ME 5th Sem Lab Manual_new

GS MVN INSTITUTE OF ENGINEERING & TECHNOLOGY

PALWAL (HARYANA)

LAB MANUAL

APPLIED NUMERICAL TECHNIQUES

AND COMPUTING LAB ( ME-319 E )

Department-CSE

H.O.D (CSE/IT) Faculty

(Mr. P.C.VASHIST) (MAHENDRA PAL SINGH)

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GS MVN INSTITUTE OF ENGINEERING & TECHNOLOGY

INDEX

S.No Document Description Document ID Page No

1 Time Table GS/ME/ANTC/TT

2 Syllabus GS/ME/ANTC/SB

3 Lab Plan GS/ME/ANTC/CP

4 Attendance Register GS/ME/ANTC/AR

Branch: - ME

Subject Name:- APPLIED NUMERICAL TECHNIQUES AND COMPUTING LAB.

Subject Code:- ME-319 E

H.O.D FACULTY

(Mr. P.C.VASHIST) (Mahendra Pal Singh)

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ME- 319 E APPLIED NUMERICAL TECHNIQUES AND COMPUTING LAB.

Sessional marks : 25 Practical marks : 25

1. Solution of Non-linear equation in single variable using the method of successive bisection.

2. Solution of Non-linear equation in single variable using the Newton Raphson, Secant, Bi_ Section and Modified Euler’s method.

3. Solution of a system of simultaneous algebraic equations using the Gaussian elimination procedure.

4. Solution of a system of simultaneous algebraic equations using the Gauss-Seidel iterative method.

5. Solution of a system of simultaneous algebraic equations using the Gauss-Seidel iterative method employing the technique of successive relaxation.

6. Numerical solution of an ordinary differential equation using the Euler’s method.

7. Numerical solution of an ordinary differential equation using the Runge-Kutta 4th order method.

8. Numerical solution of an ordinary differential equation using the Predictor-corrector method.

9. Numerical solution of a system of two ordinary differential equation using Numerical integration.

10. . Numerical solution of an elliptic boundary value problem using the method of Finite Differences.

The students will be required to carry out the following exercises, that are based on the theory course ME-311 Numerical Methods and Computing, with the help of MATLAB software / Pascal / C / C++ on personal computer.

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GS MVN INSTITUTE OF ENGINEERING & TECHNOLOGY HARYANA

Course Description

Faculty : MAHENDRA PAL SINGH

Subject Title: APPLIED NUMERICAL TECHNIQUES AND COMPUTING LAB.

Subject Code: ME-319 E

Course Objective:

Applied Numerical Techniques and Computing lab are essential part to solution of any Numerical like linear, non-linear equation. Similarly, a course in Applied Numerical Techniques and Computing is also an essential part of any numerical method. This course is intended to serve as a guideline for student to learn the basic concepts of engineering mathematics . We will be discussing how to solve numerical equations with the help of computer in C/C++ language.

Learning Outcomes: 

Introduced to the fundamental concepts numerical method. Able to solve the differential equation with the help of computer .

Able to answer the problems related to the concepts and functions of the following components:

Gauss-Seidel iterative method.

Euler’s method

Runge-Kutta 4th order method.

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GS MVN INSTITUTE OF ENGINEERING & TECHNOLOGY

HARYANA

Evaluation scheme:

Internal lab = 15 marks

Lab File = 5 marks

Attendance = 5 marks

Total internal marks = 25 marks.

M.D.U.Exam. marks = 25 marks

Total = 50 marks

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GS MVN INSTITUTE OF ENGINEERING & TECHNOLOGY

HARYANA

Weekly Lab Plan

SR. No

week Lab Lab topic Date Sign

1 1 Lab-1

1. Solution of Non-linear equation in single variable using the method of successive bisection.

22

Lab-2

2. Solution of Non-linear equation in single variable using the Newton Raphson, Secant, Bi_ Section and Modified Euler’s method.

3 3 Lab-3

3. Solution of a system of simultaneous algebraic equations using the Gaussian elimination procedure.

44

Lab-4

4. Solution of a system of simultaneous algebraic equations using the Gauss-Seidel iterative method.

55

Lab-5

5. Solution of a system of simultaneous algebraic equations using the Gauss-Seidel iterative method employing the technique of successive relaxation.

