MITCOE 2011-12 conm-submission

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

2011-12©

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;
%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)

2011-12©

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

2011-12©

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
%f',t,ea,et,es);
disp(table);
end

2011-12©

Output :

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

Enter number of significant digits: 2

Enter initial guess: 0

BY SUCCESSIVE APPROXIMATION METHOD:

0.567143290409784

2011-12©

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)

2011-12©

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

2011-12©

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.
%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)

2011-12©

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

2011-12©

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

2011-12©

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.
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

2011-12©

2011-12©

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

2011-12©

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.
%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.

2011-12©

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

2011-12©

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

2011-12©

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');

2011-12©

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

2011-12©

Assignment No: 9

Statement: Write down the Matlab Program to fit curve y = a0 + a1*x+a2x2 by using least square

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)');

2011-12©

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

2011-12©

Assignment No: 10

Statement: Write down the Matlab Program to fit curve y = a1*(xb1) by using least square

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);
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
2011-12©

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');
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)

2011-12©

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

2011-12©

Assignment No: 41

Statement: Write down the Matlab Program to fit curve y = a1 * e (b1*x) by using least square

Solution:

Input :
clc;
clear all;
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);
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;
2011-12©

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
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)

2011-12©

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

2011-12©

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

2011-12©

Assignment No: 12

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

Solution:

Input :
clc;
clear all;
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

2011-12©

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

2011-12©

Assignment No: 13

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

Solution:

Input :
clc;
clear all;
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

2011-12©

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]
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

2011-12©

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

2011-12©

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));
2011-12©

% 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);

2011-12©

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

2011-12©

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

2011-12©

Assignment No: 17

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

Solution:

Input :

clc;
clear all;
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)
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);

2011-12©

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]


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

2011-12©

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)
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);

2011-12©

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

2011-12©

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

2011-12©

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');
n=input('enter no. of segments');
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

Output :

NUMERICAL INTEGRATION BY TRAPEZOIDAL RULE

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



enter no. of segments6

Answer= 36.000000

2011-12©

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');
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);

Output :
NUMERICAL INTEGRATION BY SIMPSONS 1/3 RULE

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



Answer= 44.247402

2011-12©

Assignment No: 22

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

Solution:

Input :

clear;
clc;
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);

Output :

NUMERICAL INTEGRATION BY SIMPSONS 1/3 RULE

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



enter no.of divisions:5

Answer= 44.683772
2011-12©

Assignment No: 23

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

Solution:

Input :

clear;
clc;
while mod(n,3)~=0 % Condition for no. of segments
n=input('Enter No. of Divisions [Should be divisible by 3]: ');
end
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

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>>

2011-12©

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');
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);


2011-12©

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 No. of Divisions [(n-3) divisible by 2]: 5

Answer= 55501691.391968
>>

2011-12©

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');
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;

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

Enter a function to integrate f(x)=x^3+x-1

Answer= 68.250000

2011-12©

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');
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



Answer= -3.333333

2011-12©

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;

2011-12©

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

2011-12©

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;

2011-12©

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

2011-12©

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

2011-12©

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

2011-12©

Assignment No: 31

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

Solution:

Input:
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);

2011-12©

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

2011-12©

Assignment No: 32

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

Solution:

Input:
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
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(yans);

2011-12©

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 step size: 0.1
The value of the diff eqn at unkown x is:
0.715327926979073

2011-12©

Assignment No: 33

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

Solution:

Input:
disp('RUNGE KUTTA METHOD 4TH ORDER');
format long

eq=input('Enter the diff. eqn in x and y: ','s');
s=inline(eq); % Converts the i/p string into symbolic function

% 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(yans);

2011-12©

Output:
RUNGE KUTTA METHOD 4TH ORDER
Enter the diff. eqn in x and y: 0*x+y
Enter y: 2
Enter x: 0
2.442805141701389

2011-12©

Assignment No: 34

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

Solution:

Input:
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: ');

%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]

2011-12©

0.893399346172840
Assignment No: 35

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

Solution:

Input:
disp('RUNGE KUTTA METHOD 4TH ORDER FOR SIMLTANEOUS EQUATONS');
format long

