Re: [uml-devel] The UML scheduler

[Jeff Dike <jdike@addtoit.com>]

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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