3 4] Enter matrix for values of y: [0 0.3 0.48 0.6] Enter value of x for which to find y: 2.3 @x=2.300000, y=0.363014 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 39. Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 40. Assignment No: 16 Statement: Write down the Matlab Program for Inverse Interpolation. Solution: Input : clc; clear all; x = input('Enter the of Values of x: '); y = input('Enter the of Values of y: '); r = input('Value of y at which x is to be evaluated: '); n = length(x); % determines size of matrix. p=1; s=0; for j=1:n for i=1:n if i==j continue; end numerator=r-y(i); denominator=y(j)-y(i); v(j)=numerator/denominator; p=p*v(j); end s=s+p*x(j); p=1; end fprintf('n Value is : %f ',s) Output : Enter the of Values of x: [0 1 2 3] Enter the of Values of y: [0 1 7 25] Value of y at which x is to be evaluated: 2 Value is : 1.716138 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 41. Assignment No: 17 Statement: Write down the Matlab Program for Newton Forward Differentiation. Solution: Input : clc; clear all; x=input('Enter row matrix x : '); y=input('Enter row matrix y : '); r=input('Enter value of x at which value of function is to be calculated : '); [m,n]=size(x); p=1; h=diff(x); % Step size disp(x); disp(y); for j=1:n if (r==x(j)) p=j; end end d(1,1)=y(1,p); for j=1:(n-p) dy=diff(y); % Delta matrix disp(dy); y=dy; d(j+1)=y(1,p); % Stores p th value of delta matrix. end f=0; for k=1:n-1 fr=d(k+1)/k; f=f+((-1)^(k-1))*fr; end dx=(f/h(1)); s=sprintf('Value of dy/dx at %f is : % f',r,dx); disp (s); Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 42. Output : Enter row matrix x : [1.5 2 2.5 3 3.5 4] Enter row matrix y : [3.375 7 13.625 24 38.875 59] Enter value of x at which value of function is to be calculated : 1.5 1.5000 2.0000 2.5000 3.0000 3.5000 4.0000 3.3750 7.0000 13.6250 24.0000 38.8750 59.0000 3.6250 6.6250 10.3750 14.8750 20.1250 3.0000 3.7500 4.5000 5.2500 0.7500 0.7500 0.7500 0 0 0 Value of dy/dx at 1.500000 is : 4.750000 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 43. Assignment No: 18 Statement: Write down the Matlab Program for Newton Backward Differentiation. Solution: Input : clc; clear all; xin=input('Enter row matrix x : '); yin=input('Enter row matrix y : '); r=input('Enter value of x at which value of function is to be calculated : '); [m,n]=size(xin); p=1; h=diff(xin); % Step size y(1,n:-1:1)=yin(1,1:n); % Reversing order of matrix y so that nth value is brought 1st. x(1,n:-1:1)=xin(1,1:n); % Reversing order of matrix x so that nth value is brought 1st. disp(x) disp(y) for j=1:n if (r==x(j)) p=j; end end d(1,1)=y(1,p); for j=1:(n-p) dy=diff(y); % Delta matrix y=(-1)*dy; d(j+1)=(y(1,p)); % Stores p th value of delta matrix. disp(y); end f=0; for k=1:n-1 fr=d(k+1)/k; f=f+fr; end dx=(f/h(1)); s=sprintf('Value of dy/dx at %f is : % f',r,dx); disp (s); Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 44. Output : Enter row matrix x : [0 10 20 30 40] Enter row matrix y : [1 0.984 0.939 0.866 0.766] Enter value of x at which value of function is to be calculated : 40 40 30 20 10 0 0.7660 0.8660 0.9390 0.9840 1.0000 -0.1000 -0.0730 -0.0450 -0.0160 -0.0270 -0.0280 -0.0290 0.0010 0.0010 -2.2204e-016 Value of dy/dx at 40.000000 is : -0.011317 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 45. Assignment No: 19 Statement: Write down the Matlab Program using Trapezoidal rule(single segment) for any function. Solution: Input : clear; clc; fprintf('NUMERICAL INTEGRATION BY TRAPEZOIDAL RULE nn'); x=input('Enter a function to integrate f(x)=','s'); a=input('Enter Lower Limit: '); b=input('Enter Upper Limit: '); n=2; % No. of points y=inline(x); % Defining function h=(b-a)/n; % Step size S=0; for i=1:n-1; t=2*y(a+i*h); S=S+t; end A=h/2*(y(a)+y(b)+S); % Calculation of area fprintf('nAnswer= %fn',A); Output : NUMERICAL INTEGRATION BY TRAPEZOIDAL RULE Enter a function to integrate f(x)=4*x+2 Enter Lower Limit: 1 Enter Upper Limit: 4 Answer= 36.000000 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 