[Tccc] Jackson Network and Queueing Theory

Lachlan Andrew lachlan.andrew
Mon Nov 14 05:29:14 EST 2011


Greetings Victor,

What you say is correct, but does not address the argument that I sent
you off-list.  For the benefit of the list, I'll repeat the argument
here.

If  "P(B)=1"  is false  (as it is for many interesting queues, and in
particular for M/M/1 queues), then, as you correctly say,
  V( [P(B)=1],  [P(S)=0] ) = V(F, [P(S)=0]) = T
This contradicts your claim in the first sentence of the proof of
Theorem 1 that  V( [P(B)=1],  [P(S)=0] ) = F  for all queues in  G.


As a second point, you say 'If one lets the value of   "P(B) = 1" be
"true" ', but line 8 of the proof says that  P(B)=1  is a conclusion,
not a hypothesis.

Cheers,
Lachlan

On 14 November 2011 15:51, Prof. Victor Li <vli at eee.hku.hk> wrote:
> Dear Lachlan and Flaminio,
>
> Thanks for your comments. Let V(X, Y) stand for
> "X implies Y", a logical implication statement in general.
> Write (X, Y) = (T, F) to mean "the value of X is T (true) and
> the value of Y is F (false)". Let ?V(T, F) = F represent
> "the value of V(X, Y) is F when (X, Y) = (T, F)".
> Smilarly, ?V(F, F) = V(F, T) = V(T, T) = T.
> In fact, V(T, F) = F and V(F, F) = V(F, T) = V(T, T) = T
> correspond to the truth table values of the implication, and
> V(T, F) = F is completely determined by (X, Y) = (T, F)
> regardless of what ?X or Y means. Any other assumption
> is unnecessary for V(T, F) = F.
> Unless one is reasoning with a different logic,
> V(T, F) = F is just fine both in general and in particular
> when X and Y represent P(B) = 1 and P(S) = 0, respectively.
> If one lets the value of ? "P(B) = 1" be "true", then one must
> let the value of "P(S) = 0" be "false" ?because a queue with a.s.
> bounded waiting time is not unstable.
>
>
> Best regards,
>
> Guang-Liang and Victor



-- 
Lachlan Andrew? Centre for Advanced Internet Architectures (CAIA)
Swinburne University of Technology, Melbourne, Australia
<http://caia.swin.edu.au/cv/landrew>
Ph +61 3 9214 4837




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