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
x0=input('Enter x: '); z0=input('Enter z: ');

% 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);

2011-12©

disp('The value of the differential at unknown x is: ');
disp(zans);

2011-12©

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
The value of the diff eqn at unknown x is:
1.352724056760832

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

2011-12©

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: ');

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

2011-12©

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
The value at given x is : 0.740490

2011-12©

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
2011-12©

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

2011-12©

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
co=(xf-xi)*10000/(h*10000);
ro=(tf-ti)*10000/(k*10000);
f=inline(s);
r=k*C^2/h^2;
% define side values of matrix
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

2011-12©

disp(u);

Output:



Enter step size for x: 1


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



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

2011-12©

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
co=(xf-xi)*10000/(h*10000);
ro=(tf-ti)*10000/(k*10000);
f=inline(s);
r=h/k;
if r~=C
error('r is not equal to C');
end
% Assign side values in matrix
% 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);

2011-12©

Output:



Enter step size for x: 1


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



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

2011-12©

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

2011-12©

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);
2011-12©

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

2011-12©

so =

200 100 100 0

t1 =

75.0000 50.0000

50.0000 25.0000

2011-12©

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

2011-12©

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

2011-12©

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

2011-12©

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

2011-12©

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
2011-12©

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

2011-12©

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)
2011-12©
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
2011-12©

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

2011-12©

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

C Computer Oriented Numerical Methods
2011-12©

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

2011-12©

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
2011-12©

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
2011-12©
A

A

c(i,i)=0
x(i)=0

B

x = x’

Display A, B, C
&D

Input no. of
iterations (p)

For k=i:p

For i=1:n

x(i)=d(i)-c(i,:)*x(:,1)

Display x

2011-12©
End MITCOE Mechanical Engineering

FLOWCHART: Gauss-Seidel with Relaxation method
START

Input matrix A(a)

And B(b)

[m,n]=size (a)

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

For k = 1:n-1 E

[xyz,i]=max(abs(a(k:n,k)))

Ipr=i+k-1

NO
if ipr N=K

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

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

for i=1:n

2011-12©

Di = b(i)/a(1,i)

D=d’; c=0 M

2011-12©

M

For i=1:n

For j=1:n

C(I,j)=a(I,j)/a(I,i)

C(I,i)=0

X(i)=0

X = X’

Display a,
b, c & d

Input weighing factor (lambda)

es=0.05
ea(i)=100
set head Third Year Mechanical Engineering
2011-12©
N MITCOE Mechanical Engineering

N

Display
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
ea(i)=abs((x(i)-y)/x(i))*100

Set table

Display
table

E

end
2011-12©

Flowchart: Least square techniques – Linear fit.

2011-12©

START

Enter values of
x & y in matrix
form.

n = size of matrix x

xy(1,1) = 0

i=1

X=0

Y=0

XY=0
While i< = n
Xsqr=0

xy(1,i)=x(1,i)*y(1,i)

xsqr(1,i)=x(1,i)^2

X=X+x(1,i);

X=X+x(1,i);

XY=XY+xy(1,i)

Xsqr=Xsqr+xsqr(1,i)
i = i +1

A

2011-12©

A

Print x & 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-beta*x(1,j))^2

st(1,j)=(y(1,j)-ym)^2

j = j +1

B

2011-12©

B

xp=linspace(min(x),max(x));

yp=a0+a1*xp;

SR=sum(sr)

ST=sum(st)

r2=(ST-SR)/ST

Print Function

of best fitted line

Plot Function

& best fitted curve

END

2011-12©

Flowchart: Least square techniques – Quadratic fit.

2011-12©

START

Enter values of
x & y in matrix
form.


xy(1,1) = 0

i=1

X=0

Y=0

XY=0
sx = sum(x)
sy Xsqr=0
= 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

A

2011-12©

A

a0 = z(1)

a1 = z(2)

a2 = z(3)

Print Function

of best fitted line

xp = linspace(min(x),max(x))

yp = z(3)*(xp.*xp)+z(2)*xp+z(1)

Plot Function

& best fitted curve

END

2011-12©

Flowchart: Least square techniques – Power model.

