Computer Based Numerical & Statistical Techniques using 'C

7/31/2019 Computer Based Numerical & Statistical Techniques using 'C'

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Term Work

On

Computer Based Numerical &Statistical Techniques

(Phase II)

April 10,2012 to March 18,2012

Program no. Date Program Name Remarks

1 10-04012 Write a Program in C to implement X,R,C,P Charts

2 20-04012 Write a Program in C to find value of function UsingNewton Forward Interpolation.

3 24-04-12 Write a Program in C to find value of function UsingNewton Backward Interpolation

4 27-04-12 Write a Program in C to find value of function UsingGauss Forward Interpolation.

5 27-04-12 Write a Program in C to find value of function Using

Gauss backward Interpolation

6 01-04-12 Write a Program in C to find value of function UsingLangrange Interpolation

7 04-04012 Write a Program in C to find value of function UsingNewton Divided Difference Formula

8 08-04-12 Write a Program in C to find value of function UsingNewton Divided Difference Formula

9 11-04-12 Write a Program in C to find value of Integration using

Simpson1/3 Rule

10 15-04-12 Write a Program in C to find value of Integration usingSimpson3/8 Rule

Group Name: G1

Group Teacher: Tarun Kathuria

Signature:

Class Teacher: Tarun Kathuria

Signature:


2/35

//Program to implement X Chart

//Ashish Ruhela

//MCA 2 A

//Roll No. : 10

//Date: 10-04-12

#include

void main()

{

float x[30],cl,ucl,lcl,sum=0.0,sd,sd1;

int i,n,z;

clrscr();

printf("\nHow many samples are there:");

scanf("%d",&n);

printf("\n\nEnter the average observation each:\n\n");

for(i=0;i


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OUTPUT:

How many samples are there:25

Enter the average observation each:

For Sample 1:15.91

For Sample 2:15.99For Sample 3:15.92

For Sample 4:15.93

For Sample 5:15.98

For Sample 6:16.03

For Sample 7:15.96

For Sample 8:15.93

For Sample 9:15.96

For Sample 10:15.83

For Sample 11:15.99


For Sample 14:15.91

For Sample 15:16.05

For Sample 16:15.99

For Sample 17:15.86

For Sample 18:16.01

For Sample 19:15.98

For Sample 20:16.02

For Sample 21:16.00

For Sample 22:15.90

For Sample 23:15.86

For Sample 24:15.94

For Sample 25:15.94

Enter the value of z:(Standard Normal Variable):3

Enter the standard deviation:0.14

U.C.L=16.159100 C.L=15.950000 L.C.L=15.739100


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//Program to implement R chart

//Ashish Ruhela

//MCA 2 A

//Roll No. : 10

//Date: 10-04-12

#include

void main()

{

float r[30],cl,ucl,lcl,sum=0.0;

int i,n,d3,d4;

clrscr();


scanf("%d",&n);

printf("\nEnter the Range:\n");

for(i=0;i


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OUTPUT:


Enter the Range:

For Sample 1:0.19


For Sample 4:0.46

For Sample 5:0.47

For Sample 6:0.20

For Sample 7:0.46

For Sample 8:0.20

For Sample 9:0.21

For Sample 10:0.30

For Sample 11:0.29


For Sample 14:0.37

For Sample 15:0.31

For Sample 16:0.29

For Sample 17:0.33

For Sample 18:0.34

For Sample 19:0.28

For Sample 20:0.20

For Sample 21:0.23

For Sample 22:0.16

For Sample 23:0.32

For Sample 24:0.15

For Sample 25:0.30

Enter the value of d3:0

Enter the value of d4:2.28

U.C.L=0.598756 C.L=0.289671 L.C.L=0.000000


6/35

//Program to implement P chart

//Ashish Ruhela

//MCA 2 A

//Roll No. : 10

//Date: 10-04-12

#include

void main()

{

float x[30],y[30],cl,ucl,lcl,q,w,sum=0.0,obs=1.0;

int i,n,z;

clrscr();


scanf("%d",&n);

printf("\n\nEnter the number of defective tyres:\n\n");

for(i=0;i


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OUTPUT:


Enter the number of defective tyres:

