[Tccc] Jackson Network and Queueing Theory
Prof. Victor Li
vli
Thu Nov 17 06:48:58 EST 2011
Dear Lachlan,
Please see our replies below.
>
> On 16 November 2011 01:10, Prof. Victor Li <vli at eee.hku.hk> wrote:
>>
>> Let X represent P(B) = 1. Let Y represent P(S) = 0.
>> 1) Let the value of X be T (denote this by X = T).
>
> You are giving two definitions of X. You can't do that. Otherwise
> you could prove that black = white by saying "Let X = black. Let X =
> white. Then black = X = white".
>
> Please choose your definition of X, and then we can continue the
> discussion.
We do not understand why you are confused about our notations.
We use X and Y to represent
propositions as in propositional logic, as shown, for example, by saying
"let X represent P(B) = 1 and let Y represent P(S) = 0".
A proposition has its truth-values T and F. We use, for example,
X = T to mean "the value of X is T". Such notations do not necessarily
in any way imply that "let X represent" means "X = ".
When one uses a real-valued variable, say W, to represent a waiting time
such that W can take different values,
and uses W = 0 to mean that 0 is one of the values of W,
one does not give two definitions of W. If W can take infinitely many
different values, does one give infinitely many definitions of W?
Of course not.
Although we believe our notations are clear, to avoid any misunderstanding,
in the following discussion we shall use X : = [p] to mean
"X represents p" where p is a proposition. For example,
X : = [P(B) = 1] means "X represents P(B) = 1".
>
>> 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.
>
> Of course V(T,F)=F. I don't think I've ever disputed that (even
> though I've said many wrong things in this thread ):
>
> My argument again is:
> a) For some queues in G, (P(B)=1)=F.
> b) if X=F then V(X,F)=V(X,T)=T.
> c) Hence, there exists a queue in G, such that V(P(B)=1, P(S)=0) = T.
> d) Hence, it is not true that for all queues in G, V(P(B)=1, P(S)=0) =
> F.
>
> Please tell me which step you disagree with.
To avoid any confusion, let us follow the conventions below.
i) To determine the value
of V(X, Y), we must first specify X and Y.
ii) In any expression of the form V(A, B) = C,
A, B, and C must be truth values, rather than propositional variables.
For example, V(X, F) = F is not allowed because X is a propositional
variable.
In your argument, suppose that you mean
in a), b) , and c), X : = [P(B) = 1] and X = F;
in b), Y: = [P(S) = 0] and Y = T or F;
in c), for X : = [P(B) = 1] and Y: = [P(S) = 0]. V(F, T) = T and
V(F, F) = T;
in d) X : = [P(B) = 1], Y: = [P(S) = 0], X = T, Y = F and
V(T, F) = F.
You say that for X and Y as specified in d),
V(T, F) = F is not true for all the queues in G because
for X and Y as specified in c) V(F, T) = V(F, F) = T.
Does this mean that you are using the queue specified
in c) as a counterexample to
V(T, F) = F where X : = [P(B) = 1], Y: = [P(S) = 0]?
If the answer is yes, then such a counterexample is invalid
as explained below.
Implications (such as V(T, F) = F with X and Y we specified)
are typically used to express mathematical statements.
If one says "there is a counterexample to an implication" then the
counterexample must satisfy the assumption of the implication for
the counterexample to make sense. If an example does not
satisfy the assumption of an implication, then the example fails to be
a counterexample to the implication, and hence does not make
the implication invalid. It seems that you are using a false
counterexample to argue against an implication.
If you are not using the
queue as a counterexample,
then your argument is irrelevant to V(T, F) = F
with X : = [P(B) = 1], Y: = [P(S) = 0].
Return to your conclusion given in d).
You say that it is not true for all queues in G, V(T, F) = F
with X : = [P(B) = 1], Y: = [P(S) = 0].
Based on our explanation above,
your conclusion appears to be incorrect. The fact is,
there are queues in G that do not satisfy the assumption
X = T with X : = [P(B) = 1] of V(T, F) = F
where Y: = [P(S) = 0] with Y = F. This does not necessarily
mean that V(T, F) = F with X : = [P(B) = 1], Y: = [P(S) = 0]
is incorrect. Suppose that one wants to
use a theorem to solve a problem. But the
problem does not satisfy the assumption of the theorem.
Does this mean that the theorem is flawed? Clearly not.
The theorem is still true.
>
> You are right that my comments are about cases where V(u,z)=T. I
> agree this is irrelevant to V(T,F)=F, but it is central to the flaw in
> the first sentence of your proof: for particular statements u,z, you
> claim that V(u,z)=F for all queues, and I have argued that V(u,z)=T
> for some queues.
>
First, by our definition, both u and z are in {T, F} representing
the values of statements. Neither u nor z itself is a statement.
Second, as we have just pointed out, if you use "some queues"
as counterexamples, then they are false counterexamples. If
"some queues" are not counterexamples, then they are irrelevant
to the issue.
>> 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.
>
> Fair enough. For clarity, I propose we reserve the word "let" for
> definitions, rather to mean "if this happens to be true".
>
> Cheers,
> Lachlan
>
>
> --
> 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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Best regards,
Guang-Liang and Victor
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