Jacobi Iteration Method Using C Programming

C program to solve system of linear equations using Jacobi Iteration Method.

Program

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

/* Arrange systems of linear
   equations to be solved in
   diagonally dominant form
   and form equation for each
   unknown and define here
*/
/* In this example we are solving
   3x + 20y - z = -18
   2x - 3y + 20z = 25
   20x + y - 2z = 17
*/
/* Arranging given system of linear
   equations in diagonally dominant
   form:
   20x + y - 2z = 17
   3x + 20y -z = -18
   2x - 3y + 20z = 25
*/
/* Equations:
   x = (17-y+2z)/20
   y = (-18-3x+z)/20
   z = (25-2x+3y)/20
*/
/* Defining function */
#define f1(x,y,z)  (17-y+2*z)/20
#define f2(x,y,z)  (-18-3*x+z)/20
#define f3(x,y,z)  (25-2*x+3*y)/20

/* Main function */
int main()
{
 float x0=0, y0=0, z0=0, x1, y1, z1, e1, e2, e3, e;
 int count=1;
 clrscr();
 printf("Enter tolerable error:\n");
 scanf("%f", &e);

 printf("\nCount\tx\ty\tz\n");
 do
 {
  /* Calculation */
  x1 = f1(x0,y0,z0);
  y1 = f2(x0,y0,z0);
  z1 = f3(x0,y0,z0);
  printf("%d\t%0.4f\t%0.4f\t%0.4f\n",count, x1,y1,z1);

  /* Error */
  e1 = fabs(x0-x1);
  e2 = fabs(y0-y1);
  e3 = fabs(z0-z1);

  count++;

  /* Set value for next iteration */
  x0 = x1;
  y0 = y1;
  z0 = z1;
 }while(e1>e && e2>e && e3>e);

 printf("\nSolution: x=%0.3f, y=%0.3f and z = %0.3f\n",x1,y1,z1);

 getch();
 return 0;
}

Jacobi Iteration C Program Output

Enter tolerable error:
0.0001

Count      x   	  y      	z
1       0.8500 	 -0.9000 	1.2500
2       1.0200 	 -0.9650 	1.0300
3       1.0013 	 -1.0015 	1.0033
4       1.0004 	 -1.0000 	0.9997
5       1.0000 	 -1.0001 	1.0000

Solution: x=1.000, y=-1.000 and z = 1.000