For Sample 1:3

For Sample 2:2

For Sample 3:1For Sample 4:2

For Sample 5:1

For Sample 6:3

For Sample 7:3

For Sample 8:2

For Sample 9:1

For Sample 10:2

For Sample 11:3

For Sample 12:2

For Sample 13:2For Sample 14:1

For Sample 15:1

For Sample 16:2

For Sample 17:4

For Sample 18:3

For Sample 19:1

For Sample 20:1

Enter the value of z:3

Enter the number of observations each for 20 samples:20

U.C.L=0.301246 C.L=0.100000 L.C.L=0.000000


8/35

//Program to implement C chart

//Ashish Ruhela

//MCA 2 A

//Roll No. : 10

//Date: 10-04-12

#include

void main()

{

float x[30],y[30],cl,ucl,lcl,sum=0.0;

int i,n,z;

clrscr();

printf("\nHow many weeks are monitored:");

scanf("%d",&n);

printf("\n\nEnter the number of complaints:\n\n");

for(i=0;i


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OUTPUT:

Enter the number of complaints:

For Week 1:3

For Week 2:2

For Week 3:3For Week 4:1

For Week 5:3

For Week 6:3

For Week 7:3

For Week 8:2

For Week 9:1

For Week 10:3

For Week 11:3

For Week 12:4

For Week 13:2For Week 14:1

For Week 15:1

For Week 16:1

For Week 17:3

For Week 18:2

For Week 19:2

For Week 20:3

Enter the value of Z:3

U.C.L=6.849725 C.L=2.300000 L.C.L=0.000000


10/35

//Program to find the value using Newton Forward Method

//Ashish Ruhela

//MCA 2 A

//Roll No. : 10

//Date:20-04-12

#include

void main()

{

float x[10],y[10][10],x1,h,r=0.0,z;

int i,j,n;

clrscr();

printf("Enter the value of n:");

scanf("%d",&n);

printf("\n X Y \n");

for(i=0;i


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}

printf("\nThe value of f(%.2f)is:%.5f",x1,r);

getch();

}

int fact(int a)

{int fact=1;

while(a>=1)

{

fact=fact*a;

a--;

}

return fact;

}


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13/35

//Program to find the value using Newton Backward Method

//Ashish Ruhela

//MCA 2 A

//Roll No. : 10

//Date:24-04-12

#include

void main()

{

float x[10],y[10][10],x1,h,z,r=0.0,k;

int i,j,n;

clrscr();


scanf("%d",&n);

printf("\nX Y \n");

for(i=0;i


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r+=y[n-1][k]*z/(pow(h,k)*fact(k));

k++;

}

printf("\nThe value of f(%.2f)is:%.2f",x1,r);

getch();

}int fact(int a)

{

int fact=1;

while(a>=1)

{

fact=fact*a;

a--;

}

return fact;

}


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OUTPUT:

Enter the value of n:5

X Y

1891 46

1901 661911 81

1921 93

1931 101

**Backward difference table**

1891.00 46.00

1901.00 66.00 20.00

1911.00 81.00 15.00 -5.00

1921.00 93.00 12.00 -3.00 2.001931.00 101.00 8.00 -4.00 -1.00 -3.00

Enter the value of x:1925

The value of f(1925.00)is:99.92


16/35

//Program to find the value using Gauss Forward Method

// Ashish Ruhela

//MCA 2 A

//Roll No. : 10

//Date:27-04-1

#include

void main()

{

float x[10],y[10][10],x1,h,r=0.0,z,u,x0;

int i,j,n,index,c=0;

float fun(float,int);

clrscr();


scanf("%d",&n);

printf("\n X Y \n");for(i=0;i


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}

u=(x1-x0)/h;

printf("\nTaking the origin at %f \n",u);

index=n/2;

for(i=0;i=1)

{

fact=fact*a;

a--;

}

return fact;

}

float fun(float u,int i)

{

if(i


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OUTPUT:


X Y

21 18.4708

25 17.814429 17.1070

33 16.3432

37 15.5154

**Forward difference table**

21.00 18.47 -0.66 -0.05 -0.01 -0.00

25.00 17.81 -0.71 -0.06 -0.01

29.00 17.11 -0.76 -0.06

33.00 16.34 -0.8337.00 15.52


Taking the origin at 0.250000

The value of f(30.00)is:16.92160


19/35

//Program to find the value using Gauss Backward Method

//Ashish Ruhela

//MCA 2 A

//Roll No. : 10

//Date:27-04-12

#include

void main()

{

float x[10],y[10][10],x1,h,r=0.0,z,u,x0;

int i,j,n,index,c=1;

float fun(float,int);

clrscr();


scanf("%d",&n);