46. Assignment No: 20 Statement: Write down the Matlab Program using Trapezoidal rule(multiple segment) for any function. Solution: Input : clear; clc; fprintf('NUMERICAL INTEGRATION BY TRAPEZOIDAL RULE nn'); x=input('Enter a function to integrate f(x)=','s'); a=input('Enter Lower Limit: '); b=input('Enter Upper Limit: '); n=input('enter no. of segments'); y=inline(x); % Defining function h=(b-a)/n; % Step size S=0; for i=1:n-1; t=2*y(a+i*h); S=S+t; end A=h/2*(y(a)+y(b)+S); % Calculation of area fprintf('nAnswer= %fn',A); Output : NUMERICAL INTEGRATION BY TRAPEZOIDAL RULE Enter a function to integrate f(x)=4*x+2 Enter Lower Limit: 1 Enter Upper Limit: 4 enter no. of segments6 Answer= 36.000000 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 47. Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 48. Assignment No: 21 Statement: Write down the Matlab Program using Simpson’s 1/3rd (single segment) rule for any function. Solution: Input : clear; clc; fprintf('NUMERICAL INTEGRATION BY SIMPSONS 1/3 RULE nn'); x=input('Enter a function to integrate f(x)=','s'); a=input('Enter Lower Limit: '); b=input('Enter Upper Limit: '); n=2; % No. of segment y=inline(x); h=(b-a)/n; S=0; for i=1:n-1; if mod(i,2)==1 % Condition for even segments t=4*y(a+i*h); else t=2*y(a+i*h); end S=S+t; end A=h/3*(y(a)+y(b)+S); fprintf('nAnswer= %fn',A); Output : NUMERICAL INTEGRATION BY SIMPSONS 1/3 RULE Enter a function to integrate f(x)=exp(x) Enter Lower Limit: 0 Enter Upper Limit: 4 Answer= 44.247402 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 49. Assignment No: 22 Statement: Write down the Matlab Program using Simpson’s 1/3rd (multiple segment) rule for any function. Solution: Input : clear; clc; fprintf('NUMERICAL INTEGRATION BY SIMPSONS 1/3 RULE nn'); x=input('Enter a function to integrate f(x)=','s'); a=input('Enter Lower Limit: '); b=input('Enter Upper Limit: '); n=input(‘Enter no. of divisions: ’); y=inline(x); h=(b-a)/n; S=0; for i=1:n-1; if mod(i,2)==1 t=4*y(a+i*h); else t=2*y(a+i*h); end S=S+t; end A=h/3*(y(a)+y(b)+S); fprintf('nAnswer= %fn',A); Output : NUMERICAL INTEGRATION BY SIMPSONS 1/3 RULE Enter a function to integrate f(x)=exp(x) Enter Lower Limit: 0 Enter Upper Limit: 4 enter no.of divisions:5 Answer= 44.683772 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 50. Assignment No: 23 Statement: Write down the Matlab Program using Simpson’s 3/8th rule for any function. Solution: Input : clear; clc; fprintf('NUMERICAL INTEGRATION BY SIMPSONS 3/8 RULE nn'); x=input('Enter a function to integrate f(x)=','s'); a=input('Enter Lower Limit: '); b=input('Enter Upper Limit: '); while mod(n,3)~=0 % Condition for no. of segments n=input('Enter No. of Divisions [Should be divisible by 3]: '); end y=inline(x); % Defining function h=(b-a)/n; % Step size S=0; for i=1:n-1; if mod(i,3)==0 % Decision statement for usage of formula t=2*y(a+i*h); else t=3*y(a+i*h); end S=S+t; end A=3*h/8*(y(a)+y(b)+S); % Area calculation fprintf('nAnswer= %fn',A); Output : Enter the function: 4*x-1 Initial Value of x :1 Final Value of x :4 Enter No. of Divisions [Should be divisible by 3]: 3 Answer: 27.000000>> Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 51. Assignment No: 24 Statement: Write down the Matlab Program for Combined Simpson’s Rule. Solution: Input : clear; clc; j=1; fprintf('NUMERICAL INTEGRATION BY MULTIPLE SIMPSONS RULE nn'); x=input('Enter a function to integrate f(x)=','s'); a=input('Enter Lower Limit: '); b=input('Enter Upper Limit: '); while j==1 n=input('Enter No. of Divisions [(n-3) divisible by 2]: '); % Condition for no. of segments if mod(n-3,2)==0 j=0; end end y=inline(x); h=(b-a)/n; S=0; if n>=3 for i=1:2; t=3*y(a+i*h); S=S+t; end A=3*h/8*(y(a)+y(a+3*h)+S); end S=0; for i=4:n-1; if mod(i,2)==0 t=4*y(a+i*h); else t=2*y(a+i*h); end S=S+t; end A=A+h/3*(y(a+3*h)+y(b)+S); fprintf('nAnswer= %fn',A); Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 52. OUTPUT: NUMERICAL INTEGRATION BY MULTIPLE SIMPSONS RULE Enter a function to integrate