[CST-2] Comp. Sys. Modelling

[Timothy Hospedales]

In the answer for this question for the lower bound, they say: "Total
service demand of 215ms, and hence require 3 devices at 100ms?"  Can
anyone explain how that follows? Wouldnt you then have 300ms total
service demand?

Also, why does it matter how many devices you have at 100ms? Surely
you always get equally bad performance if any number >= 1 of devices
are operating at the same bottleneck performance?

Tim

> Hmm I just got the answers off the sample solutions he gave at the examples
> class.  I guess you can say that balanced systems with K = 1 still bound the
> throughput of the system, which gives U = 1, thus the throughput is just the
> reciprocal.
> 
> Alvin
> 
> > But that doesn't take into account N or K, which I think it needs to as
> > you'll need to queue for other customers to finish with the device
> > you're waiting for first - it's taken into account in the notes when it
> > considers the optimistic/pessimistic cases like you said, so we get  X =
> > N/(N+3) x 1/Dav or 1/Dmax.
> > 
> > There must be more to it than that, I can't believe that inverting two
> > numbers is worth 7 marks...
> > 
> > Cheerio,
> > 
> > Chris
> > do something lastminute.work
> > 
> > Chris Applegate
> > Room X6, Corpus Christi College, Cambridge, CB2 1RH
> > chris@qwghlm.co.uk / www.qwghlm.co.uk / [Redacted by SRCF sysadmins on request]
> > ICQ 41706821          PGP key available on request
> > 
> > 
> > 
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