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    SubjectMeasured overhead of timer interrupts
    > > Do you really want to spend 1% CPU servicing timer interrupts just so
    > > _occasionally_ a program can get a short sleep to 2ms accuracy?
    > Did you measure the difference? I believe I tried to actually measure
    > that and failed.

    Ok, I just measured it, and it is more than I expected. The code is
    attached at the end. (You can use it for measuring other interrupt
    overheads too).

    If I disable interrupts using cli(),
    I consistently get:

    low: 0000002e, high 0000002e, total 1b6b0b00, lost 00000000
    Percentage cycles lost: 0.000000%

    With interrupts enabled, network interface disabled, task run as top
    priority SCHED_FIFO, and I checked only timer interrupts occur during
    the test:

    At 100Hz, on a single 300MHz Pentium II, 440LX:

    low: 0000002e, high 0000a0e7, total 1b758d4f, lost 000a824f
    Percentage cycles lost: 0.149498%
    low: 0000002e, high 0000a08c, total 1b7528ee, lost 000a1dee
    Percentage cycles lost: 0.143928%
    low: 0000002e, high 00009fdb, total 1b749e93, lost 00099393
    Percentage cycles lost: 0.136250%
    low: 0000002e, high 00009ff9, total 1b7517b1, lost 000a0cb1
    Percentage cycles lost: 0.142971%
    low: 0000002e, high 0000a0e0, total 1b74c678, lost 0009bb78
    Percentage cycles lost: 0.138464%
    low: 0000002e, high 0000a18f, total 1b7533b8, lost 000a28b8
    Percentage cycles lost: 0.144526%
    low: 0000002e, high 0000a0f8, total 1b749561, lost 00098a61
    Percentage cycles lost: 0.135739%
    low: 0000002e, high 0000a441, total 1b752bbc, lost 000a20bc
    Percentage cycles lost: 0.144083%
    low: 0000002e, high 0000a156, total 1b7526b8, lost 000a1bb8
    Percentage cycles lost: 0.143805%
    low: 0000002e, high 0000a0b1, total 1b75194e, lost 000a0e4e
    Percentage cycles lost: 0.143061%

    This counts the overhead of timer interrupts and anything that happens
    in return path of timer interrupts.

    Summary: Single 300MHz Pentium II on 440LX chipset.
    Average timer interrupt overhead at 100Hz: 0.14225%
    Average cycles per timer interrupt: 41240.6
    Average time per timer interrupt: 137.47

    Extrapolating to 1024Hz, expect 1.4566% time to go servicing the timer

    Quite how this translates to "reasonable 486+ its definitely a non
    issue" I can't see. Maybe 486s are faster than Pentium IIs?

    -- Jamie

    /* Measure overhead taken by kernel from a running process.
    x86 with rdtsc and cmov instructions required. Version 1.

    Copyright (C) 1999 Jamie Lokier.

    This program is free software; you can redistribute it and/or modify
    it under the terms of the GNU General Public License as published by
    the Free Software Foundation; either version 2 of the License, or (at
    your option) any later version.

    This program is distributed in the hope that it will be useful,
    but WITHOUT ANY WARRANTY; without even the implied warranty of
    GNU General Public License for more details.

    You should have received a copy of the GNU General Public License
    along with this program; if not, write to the Free Software
    Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA */

    #define LOOP 10000000 /* You may have to reduce this to avoid overflow. */

    /* Define CLI to disable interrupts. sync() your disks first! */
    /* Define SCHED to run at high priority. sync() your disks first! */

    #include <stdio.h>
    #include <sys/io.h>
    #include <sched.h>
    int main ()
    unsigned int dummy, low, high, total;

    #ifdef SCHED
    struct sched_param param;
    memset (&param, 0, sizeof (param));
    param.sched_priority = sched_get_priority_max (SCHED_FIFO);
    sched_setscheduler (getpid (), SCHED_FIFO, &param);

    #ifdef CLI
    iopl (3); __asm__ volatile ("cli");

    __asm__ volatile ("
    pushl %%ebp
    0: movl %%eax,%%ebp
    subl %%eax,%%ebp
    cmpl %%ebp,%%esi
    cmovbl %%ebp,%%esi
    cmpl %%ebp,%%edi
    cmoval %%ebp,%%edi
    decl %%ecx
    cmovel %%ecx,%%esi
    cmovel %%ebx,%%edi
    cmovel %%eax,%%ebx
    cmpl %4,%%ecx
    jnz 0b
    subl %%ebx,%%eax
    popl %%ebp"
    : "=c" (dummy), "=S" (low), "=D" (high), "=a" (total)
    : "i" (-LOOP), "c" (50), "b" (0xffffffff)
    : "ebx", "edx");

    #ifdef CLI
    __asm__ volatile ("sti"); iopl (0);

    low = -low;
    high = -high;
    printf ("low: %08x, high %08x, total %08x, lost %08x\n"
    "Percentage cycles lost: %0.6f%%\n",
    low, high, total, (total - low * LOOP),
    (total - low * (double) LOOP) / total * 100);

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      Last update: 2005-03-22 13:53    [W:0.026 / U:42.156 seconds]
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