Re: bug in calcru()

From: Chris Landauer (cal_at_rush.aero.org)
Date: 01/26/05

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    Date: Wed, 26 Jan 2005 13:05:59 -0800 (PST)
To: ambrisko@ambrisko.com

    hihi, doug -

    > Doug Ambrisko <ambrisko@ambrisko.com> wrote
    > ...
    > The assumption with this calculation is that st & it tend to be
    > small compared to tt so the 1024 X shouldn't overflow much.
    > ...
    > cal@aero.org wrote:
    > | ...but i'm a little worried that the 1024 multiplications aren't
    > | large enough when tt gets really large
    > | > Doug Ambrisko <ambrisko@ambrisko.com> wrote
    > | > ...
    > | > /* Subdivide tu. try to becareful of overflow */
    > | > su = tu * (st * 1024 / tt) / 1024;
    > | > iu = tu * (it * 1024 / tt) / 1024;
    > | > uu = tu - (su + iu);
    > | > ...

    i'm not so worried about the overflow limit (that's what the mathematical
    analysis is intended to discover, and i assume that the bound is large enough
    to ignore the issue - that is the really clever part about computing su and iu
    first instead of uu), but the underflow - if st and it are small enough and tt
    is large enough, these equations produce 0 for both su and iu (and the
    reported percentage will rightly be 0.00%, but i want to see the rest of the
    detail for my time models)

    what i think i'll do is try out my set of if-protected equations, but with
    your choice of computing su and iu - that ought to avoid both problems

    #define TOOBIG 45254834 /* 32*sqrt(2)*10^6 */
            if (tt >= TOOBIG && tu >= tt)
                    {
                    u_int64_t q, r;
                    q = tu / tt;
                    r = tu % tt;
                    su = (q * st) + (r * st) / tt;
                    iu = (q * it) + (r * it) / tt;
                    }
            else {
                    su = (tu * st) / tt;
                    iu = (tu * it) / tt;
                    }
            uu = tu - (su + iu);

    of course, JUST using the lines

            su = (tu * st) / tt;
            iu = (tu * it) / tt;
            uu = tu - (su + iu);

    will help a lot already - perhaps this much should be done right now
    (according to the usual committment protocols) while we analyze the various
    alternative suggestions for improvement (or are there high-utilization
    processes that are more system time than user time?)

    general question - is it (interrupt time) where the unreported time is when
    user and system percentages add up to much less than 100%? i assume that the
    monotonicity rules are what makes programs report over 100% utilization (which
    i get occasionally, but only on very short programs)

    more later,
    cal

    Chris Landauer
    Aerospace Integration Science Center
    The Aerospace Corporation
    cal@aero.org

    PS - i'm sorry i don't speak "patch" well enough yet, but i will eventually

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