Re: [PATCH 2/2] x86/tsc_sync: Add synchronization overhead to tsc adjustment

From: Waiman Long
Date: Mon Apr 25 2022 - 09:21:10 EST


On 4/22/22 06:41, Thomas Gleixner wrote:
On Mon, Apr 18 2022 at 11:41, Waiman Long wrote:
On 4/3/22 06:03, Thomas Gleixner wrote:
[    0.008815] TSC ADJUST compensate: CPU36 observed 86056 warp
(overhead 150). Adjust: 86206
[    0.008815] TSC ADJUST compensate: CPU54 observed 86148 warp
(overhead 158). Adjust: 86306

Also if the compensation value is at the upper end and the real overhead
is way lower then the validation run might end up with the opposite
result. I'm a bit worried about this variation.
I also have a little concern about that. That is why I add patch 1 to
minimize as much external interference as possible. For the TSC
adjustment samples that I got so far, I have never seen one that need a
2nd adjustment to go backward.
I did some experiments and noticed that the boot time overhead is
different from the overhead when doing the sync check after boot
(offline a socket and on/offline the first CPU of it several times).

During boot the overhead is lower on this machine (SKL-X), during
runtime it's way higher and more noisy.

The noise can be pretty much eliminated by running the sync_overhead
measurement multiple times and building the average.

The reason why it is higher is that after offlining the socket the CPU
comes back up with a frequency of 700Mhz while during boot it runs with
2100Mhz.

Sync overhead: 118
Sync overhead: 51 A: 22466 M: 22448 F: 2101683
One explanation of the sync overhead difference (118 vs 51) here is whether the lock cacheline is local or remote. My analysis the interaction between check_tsc_sync_source() and check_tsc_sync_target() is that real overhead is about locking with remote cacheline (local to source, remote to target). When you do a 256 loop of locking, it is all local cacheline. That is why the overhead is lower. It also depends on if the remote cacheline is in the same socket or a different socket.

Sync overhead: 178
Sync overhead: 152 A: 22477 M: 67380 F: 700529

Sync overhead: 212
Sync overhead: 152 A: 22475 M: 67380 F: 700467

Sync overhead: 153
Sync overhead: 152 A: 22497 M: 67452 F: 700404

Can you try the patch below and check whether the overhead stabilizes
accross several attempts on that copperlake machine and whether the
frequency is always the same or varies?
Yes, I will try that experiment and report back the results.

Independent of the outcome on that, I think have to take the actual CPU
frequency into account for calculating the overhead.

Assuming that the clock frequency remains the same during the check_tsc_warp() loop and the sync overhead computation time, I don't think the actual clock frequency matters much. However, it will be a different matter if the frequency does change. In this case, it is more likely the frequency will go up than down. Right? IOW, we may underestimate the sync overhead in this case. I think it is better than overestimating it.

Cheers,
Longman


Thanks,

tglx
---
--- a/arch/x86/kernel/tsc_sync.c
+++ b/arch/x86/kernel/tsc_sync.c
@@ -446,7 +446,8 @@ void check_tsc_sync_target(void)
unsigned int cpu = smp_processor_id();
cycles_t cur_max_warp, gbl_max_warp;
cycles_t start, sync_overhead;
- int cpus = 2;
+ u64 ap1, ap2, mp1, mp2;
+ int i, cpus = 2;
/* Also aborts if there is no TSC. */
if (unsynchronized_tsc())
@@ -514,6 +515,23 @@ void check_tsc_sync_target(void)
arch_spin_lock(&sync.lock);
arch_spin_unlock(&sync.lock);
sync_overhead = rdtsc_ordered() - start;
+ pr_info("Sync overhead: %lld\n", sync_overhead);
+
+ sync_overhead = 0;
+ rdmsrl(MSR_IA32_APERF, ap1);
+ rdmsrl(MSR_IA32_MPERF, mp1);
+ for (i = 0; i < 256; i++) {
+ start = rdtsc_ordered();
+ arch_spin_lock(&sync.lock);
+ arch_spin_unlock(&sync.lock);
+ sync_overhead += rdtsc_ordered() - start;
+ };
+ rdmsrl(MSR_IA32_APERF, ap2);
+ rdmsrl(MSR_IA32_MPERF, mp2);
+ ap2 -= ap1;
+ mp2 -= mp1;
+ pr_info("Sync overhead: %lld A: %llu M: %llu F: %llu\n", sync_overhead >> 8,
+ ap2, mp2, mp2 ? div64_u64((cpu_khz * ap2), mp2) : 0);
/*
* If the warp value of this CPU is 0, then the other CPU