Re: [uml-devel] The UML scheduler

[Young Koh <young.koh@gmail.com>]

Thank you for your reply. I have one following question.

in native Linux, a timer tick goes to the kernel, which has the
previliege to save the current context and switch the running process.
But UML kernel is just another user level process, even though it is
tracing the application processes. in UML case, when a timer tick goes
to the UML kernel, how does it stop the running process, which is
another process?

it seems it's a primitive quesion, but it will help me a lot to
understand more about UML. Thank you!

On 5/16/05, Jeff Dike <jdike@addtoit.com> wrote:
> On Mon, May 16, 2005 at 10:30:33AM -0400, Young Koh wrote:
> > suppose there are 2 processes on top of UML and 3 host processes.
> > then, the host kernel sees 5 processes total, so, each of the process
> > will get 20% of the CPU time? is that right?
> 
> In tt mode, there will be 5 processes, but it will be difficult for the 2 UML
> processes to each get 20% of the time, since only one of them will be runnable
> at a time on the host.  This is still unfair because each of them will have
> a higher dynamic priority than the 3 host processes because they are each only
> runnable half the time.
> 
> In skas mode, there is only one host process for all of the UML processes.
> 
> > Or, UML is somehow viewed as a group of processes and UML get 25% of
> > the CPU and each UML got 12.5% of the CPU time? if so, how is it
> > possible?
> 
> In skas mode, this is true.
> 
> > one more question, when the UML kernel wants to schedule its
> > processes, how does it preempt UML processes?
> 
> The same way as any other architecture.  A timer tick comes in, and the
> timer handler calls schedule().
> 
> > this way of scheduling and preemption will be different from TT mode
> > and SKAS mode? if so, how?
> 
> No, exactly the same.
> 
>                                Jeff
>


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