6 6 Lab-66. Numerical solution of an ordinary differential equation using the Euler’s method.

7 7 Lab-7 7. Numerical solution of an ordinary differential equation using the Runge-Kutta 4th order method.

8 8 Lab-88. Numerical solution of an ordinary differential equation using the Predictor-corrector method.

9 9 Lab-99. Numerical solution of a system of two ordinary differential equation using Numerical integration.

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10 10 Lab-1010. Numerical solution of an elliptic boundary value problem using the method of Finite Differences.

/*PROGRAM 1 BISECTION METHOD*/

#include<stdio.h>

#include<conio.h

#include<math.h>

float f(float x){return (x*x*x+x*x-3*x-3);}void bisect(float *x,float a,float b,int *itr){*x=(a+b)/2;++(*itr);printf("Iteration no %3d x=%7.5f\n",*itr,*x);getch ();}main(){int itr=0,maxitr;float x,a,b,aerr,x1;clrscr();printf("enter the value of a,b\nallowed error\nmax iteration\n");scanf("%f%f%f%d",&a,&b,&aerr,&maxitr);bisect(&x,a,b,&itr);do{if(f(a)*f(x)<0)b=x;elsea=x;bisect(&x1,a,b,&itr);if(fabs(x1-x)<aerr){

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printf("after %d iteration, root<169>=%6.4f\n",itr,x1);return 0;}x=x1;}while(itr<maxitr);printf("solution does not converger iteration not sufficient");getch ();return 1;}

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OUTPUT -1 BISECTION METHOD

Enter the value of a,b,allowed error, maximum iteration.

1

2

.00000000005

18

Iteration no 1*=1.50000

Iteration no 2*=1.75000

Iteration no 3*=1.62500

Iteration no 4*=1.68750

Iteration no 5*=1.71875

Iteration no 6*=1.73438

Iteration no 7*=1.72656

Iteration no 8*=1.73047

Iteration no 9*=1.73242

Iteration no 10*=1.73145

Iteration no 11*=1.73193

Iteration no 12*=1.73218

Iteration no 13*=1.73206

Iteration no 14*=1.73199

Iteration no 15*=1.73203

Iteration no 16*=1.73204

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Iteration no 17*=1.73205

Iteration no 18*=1.73205

/*PROGRAM 2 SECANT METHOD*/

#include<stdio.h>#include<conio.h>#include<math.h>float f(float x){return (cos(x)-x*exp(x));}void main(){float x1,x2,x3,y1,y2,y3,acc;clrscr();printf("enter the value of initial auess x1=");scanf("%f",&x1);printf("\n enter the value of another guess x2=");scanf("%f",&x2);printf("enter the calue of accuracy=");scanf("%f",&acc);y1=f(x1);y2=f(x2);while(fabs(x2-x1)>acc){x3=(((x2*y1)-(x1*y2))/(y1-y2));y3=f(x3);x1=x2;y1=y2;x2=x3;y2=y3;}printf("\n the root x3=%f",x3);getch ();}

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OUTPUT-2 SECANT METHOD

Enter the value of x1

And x2

X1=0

X2=1

Accuracy =0.001

The root x3=0.517757

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/*PROGRAM 3 GAUSSIAN ELIMINATION PROCEDURE*/#include<stdio.h>#include<conio.h>#include<math.h>#define n 3void main(){float a[n][n+1],x[n],t,s;int i,j,k;clrscr();printf("enter the elements of the augmented matrix rowwise\n");for(i=0;i<n;i++)for(j=0;j<n+1;j++)scanf("%f",&a[i][j]);for(j=0;j<n-1;j++)for(i=j+1;i<n;i++){t=a[i][j]/a[j][j];for(k=0;k<n+1;k++)a[i][k]-=a[j][k]*t;}printf("the upper triangular matrix is\n");for(i=0;i<n;i++){for(j=0;j<n+1;j++)printf("%8.4f",a[i][j]);printf("\n");}for(i=n-1;i>=0;i--){s=0;for(j=i+1;j<n;j++)s+=a[i][j]*x[j];x[i]=(a[i][n]-s)/a[i][i];}printf("the solution is:-\n");

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for(i=0;i<n;i++)printf("x[%3d]=%7.4f\n",i+1,x[i]);getch();}

