Re: [ck] Re: is RSDL an "unfair" scheduler too?

[jos poortvliet]

Op Saturday 17 March 2007, schreef Ingo Molnar:
> * Con Kolivas <kernel@xxxxxxxxxxx> wrote:
> > Despite the claims to the contrary, RSDL does not have _less_
> > heuristics, it does not have _any_. It's purely entitlement based.
>
> RSDL still has heuristics very much, but this time it's hardcoded into
> the design! Let me demonstrate this via a simple experiment.
>
> in the vanilla scheduler, the heuristics are ontop of a fairly basic
> (and fast) scheduler, they are plain visible and thus 'optional'. In
> RSDL, the heuristics are still present but more hidden and more
> engrained into the design.
>
> But it's easy to demonstrate this under RSDL: consider the following two
> scenarios, which implement precisely the same fundamental computing
> workload (everything running on the same, default nice 0 level):
>
> 1) a single task runs almost all the time and sleeps about 1 msec every
>    100 msecs.
>
>    [ run "while N=1; do N=1; done &" under bash to create such a
>      workload. ]
>
> 2) tasks are in a 'ring' where each runs for 100 msec, sleeps for 1
>    msec and passes the 'token' around to the next task in the ring. (in
>    essence every task will sleep 9900 msecs before getting another run)
>
>    [ run http://redhat.com/~mingo/scheduler-patches/ring-test.c to
>      create this workload. If the 100 tasks default is too much for you
>      then you can run "./ring-test 10" - that will show similar effects.
>    ]
>
> Workload #1 uses 100% of CPU time. Workload #2 uses 99% of CPU time.
> They both do in essence the same thing.
>
> if RSDL had no heuristics at all then if i mixed #1 with #2, both
> workloads would get roughly 50%/50% of the CPU, right? (as happens if i
> mix #1 with #1 - both CPU-intense workloads get half of the CPU)

Isn't RSDL fair to each task? So each of the 101 tasks (1 and 2) gets an equal 
share... That doesn't sound unfair to me. if the current scheduler manages to 
give the single task 10 times more cpu to task one, that wouldn't be fair. 
Unless you want to be fair to a single process, no matter how many threads.

> in reality, in the 'ring workload' case, RSDL will only give about _5%_
> of CPU time to the #1 CPU-intense task, and will give 95% of CPU time to
> the #2 'ring' of tasks. So the distribution of timeslices is
> significantly unfair!
>
> Why? Because RSDL still has heuristics, just elsewhere and more hidden:
> in the "straightforward CPU intense task" case RSDL will 'penalize' the
> task by depleting its quota for running nearly all the time, in the
> "ring of tasks" case the 100 tasks will each run near their priority
> maximum, fed by 'major epoch' events of RSDL, thus they get 'rewarded'
> for seemingly sleeping alot and spreading things out. So RSDL has
> fundamental unfairness built in as well - it's just different from the
> vanilla scheduler.

I don't see RSDL having heuristics here - it just gives each task an equal 
share of CPU. The single task gets 1/101th of cpu, equal to all others...

> 	Ingo

I guess I just don't get it, would the current kernel give 50% cpu to the 
single thread, and 0,5% to each of the 100 other threads?!? How would it do 
that? Does it schedule a process with several threads equal to a process 
having 1 thread, so each gets an equal share of cpu?

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