How to measure i386 timer interrupt overhead

From: Chuck Ebbert
Date: Mon Dec 06 2004 - 22:11:55 EST


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
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