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}

u=(x1-x0)/h;

printf("\nTaking the origin at %f \n",x0);

index=n/2;

for(i=0;i=1)

{

fact=fact*a;

a--;

}

return fact;

}

float fun(float u,int i)

{

if(i


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OUTPUT:


X Y

12500 111.80339912510 111.848111

12520 111.892806

12530 111.937483

**Backward difference table**

12500.00 111.80

12510.00 111.85 0.04

12520.00 111.89 0.04 -0.00

12530.00 111.94 0.04 -0.00 0.00


Taking the origin at 12520.000000

The value of f(12516.00)is:111.87492


22/35

//Program to find the value using Lagrange's Method

//Ashish Ruhela

//MCA 2 A

//Roll No. : 10

//Date:01-05-12

void main()

{

float x[30],y[30],x1,sum=0,s1,s2;

int n,i,j;

clrscr();

printf("\n Enter the number of terms:");

scanf("%d",&n);

printf("\n\nEnter the values of x:");

for(i=0;i


23/35

OUTPUT:

Enter the number of terms:4

Enter the values of x:7

8

9

10Enter the values of y:3

1

1

9

Enter the value of x,for which you find the value of y:9.5

Value of y is 3.625000


24/35

//Program to find the value using Newton divided difference formula

//Ashish Ruhela

//MCA 2 A

//Roll No. : 10

//Date:04-04-12

#include

#include

void main()

{

float x[10],y[10][10],x1,z,r=0.0;

int i,j,n,k=0,m=0;

clrscr();


scanf("%d",&n);



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z=1.0;

for(j=0;j


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OUTPUT:


X Y

4 48

5 1007 294

10 900

11 1210

13 2028

**Divided difference table**

4.00 48.00 52.00 15.00 1.00 0.00 0.00

5.00 100.00 97.00 21.00 1.00 0.00

7.00 294.00 202.00 27.00 1.0010.00 900.00 310.00 33.00

11.00 1210.00 409.00

13.00 2028.00

Enter the value at which you want to find the function:8

The value of f(8.000) is: 448.000


27/35

//Program to evaluate the given function using Trapezoidal rule

//Ashish Ruhela

//MCA 2 A

//Roll No. : 10

//Date: 08-05-12

#define f(x) ((x)*(x)*(x))

void main()

{

int ub,lb,j=0;

float x[20],y[20],h,n,i,res;

aa:

printf("\nEnter the limit:\n");

printf("\nEnter the upper limit:");

scanf("%d",&ub);

printf("Enter the lower limit:");scanf("%d",&lb);

if(ub


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OUTPUT:

Enter the limit:

Enter the upper limit:1

Enter the lower limit:0

Enter the no. of subintervals:5

x f(x)

0.00 0.000

0.20 0.008

0.40 0.064

0.60 0.216

0.80 0.512

1.00 1.000

Result is:0.260000


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//Program to Evaluate the given function using Simpson's 1/3 rule

//Ashish Ruhela

//MCA 2 A

//Roll No. : 10

//Date: 11-05-12

#define f(x) (1/(1+(x)*(x)))

void main()

{

int ub,lb,j=0,k;

float x[20],y[20],h,n,res,eve,odd,i;

aa:



scanf("%d",&ub);


if(ub


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}

res=h/3*(res+eve+odd);

printf("\n\nResult is:%f",res);

getch(); }


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OUTPUT:

Enter the limit:




x f(x)

0.00 1.000

1.00 0.500

2.00 0.200

3.00 0.100

4.00 0.0595.00 0.038

6.00 0.027

Result is:1.366174


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//Program to Evaluate the given function using Simpson's 3/8 rule

//Ashish Ruhela

//MCA 2 A

//Roll No. : 10

//Date: 15-05-12

#define f(x) (1/(1+(x)*(x)))

void main()

{

int ub,lb,j=0,k;

float x[20],y[20],h,n,res,eve,odd,i;

aa:



scanf("%d",&ub);


if(ub


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}

res=3*h/8*(res+eve+odd);

printf("\n\nResult is:%f",res);

getch(); }


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OUTPUT:

Enter the limit:




x f(x)

0.00 1.000

1.00 0.500

2.00 0.200

3.00 0.100

4.00 0.059

5.00 0.0386.00 0.027

Result is:1.357081

Documents

Computer Based Numerical & Statistical Techniques using 'C