OUTPUT- 3 GAUSSIAN ELIMINATION PROCEDURE

Enter the elements of the augmented matrix rowwise

2 1 1 10

3 2 3 18

1 4 9 16

The upper tringular matrix is

2.0000 1.0000 1.0000 10.0000

0.0000 0.5000 1.5000 3.0000

0.0000 0.0000 2.0000 10.0000

The solution is

X[1]=7.0000

X[2]=9.0000

X[3]=5.0000

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/*PROGRAM 4 REGULA FALSI METHOD*/#include<stdio.h>#include<conio.h>#include<math.h>float f(float x){return (x*x*x+x*x-3*x-3);}void regula(float *x,float x0,float x1,float fx0,float fx1,int *itr){*x=x0-((x1-x0)/(fx1-fx0))*fx0;++(*itr);printf("Iteration no %3d x=%7.5f\n",*itr,*x);getch ();}main(){int itr=0,maxitr;float x0,x1,x2,x3,aerr;printf("enter the value of x0,x1\nallowed error\nmax iteration\n");scanf("%f%f%f%d",&x0,&x1,&aerr,&maxitr);regula(&x2,x0,x1,f(x0),f(x1),&itr);do{if(f(x0)*f(x2)<0)x1=x2;elsex0=x2;regula(&x3,x0,x1,f(x0),f(x1),&itr);if(fabs(x3-x2)<aerr){printf("after %d iteration, root=6.5f\n",itr,x3);getch ();return 0;}x2=x3;}

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while(itr<maxitr);printf("solution does not converges iteration not sufficient");return 1;}

OUTPUT- 4 REGULA FALSI METHOD

Enter the value of x0,y0,x1,allowed error, maximum iteration.

1

2

.00000000006

10

Iteration no 1*=1.57143

Iteration no 2*=1.70541

Iteration no 3*=1.72788

Iteration no 4*=1.73140

Iteration no 5*=1.73195

Iteration no 6*=1.73204

Iteration no 7*=1.73205

Iteration no 8*=1.73205

Iteration no 9*=1.73205

Iteration no 10*=1.73205

After 10 iteration, root =1.73205

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/*PROGRAM 5 RUNGE KUTTA 4TH ORDER METHOD*/

#include<stdio.h>#include<conio.h>#include<math.h>float f(float x,float y){return (x+y*y);}int main(){float x0,y0,h,xn,x,y,k1,k2,k3,k4,k;clrscr();printf("enter the value of x0,y0,h,xn");scanf("%f%f%f",&x0,&y0,&h,&x);x=x0;y=y0;printf("\n\n the execution of the program to the given problem\n");while(1) break;{if(x==xn)k1=h*f(x,y);k2=h*f(x+h/2,y+k1/2);k3=h*f(x+h/2,y+k2/2);k4=h*f(x+h,y+k3);k=(k1+(k2+k3)*2+k4)/6;x+=h;y+=k;printf("when x=%8.4f,then y=%8.4f\n\n",x,y);}getch ();return 0;}

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OUTPUT- 5 RUNGE KUTTA 4TH ORDER METHOD

Enter the vallue of x0, yo, h, xn 0.0 1.0 0.2 0.2

When x=0.2000 y=1.2736

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/*PROGRAM -6 NEWTON RAPSHON METHOD*/#include<stdio.h>#include<conio.h>#include<math.h>float f(float x){return (x*log10(x)-1.2);}float df(float x){return (log10(x)+0.43429);}main(){int itr=0,maxitr;float x0,x1,h,aerr;clrscr();printf("enter the value of x0\nallowed error\nmax iteration\n");scanf("%f%f%d",&x0,&aerr,&maxitr);for(itr=1;itr<=maxitr;++itr){h=f(x0)/df(x0);x1=x0-h;printf("\nafter %d iteration x=%9.6f" ,itr,x1);getch ();if(fabs(h)<aerr){printf("\n\n\nafter %d iteration, root=%6.5f\n",itr,x1);getch ();return 0;}x0=x1;}printf("solution does not converger iteration not sufficient");getch ();return 1;}

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OUTPUT- 6 NEWTON RAPSHON METHODEnter the value of x0,allowed error, maximum iteration.

