dfb2146430
This commit was SVN r21132.
197 строки
6.4 KiB
C
197 строки
6.4 KiB
C
/* -*- C -*-
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*
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* Copyright (c) 2008 Los Alamos National Security, LLC. All rights reserved.
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*
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* $COPYRIGHT$
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*
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* Additional copyrights may follow
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*
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* $HEADER$
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*
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*/
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#include <stdio.h>
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#include <stdbool.h>
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#include <sys/types.h>
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#include <unistd.h>
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#include <stdlib.h>
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#include <time.h>
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#include <mpi.h>
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int main(int argc, char* argv[])
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{
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int msg;
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int rank, size, my_twin;
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int ppn, my_node;
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struct timeval tv;
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unsigned long my_timestamp[2];
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long *timestamps;
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int i, maxrank;
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unsigned long maxsec, maxusec, minutes, seconds;
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unsigned long start_sec, start_usec;
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float fsecs;
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int nnodes;
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bool odd_nnodes;
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bool recvit;
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char *ppnstr;
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if (argc < 3) {
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fprintf(stderr, "start times must be provided\n");
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return 1;
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}
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ppnstr = getenv("OMPI_COMM_WORLD_LOCAL_SIZE");
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ppn = strtol(ppnstr, NULL, 10);
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start_sec = strtol(argv[1], NULL, 10);
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start_usec = strtol(argv[2], NULL, 10);
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MPI_Init(NULL, NULL);
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MPI_Comm_rank(MPI_COMM_WORLD, &rank);
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MPI_Comm_size(MPI_COMM_WORLD, &size);
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/* this program requires that the size be an integer multiple of ppn */
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if (0 != (size % ppn)) {
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if (0 == rank) {
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fprintf(stderr, "The number of procs must be an integer multiple of the ppn\n"
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"Given: num_procs %d ppn %d\n", size, ppn);
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MPI_Abort(MPI_COMM_WORLD, 1);
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} else {
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goto cleanup;
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}
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}
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/* see how many nodes we have */
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nnodes = size / ppn;
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odd_nnodes = false;
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if (0 != (nnodes % 2)) {
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/* we have an odd # of nodes */
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odd_nnodes = true;
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}
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/* compute the rank of the rank with which I am to exchange a message.
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* Per requirements, this proc must be on another node. To accomplish
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* this with max efficiency, we take advantage of knowing that the ppn
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* on every node will be the same. We therefore pair up the nodes, and
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* pair up the procs on each node, so that only one connection is setup
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* for each proc. We also want to ensure that the node pairs are
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* "neighboring" - i.e., that they hopefully share a switch so that the
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* hop count of sending the messages is minimized.
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*/
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/* first, determine if my node is odd or even */
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my_node = rank / ppn;
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if (0 != (my_node % 2)) {
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/* compute my twin's rank - as I am an odd numbered node, my
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* twin will be on the node below me. Thus, its rank will be
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* my rank - ppn
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*/
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my_twin = rank - ppn;
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/* if I am an odd numbered node, then I will receive first */
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MPI_Recv(&msg, 1, MPI_INT, my_twin, 1, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
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/* receive the return message so that we meet the stated requirement
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* that -every- proc send a message
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*/
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MPI_Send(&msg, 1, MPI_INT, my_twin, 1, MPI_COMM_WORLD);
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} else {
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/* compute my twin's rank - as I am an even numbered node, my
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* twin will be on the node above me. Thus, its rank will be
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* my rank + ppn
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*/
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my_twin = rank + ppn;
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/* if we have an odd number of nodes, then the last node will be
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* even and will have no one above them. In this case, we wrap around
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* and ask that node=0 take the additional connections
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*/
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recvit = true;
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if (my_twin >= size) {
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my_twin = my_twin - size;
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recvit = false;
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}
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/* I am an even numbered node, so I send first */
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MPI_Send(&msg, 1, MPI_INT, my_twin, 1, MPI_COMM_WORLD);
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/* now receive the reply so my twin also meets the requirement - but only
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* if we don't have an odd number of nodes. If we have an odd number of
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* nodes, then the node=0 procs will already have met their requirement
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*/
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if (recvit) {
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MPI_Recv(&msg, 1, MPI_INT, my_twin, 1, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
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}
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}
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/* if we have an odd number of nodes and I am on node=0, then I have
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* to take the extra recv
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*/
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if (odd_nnodes && 0 == my_node) {
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my_twin = size - ppn + rank;
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MPI_Recv(&msg, 1, MPI_INT, my_twin, 1, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
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}
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/* get a completion time stamp */
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gettimeofday(&tv, NULL);
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my_timestamp[0] = tv.tv_sec;
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my_timestamp[1] = tv.tv_usec;
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/* THIS COMPLETES THE OFFICIAL TIMING POINT */
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/* Gather to get all the timestamps to rank 0 */
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timestamps = NULL;
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if (0 == rank) {
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timestamps = malloc(2 * size * sizeof(unsigned long));
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if (NULL == timestamps) {
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MPI_Abort(MPI_COMM_WORLD, 1);
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}
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}
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MPI_Gather(&my_timestamp, 2, MPI_LONG,
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timestamps, 2, MPI_LONG, 0, MPI_COMM_WORLD);
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if (0 == rank) {
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/* The "timestamps" array will now have everyone's timestamp
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(i.e., rank 0's timestamp will be in pos 0 & 1,, rank 1's timestamp
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will be in 2 & 3, ...etc. */
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/* find the maximum timestamp */
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maxsec = start_sec;
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maxusec = start_usec;
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maxrank = -1;
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for (i=0; i < 2*size; i+=2) {
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if (timestamps[i] < maxsec) {
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continue;
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}
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if (timestamps[i] == maxsec &&
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timestamps[i+1] < maxusec) {
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continue;
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}
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maxsec = timestamps[i];
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maxusec = timestamps[i+1];
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maxrank = i/2;
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}
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free(timestamps);
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/* subtract starting time to get time in microsecs for test */
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maxsec = maxsec - start_sec;
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if (maxusec >= start_usec) {
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maxusec = maxusec - start_usec;
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} else {
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maxsec--;
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maxusec = 1000000 - start_usec + maxusec;
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}
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/* pretty-print the result */
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seconds = maxsec + (maxusec / 1000000l);
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minutes = seconds / 60l;
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seconds = seconds % 60l;
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if (0 == minutes && 0 == seconds) {
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fsecs = ((float)(maxsec)*1000000.0 + (float)maxusec) / 1000.0;
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fprintf(stderr, "Time test was completed in %8.2f millisecs\nSlowest rank: %d\n",
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fsecs, maxrank);
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} else {
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fprintf(stderr, "Time test was completed in %3lu:%02lu min:sec\nSlowest rank: %d\n",
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minutes, seconds, maxrank);
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}
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}
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cleanup:
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/* this completes the test */
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MPI_Finalize();
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return 0;
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}
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