f(x)=x^0.1*(1.2-x)*(1-exp(20*(x-1))) Enter Lower Limit: 0 Enter Upper Limit: 2 Enter No. of Divisions [(n-3) divisible by 2]: 5 Answer= 55501691.391968 >> Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 53. Assignment No: 25 Statement: Write down the Matlab Program for Gauss-Legendre 2-pt method. Solution: Input : clear; clc; fprintf('NUMERICAL INTEGRATION BY GAUSS-LEGENDRE 2-POINT FORMULA nn'); x=input('Enter a function to integrate f(x)=','s'); a=input('Enter Lower Limit: '); b=input('Enter Upper Limit: '); f=inline(x); % Defining function c=(b-a)/2; % Constants d=(b+a)/2; % Constants x1=c/sqrt(3)+d; x2=-c/sqrt(3)+d; y1=f(x1); y2=f(x2); A=(y1+y2)*c; fprintf('nAnswer= %fn',A); OUTPUT: NUMERICAL INTEGRATION BY GAUSS-LEGENDRE 2-POINT FORMULA Enter a function to integrate f(x)=x^3+x-1 Enter Lower Limit: 1 Enter Upper Limit: 4 Answer= 68.250000 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 54. Assignment No: 26 Statement: Write down the Matlab Program using Gauss Legendre 3-pt rule for any function. Solution: Input : clear; clc; fprintf('NUMERICAL INTEGRATION BY GAUSS-LEGENDRE 3-POINT FORMULA nn'); x=input('Enter a function to integrate f(x)=','s'); a=input('Enter Lower Limit: '); b=input('Enter Upper Limit: '); f=inline(x); % Defining function c=(b-a)/2; d=(b+a)/2; x1=c*sqrt(3/5)+d; x2=-c*sqrt(3/5)+d; x3=d; y1=f(x1); y2=f(x2); y3=f(x3); A=(5/9*y1+5/9*y2+8/9*y3)*c; fprintf('n Answer= %fn',A); Output : NUMERICAL INTEGRATION BY GAUSS-LEGENDRE 3-POINT FORMULA Enter a function to integrate f(x)=x^2-5*x+2 Enter Lower Limit: 3 Enter Upper Limit: 5 Answer= -3.333333 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 55. Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 56. Assignment No: 27 Statement: Write down the Matlab Program using Double integration by trapezoidal rule for any function. Solution: Input : clear; clc; % Taking Input fprintf('DOUBLE INTEGRATION BY TRAPEZOIDAL RULE nn'); xy=input('Enter a function to integrate f(x,y)=','s'); ax=input('Enter Lower Limit of x: '); bx=input('Enter Upper Limit of x: '); ay=input('Enter Lower Limit of y: '); by=input('Enter Upper Limit of y: '); nx=input('No. of intervals for integration w.r.t. x: '); ny=input('No. of intervals for integration w.r.t. y: '); % Defining the function f=inline(xy); % Main Calculations h=(bx-ax)/nx; k=(by-ay)/ny; an=0; for i=0:nx-1 for j=0:ny-1 tr=f(ax+i*h,ay+j*k)+f(ax+i*h,ay+(j+1)*k)+f(ax+(i+1)*h,ay+(j+1)*k)+f(ax+(i+1)* h,ay+j*k); an=an+tr; end end A=h*k/4*an; fprintf('nAnswer= %fn',A); Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 57. Output : DOUBLE INTEGRATION BY TRAPEZOIDAL RULE Enter a function to integrate f(x,y)=x+y Enter Lower Limit of x: 0 Enter Upper Limit of x: 2 Enter Lower Limit of y: 1 Enter Upper Limit of y: 3 No. of intervals for integration w.r.t. x: 2 No. of intervals for integration w.r.t. y: 2 Answer= 12.000000 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 58. Assignment No: 28 Statement: Write down the Matlab Program using double integration by Simpson’s 1/3rd rule for any function. Solution: Input : clear; clc; % Taking Input fprintf('DOUBLE INTEGRATION BY SIMPSONS 1/3rd RULE nn'); xy=input('Enter a function to integrate f(x,y)=','s'); ax=input('Enter Lower Limit of x: '); bx=input('Enter Upper Limit of x: '); ay=input('Enter Lower Limit of y: '); by=input('Enter Upper Limit of y: '); nx=3; ny=3; while mod(nx,2)~=0 || mod(ny,2)~=0 nx=input('No. of intervals for integration w.r.t. x (Should be even): '); ny=input('No. of intervals for integration w.r.t. y (Should be even): '); end % Defining the function f=inline(xy); % Main Calculations h=(bx-ax)/nx; k=(by-ay)/ny; an=0; for i=0:2:nx-1 for j=0:2:ny-1 tr1=f(ax+i*h,ay+j*k)+f(ax+i*h,ay+(j+2)*k)+f(ax+(i+2)*h,ay+(j+2)*k)+f(ax+(i+2) *h,ay+j*k); tr2=f(ax+i*h,ay+(j+1)*k)+f(ax+(i+1)*h,ay+(j+2)*k)+f(ax+(i+2)*h,ay+(j+1)*k)+f( ax+(i+1)*h,ay+j*k); tr3=f(ax+(i+1)*h,ay+(j+1)*k); an=an+tr1+4*tr2+16*tr3; end end A=h*k/9*an; fprintf('nAnswer= %fn',A); Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 