An example using mpich in a cluster (updated 10.2016)

Here is a C program that shows an example of setting up a different task for each processor core in a cluster of computers. It's very simple but shows the minimum needed to be useful.





/*

domath.c - test of mpich on multi-processor system
k theis 3/31/2016

to run:
mpicc -o domath domath.c -lm
mv domath /mnt/nfs/dist/domath
cd /mnt/nfs/dist/
mpiexec -n 6 -f machinefile ./domath

make sure machinefile exists in /mnt/nfs/dist

*/



#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <mpi.h>

#define PI 3.141592654

int do_sin(void){
double n, res;
FILE *sout;
int err;
    sout = fopen("sin.txt","w");
    if (sout == NULL) return 1;
    fprintf(sout,"Sin Table\n\n");
    err = system("/bin/uname -a > sin_s.txt");    // show machine this is on
    n = 0.0;
    while (n <= PI/2.0){
        res = sin(n);
        fprintf(sout,"\nN: %f    Res: %f",n*(90.0/(PI/2.0)),res);
        n += 1.0/90.0;
    }
    fprintf(sout,"\n");
    fflush(sout);
    fclose(sout);
    return 0;
}

int do_cos(void){
double n, res;
FILE *cout;
int err;
    cout = fopen("cos.txt","w");
    if (cout == NULL) return 1;
    fprintf(cout,"Cos Table\n\n");
    err = system("/bin/uname -a > cos_s.txt");    // show machine this is on
    n = 0.0;
    while (n <= PI/2.0){
        res = cos(n);
        fprintf(cout,"\nN: %f    Res: %f",n*(90.0/(PI/2.0)),res);
        n += 1.0/90.0;
    }
    fprintf(cout,"\n");
    fflush(cout);
    fclose(cout);
    return 0;
}

int do_tan(void){
double n, res;
FILE *tout;
int err;
    tout = fopen("tan.txt","w");
    if (tout == NULL) return 1;
    fprintf(tout,"Tan Table\n\n");
    err = system("/bin/uname -a > tan_s.txt");    // show machine this is on
    n = 0.0;
    while (n <= PI/2.0){
        res = tan(n);
        fprintf(tout,"\nN: %f    Res: %f",n*(90.0/(PI/2.0)),res);
        n += 1.0/90.0;
    }
    fprintf(tout,"\n");
    fflush(tout);
    fclose(tout);
    return 0;
}

int do_log10(void){
double n, res;
FILE *l10out;
int err;
    l10out = fopen("log10.txt","w");
    if (l10out == NULL) return 1;
    fprintf(l10out,"Log Table Base 10\n\n");
    err = system("/bin/uname -a > log10_s.txt");    // show machine this is on
    n = 0.0;
    while (n <= 100.0){
        res = log10(n);
        fprintf(l10out,"\nN: %f    Res: %f",n,res);
        n += 0.5;
    }
    fprintf(l10out,"\n");
    fflush(l10out);
    fclose(l10out);
    return 0;
}


int do_log(void){
double n, res;
FILE *lout;
int err;
    lout = fopen("log.txt","w");
    if (lout == NULL) return 1;
    fprintf(lout,"Log Table\n\n");
    err = system("/bin/uname -a > log_s.txt");    // show machine this is on
    fprintf(lout,"\nsys err = %d",err);
    n = 0.0;
    while (n <= 100.0){
        res = log(n);
        fprintf(lout,"\nN: %f    Res: %f",n,res);
        n += 0.5;
    }
    fprintf(lout,"\n");
    fflush(lout);
    fclose(lout);
    return 0;
}


int main(int argc, char **argv){
int my_id, root_process, ierr, num_procs;
MPI_Status status;

    ierr = MPI_Init(&argc, &argv);
    ierr = MPI_Comm_rank(MPI_COMM_WORLD, &my_id);
    ierr = MPI_Comm_size(MPI_COMM_WORLD, &num_procs);

    if (my_id == 0) printf("\n# of Processes: %d",num_procs);

    if (my_id == 1) do_sin();
    if (my_id == 2) do_cos();
    if (my_id == 3) do_tan();
    if (my_id == 4) do_log10();
    if (my_id == 5) do_log();
    if (my_id == 0) printf("\n");
    ierr = MPI_Finalize();
    exit(0);
}