Per popular demand. Here are a few numbers for small thread counts
running the sched_yield_test benchmark on a 2-way SMP with the following
characteristics.
model name : Pentium III (Katmai)
stepping : 3
cpu MHz : 551.266
cache size : 512 KB
I compare 2.4.1-pre8 kernels (vanilla, table/prio scheduler and
multiqueue).
#T : van prio MQ
----------------------
1 : 0.591 0.582 0.750
2 : 0.295 0.293 0.377
3 : 2.091 2.373 1.010
4 : 1.894 1.783 1.558
5 : 1.949 1.794 1.591
6 : 2.003 1.803 1.605
7 : 2.050 1.805 1.654
8 : 2.118 1.816 1.676
9 : 2.174 1.811 1.708
10 : 2.235 1.821 1.744
11 : 2.304 1.823 1.780
12 : 2.365 1.831 1.863
13 : 2.427 1.829 1.870
14 : 2.494 1.841 1.950
15 : 2.578 1.839 1.959
16 : 2.691 1.865 2.043
17 : 2.804 1.855 2.041
18 : 2.893 1.873 2.127
19 : 3.001 1.851 2.079
20 : 3.098 1.878 2.182
21 : 3.191 1.851 2.178
22 : 3.263 1.884 2.233
23 : 3.332 1.850 2.231
24 : 3.403 1.901 2.272
25 : 3.472 1.865 2.251
26 : 3.540 1.923 2.305
27 : 3.604 1.872 2.295
28 : 3.680 1.900 2.333
29 : 4.204 1.883 2.329
30 : 4.256 1.944 2.358
31 : 3.875 1.936 2.325
32 : 4.476 1.953 2.339
Hubertus Franke
Enterprise Linux Group (Mgr), Linux Technology Center (Member Scalability)
, OS-PIC (Chair)
email: frankeh@us.ibm.com
(w) 914-945-2003 (fax) 914-945-4425 TL: 862-2003
Andrea Arcangeli <andrea@suse.de>@lists.sourceforge.net on 01/18/2001
08:30:41 PM
Sent by: lse-tech-admin@lists.sourceforge.net
To: Mike Kravetz <mkravetz@sequent.com>
cc: lse-tech@lists.sourceforge.net, linux-kernel@vger.kernel.org
Subject: Re: [Lse-tech] Re: multi-queue scheduler update
On Thu, Jan 18, 2001 at 04:52:25PM -0800, Mike Kravetz wrote:
> was less than the number of processors. I'll give the tests a try
> with a smaller number of threads. I'm also open to suggestions for
OK!
> what benchmarks/test methods I could use for scheduler testing. If
> you remember what people have used in the past, please let me know.
It was this one IIRC (it spawns threads calling sched_yield() in loop).
/*
Tester for the kernel's speed in scheduling.
(C) 1999 / Willy Tarreau <willy@meta-x.org>
Modified by Davide Libenzi <davidel@maticad.it>
You can do whatever you want with this program, but I'm not
responsible for any misuse. Be aware that it can heavily load
a host. As it is multithreaded, it might take advantages of SMP.
It basically creates a growing amount of threads and measures
their cumulative work (i.e. loop iterations/second). The output
is easily useable by gnuplot.
To compile, you need libpthread :
gcc -O2 -fomit-frame-pointer -o threads threads.c -lpthread
Output on stdout is :
<nb_threads> <average_work> <zero_work_threads> <std_deviation>
*/
#include <stdio.h>
#include <pthread.h>
#include <signal.h>
#include <unistd.h>
#include <time.h>
#define MAXTHREADS 450
#define MEASURE_TIME 60
pthread_t thr[MAXTHREADS];
int nbthreads = MAXTHREADS;
int measure_time = MEASURE_TIME;
volatile actthreads = 0;
long long int totalwork[MAXTHREADS];
volatile int stop = 0,
start = 0,
count = 0;
void oneatwork(int thr)
{
int i;
while (!start) /* don't disturb pthread_create() */
usleep(10000);
actthreads++;
while (!stop)
{
if (count)
totalwork[thr]++;
syscall(158); /* sys_sched_yield() */
}
actthreads--;
pthread_exit(0);
}
main(int argc, char **argv)
{
int i,
err,
avgwork,
thrzero;
long long int value,
avgvalue;
double sqrdev;
time_t ts,
te;
if (argc < 3)
{
printf("usage: %s threads time\n", argv[0]);
exit(1);
}
nbthreads = atoi(argv[1]);
measure_time = atoi(argv[2]);
start = 0;
count = 0;
stop = 0;
actthreads = 0;
thrzero = 0;
value = 0;
sqrdev = 0.0;
fprintf(stderr, "\nCreating %d threads ...", nbthreads);
for (i = 0; i < nbthreads; i++)
{
if ((err = pthread_create(&thr[i], NULL, (void *) &oneatwork, (void
*) i)) != 0)
{
fprintf(stderr, "thread %d pthread_create=%d -> ", i, err);
perror("");
exit(1);
}
pthread_detach(thr[i]);
}
for (i = 0; i < nbthreads; i++)
totalwork[i] = 0;
fprintf(stderr, " OK !\nWaiting for all threads to start ...");
start = 1;
while (actthreads != nbthreads)
usleep(10000); /* waiting for a bit of stability */
fprintf(stderr, "Go !\n");
count = 1;
time(&ts);
sleep(measure_time);
count = 0;
stop = 1;
time(&te);
for (i = 0; i < nbthreads; i++)
{
value += totalwork[i];
if (totalwork[i] == 0)
++thrzero;
}
avgvalue = value / nbthreads;
value /= (int) difftime(te, ts);
avgwork = (int) (value / nbthreads);
for (i = 0; i < nbthreads; i++)
{
double difvv = (double) (totalwork[i] - avgvalue);
sqrdev += difvv * difvv;
}
while (actthreads > 0)
usleep(10000);
printf("%d\t\t%lld\t\t%d\t\t%d\t\t%f\n", nbthreads, value, avgwork,
thrzero,
sqrdev / ((double) nbthreads * avgvalue * avgvalue));
exit(0);
}
Andrea
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