2011-12©

START

Enter values of
x & y in matrix
form.


xy(1,1) = 0

y(1,1) = 0
i=1

X=0

Y=0

XY=0

While i< = n
Xsqr=0

y(1,i)=log10(ya(1,i))
x(1,i)=log10(xa(1,i))
xy(1,i)=x(1,i)*y(1,i)
xsqr(1,i)=x(1,i)^2

X=X+x(1,i);
X=X+x(1,i);
XY=XY+xy(1,i)
Xsqr=Xsqr+xsqr(1,i)

i = i +1

A

2011-12©

A

Print x & y

beta=(n*XY-Y*X)/(n*Xsqr-X^2)


alpha=10^(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

st(1,j)=(y(1,j)-ym)^2

j = j +1

B

2011-12©

B

SR=sum(sr)

ST=sum(st)

r2=(ST-SR)/ST

Print Function

Of best fitted curve


yp = (xp.^beta)*alpha

Plot Function

& best fitted curve

END

2011-12©

Flowchart: Least square techniques – Exponential fit.

2011-12©

START

Enter values of
x & ya in matrix
form.


xy(1,1) = 0

y(1,1) = 0
i=1

X=0

Y=0

XY=0
While i< = n
Xsqr=0

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);

X=X+x(1,i);

XY=XY+xy(1,i)

Xsqr=Xsqr+xsqr(1,i)
i = i +1

A

2011-12©

A

Print x & ya

a1=(n*XY-Y*X)/(n*Xsqr-X^2)


alpha=exp(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

st(1,j)=(y(1,j)-ym)^2

j = j +1

B

2011-12©

B


yp= alpha*exp(a1*xp)

SR=sum(sr)

ST=sum(st)

r2=(ST-SR)/ST

Print Function

Of best fitted curve

Plot Function

& best fitted curve

END

2011-12©

Flowchart: Lagrange interpolation.

START

Enter values of x & y
in matrix form.
& Enter value of u

n = length of matrix x

p=1; s=0

dx = x(i+1) – x(i)

For i = 1 to n

p = y(i)

For j = 1 to n

B
If i=~j End

p=p*(u-x(j))/(x(i)-x(j))

A

2011-12©

A
B

s = s+p

Print Y

END

2011-12©

Flowchart: N-G forward difference interpolation.

START

in matrix form &
Enter value of X

d(1,1) = y(1,1)

dx = x(i+1) – x(i)

For j = 1 to (n-1)

dy = y(j+1) – y(j)

Display dy

d(j+1) = dy(1)

y = dy

A

2011-12©

A

alpha = (X-x(1))/dx(1)

a(1,1)=1

prod=1

For k = 1 to (n-2)

prod = prod * (alpha-k+1)

a(k+1) = prod

func = 0

For i = 1 to (n-1)

fx = a(i) * d(i) / (factorial(i-1))

func=func+fx

B

2011-12©

B

Print Y

END

2011-12©

Flowchart: N-G backward difference interpolation.

START

Enter values of x & y in
matrix form.
& Enter value of X


d(1,1) = y(n)

dx = x(i+1) – x(i)

newx(1,n:-1:1) = x(1,1:n)

newy(1,n:-1:1) = y(1,1:n)

For j = 1 to (n-1)

dy = y(j+1) – y(j)

B

Display dy

A

2011-12©

A

B
d(j+1) = dy(1)

y = dy

alpha = (x(n)-X)/dx(1)

a(1,1)=1

prod=1

For k = 1 to (n-2)

prod = prod * (alpha-k+1)

a(k+1) = prod

func = 0

C

2011-12©

C

For i = 1 to (n-1)

fx = a(i) * d(i) / (factorial(i-1))

func=func+fx

Print Y

END

2011-12©

FLOWCHART:Hermite Interpolation

START

Accept values of x,f(x) and f’(x)

accept unknown x

n=size(x) suma=0

sumb=0 sum=0

for i=1 to n

pro=1 pro1=1

for j=1 to n

if i≠j

pro=pro*(xu-x(j))/(x(i)-x(j))
pro1=pro1*(x(i)-x(j))

