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From: Chuck Ebbert <76306.1226@compuserve.com>
To: linux-kernel <linux-kernel@vger.kernel.org>
Subject: How to measure i386 timer interrupt overhead
Date: Mon, 6 Dec 2004 22:05:42 -0500	[thread overview]
Message-ID: <200412062209_MC3-1-902F-B102@compuserve.com> (raw)

If this program actually works, then the SMP timer interrupt on this
system takes 1200-1420 cycles, with 448 of 945 taking 1200-1219 cycles.
(All the normal interrupts are bound to CPU 0.)

Is this a valid test program?

$ taskset 2 ./t 6800 332490000 | sort -g | uniq -c | grep -v '^      [1-9]'
Elasped time: 945 ticks (assuming Hz = 1000)
   9731 0
     35 10
    224 1200
    204 1210
     16 1220
     14 1270
     65 1300
    131 1310
     33 1320
     14 1370
     20 1380
     22 1420

$ cat t.c
/* Measure i386 timer overhead by timing loops with TSC and
 * subtracting out the minimum time.  Rounds to granularity 10.
 * (Run this on an otherwise idle system.)
 */
#include <stdio.h>
#define HZ              1000
#define NUM_TESTS       (11000 + 1)
unsigned long long t0, t1, t2, t3;
unsigned long long sample[NUM_TESTS], min = 1000000;
unsigned int loop_iters, tsc_per_tick, i, j;

main (int argc, char **argv) {
        if (argc < 3) {
                fprintf(stderr, "Usage: %s <loop_size> <CPU Hz>\n", argv[0]);
                exit(1);
        }
        loop_iters   = atoi(argv[1]);
        tsc_per_tick = atoll(argv[2]) / HZ;
restart:
        asm volatile ("rdtsc" : "=A"(t0));
        for (i = 0; i < NUM_TESTS; i++) {
                asm volatile ("rdtsc" : "=A"(t1));
                for (j = 0; j < loop_iters; j++)
                        /* empty */;
                asm volatile ("rdtsc" : "=A"(t2));
                sample[i] = t2 - t1;
        }
        asm volatile ("rdtsc" : "=A"(t3));
        if (t3 < t0)
                goto restart;
        for (i = 1; i < NUM_TESTS; i++) { /* skip first sample */
                if (min > sample[i])
                        min = sample[i];
        }
        for (i = 1; i < NUM_TESTS; i++)
                printf("%llu\n", (sample[i] - min) / 10 * 10);
        fprintf(stderr, "Elasped time: %d ticks (assuming Hz = %d)\n",
                        (int)((t3 - t0) / tsc_per_tick), HZ);
}

--Chuck Ebbert  06-Dec-04  22:03:04

             reply	other threads:[~2004-12-07  3:10 UTC|newest]

Thread overview: 2+ messages / expand[flat|nested]  mbox.gz  Atom feed  top
2004-12-07  3:05 Chuck Ebbert [this message]
2004-12-07 11:36 ` Jan Engelhardt

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