2

.000000005

10

Iteration no 1*=2.813170

Iteration no 2*=2.741109

Iteration no 3*=2.740646

Iteration no 4*=2.740646

Iteration no 5*=2.740646

Iteration no 6*=2.740646

Iteration no 7*=2.740646

Iteration no 8*=2.740646

Iteration no 9*=2.740646

Iteration no 10*=2.740646

Iterations not sufficient , solution does not converge.

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/*PROGRAM-7 GAUSS JORDAN MEHTOD*/

#include<stdio.h>#include<conio.h>#include<math.h>#define N 3 main(){float a[N][N+1],t;int i,j,k,n;clrscr();printf("enter the elements of the augmented matrix rowwise\n");for(i=0;i<N;i++)for(j=0;j<N+1;j++)scanf("%f",&a[i][j]);for(j=0;j<N;j++)for(i=0;i<N;i++)if(i!=j){t=a[i][j]/a[j][j];for(k=0;k<N+1;k++)a[i][k]-=a[j][k]*t;}printf("the diagonal matrix is\n");for(i=0;i<N;i++){for(j=0;j<N+1;j++)printf("%9.4f",a[i][j]);printf("\n");}printf("the solution is:\n");for(i=0;i<N;i++)printf("x[%3d]=%7.4f\n",i+1,a[i][N]/a[i][i]);getch ();return 0;}

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OUTPUT-7 GAUSS JORDAN METHODenter the elements of the augmented matrix rowwise

2 1 1 10

3 2 3 18

1 4 9 16

the diagonal matrix is

2.0000 0.0000 0.0000 14.0000

0.0000 0.5000 0.0000 -4.5000

0.0000 0.0000 -2.0000 -10.0000

the solution is:

x[ 1]= 7.0000

x[ 2]=-9.0000

x[ 3]= 5.0000

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/*PROGRAM-8 SIMPSON 1/3 MEHTOD*/ #include<stdio.h>#include<conio.h>#include<math.h>float f(float x){return (1+x*x);}int main(){int i,n;float x0,xn,h,s;clrscr();printf("enter the value of x0,xn,n\n");scanf("%f%f%d",&x0,&xn,&n);h=(xn-x0)/n;s=f(x0)+f(xn)+4*f(x0+h);for(i=3;i<=n-1;i+=2)s+=4*f(x0+i*h)+2*f(x0+(i-h)*h);printf("the value of integral is %f",(h*s)/3);getch();return 0;}

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OUTPUT-8 SIMPSON 1/3 MEHTOD enter the value of x0,xn,n066the value of integral is 78.000000

23

/*PROGRAM-9 TRAPEZOIDAL MEHTOD*/

#include<stdio.h>#include<conio.h>#include<math.h>

float f(float x){return(1+x*x);}int main(int argc,char*argv[]){int i,n;float x0,xn,h,s;clrscr();printf("enter the value of x0,xn,n\n");scanf("%f%f%d",&x0,&xn,&n);h=(xn-x0)/n;s=f(x0)+f(xn);for(i=1;i<=n-1;++i)s+=2*f(x0+i*h);printf("the value of integral is %f\n",(h/2)*s);getch();return 0;}

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OUTPUT-9 TRAPEZOIDAL MEHTOD

enter the value of x0,xn,n

0

6

6

the value of integral is 79.000000

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/*PROGRAM-10 MATRIX MULTIPLICATION */

#include<stdio.h>#include<conio.h>void main(){int a[4][4],b[4][4],p[4][4],i,j,k,m,n;clrscr();printf("enter the elements of first matrix\n");for(i=1;i<=2;i++)for(j=1;j<=2;j++)scanf("%d",&a[i][j]);printf("enter the elements of second matrix\n");for(i=1;i<=2;i++)for(j=1;j<=2;j++)scanf("%d",&b[i][j]);for(i=1;i<=2;i++){for(j=1;j<=2;j++){p[i][j]=0;for(k=1;k<=2;k++)p[i][j]+=a[i][k]*b[k][j];}}printf("resultant matrix is\n");for(i=1;i<=2;i++){for(j=1;j<=2;j++)printf("%4d",p[i][j]);printf("\n");}getch();}

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OUTPUT-10 MATRIX MULTIPLICATION enter the elements of first matrix

10

20

30

40

enter the elements of second matrix

10

20

30

40

resultant matrix is

700 1000

1500 2200

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