From: Peter Zijlstra <peterz@infradead.org> mainline inclusion from mainline-v6.19-rc1 commit 9359d9785d85bb53f1ff1738a59aeeec4b878906 category: bugfix bugzilla: https://gitee.com/openeuler/kernel/issues/ICDF44 Reference: https://git.kernel.org/pub/scm/linux/kernel/git/torvalds/linux.git/commit/?i... -------------------------------- I always end up having to re-read these emails every time I look at this code. And a future patch is going to change this story a little. This means it is past time to stick them in a comment so it can be modified and stay current. Signed-off-by: Peter Zijlstra (Intel) <peterz@infradead.org> Link: https://lkml.kernel.org/r/20200506143506.GH5298@hirez.programming.kicks-ass.... Link: https://lkml.kernel.org/r/20200515103844.GG2978@hirez.programming.kicks-ass.... Link: https://patch.msgid.link/20251106111603.GB4068168@noisy.programming.kicks-as... Conflicts: kernel/sched/fair.c [Conflicts in nr_queued and nr_running in `task_tick_core()`, so keep nr_running.] Signed-off-by: Zicheng Qu <quzicheng@huawei.com> --- kernel/sched/fair.c | 181 ++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 181 insertions(+) diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c index d0e0e66da95f..70b49b96c117 100644 --- a/kernel/sched/fair.c +++ b/kernel/sched/fair.c @@ -14505,6 +14505,187 @@ static inline void task_tick_core(struct rq *rq, struct task_struct *curr) resched_curr(rq); } +/* + * Consider any infeasible weight scenario. Take for instance two tasks, + * each bound to their respective sibling, one with weight 1 and one with + * weight 2. Then the lower weight task will run ahead of the higher weight + * task without bound. + * + * This utterly destroys the concept of a shared time base. + * + * Remember; all this is about a proportionally fair scheduling, where each + * tasks receives: + * + * w_i + * dt_i = ---------- dt (1) + * \Sum_j w_j + * + * which we do by tracking a virtual time, s_i: + * + * 1 + * s_i = --- d[t]_i (2) + * w_i + * + * Where d[t] is a delta of discrete time, while dt is an infinitesimal. + * The immediate corollary is that the ideal schedule S, where (2) to use + * an infinitesimal delta, is: + * + * 1 + * S = ---------- dt (3) + * \Sum_i w_i + * + * From which we can define the lag, or deviation from the ideal, as: + * + * lag(i) = S - s_i (4) + * + * And since the one and only purpose is to approximate S, we get that: + * + * \Sum_i w_i lag(i) := 0 (5) + * + * If this were not so, we no longer converge to S, and we can no longer + * claim our scheduler has any of the properties we derive from S. This is + * exactly what you did above, you broke it! + * + * + * Let's continue for a while though; to see if there is anything useful to + * be learned. We can combine (1)-(3) or (4)-(5) and express S in s_i: + * + * \Sum_i w_i s_i + * S = -------------- (6) + * \Sum_i w_i + * + * Which gives us a way to compute S, given our s_i. Now, if you've read + * our code, you know that we do not in fact do this, the reason for this + * is two-fold. Firstly, computing S in that way requires a 64bit division + * for every time we'd use it (see 12), and secondly, this only describes + * the steady-state, it doesn't handle dynamics. + * + * Anyway, in (6): s_i -> x + (s_i - x), to get: + * + * \Sum_i w_i (s_i - x) + * S - x = -------------------- (7) + * \Sum_i w_i + * + * Which shows that S and s_i transform alike (which makes perfect sense + * given that S is basically the (weighted) average of s_i). + * + * Then: + * + * x -> s_min := min{s_i} (8) + * + * to obtain: + * + * \Sum_i w_i (s_i - s_min) + * S = s_min + ------------------------ (9) + * \Sum_i w_i + * + * Which already looks familiar, and is the basis for our current + * approximation: + * + * S ~= s_min (10) + * + * Now, obviously, (10) is absolute crap :-), but it sorta works. + * + * So the thing to remember is that the above is strictly UP. It is + * possible to generalize to multiple runqueues -- however it gets really + * yuck when you have to add affinity support, as illustrated by our very + * first counter-example. + * + * Luckily I think we can avoid needing a full multi-queue variant for + * core-scheduling (or load-balancing). The crucial observation is that we + * only actually need this comparison in the presence of forced-idle; only + * then do we need to tell if the stalled rq has higher priority over the + * other. + * + * [XXX assumes SMT2; better consider the more general case, I suspect + * it'll work out because our comparison is always between 2 rqs and the + * answer is only interesting if one of them is forced-idle] + * + * And (under assumption of SMT2) when there is forced-idle, there is only + * a single queue, so everything works like normal. + * + * Let, for our runqueue 'k': + * + * T_k = \Sum_i w_i s_i + * W_k = \Sum_i w_i ; for all i of k (11) + * + * Then we can write (6) like: + * + * T_k + * S_k = --- (12) + * W_k + * + * From which immediately follows that: + * + * T_k + T_l + * S_k+l = --------- (13) + * W_k + W_l + * + * On which we can define a combined lag: + * + * lag_k+l(i) := S_k+l - s_i (14) + * + * And that gives us the tools to compare tasks across a combined runqueue. + * + * + * Combined this gives the following: + * + * a) when a runqueue enters force-idle, sync it against it's sibling rq(s) + * using (7); this only requires storing single 'time'-stamps. + * + * b) when comparing tasks between 2 runqueues of which one is forced-idle, + * compare the combined lag, per (14). + * + * Now, of course cgroups (I so hate them) make this more interesting in + * that a) seems to suggest we need to iterate all cgroup on a CPU at such + * boundaries, but I think we can avoid that. The force-idle is for the + * whole CPU, all it's rqs. So we can mark it in the root and lazily + * propagate downward on demand. + */ + +/* + * So this sync is basically a relative reset of S to 0. + * + * So with 2 queues, when one goes idle, we drop them both to 0 and one + * then increases due to not being idle, and the idle one builds up lag to + * get re-elected. So far so simple, right? + * + * When there's 3, we can have the situation where 2 run and one is idle, + * we sync to 0 and let the idle one build up lag to get re-election. Now + * suppose another one also drops idle. At this point dropping all to 0 + * again would destroy the built-up lag from the queue that was already + * idle, not good. + * + * So instead of syncing everything, we can: + * + * less := !((s64)(s_a - s_b) <= 0) + * + * (v_a - S_a) - (v_b - S_b) == v_a - v_b - S_a + S_b + * == v_a - (v_b - S_a + S_b) + * + * IOW, we can recast the (lag) comparison to a one-sided difference. + * So if then, instead of syncing the whole queue, sync the idle queue + * against the active queue with S_a + S_b at the point where we sync. + * + * (XXX consider the implication of living in a cyclic group: N / 2^n N) + * + * This gives us means of syncing single queues against the active queue, + * and for already idle queues to preserve their build-up lag. + * + * Of course, then we get the situation where there's 2 active and one + * going idle, who do we pick to sync against? Theory would have us sync + * against the combined S, but as we've already demonstrated, there is no + * such thing in infeasible weight scenarios. + * + * One thing I've considered; and this is where that core_active rudiment + * came from, is having active queues sync up between themselves after + * every tick. This limits the observed divergence due to the work + * conservancy. + * + * On top of that, we can improve upon things by moving away from our + * horrible (10) hack and moving to (9) and employing (13) here. + */ + /* * se_fi_update - Update the cfs_rq->min_vruntime_fi in a CFS hierarchy if needed. */ -- 2.34.1