On Tue, Aug 20, 2024 at 10:50 AM Steven Rostedt <rostedt@xxxxxxxxxxx> wrote: > > On Tue, 20 Aug 2024 15:15:55 +0530 > Srikar Dronamraju <srikar@xxxxxxxxxxxxx> wrote: > > > * Suleiman Souhlal <suleiman@xxxxxxxxxx> [2024-08-06 20:11:57]: > > > > > When steal time exceeds the measured delta when updating clock_task, we > > > currently try to catch up the excess in future updates. > > > However, this results in inaccurate run times for the future clock_task > > > measurements, as they end up getting additional steal time that did not > > > actually happen, from the previous excess steal time being paid back. > > > > > > For example, suppose a task in a VM runs for 10ms and had 15ms of steal > > > time reported while it ran. clock_task rightly doesn't advance. Then, a > > > different task runs on the same rq for 10ms without any time stolen. > > > Because of the current catch up mechanism, clock_sched inaccurately ends > > > up advancing by only 5ms instead of 10ms even though there wasn't any > > > actual time stolen. The second task is getting charged for less time > > > than it ran, even though it didn't deserve it. > > > In other words, tasks can end up getting more run time than they should > > > actually get. > > > > > > So, we instead don't make future updates pay back past excess stolen time. > > In other words, If one task had more time stolen from it than it had run, > the excess time is removed from the next task even though it ran for its > entire slot? > > I'm curious, how does a task get queued on the run queue if 100% of it's > time was stolen? That is, how did it get queued if the vCPU wasn't running? The scenario seems plausible to me, like say several tasks were already queued on the RQ (overloaded RQ) before any time stealing happened. Then vCPU is preempted thus stealing time. When vCPU returns to execution, the running task runs without the clock_task advancing and then goes to sleep thus freeing the RQ for other tasks (that were previously queued). Or did I miss something? thanks, - Joel
