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(17) MPI Numerical Integration Reduction Example Previous Top Next

Numerical Integration Example using a MPI Reduction Function

where p = # of processes
n = number of intervals per process
a = lower limit of integration
b = upper limit of integration
h = (b-a)/(n*p)
aij = a +[ i*n +j]h


/* C Example */
#include <mpi.h>
#include <math.h>
#include <stdio.h>
float fct(float x)
{
return cos(x);
}
/* Prototype */
float integral(float a, int n, float h);
void main(argc,argv)
int argc;
char *argv[];
{
/***********************************************************************
* *
* This is one of the MPI versions on the integration example *
* It demonstrates the use of : *
* *
* 1) MPI_Init *
* 2) MPI_Comm_rank *
* 3) MPI_Comm_size *
* 4) MPI_Reduce * *
* 5) MPI_Finalize *
* *
***********************************************************************/
int n, p, i, j, ierr,num;
float h, result, a, b, pi;
float my_a, my_range;

int myid, source, dest, tag;
MPI_Status status;
float my_result;

pi = acos(-1.0); /* = 3.14159... */
a = 0.; /* lower limit of integration */
b = pi*1./2.; /* upper limit of integration */
n = 100000; /* number of increment within each process */

dest = 0; /* define the process that computes the final result */
tag = 123; /* set the tag to identify this particular job */

/* Starts MPI processes ... */

MPI_Init(&argc,&argv); /* starts MPI */
MPI_Comm_rank(MPI_COMM_WORLD, &myid); /* get current process id */
MPI_Comm_size(MPI_COMM_WORLD, &p); /* get number of processes */

h = (b-a)/n; /* length of increment */
num = n/p; /* number of intervals calculated by each process*/
my_range = (b-a)/p;
my_a = a + myid*my_range;
my_result = integral(my_a,num,h);

printf("Process %d has the partial result of %f\n", myid,my_result);

/* Use an MPI sum reduction to collect the results */
MPI_Reduce(&my_result, &result,1,MPI_REAL,MPI_SUM,0,MPI_COMM_WORLD);

MPI_Finalize(); /* let MPI finish up ... */
}
float integral(float a, int n, float h)
{
int j;
float h2, aij, integ;

integ = 0.0; /* initialize integral */
h2 = h/2.;
for (j=0;j<n;j++) { /* sum over all "j" integrals */
aij = a + j*h; /* lower limit of "j" integral */
integ += fct(aij+h2)*h;
}
return (integ);
}




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Numerical_integration_example2.src  last modified Feb 14, 2011 Introduction Table of Contents
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