[CST-2] CSM 98.9.3?

[Phebe Mann]

I can't claim that this is a good answer, but that's what I've written
down


(b) In outline for the M/M/1 case:

consider a brief interval of time, \delta_t, sufficiently small
that we don't have to worry about multiple jobs completing during the
interval.  The probability that the server is busy during this interval
is simply the utilization, \rho = \lambda / \mu.  If the server is
busy then a job will complete with probability \mu * \delta_t.
If the server is idle then evidently no job will complete.  Therefore
overall we'd expect \rho*\mu*\delta_t jobs to complete, i.e.
\lambda * \delta_t.  This is the same as the number of jobs we'd
expect to arrive during that interval, hence the departure process
is the same as the arrivals process.

HTH
Phebe

On Sun, 27 May 2001, garan wrote:

> Anyone got a agood answer I can look at to the proof part and the rest, I
> am rubbish.
>
> Garan
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> http://thor.cam.ac.uk/~gwlj2
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