59. Output : DOUBLE INTEGRATION BY SIMPSONS 1/3rd RULE Enter a function to integrate f(x,y)=x-y+1 Enter Lower Limit of x: 6 Enter Upper Limit of x: 14 Enter Lower Limit of y: 1 Enter Upper Limit of y: 5 No. of intervals for integration w.r.t. x (Should be even): 4 No. of intervals for integration w.r.t. y (Should be even): 4 Answer= 256.000000 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 60. Assignment No: 29 Statement: Write down the Matlab Program for Euler Method. Solution: Input : clc; clear all; dydx=input('Emter A Function dy/dx : ','s'); x0=input('Enter The Initial Value of x :'); y0=input('Enter The Initial Value of y :'); xf=input('Enter Value of "x" At Which Value of "y" Is To Be Found: '); h=input('Enter Step Size :'); f=inline(dydx); % Defining function n=(xf-x0)/h; for i=1:n y(i) = y0 + h*(f(x0,y0)); % Evaluating function at given x & y y0 = y(i); x0 = x0 + h; end s=sprintf('n Value of y At x = %f Is : %f',xf,y(n)); disp(s); Output : Enter A Function dy/dx : (x+y)/((y^2)-(sqrt(x*y))) Enter The Initial Value of x :1.3 Enter The Initial Value of y :2 Enter Value of "x" At Which Value of "y" Is To Be Found: 1.8 Enter Step Size :.05 Value of y At x = 1.800000 Is : 2.578164 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 61. Assignment No: 30 Statement: Write down the Matlab Program for Heun’s method. Solution: Input : clc; clear all; disp('HEUNS METHOD'); format long; dydx=input('nEnter The Function dy/dx : ','s'); x0=input('Enter The Initial Value of x: '); y0=input('Enter Initial Value of y: '); h=input('Enter step size: '); xf=input('Enter Value of x For Which y Is To Be Found: '); fprintf('n'); f=inline(dydx); n=(xf-x0)/h; for i=1:n yf = y0 + h*f(x0,y0); yff = y0 + h*(f(x0,y0) + f(x0+h,yf))/2; y0 = yff; x0 = x0 + h; s = sprintf('Value y = %f At x%d',yff,i); disp(s); end Output : HEUNS METHOD Enter The Function dy/dx : 4*exp(.8*x) - .5*y Enter The Initial Value of x: 0 Enter Initial Value of y: 2 Enter step size: 1 Enter Value of x For Which y Is To Be Found: 4 Value y = 6.701082 At x1 Value y = 16.319782 At x2 Value y = 37.199249 At x3 Value y = 83.337767 At x4 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 62. Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 63. Assignment No: 31 Statement: Write down the Matlab Program for Modified Euler method. Solution: Input: clc; clear all; % Clears the workspace disp('MODIFIED EULER METHOD'); format long % Take the input from user eq=input('nEnter the diff. eqn in x and y: ','s'); s=inline(eq); y0=input('Enter y: '); x0=input('Enter x: '); xu=input('Enter unknown x: '); acc=input('Enter accuracy required: '); % Calculatoins h=(xu-x0)/2;n=2; for i=1:n x1=x0+h; y1=y0+h*s(x0,y0); y1n=y0+(h/2)*(s(x0,y0)+s(x1,y1)); dy=abs(y1-y1n); while dy>acc y1=y1n; y1n=y0+(h/2)*(s(x0,y0)+s(x1,y1)); dy=abs(y1-y1n); end x0=x1; y0=y1n; end % Prints the answer disp('The value of the diff eqn at unkown x is: '); disp(y1n); Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 64. Output: MODIFIED EULER METHOD Enter the diff. eqn in x and y: sqrt(x+y) Enter y: 2.2 Enter x: 1 Enter unknown x: 1.2 Enter accuracy required: 0.0001 The value of the diff eqn at unkown x is: 2.573186212370175 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 65. Assignment No: 32 Statement: Write down the Matlab Program for Runge-Kutta 2nd order method. Solution: Input: clc; clear all; % Clears the workspace disp('RUNGE KUTTA METHOD 2ND ORDER'); format long % Takes the input from user eq=input('Enter the diff. eqn in x and y: ','s'); s=inline(eq); % Converts the i/p string into symbolic function y0=input('Enter y: '); x0=input('Enter x: '); xu=input('Enter unknown x: '); h=input('Enter step size: '); n=(xu-x0)/h; for i=1:n+1 x1=x0+h; y1=y0+h*s(x0,y0); c1=h*s(x0,y0); c2=h*s(x1,y1); c=(c1+c2)/2; yans=y0+c; y0=yans; x0=x1; end % Prints the answer disp('The value of the diff eqn at unkown x is: '); disp(yans); Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 66. Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 67. Output: RUNGE KUTTA METHOD 2ND ORDER Enter the diff. eqn in x and y: -(y+x*y^2) Enter y: 1 Enter x: 0 Enter unknown x: 0.3 Enter step size: 0.1 The value of the diff eqn at unkown x is: 0.715327926979073 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 68. Assignment No: 33 Statement: Write down the Matlab Program for Runge-Kutta 4th order method. Solution: Input: clc; clear all; % Clears the workspace disp('RUNGE KUTTA METHOD 4TH ORDER'); format long % Takes the input from user eq=input('Enter the diff. eqn in x and y: ','s'); s=inline(eq); % Converts the i/p string into symbolic function y0=input('Enter y: '); x0=input('Enter x: '); xu=input('Enter unknown x: '); h=input('Enter step size: '); % Calculation n=(xu-x0)/h; for i=1:n x1=x0+h; y1=y0+h*s(x0,y0); c1=h*s(x0,y0); c2=h*s((x0+(h/2)),(y0+(c1/2))); c3=h*s((x0+(h/2)),(y0+(c2/2))); c4=h*s(x1,(y0+c3)); c=(c1+2*c2+2*c3+c4)/6; yans=y0+c; y0=yans; x0=x1; end % Prints the answer disp('The value of the diff eqn at unkown x is: '); disp(yans); Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 69. Output: RUNGE KUTTA METHOD 4TH ORDER Enter the diff. eqn in x and y: 0*x+y Enter y: 2 Enter x: 0 Enter unknown x: 0.2 Enter step size: 0.1 The value of the diff eqn at unkown x is: 2.442805141701389 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 70. Assignment No: 34 Statement: Write down the Matlab Program for Milne’s correct prediction method . Solution: Input: clc; clear all; % Clears the workspace disp('MILNE PREDICTION'); format long % Take the input from user eq=input('Enter the 1st diff. eqn in x, y: ','s'); s=inline(eq); y=input('Enter y: '); x=input('Enter x: '); xu=input('Enter unknown x: '); h=input('Enter step size: '); %calculation n=(xu-x(4))/h; f1=s(x(2),y(2)); f2=s(x(3),y(3)); f3=s(x(4),y(4)); for i=1:n+1 y4pr=y(1)+(4*h/3)*(2*f1-f2+2*f3); f4pr=s(xu-h*(n-i),y4pr); y4cr=y(3)+(h/3)*(f2+4*f3+f4pr); if y4pr~=y4cr y4pr=y4cr; y4=y4cr; end f4=s(xu-h*(n-i),y4); f1=f2;f2=f3;f3=f4; y(1)=y(2); y(3)=y(4); yans=y4cr; end disp('The value of the diff eqn at unkown x is: '); disp(yans); Output: MILNE PREDICTION Enter the 1st diff. eqn in x, y: x-y+1 Enter y: [0;0.1951;0.3812;0.5591] Enter x: [1;1.1;1.2;1.3] Enter unknown x: 1.5 Enter step size: 0.1 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 71. The value of the diff eqn at unkown x is: 0.893399346172840 Assignment No: 35 Statement: Write down the Matlab Program for Runge-Kutta simultaneous method. Solution: Input: clc; clear all; % Clears the workspace disp('RUNGE KUTTA METHOD 4TH ORDER FOR SIMLTANEOUS EQUATONS'); format long % Takes the input from user eq=input('Enter the 1st diff. eqn in x, y, z: ','s'); eq1=input('Enter the 2nd diff. eqn in x, y, z: ','s'); s=inline(eq,'x','y','z'); % Converts the i/p string into symbolic function s1=inline(eq1,'x','y','z'); % Converts the i/p string into symbolic function y0=input('Enter y: '); x0=input('Enter x: '); z0=input('Enter z: '); xu=input('Enter unknown x: '); h=input('Enter step size: '); % Calculation n=(xu-x0)/h; for i=1:n x1=x0+h; c1=h*s(x0,y0,z0); d1=h*s1(x0,y0,z0); c2=h*s((x0+(h/2)),(y0+(c1/2)),(z0+(d1/2))); d2=h*s1((x0+(h/2)),(y0+(c1/2)),(z0+(d1/2))); c3=h*s((x0+(h/2)),(y0+(c2/2)),(z0+(d2/2))); d3=h*s1((x0+(h/2)),(y0+(c2/2)),(z0+(d2/2))); c4=h*s(x1,(y0+c3),(z0+d3)); d4=h*s1(x1,(y0+c3),(z0+d3)); c=(c1+2*c2+2*c3+c4)/6; d=(d1+2*d2+2*d3+d4)/6; yans=y0+c; zans=z0+d; y0=yans; z0=zans; x0=x1; end % Prints the answer disp('The value of the diff eqn at unknown x is: '); disp(yans); Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 72. disp('The value of the differential at unknown x is: '); disp(zans); Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 73. Output: RUNGE KUTTA METHOD 4TH ORDER FOR SIMLTANEOUS EQUATONS Enter the 1st diff. eqn in x, y, z: x+y*z Enter the 2nd diff. eqn in x, y, z: x^2-y^2 Enter y: 1 Enter x: 0 Enter z: 0.5 Enter unknown x: 1.2 Enter step size: 1.2 The value of the diff eqn at unknown x is: 1.352724056760832 The value of the differential at unknown x