On Sun, Aug 10, 2008 at 12:21 PM, Mulyadi Santosa <mulyadi.santosa@xxxxxxxxx> wrote: > Just trying to help the best I can... > > > On Mon, Aug 4, 2008 at 2:11 AM, Sukanto Ghosh > <sukanto.cse.iitb@xxxxxxxxx> wrote: >> 1. What is 'fair_clock' ? >> (Ingo says it is 'CPU time a runnable task would have fairly gotten, >> had it been runnable during that *time* ') >> >> How is this calculated ? >> >> (my understanding: suppose there are 2 runnable tasks for 4ms and in >> the next 4ms there are 4 runnable tasks, then fair_clock would be >> incremented by 3ms in this 8ms period ? >> >> If this is correct then, what is *that time* ? is it the 8ms period ? ) > > hard to say. My own understanding is like this: > > CFS learns that there are now 2 runnable tasks, each has 4 ms time > slice (let's calle the tasks A and B). Assuming none of them were > blocking or doing yield(), they use 4ms fully...thus 8ms is the total > duration of the 1st iteration. > > > Then on 2nd iteration, comes 2 new runnable task. Since previously > total interval is 8ms, making now to 16ms won't make the tasks run > fair. Thus, I think what CFS does is creating somekind of "deadline". > each tasks could still own 4 ms time slice, but will be preempted > after 2ms. That way, A will run for 2ms, B, C and then D also runs for > 2ms. Then they will spent the rest of their 2ms time slice for 3rd > iteration. > > I think this is what "fair clock" means... > fair_clock is a variable associated with every runqueue/rbtree. so in your example what do you think would be the value of the fair_clock for this runqueue ? >> 4. When is a running task preempted ? Or, how is the length of the >> timeslices calculated ? > > well, process is preempted when it is blocking...or exhaust it's time > slice...or entering what I call "scheduling point" i.e going to user > space from kernel space after syscall. > > But in CFS context, i think it is also preempted when its "fair clock" > is expired.... > See, if the 'rq->fair_clock - p->wait_runtime' is used as an index into the rbtree, the timeslice for the leftmost task (pl) will be equal to (pl->wait_runtime - pn->wait_runtime)/2 (pn being the next to the leftmost task) just before it was scheduled (as rq->fairclock is same for both of them). But what if a task is added to the runqueue/rbtree in the meantime. -- Regards, Sukanto Ghosh -- To unsubscribe from this list: send an email with "unsubscribe kernelnewbies" to ecartis@xxxxxxxxxxxx Please read the FAQ at http://kernelnewbies.org/FAQ
