Re: [PATCH v5 2/3] x86/bus_lock: Handle #DB for bus lock

[Fenghua Yu <fenghua.yu@intel.com>]

Hi, Thomas,

On Sat, Mar 20, 2021 at 01:42:52PM +0100, Thomas Gleixner wrote:
> On Fri, Mar 19 2021 at 22:19, Fenghua Yu wrote:
> > On Fri, Mar 19, 2021 at 10:30:50PM +0100, Thomas Gleixner wrote:
> >> > +	if (sscanf(arg, "ratelimit:%d", &ratelimit) == 1 && ratelimit > 0) {
> >> > +		bld_ratelimit = ratelimit;
> >> 
> >> So any rate up to INTMAX/s is valid here, right?
> >
> > Yes. I don't see smaller limitation than INTMX/s. Is that right?
> 
> That's a given, but what's the point of limits in that range?
> 
> A buslock access locks up the system for X cycles. So the total amount
> of allowable damage in cycles per second is:
> 
>    limit * stall_cycles_per_bus_lock
> 
> ergo the time (in seconds) which the system is locked up is:
> 
>    limit * stall_cycles_per_bus_lock / cpufreq
> 
> Which means for ~INTMAX/2 on a 2 GHz CPU:
> 
>    2 * 10^9 * $CYCLES  / 2 * 10^9  = $CYCLES seconds
> 
> Assumed the inflicted damage is only 1 cycle then #LOCK is pretty much
> permanently on if there are enough threads. Sure #DB will slow them
> down, but it still does not make any sense at all especially as the
> damage is certainly greater than a single cycle.
> 
> And because the changelogs and the docs are void of numbers I just got
> real numbers myself.
> 
> With a single thread doing a 'lock inc *mem' accross a cache line
> boundary the workload which I measured with perf stat goes from:
> 
>      5,940,985,091      instructions              #    0.88  insn per cycle         
>        2.780950806 seconds time elapsed
>        0.998480000 seconds user
>        4.202137000 seconds sys
> to
> 
>      7,467,979,504      instructions              #    0.10  insn per cycle         
>        5.110795917 seconds time elapsed
>        7.123499000 seconds user
>       37.266852000 seconds sys
> 
> The buslock injection rate is ~250k per second.
> 
> Even if I ratelimit the locked inc by a delay loop of ~5000 cycles
> which is probably more than what the #DB will cost then this single task
> still impacts the workload significantly:
> 
>      6,496,994,537      instructions              #    0.39  insn per cycle         
>        3.043275473 seconds time elapsed
>        1.899852000 seconds user
>        8.957088000 seconds sys
> 
> The buslock injection rate is down to ~150k per second in this case.
> 
> And even with throttling the injection rate further down to 25k per
> second the impact on the workload is still significant in the 10% range.

Thank you for your insight!

So I can change the ratelimit to system wide and call usleep_range()
to sleep: 
               while (!__ratelimit(&global_bld_ratelimit))
                       usleep_range(1000000 / bld_ratelimit,
                                    1000000 / bld_ratelimit);

The max bld_ratelimit is 1000,000/s because the max sleeping time is 1 usec.
The min bld_ratelimit is 1/s.

> 
> And of course the documentation of the ratelimit parameter explains all
> of this in great detail so the administrator has a trivial job to tune
> that, right?

I will explain how to tune the parameter in buslock.rst doc.

> 
> >> > +	case sld_ratelimit:
> >> > +		/* Enforce no more than bld_ratelimit bus locks/sec. */
> >> > +		while (!__ratelimit(&get_current_user()->bld_ratelimit))
> >> > +			msleep(1000 / bld_ratelimit);
> 
> For any ratelimit > 1000 this will loop up to 1000 times with
> CONFIG_HZ=1000.
> 
> Assume that the buslock producer has tons of threads which all end up
> here pretty soon then you launch a mass wakeup in the worst case every
> jiffy. Are you sure that the cure is better than the disease?

if using usleep_range() to sleep, the threads will not sleep and wakeup,
right? Maybe I can use msleep() for msec (bigger bld_ratelimit) and
usleep_range() for usec (smaller bld_ratelimit)?

Even if there is mass wakeup, throttling the threads can avoid the system
wide performance degradation (e.g. 7x slower dd command in another user).
Is that a good justification for throttling the threads?

> 
> > If I split this whole patch set into two patch sets:
> > 1. Three patches in the first patch set: the enumeration patch, the warn
> >    and fatal patch, and the documentation patch.
> > 2. Two patches in the second patch set: the ratelimit patch and the
> >    documentation patch.
> >
> > Then I will send the two patch sets separately, you will accept them one
> > by one. Is that OK?
> 
> That's obviously the right thing to do because #1 should be ready and we
> can sort out #2 seperately. See the conversation with Tony.

Thank you for picking up the first patch set!

-Fenghua