[Tccc] Jackson Network and Queueing Theory
Prof. Victor Li
vli
Tue Nov 15 09:10:42 EST 2011
Dear Lachlan,
For convenience of discussion, we shall continue using the
notations in our previous e-mail.
Let X represent P(B) = 1.
Let Y represent P(S) = 0.
To see V(T, F) = F, one may proceed as follows:
1) Let the value of X be T (denote this by X = T).
2) Find a value (denoted by z) in {T, F} for Y,
such that Y = z is compatible with X = T.
3) Since z = F, one sees V(T, F) = F.
If one does not follow the above line of reasoning,
then one may not be able to see V(T, F) = F.
However, the inability to see V(T, F) = F does not
necessarily justify one's denial of V(T, F) = F.
In your comments based on an M/M/1 queue,
there seems to be a
misunderstanding about V(T, F) = F.
To see V(T, F) = F, one assumes X = T.
Even if X = T does not hold, V(T, F) = F
is still correct. Moreover,
your comment concerns
only values (X, Y) = (u, z)
where u and z are in {T, F} such that
V(u, z) = T. This makes the comment
irrelevant to V(T, F) = F.
As to the second point raised in your e-mail,
we would like to address the difference between
the contexts of 'If one lets the value
of "P(B) = 1" be "true" ' in our previous e-mail
and the context of "Thus P(B) = 1 is true"
in line 8 of the proof.
In the former we discuss V(T, F) = F.
In the latter we prove P(B) = 1 for a stable queue
as shown by the first 8 lines of the proof.
Best regards,
Guang-Liang and Victor
-----Original Message-----
From: Lachlan Andrew
Sent: Monday, November 14, 2011 6:29 PM
To: Prof. Victor Li
Cc: Flaminio Borgonovo ; tccc at lists.cs.columbia.edu ; glli at eee.hku.hk
Subject: Re: [Tccc] Jackson Network and Queueing Theory
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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