L(I,1)=pro

L’(I,1)=pro1

for k=1 to n

2011-12©

A B

suma=suma+(1-2*(xu-x(k))*dL(k))*((L(k))^2)*fx(k)

A sumb=sumb+(xu-x(k))*((L(k))^2)*dfx(k)

sumf=suma+sumb

Display sumf

END

2011-12©

FLOWCHART: Cubic Spline Interpolation

START

Accept pts. X1,y1

Accept unknown xg

m1=size(x1) n=m(1,2)

x=x1’ y=y1’

plot curve(x,y)

assigning pts. In yy betwn x and y with 0:0.01:100

plot (x,y) with intermediate yy

defining f’’(x)=0 for end pts.

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))

(solving for first row)

If n>3 else

For l=2:n-2

Solve Subsequent rows till n-2 by above method

2011-12©

A D
D

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))
(solving the last row)

Finding other values of f’’(x)
By using gauss Solution for (A,B)
Storing in N

For i=1 to n-1

M(i+1)=N(i)
(assigning the values to f’’(x) from N)

Define function for cubic spline for each interval
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)

xn(1:1000)=0
Yn(1:1000)=0

For i=1 to n-1

J=1
dx=(x(i+1)-x(i))/1000

for k=x(i):dx:x(i+1)

2011-12©

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

C B
C B

else
if xg>=x(i) and xg<=x(i+1)

yg=f(xg,M(i),M(i+1),x(i),x(i+1),y(i),y(i+1))

plot (xn,yn)

display xg,yg

END

2011-12©

START

in matrix form.
& Enter value of r


p=1,s=0

for j=1 to n

for i= 1 to n

if i=j

numerator= r-y(i)

denominator=y(j)-y(i)

v(j)=numerator/denominator

p=p*v(j)
s=s+p*x(j)
p=1

print s

End

2011-12©

Flowchart: Newton Forward differentiation.

START

in matrix form.
& Enter value of X


p=1

h = x(i+1) – x(i)

For j = 1 to n

If r=~x(j)
End

p=j

d(1,1)=y(1,p)

A

2011-12©

A

For j = 1 to (n-p)

dy = y(j+1) – y(j)

Display dy

y = dy

d(j+1)=y(1,p)

f=0

For k = 1 to (n-1)

fr=d(k+1)/k

f=f+((-1)^(k-1))*fr

B

2011-12©

START

in matrix form.

& Enter value of X


p=1

h = x(i+1) – x(i)

x(1,n:-1:1)=xin(1,1:n);

y(1,n:-1:1)=yin(1,1:n)

For j = 1 to n

If r=~x(j)
End

p=j

d(1,1)=y(1,p)

A

2011-12©

A

For j = 1 to (n-p)

dy = y(j+1) – y(j)

Display dy

y = dy*(-1)

d(j+1)=y(1,p)

f=0

For k = 1 to (n-1)

fr=d(k+1)/k

f=f+fr

B

2011-12©

FLOWCHART: Trapezoidal (single segment) method

START

print statement

Read function(x)
Input lower limit(a)
Upper limit(b)

y=inline(x)
h=(b-a)/n
s=0,n=1

for i=1:n-1

t=2*y(a+i*h)
s=s+t

A=h*(y(a)+y(b)+s)

print A

END

2011-12©

FLOWCHART: Trapezoidal(multiple segment) method

START

print statement

Read function(x)
Input lower limit(a)
Upper limit(b)
No. of divisions(n)

y=inline(x)
h=(b-a)/n
s=0

for i=1:n-1

t=2*y(a+i*h)
s=s+t

A=h*(y(a)+y(b)+s)

print A

END

2011-12©

FLOWCHART: Simpson’s One Third Rule(single segment) for Integration

Start

print
statement

Read function(x) & input lower limit(a) &
input upper limit (b)

y=inline (x)

h=(b-a)/n

s=0,n=2

for i=1:n-1

if
mod(I,2)==1

M

2011-12©

FLOWCHART: Simpson’s One Third Rule(multiple segment) for Integration
Start

print
statement

input upper limit (b) & no. of divisions n

while
mod(n,2)~=0

input no. of division
divisible by 2

y=inline (x)

