On Fri, 25 Aug 2000, Stuart MacDonald wrote:
> From: "Richard B. Johnson" <root@chaos.analogic.com>
> > On Fri, 25 Aug 2000, Stuart MacDonald wrote:
> >
> > > From: "Chris Swiedler" <ceswiedler@mindspring.com>
> > > > > Let's say the minimum time is 50 cycles:
> > > > >
> > > > > Process A last_cpu = 1
> > > > > Process B last_cpu = 1
> > > > > Process C last_cpu = 1
> > > > >
> > > > > Process C runs for 200 cycles on CPU 1
> > > > > Process C last_cpu = 1
> > > > > Process A runs for 300 cycles on CPU 2
> > > > > Process A last_cpu = 2
> > > > >
> > > > > Process C is running on CPU 1
> > > > > Process C last_cpu = 1
> > > > > Process B runs for 15 cycles on CPU 2 but is interrupted
> > > > > Process B last_cpu = 1 (unaltered)
> > > > >
> > > > > Here we have:
> > > > > Process A last_cpu = 2
> > > > > Process B last_cpu = 1
> > > > > Process C last_cpu = 1
> > > > > C is currenty running on 1
> > > > > Scheduler needs to pick a process for 2
> > > > > A runs on 2
> > > > >
> > > > > C is starved
> > > >
> > > > ??? I don't see how C is starved. C and B have an equal chance of
> being
> > > > scheduled for CPU 1 (barring other factors). Certainly, C won't be
> starved
> > > > in an extreme sense, because we're only adjusting the goodness(), and
> so
> > > > eventually it will be scheduled again.
> > >
> > > Sorry, typo. B is starved. C is already running on 1
> > > and has 185 cycles left.
> > >
> > > Also, I meant starved in that even though B is the
> > > process time-affinity scheduling should choose, it
> > > won't get chosen.
> > >
> > > ..Stu
> > >
> >
> > But this is a 'Unix' system, not VMS! A task that gets interrupted
> > will get the CPU back as soon as the ISR is complete. Since you can't
> > schedule in an interrupt, this rule is absolute. That is true even
> > if a "bottom-half" is queued within the ISR. The bottom-half runs
> > after somebody has either given up the CPU, or has it stolen from
> > them via a context-switch.
> >
> > Are you saying that a task will "switch CPUs" as a result of an
> > interrupt? I don't think that this is allowed to happen because
> > some task that's interrupted isn't going to get interrupted while
> > it's interrupted! Some other task might, but not this one.
>
> Now I know I'm missing something. According to the above two
> paragraphs, "time-affinity" scheduling is already the implicit
> behaviour.
>
I think, as you have already observed, is the implicit behavior.
> But Chris Swiedler was arguing for it.
>
> I was just trying to introduce a counter-example to Chris' proposed
> changes.
>
> Did I misunderstand Chris' original proposal?
>
I don't know. I think there may be some confusion about how Linux/Unix
operates.
In VAX/VMS, you have the problem Chris cited because every interrupt
causes a context-switch. The task that was interrupted is simply put
into the queue of runnable tasks. Also, any time some task makes a
system-call, it also loses out because it is also put into the queue
and won't get the CPU again until the kernel has completed whatever
function the user requested.
This was a nice scheduling idea for slow I/O bound processes like
in the old VAXen days. This was before RAM cache, prefetch buffers,
etc. Also, you didn't have two or more CPUs ready to pounce at any
instant.
Unix, and Unix variants perform kernel functions on behalf of, and
in the context of, the caller. You don't lose the CPU just because
you called getpid(). However, when the kernel can't do something
right away (it's I/O bound), you sleep and somebody else gets the
CPU. Things done in the kernel are done with kernel privs, but
they are done in the context of a caller.
Interrupts just save the caller's registers, load kernel registers,
execute an ISR with kernel privs, reload caller's registers, then
return to the interrupted task.
Cheers,
Dick Johnson
Penguin : Linux version 2.2.15 on an i686 machine (797.90 BogoMips).
"Memory is like gasoline. You use it up when you are running. Of
course you get it all back when you reboot..."; Actual explanation
obtained from the Micro$oft help desk.
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