Subject: RE: [manet] Question regarding Mobility

From: "Dearlove, Chris (UK)" <[email-masked]>
Date: Wed, 26 Nov 2003 16:40:41 +0000

> Actually, we only need to consider a single node, and show that it is
> uniformly distributed in the square.  Thus, even though Philippe's
> explanation is accurate, I don't think it is necessary to talk about
> a Poisson distribution.   Allowing the single node to travel straight
> through boundaries, we only need to show that it is uniformly
> distributed in whatever "clone" of the square it happens to be in
> at any time.   I think this follows simply because the node is
> initially placed uniformly, and its movements are independent of
> its location.

It doesn't travel through walls, it bounces off them, so we need to
apply some reflections, and we need basic symmetry properties (left/right,
up/down).

Otherwise consider a node which always picks its next destination a
short distance away to the right, regardless of position. It will keep
hitting the right wall, bouncing, doing the same again and always be
close to the right wall, i.e. not uniform.

With this minor tweak, I otherwise agree with you. (I'm assuming you
are talking about the billiard ball model. The torus model of course
is exactly as you say - with however the issue of what a jump means.)

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From: "Richard Ogier" <[email-masked]>
Date: Wed, 26 Nov 2003 10:10:39 -0800

Chris,

> close to the right wall, i.e. not uniform.

Good point.  But I am talking about the probability distribution at any
point in time.  Even if the node never moves, it's position is still
uniform because it was placed that way.  If you want a single node
to eventually travel near all points of the square, that is a different
problem, but I agree that this is a desirable property.
This property will be achieved if the node chooses a new random
direction every T seconds.

Regarding the application of reflections in the "clone" argument (if that's
what you meant), when I said that we need to show the node position
is uniformly distributed in whatever "clone" it happens to be in, the
orientation (reflection) of the clone should not matter, since if the
node is uniform for any reflection of the square, it is uniform for the
original square.  Also, it would be more accurate to say that the
fractional part (i..e, modulo 1) of the node position is uniform in the
unit square.

Richard