Subject: Re: various questions about Generic QoS Socket Interface

From: [email-masked] (Mark Garrett)
Date: Thu, 18 Aug 1994 09:30:51 -0400

Zheng,

> By average I mean AV(t)=A(0, t)/t, where A(0, t) is the total number of bits 
> that entered into the network during (0, t). 

> Yes, I disagree. R is not the upper bound of the average (as defined above).
> Just imagine that a traffic goes "greedy" (i.e. a bursts of back-to-back
> packets) at t=0. In this case, the upper bound is the speed of the burst
> which is only constrained by the outgoing interface/link. But as long as
> the burst size is smaller than B, it has no effects on R.
> AV(t) can move above or below R. This is why I said "close-to-average".
> Informally, we can say that R is the rate after smoothing B size bursts.

OK, we agree on what average means.  So, to be more precise, the upper
bound on the average is (Rt+B)/t = R+B/t.  Now, for the greedy source
you're right, then R is slightly more than average, by B/t.  However
there is no statistical multiplexing gain to be had with greedy sources.
You allocate them R and they get R.  I think it's safe to say that if
B/t is not insignificant compared to R, then we're talking about a short
datagram, and not something you should be allocating rates to at all.

In general, however, a source is not always greedy, and then you can
have statistical multiplexing.  It was this case I was thinking of.
Here R+B/t (which I considered to be approximately R) is an upper bound
on the average, and the lower bound is zero, so the average could be
anywhere in between.

-Mark

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From: Zheng Wang <[email-masked]>
Date: Thu, 18 Aug 94 15:23:54 +0100

>datagram, and not something you should be allocating rates to at all.

Note that if a source sends a burst of packets at t=0, we have
B/t=line speed (i.e. the highest speed at which an application
can transmit, which can be very high), so AV_max >> R.

Cheers
Zheng