In need of Fixed-point algorithm

Question

hi peeps,
Im in need ( a very urgent need ) of a fix-point algorithm , with explanations if possible. Can anybody point me to the right direction?
thanks....

Answer 1

See if this helps you,
To find a solution to p=g(p)given an initial approximation p0.
Input: Initial approximation p0; tolerance TOL; maximum number of
     iterations N0.
Output: Approximate solution p or message of failure.
Step 1: Set i=1;
Step 2: While i<=N0 do Steps 3-6.
Step 3: Set p=g(p0). (Compute p.)
Step 4: If |p-p0| < TOL then
OUTPUT (p); (Procedure completed successfully.)
STOP.
Step 5: Set i=i+1.
Step 6: Set p0=p. (Update p0.)
Step 7: OUTPUT('Method failed after N0 iterations, N0=', N0);
(procedure completed unsuccessfully.)
STOP.
Fixed-Point Iteration Code in C Language
* Fixed-Point Iteration Code in C Language
* To find a solution to p = g(p) given an
* initial approximation p0
* INPUT: initial approximation; tolerance TOL;
* maximum number of iterations NO.
* OUTPUT: approximate solution p or
* a message that the method fails.
#include<stdio.h>
#include<math.h>
#define true 1
#define false 0
main()
double TOL,P0,P;
int I,NO,FLAG,OK;
FILE *OUP[1];
void INPUT(int *, double *, double *, int *);
void OUTPUT(FILE **, int *);
double absval(double);
double G(double );
INPUT(&OK, &P0, &TOL, &NO);
if (OK)
OUTPUT(OUP, &FLAG);
/* STEP 1 */
I = 1; OK = true;
/* STEP 2 */
while((I<=NO) && OK)
/* STEP 3 */
/* compute P(I) */
P = G(P0);
if (FLAG == 2)
fprintf(*OUP, "%3d %15.8e\n", I, P);
/* STEP 4 */
if (absval(P-P0) < TOL)
/* procedure completed successfully */
fprintf(*OUP, "\nApproximate solution P = %12.8f\n", P);
fprintf(*OUP, "Number of iterations = %3d", I);
fprintf(*OUP, " Tolerance = %14.8e\n",TOL);
OK = false;
else
/* STEP 5 */
I++;
/* STEP 6 */
/* update P0 */
P0 = P;
if (OK)
/* STEP 7 */
/* procedure completed unsuccessfully */
fprintf(*OUP, "\nIteration number %3d", NO);
fprintf(*OUP, " gave approximation %12.8f\n", P);
fprintf(*OUP, "not within tolerance %14.8e\n",TOL);
fclose(*OUP);
/* Change function G for a new problem */
double G(double X)
double g;
g = sqrt(10.0 / (4.0 + X));
return g;
void INPUT(int OK, double P0, double TOL, int NO)
char AA;
printf("This is the Fixed-Point Method.\n");
printf("Has the function G been created in the program immediately preceding\n");
printf("the INPUT function?\n");
printf("Enter Y or N\n");
scanf("%c",&AA);
if ((AA == 'Y') || (AA == 'y'))
*OK = false;
printf("Input initial approximation\n");
scanf("%lf",P0);
while(!(*OK))
printf("Input tolerance\n");
scanf("%lf", TOL);
if (*TOL <= 0.0)
printf("Tolerance must be positive\n");
else
*OK = true;
*OK = false;
while (!(*OK))
printf("Input maximum number of iterations - no decimal point\n");
scanf("%d", NO);
if (*NO <= 0)
printf("Must be positive integer\n");
else
*OK = true;
else
printf("The program will end so that the function G can be created\n");
*OK = false;
void OUTPUT(FILE **OUP, int *FLAG)
char NAME[30];
printf("Select output destination\n");
printf("1. Screen\n");
printf("2. Text file\n");
printf("Enter 1 or 2\n");
scanf("%d", FLAG);
if (*FLAG == 2)
printf("Input the file name in the form - drive:name.ext\n");
printf("For example: A:OUTPUT.DTA\n");
scanf("%s", NAME);
*OUP = fopen(NAME, "w");
else
*OUP = stdout;
printf("Select amount of output\n");
printf("1. Answer only\n");
printf("2. All intermeditate approximations\n");
printf("Enter 1 or 2\n");
scanf("%d", FLAG);
fprintf(*OUP, "FIXED-POINT METHOD\n");
if (*FLAG == 2)
fprintf(*OUP, " I P\n");
/* Absolute Value Function */
double absval(double val)
if (val >= 0)
return val;
else
return -val;