h=(b-a)/n

s=0

for i=1:n-1

if
mod(I,2)==1
2011-12©
M

FLOWCHART: Simpson’s Three Eight Rule for Integration
Start

print
statement

input upper limit (b) & no. of divisions n

while
mod(n,3)~=0

input no. of division
divisible by 3

y=inline (x)

h=(b-a)/n

s=0

for i=1:n-1

if
mod(I,3)==1
2011-12©
M

FLOWCHART: Gauss-Legendre 2-pt method

Start

Print NUMERICAL INTEGRATION BY
GAUSS LEGEDRE 2-POINT FORMULA

Read function (x)
Input lower limit a & upper limit b

f=inline(x)
c=(b-a)/2
d=(b+a)/2
x1=c/sqrt(3)+d
x2=-c/sqrt(3)+d
y1=f(x1)
y2=f(x2)
A=(y1+y2)*c

Print A

End

2011-12©

FLOWCHART: Gauss-Legendre 3-pt method

Start

Print NUMERICAL INTEGRATION BY
GAUSS LEGEDREN 3-POINT FORMULA

Read function (x)
Input lower limit a & upper limit b

f=inline(x)
c=(b-a)/2
d=(b+a)/2
x1=c/sqrt(3)+d
x2=-c/sqrt(3)+d
x3=d
y1=f(x1)
y2=f(x2)
y=f(x3)
A=(5/9*y1+5/9*y2+8/9*y3)*c

Print A

End

2011-12©

FLOWCHART: Double Integration by trapezoidal method

START

print statement

Read function(xy)
Input lower limit of x(ax)
Input lower limit of y(ay)
Upper limit of x(bx)
Upper limit of y(by)

Input nx
Input ny

while
mod(nx,2)~=0

input no. of intervals wrt x(nx)
no. of intervals wrt (ny)

f= inline(xy)

h=(bx-ax)/nx
k=(by-ay)/ny
an=0

for i=0:nx-1 Third Year Mechanical Engineering
2011-12©

B A

A
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;

A=h*k*an/4

B
Print A

END

2011-12©

FLOWCHART: Double Integration by Simpson’s One Third method

START

print statement

Read function(xy)
Input lower limit of x(ax)
Input lower limit of y(ay)
Upper limit of x(bx)
Upper limit of y(by)

nx=3
ny=3

while
mod(nx,2)~=0

input no. of intervals wrt x(nx)
no. of intervals wrt (ny)

f= inline(xy)

h=(bx-ax)/nx
k=(by-ay)/ny
an=0

for i=0:2:nx-1 Third Year Mechanical Engineering
2011-12©

B A

A

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(a
x+(i+1)*h,ay+j*k)
tr3=f(ax+(i+1)*h,ay+(j+1)*k)
an=an+tr1+4*tr2+16*tr3

A=h*k/9*an

B
Print A

END

2011-12©

FLOWCHART: Euler method

START

Accept the function f(x,y)

Get inputs of initial values x0,y0

Get step size h and unknown xu

n= (xu-x0)/h

for i=1 to n

x1=x0+h

y1=y0+h*f(x0,y0)

x0=x1

y0=y1

Displaying y1

END

2011-12©

START

Accept function

Accept x0,y0,h
and unknown x

n = (x-x0)/h

for i= 1 to n

yf = y0 + h*f(x0,y0)

yff = y0 + h*(f(x0,y0) + f(x0+h,yf))/2

y0 = yff;

x0 = x0 + h

display yff

END

2011-12©

FLOWCHART: Modified Euler method

START



n=2 and unknown xu

Get accuracy

h= (xu-x0)/n

for i=1 to n

x1=x0+h

y1=y0+h*f(x0,y0)

ynew=y0+(h/2)*(f(x0,y0)+f(x1,y1)

diff=|y1-ynew|

while diff>accuracy

A

y1=ynew

ynew= y0+(h/2)*(f(x0,y0)+f(x1,y1)

diff=|y1-ynew|

B

2011-12©

FLOWCHART: Runge-Kutta2nd order method

START


Get inputs of initial values x,y


n= (xu-x0)/h m =size(x)

for i=1 to n

ya=y3+(h/24) (-9*f(x0,y0) +37*f(x1,y1)- 59*f(x2,y2) +55*f(x3,y3))
ya1=y3+ (h/24)(f(x1,y1)-5*f(x2,y2)+19*f(x3,y3)+9*f(x3+h,ya)

while ya1=!ya

ya=ya1
ya1=y3+(h/24)*(f(x1,y1)-5*f(x2,y2)+19*f(x3,y3)+ 9*f(x3+h,ya);

for i=1: m

A

x(j)=x(j+1);
y(j)=y(j+1);

