Re: [o(1) sched J0] higher priority smaller timeslices, in fact

From: Davide Libenzi (davidel@xmailserver.org)
Date: Wed Jan 16 2002 - 18:13:45 EST


On Wed, 16 Jan 2002, Alexei Podtelezhnikov wrote:

> man nice helped. Thanks!
>
> On Wed, 16 Jan 2002, Davide Libenzi wrote:
>
> > On Wed, 16 Jan 2002, Alexei Podtelezhnikov wrote:
> >
> > >
> > > The comment and the actual macros are inconsistent.
> > > positive number * (19-n) is a decreasing function of n!
> >
> > # man nice
> >
> >
> > > + * The higher a process's priority, the bigger timeslices
> > > + * it gets during one round of execution. But even the lowest
> > > + * priority process gets MIN_TIMESLICE worth of execution time.
> > > + */
> > >
> > > -#define NICE_TO_TIMESLICE(n) (MIN_TIMESLICE + \
> > > - ((MAX_TIMESLICE - MIN_TIMESLICE) * (19 - (n))) / 39)
> > > +#define NICE_TO_TIMESLICE(n) (MIN_TIMESLICE + \
> > > + ((MAX_TIMESLICE - MIN_TIMESLICE) * (19-(n))) / 39)
> > >
> > > I still suggest a different set as faster and more readable at least to
> > > me. Just two operations instead of 4!
> >
> > this seems quite readable to me, it's the equation at page 1 of any know
> > linear geometry book.

and this macro gets called about every 80ms, that is nothing. try to run a
cycle counter between the two implementation, get the time difference
using the CPU speed and weight it with 80ms. you'll get a percent that
compared to that, the probability of having snow in Miami in August is a
big number :-)

- Davide

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