Messages in this thread | | | From | Youssef Esmat <> | Date | Fri, 30 Sep 2022 10:44:39 -0500 | Subject | Re: Sum of weights idea for CFS PI |
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Hi Everyone!
I am not sure we should care about A's sleeping pattern. The case we care about is when A is running or wants to run but can't because it is blocked on C. In that case C should get the weight of A as if A was running.
Ideally this is also a temporary boost since critical sections should be relatively small, so erring on the side of giving C slightly more runtime would be safe I think.
Thanks, Youssef
On Fri, Sep 30, 2022 at 8:49 AM Qais Yousef <qais.yousef@arm.com> wrote: > > Hi Joel > > I'm interested in the topic, if I can be CCed in any future discussions I'd > appreciate it :) > > On 09/29/22 16:38, Joel Fernandes wrote: > > Hi Peter, all, > > > > Just following-up about the idea Peter suggested at LPC22 about sum of weights > > to solve the CFS priority inversion issues using priority inheritance. I am not > > sure if a straight forward summation of the weights of dependencies in the > > chain, is sufficient (or may cause too much unfairness). > > > > I think it will work if all the tasks on CPU are 100% in utilization: > > > > Say if you have 4 tasks (A, B, C, D) running and each one has equal > > weight (W) except for A which has twice the weight (2W). > > So the CPU bandwidth distribution is (assuming all are running): > > A: 2 / 5 > > B, C. D: 1 / 5 > > > > Say out of the 4 tasks, 3 of them are a part of a classical priority > > inversion scenario (A, B and C). > > > > Say now A blocks on a lock and that lock's owner C is running, however now > > because A has blocked, B gets 1/3 bandwidth, where as it should have been > > limited to 1/5. To remedy this, say you give C a weight of 2W. B gets 1/4 > > bandwidth - still not fair since B is eating away CPU bandwidth causing the > > priority inversion we want to remedy. > > > > The correct bandwidth distribution should be (B and D should be unchanged): > > B = 1/5 > > D = 1/5 > > > > C = 3/5 > > > > This means that C's weight should be 3W , and B and D should be W each > > as before. So indeed, C's new weight is its original weight PLUS the > > weight of the A - that's needed to keep the CPU usage of the other > > tasks (B, D) in check so that C makes forward progress on behalf of A and the > > other tasks don't eat into the CPU utilization. > > > > However, I think this will kinda fall apart if A is asleep 50% of the time > > (assume the sleep is because of I/O and unrelated to the PI chain). > > > > Because now if all were running (and assume no PI dependencies), with A being > > 50%, the bandwidth of B, C and D each would be divided into 2 components: > > > > a. when A is running, it would be as above. > > b. but if A was sleeping, B, C, and D would get 1/3. > > > > So on average, B, C and D get: (1/3 + 1/5) / 2 = 8/30. This gives A about 6/30 > > or 1/5 bandwidth. > > The average metric is interesting one. It can be confusing to reason about too. > > I think we have 3 events to take into account here, not 2: > > a. when A is running and NOT blocked on C. > b. when A is running and BLOCKED on C. > c. A is sleeping. > > This means A, B, C and D's shares will be: > > A , B , C , D > a. 2/5, 1/5, 1/5, 1/5 > b. - , 3/5, 1/5, 1/5 > c. - , 1/3, 1/3, 1/3 > > Since A is sleeping for 50%, I don't think we can assume equal distribution for > the 3 events (can't just divide by 3). > > I believe we can assume that > > a. occurs 25% of the time > b. occurs 25% of the time > c. occurs 50% of the time > > I *think* this should provide something more representative. > > > > > But now say A happen to block on a lock that C is holding. You would boost C to > > weight 3W which gives it 3/5 (or 18/30) as we saw above, which is more than what > > C should actually get. > > > > C should get (8/30 + 6/30 = 14/30) AFAICS. > > > > Hopefully one can see that a straight summation of weights is not enough. It > > needs to be something like: > > > > C's new weight = C's original weight + (A's weight) * (A's utilization) > > > > Or something, otherwise the inherited weight may be too much to properly solve it. > > > > Any thoughts on this? You mentioned you had some notes on this and/or proxy > > execution, could you share it? > > I assume we'll be using rt-mutex inheritance property to handle this? If this > was discussed during a talk, I'd appreciate a link to that. > > In the past in OSPM conference we brought up an issue with performance > inversion where a task running on a smaller (slower to be more generic) CPU is > holding the lock and causing massive delays for waiters. This is an artefact of > DVFS. For HMP, there's an additional cause due to the unequal capacities of the > CPUs. > > Proxy execution seems to be the nice solution to all of these problems, but > it's a long way away. I'm interested to learn how this inheritance will be > implemented. And whether there are any userspace conversion issues. i.e: do > we need to convert all locks to rt-mutex locks? > > > Thanks > > -- > Qais Yousef
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