B

2011-12©

FLOWCHART: Runge-Kutta 4th order method

START




n= (xu-x0)/h

for i=1 to n

x1=x0+h

y1=y0+h*f(x0,y0)

x0=x1

y0=y1

A 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;

yanswer=y0+c

y0=yanswer

x0=x1

B

2011-12©

FLOWCHART: Milne’s prediction-correction order method

START


Get inputs of initial values (x0,y0),

(x1,y1), (x2,y2), (x3,y3)


n= (xu-x0)/h

f1=f(x1,y1),f2=f(x2,y2),

f3=f(x3,y3)

for i=1 to n

y4pr=y(1)+(4*h/3)*(2*f1-f2+2*f3);

f4pr=f(xu-h*(n-i),y4pr);

y4cr=y(3)+(h/3)*(f2+4*f3+f4pr)

A if y4pr≠p4cr

B

2011-12©

B

y4pr=y4cr

y4=y4cr

A f4=f(xu-h*(n-i),y4)
f1=f2;f2=f3;f3=f4
y0=y1
y2=y3
yans=y4cr

yanswer=y0+c

y0=yanswer

x0=x1

Displaying yans

END

2011-12©

FLOWCHART: Runge-Kutta for simultaneous equation

START

Accept the function f(x,y,z)

Accept the function g(x,y,z)

Get inputs of initial values x0,y0,z0


n= (xu-x0)/h

for i=1 to n

x1=x0+h

y1=y0+h*f(x0,y0)

x0=x1

y0=y1

A c1=h*f(x0,y0,z0)
d1=h*g(x0,y0,z0)
c2=h*g((x0+(h/2)),(y0+(c1/2)),(z0+(d1/2)))
d2=h*sg((x0+(h/2)),(y0+(c1/2)),(z0+(d1/2)))
c3=h*f((x0+(h/2)),(y0+(c2/2)),(z0+(d2/2)))
d3=h*g((x0+(h/2)),(y0+(c2/2)),(z0+(d2/2)))
c4=h*f(x1,(y0+c3),(z0+d3))
d4=h*g (x1,(y0+c3),(z0+d3))

B

2011-12©

B

c=(c1+2*c2+2*c3+c4)/6
d=(d1+2*d2+2*d3+d4)/6
yans=y0+c
zans=z0+d

y0=yans

A z0=zans

x0=x1

Displaying yans

Displaying zans

END

2011-12©

FLOWCHART: Adam Bashforth order method

START


Get inputs of initial values x,y


n= (xu-x0)/h m =size(x)

for i=1 to n

ya=y3+(h/24) (-9*f(x0,y0) +37*f(x1,y1)- 59*f(x2,y2) +55*f(x3,y3))
ya1=y3+ (h/24)(f(x1,y1)-5*f(x2,y2)+19*f(x3,y3)+9*f(x3+h,ya)

while ya1=!ya

ya=ya1
ya1=y3+(h/24)*(f(x1,y1)-5*f(x2,y2)+19*f(x3,y3)+ 9*f(x3+h,ya);

for i=1: m

A

x(j)=x(j+1);
y(j)=y(j+1);

