Re: Adjustment of spinlock sleep delays

Rod Taylor <rbt(at)rbt(dot)ca> writes:
> How about (round to nearest 10msec):

> time =3D oldtime + oldtime / 2 + oldtime * rand()

> while (time > 1 second)
> time =3D time - 0.80sec

> This would stagger the wakeup times, and ensure a larger number of
> retries -- but the times should be large enough after the first few
> tries (larger than 200msec) that further backoff won't be required.

But after the first few tries the sleep time would always exceed
200msec, so there would be a *maximum* of 60*5 = 300 tries before
failing --- probably a lot less, like about 120 on average.

The random component should already help to scatter the wakeups pretty
well, so I'm thinking about just

if (oldtime > 1 sec)
time = 10msec
else
time = oldtime + oldtime * rand()

ie random growth of a maximum of 2x per try, and reset to minimum delay
when you get past 1 sec. This would guarantee at least as many tries
as I'm getting currently with the deterministic algorithm (which is
effectively this if rand() always returned 1).

regards, tom lane