is: -0.775714711925248 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 74. Assignment No: Statement: Write down the Matlab Program for Adams Bashforth. Solution: Input: clear; clc; % Clears the work space % Get the input from user g=input('Enter the function dy/dx: ','s'); x=input('Enter values of x: '); y=input('Enter values of y: '); xg=input('Enter x at which value is to be found: '); h=input('Enter step size: '); f=inline(g); % Convert the input string into a symbolic function m=size(x); % Calculate the size of matrix x % Main calculation n=(xg-x(4))/h; for i=1:n ya=y(4)+(h/24)*(-9*(f(x(1),y(1)))+(37*(f(x(2),y(2))))- (59*(f(x(3),y(3))))+(55*(f(x(4),y(4))))); ya1=y(4)+(h/24)*((f(x(2),y(2)))- (5*(f(x(3),y(3))))+(19*(f(x(4),y(4))))+(9*f(x(4)+h,ya))); while(ya1~=ya) ya=ya1; ya1=y(4)+(h/24)*((f(x(2),y(2)))- (5*(f(x(3),y(3))))+(19*(f(x(4),y(4))))+(9*f(x(4)+h,ya))); end for j=1:m-1 x(j)=x(j+1); y(j)=y(j+1); end x(m)=x(m)+h; y(4)=ya1; end fprintf('The value at given x is : %f n',ya1); % Prints the answer Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 75. OUTPUT: Enter the function dy/dx: 1+x*y^2 Enter values of x: [0 0.1 0.2 0.3] Enter values of y: [0.2 0.3003 0.4022 0.5075] Enter x at which value is to be found: 0.5 Enter step size: 0.1 The value at given x is : 0.740490 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 76. Assignment No: 36 Statement: Write down the Matlab Program for Parabolic method. Solution: Input : % Program for Parabollic Equation (Schmidt Method) clear all; clc; a=1; b=1; % input xi=input('Enter initial value of x: '); xf=input('Enter final value of x: '); while a==1 h=input('Enter step size for x: '); co=(xf-xi)*10000/(h*10000); if mod(co,1)==0 a=0; end end ti=input('Enter initial value of t: '); tf=input('Enter final value of t: '); while b==1 k=input('Enter step size for t: '); ro=(tf-ti)*10000/(k*10000); if mod(ro,1)==0 b=0; end end s=input('For all values of x at t=0, u(x)=','s'); f=inline(s); C=input('Enter value of C: '); r=k/h^2*C^2; % Assign side values in matrix u(1,2:co+2)=xi:h:xf; u(2:ro+2,1)=ti:k:tf; u(2:ro+2,2)=input('Enter constant value of u for x=xi: '); u(2:ro+2,co+2)=input('Enter constant value of u for x=xf: '); % Assign central values in matrix by finding them for i=3:co+1 u(2,i)=f(u(1,i)); end for i=3:ro+2 for j=3:co+1 u(i,j)=r*u(i-1,j-1)+(1-2*r)*u(i-1,j)+r*u(i-1,j+1); end end % display output Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 77. disp(u); Output: Enter initial value of x: 0 Enter final value of x: 1 Enter step size for x: 0.2 Enter initial value of t: 0 Enter final value of t: 0.006 Enter step size for t: 0.002 For all values of x at t=0, u(x)=sin(pi*x) Enter value of C: 1 Enter constant value of u for x=xi: 0 Enter constant value of u for x=xf: 0 0 0 0.2000 0.4000 0.6000 0.8000 1.0000 0 0 0.5878 0.9511 0.9511 0.5878 0 0.0020 0 0.5766 0.9329 0.9329 0.5766 0 0.0040 0 0.5655 0.9151 0.9151 0.5655 0 0.0060 0 0.5547 0.8976 0.8976 0.5547 0 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 78. Assignment No: 37 Statement: Write down the Matlab Program for Crank-Nicholeson method. Solution: Input : % Crank Nicoleson clear all; clc; a=1; b=1; c=1; % input xi=input('Enter initial value of x: '); xf=input('Enter final value of x: '); h=input('Enter step size for x: '); co=(xf-xi)*10000/(h*10000); ti=input('Enter initial value of t: '); tf=input('Enter final value of t: '); k=input('Enter step size for t: '); ro=(tf-ti)*10000/(k*10000); s=input('For all values of x at t=0, u(x)=','s'); f=inline(s); C=input('Enter value of C: '); r=k*C^2/h^2; % define side values of matrix u(1,2:co+2)=xi:h:xf; u(2:ro+2,1)=ti:k:tf; u(2:ro+2,2)=input('Enter constant value of u for x=xi: '); u(2:ro+2,co+2)=input('Enter constant value of u for x=xf: '); for i=3:co+1 u(2,i)=f(u(1,i)); end ui=u; k=1; % find central values of matrix while c==1 && k<=1000 ui=u; for i=2:ro+1 for j=3:co+1 %u(i+1,j)=r/(2*(1+r))*(u(i+1,j-1)+u(i+1,j+1)+u(i,j-1)-2*u(i,j)- u(i,j+1))+u(i,j)/(1+r); u(i+1,j)=1/4*(u(i+1,j-1)+u(i+1,j+1)+u(i,j-1)+u(i,j+1)); end