B

2011-12©

FLOWCHART:PARABOLIC EQUATION START

accept x0,xn,t0,tn

accept h,k,C

accept boundary value (b)

define function u

x=x0:h:xf t=t0:k:tf

x=x’ t=t’

n=(xf-x0)/h m=(tf-t0)/k

Define matrix

A(:,1)=b A(:,n)=b

r=k/h^2*C^2

for i=2 to n

A(m+1,i)=u(x(i))

for j=m:-1:1

for i=2 to n

A B C
2011-12©

FLOWCHART:Crank-Nicolson

START

`Accept xi,xf,h,ti,tf,k

Accept function

Accept value of c

co=(xf-xi)/h ro=(tf-ti)/k

r=k*C^2/h^2 a=1,b=1.c=1

u(1,2:co+2)=xi:h:xf
u(2:ro+2,1)=ti:k:t

Accept const values for:

u(2:ro+2,2)

u(2:ro+2,2)

for i=3 to co+1

u(2,i)=f(u(1,i))

ui=u;

k=1

2011-12©

B

B

while c==1 and k<=1000

ui=u;

for i=2 to ro+1

for j=3 to co+1

u(i+1,j)=1/4*(u(i+1,j-1)+u(i+1,j+1)+u(i,j-1)+u(i,j+1))

k=k+1

uf=(u-ui)./u

if max(max(uf))<=0.001

c=0

display u

END
2011-12©

FLOWCHART:Hyperbolic Equation
Start

Input Initial value of x (xi)
Final value of x (xf)
Step size for x (h)

co=(xf-xi)*10000/(h*10000)

Input Initial value of t (ti)
Final value of t (tf)
Step size for t (k)

ro=(tf-ti)*10000/(k*10000)

Input all values of u(x) (s)

r=h/k

If r~=C YES
error

NO
A 2011-12©

A

u(1,2:co+2)=xi:h:xf
u(2:ro+2,1)=ti:k:tf

Input Constant value for xi
(u(2:ro+2,2))
Input Constant value for xf
(u(2:ro+2,co+2)

For i=3:co+1

u(2,i)=f(u(1,i))

For i=3:co+1

u(3,i)=(u(2,i-1)+u(2,i+1))/2

A For i=4:ro+2

For j=3:co+1
B

C 2011-12©

FLOWCHART:Elliptical Equation
Start

Input Temperature of
Upper surface (u)
Left surface (l)
Right surface (r)
Lower surface (b)
No. of elements in a row (cs)

n=cs-1
an(n,n)=0

A For i=1:n^2

B For
j=i:n^2

YES
an(i,j)=4
IF i==j

NO
M

IF YES
mod(i,n)==1 (i,j)=-1
&& j==i+1

M
NO Third Year Mechanical Engineering
C 2011-12©

C

YES
IF j==i-n an(i,j)=-1
&& j>0

NO M

IF j==i+n
YES
an(i,j)=-1
&& j<=n^2

NO M

mod(i,n)==0 YES an(i,j)=-1
&& j==i-1

NO M

IF NO
mod(i,n)>1 B
&& j==i+1
|| j==i-1 )

YES

an(i,j)=-1
M

B

A

so(n)=0

D Computer Oriented Numerical Methods
2011-12©

D

E For i=1:n^2

YES
If i==1 so(i)=u+l

NO
N

YES
If i>1 so(i)=u
&& i<n

NO
N

If YES
so(i)=u+r
i==n

NO N

If mod(i,n)==1
YES
so(i)=l
&& i>n &&
i<=n^2-n

NO N

mod(i,n)>1 && YES
mod(i,n)<n && so(i)=0
i>n && i<=n^2-n

NO N
F 2011-12©

F

YES
If mod(i,n)==0
&& i>n && so(i)=r
i<=n^2-n

NO N

YES
If i==n^2- so(i)=l+b
n+1

NO N

If i>n^2-n+1 YES
so(i)=b
&& i<n^2

NO N

NO
If
i==n^2 E

YES

so(i)=b+r
N

E

Display
an & so

G Third Year Mechanical Engineering
2011-12©

G

an1=an
clear an
t=GaussSoln(an1,so,n^2)
k=1

For i=1:n

For j=1:n

t1(i,j)=t(k)
k=k+1

Display t1

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)

axis ij
axis ([ 0 n+1 0 n+1])
2011-12©
End MITCOE Mechanical Engineering

MITCOE 2011-12 conm-submission

More Related Content

What's hot

Viewers also liked

Similar to MITCOE 2011-12 conm-submission

Recently uploaded

MITCOE 2011-12 conm-submission