Messages in this thread | | | Date | Sat, 13 Apr 2013 11:18:00 -0700 | From | Josh Triplett <> | Subject | Re: [PATCH tip/core/rcu 6/7] rcu: Drive quiescent-state-forcing delay from HZ |
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On Fri, Apr 12, 2013 at 11:38:04PM -0700, Paul E. McKenney wrote: > On Fri, Apr 12, 2013 at 04:54:02PM -0700, Josh Triplett wrote: > > On Fri, Apr 12, 2013 at 04:19:13PM -0700, Paul E. McKenney wrote: > > > From: "Paul E. McKenney" <paulmck@linux.vnet.ibm.com> > > > > > > Systems with HZ=100 can have slow bootup times due to the default > > > three-jiffy delays between quiescent-state forcing attempts. This > > > commit therefore auto-tunes the RCU_JIFFIES_TILL_FORCE_QS value based > > > on the value of HZ. However, this would break very large systems that > > > require more time between quiescent-state forcing attempts. This > > > commit therefore also ups the default delay by one jiffy for each > > > 256 CPUs that might be on the system (based off of nr_cpu_ids at > > > runtime, -not- NR_CPUS at build time). > > > > > > Reported-by: Paul Mackerras <paulus@au1.ibm.com> > > > Signed-off-by: Paul E. McKenney <paulmck@linux.vnet.ibm.com> > > > > Something seems very wrong if RCU regularly hits the fqs code during > > boot; feels like there's some more straightforward solution we're > > missing. What causes these CPUs to fall under RCU's scrutiny during > > boot yet not actually hit the RCU codepaths naturally? > > The problem is that they are running HZ=100, so that RCU will often > take 30-60 milliseconds per grace period. At that point, you only > need 16-30 grace periods to chew up a full second, so it is not all > that hard to eat up the additional 8-12 seconds of boot time that > they were seeing. IIRC, UP boot was costing them 4 seconds. > > For HZ=1000, this would translate to 800ms to 1.2s, which is nowhere > near as annoying.
That raises two questions, though. First, who calls synchronize_rcu() repeatedly during boot, and could they call call_rcu() instead to avoid blocking for an RCU grace period? Second, why does RCU need 3-6 jiffies to resolve a grace period during boot? That suggests that RCU doesn't actually resolve a grace period until the force-quiescent-state machinery kicks in, meaning that the normal quiescent-state mechanism didn't work.
> > Also, a comment below. > > > > > --- a/kernel/rcutree.h > > > +++ b/kernel/rcutree.h > > > @@ -342,7 +342,17 @@ struct rcu_data { > > > #define RCU_FORCE_QS 3 /* Need to force quiescent state. */ > > > #define RCU_SIGNAL_INIT RCU_SAVE_DYNTICK > > > > > > -#define RCU_JIFFIES_TILL_FORCE_QS 3 /* for rsp->jiffies_force_qs */ > > > +#if HZ > 500 > > > +#define RCU_JIFFIES_TILL_FORCE_QS 3 /* for jiffies_till_first_fqs */ > > > +#elif HZ > 250 > > > +#define RCU_JIFFIES_TILL_FORCE_QS 2 > > > +#else > > > +#define RCU_JIFFIES_TILL_FORCE_QS 1 > > > +#endif > > > > This seems like it really wants to use a duration calculated directly > > from HZ; perhaps (HZ/100)? > > Very possibly to the direct calculation, but HZ/100 would get 10 ticks > delay at HZ=1000, which is too high -- the value of 3 ticks for HZ=1000 > works well. But I could do something like this: > > #define RCU_JIFFIES_TILL_FORCE_QS (((HZ + 199) / 300) + ((HZ + 199) / 300 ? 0 : 1)) > > Or maybe a bit better: > > #define RCU_JTFQS_SE ((HZ + 199) / 300) > #define RCU_JIFFIES_TILL_FORCE_QS (RCU_JTFQS_SE + (RCU_JTFQS_SE ? 0 : 1)) > > This would come reasonably close to the values shown above. Would > this work for you?
I'd argue that if you need something that complex, you should just explicitly write it as a step function:
#define RCU_JIFFIES_TILL_FORCE_QS (1 + (HZ > 250) + (HZ > 500))
- Josh Triplett
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