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Modify MPICH program

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Old   May 17, 2011, 08:31
Default Modify MPICH program
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Monika
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Hello,I have one MPICH program:
/************************************************** ****************************
* FILE: mpi_array.c
* DESCRIPTION:
* MPI Example - Array Assignment - C Version
* This program demonstrates a simple data decomposition. The master task
* first initializes an array and then distributes an equal portion that
* array to the other tasks. After the other tasks receive their portion
* of the array, they perform an addition operation to each array element.
* They also maintain a sum for their portion of the array. The master task
* does likewise with its portion of the array. As each of the non-master
* tasks finish, they send their updated portion of the array to the master.
* An MPI collective communication call is used to collect the sums
* maintained by each task. Finally, the master task displays selected
* parts of the final array and the global sum of all array elements.
* NOTE: the number of MPI tasks must be evenly disible by 4.
* AUTHOR: Blaise Barney
* LAST REVISED: 04/13/05
************************************************** **************************/
#include "mpi.h"
#include <stdio.h>
#include <stdlib.h>
#define ARRAYSIZE 16000000
#define MASTER 0

float data[ARRAYSIZE];

int main (int argc, char *argv[])
{
int numtasks, taskid, rc, dest, offset, i, j, tag1,
tag2, source, chunksize;
float mysum, sum;
float update(int myoffset, int chunk, int myid);
MPI_Status status;

/***** Initializations *****/
MPI_Init(&argc, &argv);
MPI_Comm_size(MPI_COMM_WORLD, &numtasks);
if (numtasks % 4 != 0) {
printf("Quitting. Number of MPI tasks must be divisible by 4.\n");
MPI_Abort(MPI_COMM_WORLD, rc);
exit(0);
}
MPI_Comm_rank(MPI_COMM_WORLD,&taskid);
printf ("MPI task %d has started...\n", taskid);
chunksize = (ARRAYSIZE / numtasks);
tag2 = 1;
tag1 = 2;

/***** Master task only ******/
if (taskid == MASTER){

/* Initialize the array */
sum = 0;
for(i=0; i<ARRAYSIZE; i++) {
data[i] = i * 1.0;
sum = sum + data[i];
}
printf("Initialized array sum = %e\n",sum);

/* Send each task its portion of the array - master keeps 1st part */
offset = chunksize;
for (dest=1; dest<numtasks; dest++) {
MPI_Send(&offset, 1, MPI_INT, dest, tag1, MPI_COMM_WORLD);
MPI_Send(&data[offset], chunksize, MPI_FLOAT, dest, tag2, MPI_COMM_WORLD);
printf("Sent %d elements to task %d offset= %d\n",chunksize,dest,offset);
offset = offset + chunksize;
}

/* Master does its part of the work */
offset = 0;
mysum = update(offset, chunksize, taskid);

/* Wait to receive results from each task */
for (i=1; i<numtasks; i++) {
source = i;
MPI_Recv(&offset, 1, MPI_INT, source, tag1, MPI_COMM_WORLD, &status);
MPI_Recv(&data[offset], chunksize, MPI_FLOAT, source, tag2,
MPI_COMM_WORLD, &status);
}

/* Get final sum and print sample results */
MPI_Reduce(&mysum, &sum, 1, MPI_FLOAT, MPI_SUM, MASTER, MPI_COMM_WORLD);
printf("Sample results: \n");
offset = 0;
for (i=0; i<numtasks; i++) {
for (j=0; j<5; j++)
printf(" %e",data[offset+j]);
printf("\n");
offset = offset + chunksize;
}
printf("*** Final sum= %e ***\n",sum);

} /* end of master section */



/***** Non-master tasks only *****/

if (taskid > MASTER) {

/* Receive my portion of array from the master task */
source = MASTER;
MPI_Recv(&offset, 1, MPI_INT, source, tag1, MPI_COMM_WORLD, &status);
MPI_Recv(&data[offset], chunksize, MPI_FLOAT, source, tag2,
MPI_COMM_WORLD, &status);

mysum = update(offset, chunksize, taskid);

/* Send my results back to the master task */
dest = MASTER;
MPI_Send(&offset, 1, MPI_INT, dest, tag1, MPI_COMM_WORLD);
MPI_Send(&data[offset], chunksize, MPI_FLOAT, MASTER, tag2, MPI_COMM_WORLD);

MPI_Reduce(&mysum, &sum, 1, MPI_FLOAT, MPI_SUM, MASTER, MPI_COMM_WORLD);

} /* end of non-master */


MPI_Finalize();

} /* end of main */


float update(int myoffset, int chunk, int myid) {
int i;
float mysum;
/* Perform addition to each of my array elements and keep my sum */
mysum = 0;
for(i=myoffset; i < myoffset + chunk; i++) {
data[i] = data[i] + i * 1.0;
mysum = mysum + data[i];
}
printf("Task %d mysum = %e\n",myid,mysum);
return(mysum);
}


Modify the computation of pi problem to
read in the number of intervals on one processor and broadcast the value
to the other processors.
Use a collective operation to get the solution rather than the send and
receive operations.
Please help me ...for tomorrow
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