end k=k+1; uf=(u-ui)./u; if max(max(uf))<=0.001 c=0; end end Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 79. disp(u); Output: Enter initial value of x: 0 Enter final value of x: 3 Enter step size for x: 1 Enter initial value of t: 0 Enter final value of t: .3 Enter step size for t: .1 For all values of x at t=0, u(x)=x^2 Enter value of C: 1 Enter constant value of u for x=xi: 0 Enter constant value of u for x=xf: 0 0 0 1.0000 2.0000 3.0000 0 0 1.0000 4.0000 0 0.1000 0 1.1333 0.5333 0 0.2000 0 0.2178 0.3378 0 0.3000 0 0.1046 0.0806 0 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 80. Assignment No: 38 Statement: Write down the Matlab Program for Hyperbolic method. Solution: Input : % Program to solve Hyperbolic Partial Differential Equation clear all; clc; a=1; b=1; % input xi=input('Enter initial value of x: '); xf=input('Enter final value of x: '); h=input('Enter step size for x: '); co=(xf-xi)*10000/(h*10000); ti=input('Enter initial value of t: '); tf=input('Enter final value of t: '); k=input('Enter step size for t: '); ro=(tf-ti)*10000/(k*10000); s=input('For all values of x at t=0, u(x)=','s'); f=inline(s); C=input('Enter value of C: '); r=h/k; if r~=C error('r is not equal to C'); end % Assign side values in matrix u(1,2:co+2)=xi:h:xf; u(2:ro+2,1)=ti:k:tf; u(2:ro+2,2)=input('Enter constant value of u for x=xi: '); u(2:ro+2,co+2)=input('Enter constant value of u for x=xf: '); % Assign unknown values in matrix for i=3:co+1 u(2,i)=f(u(1,i)); end for i=3:co+1 u(3,i)=(u(2,i-1)+u(2,i+1))/2; end for i=4:ro+2 for j=3:co+1 u(i,j)=u(i-1,j-1)+u(i-1,j+1)-u(i-2,j); end end % display output disp(u); Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 81. Output: Enter initial value of x: 0 Enter final value of x: 4 Enter step size for x: 1 Enter initial value of t: 0 Enter final value of t: 2.5 Enter step size for t: 0.5 For all values of x at t=0, u(x)=(x^2)*(2-x) Enter value of C: 2 Enter constant value of u for x=xi: 0 Enter constant value of u for x=xf: 0 0 0 1.0000 2.0000 3.0000 4.0000 0 0 1.0000 0 -9.0000 0 0.5000 0 0 -4.0000 0 0 1.0000 0 -5.0000 0 5.0000 0 1.5000 0 0 4.0000 0 0 2.0000 0 9.0000 0 -1.0000 0 2.5000 0 0 4.0000 0 0 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 82. Assignment No: 39 Statement: Write down the Matlab Program for Elliptical method. Solution: Input : clear all; clc; % take user input u=input('Temperature of upper surface: '); l=input('Temperature of left surface: '); r=input('Temperature of right surface: '); b=input('Temperature of lower surface: '); cs=input('No. of elements in a row: '); n=cs-1; % Create a equation matrix an(n,n)=0; for i=1:n^2 for j=1:n^2 if i==j an(i,j)=4; elseif mod(i,n)==1 && j==i+1 an(i,j)=-1; elseif j==i-n && j>0 an(i,j)=-1; elseif j==i+n && j<=n^2 an(i,j)=-1; elseif mod(i,n)==0 && j==i-1 an(i,j)=-1; elseif mod(i,n)>1 && ( j==i+1 || j==i-1 ) an(i,j)=-1; end end end so(n)=0; for i=1:n^2 if i==1 so(i)=u+l; elseif i>1 && i<n so(i)=u; elseif i==n so(i)=u+r; elseif mod(i,n)==1 && i>n && i<=n^2-n so(i)=l; elseif mod(i,n)>1 && mod(i,n)<n && i>n && i<=n^2-n so(i)=0; elseif mod(i,n)==0 && i>n && i<=n^2-n so(i)=r; elseif i==n^2-n+1 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 83. so(i)=l+b; elseif i>n^2-n+1 && i<n^2 so(i)=b; elseif i==n^2 so(i)=b+r; end end an so an1=an; clear an; % solve the matrix t=GaussSoln(an1,so,n^2); k=1; % interpret the answers for i=1:n for j=1:n t1(i,j)=t(k); k=k+1; end end t1 hold off; % plot the answers for i=1:n for j=1:n scatter(i,j,80,[0.5 0 0],'filled'); s=sprintf('n %1.2f',(t1(i,j))); text(j,i,s); hold on; end end axis ij; axis ([ 0 n+1 0 n+1]); hold off; GaussSoln: function Soln=GaussSoln(x,y,n1) A1=x; B=y; n=n1; clear x; clear y; % Check the conditions if det(A1)==0 disp('Either no solution or infinitely many solutions.'); else % forward elimination A=A1; A(:,n+1)=B(1:n); for i=1:n-1 for j=i:n-1 fac=A(j+1,i)/A(i,i); Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 84. fac_mat=fac*A(i,:); A(j+1,:)=A(j+1,:)-fac_mat; end end i=0;j=0; % Back substitution if A(n,n)==0 an(n)=0; else an(n)=A(n,n+1)/A(n,n); end for i=n-1:-1:1 for j=n:-1:1 x(j)=an(j)*A(i,j); end y=sum(x); if y==0 an(i)=0; else an(i)=(A(i,n+1)-y)/A(i,i); end end end % answer Soln=an; Output: Temperature of upper surface: 100 Temperature of left surface: 100 Temperature of right surface: 0 Temperature of lower surface: 0 No. of elements in a row: 3 an = 4 -1 -1 0 -1 4 0 -1 -1 0 4 -1 0 -1 -1 4 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 85. so = 200 100 100 0 t1 = 75.0000 50.0000 50.0000 25.0000 Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 86. Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 87. FLOWCHART: Newton Raphson method START Set v=1500 vr=2500 g=9.81 m=2,00,000 uf=300 ` Read function (a) Derivative of function(b) ft=inline(a) dft=inline(b) Read significant digits ‘n’ epsilon_s= (0.5*10^(2-n)) epsilon_a=100 Input initial guess td=tx Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 88. A A while epsilon_a >=epsilon_s tnew= td-(ft(td)/dft(td)) epsilon_a= abs((tnew-td)/tnew)*100 td=tnew print error,tnew print tnew END Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 89. FLOWCHART: Modified Newton Raphson method START Read function (a) Derivative of function(b) Second derivative (c) t0=0 f=inline(a) df=inline(b) ddf=inline(c) Read significant digits ‘n’ epsilon_s= (0.5*10^(2-n)) epsilon_a=100 tr=fzero(inline ft) disp tr Input initial guess print head disp head Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 90. A A while (1) tnew=told-((fx(told)*dfx(told)/((dfx(told)^2)-(fx(told)*d2fx(told))))) err=abs((tnew-told)/tnew)*100 epsilon_t=abs((tr-tnew)/tr)*100 told=tnew disp table NO if err<=epsilon_s YES print tnew END Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 91. FLOWCHART: Successive Approximation START Read function (g) f=inline(g) Read significant digits ‘n’ Input initial guess epsilon_s= (0.5*10^(2-n)) epsilon_a=100 tr=fzero(inline (g)) disp tr set abcd disp abcd set head Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 92. A A disp head while ea>=es temp=t t=f(t) ea=abs((t-temp)/t)*100 et=abs((tr-t)/tr)*100 disp table END Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 93. FLOWCHART: Gauss-Naïve Elimination method Start Input matrices A&B [m,n]=size [A] NO Print “Matrix If m~=n must be square!” YES For k=1:n-1 A For i=k+1:n Factor a(i,k)/a(k,k) 9i,k For j=k:n a(I,j)=a(I,j)-factor a(k,j) b(i)=b(i)- factor*b(k) Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering M
  • 94. M Display A & B For i=n:-1:1 x(i) = b(i) / a(I,i) for j=1:i-1 b(j) = b(j) - x(i)*a(j,i) Display values of x A End Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 95. FLOWCHART: Gauss with Partial Pivoting method Start Input matrices A & B [m,n] = size (a) NO If m~=n Print “Matrix must be square!” YES E A For k=1:n-1 [xyz,i]=max(abs(a(k:n,k))) ipr=i+k-1; if ipr~=k Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 96. B a([k,ipr],:)=a([ipr,k],:) b([k,ipr],:)=b([ipr,k],:) For i=k+1:n B factor=a(i,k)/a(k,k) For j=k:n a(i,j)=a(i,j)-(factor*(a(k,j))) b(i)=b(i)-factor*(b(k)) A Display A & B D for i=n:-1:1 Third Year Mechanical Engineering C Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 97. C x(i)=b(i)/a(i,i) For j=1:i-1 b(j)=b(j)-x(i)*a(j,i) D Display x E End Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 98. FLOWCHART: Thomas Algorithm START Input matix e,f,g,r n=length (e) for k=1:n factor =e(k)/f(k) f(k+1)=f(k+1)-xg(k) r(k+1)= r(k+1)-factor*r(k) x(n+1)=r(n+1)/f(n+1) for k=n:1 x(k)=r(k)-g(k)*x(k+1)/k display end Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 99. Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering
  • 100. FLOWCHART: Gauss-Seidel without Relaxation method Start Input matrices A&B [m,n]=size(a) NO Print “Matrix Must if (m~=n) Be Square!” YES For i= m:n d(i)=b(i)/a(i,i) d=d’ c=a For i=1:n B For j=1:n c(i,j)=a(i,j)/a(i,i) Third Year Mechanical Engineering Computer Oriented Numerical Methods 2011-12© MITCOE Mechanical Engineering A