Re: [ns] average end-to-end delay per packet

[Sarafidis Iordanis <eem2is@surrey.ac.uk>]

As u can see it is a bit difficult to play with mean values in statistics :)
You have to take into account what sort of distribution is giving you this
mean values that u have! If your packets' length "x" follows  the
exponential distribution (which is the most probable, since u talk about
average length) and the mean length of the packet is E(x)=4kb, then you have
to know that: var(x)=E(x)^2 (based on the known in statistics formula:
var(x)=E(x^2)-E(x)^2).
now the total delay is: mean time to wait for other queued packets(if any)
+mean time of server to become free from present job (if any)+meam time for
this message to be send + time to travel in the link (asume 10ms)
So u have also to take into account two other factors (I have the feeling
that u overload the link):
m = rate of packets send away from the node
l = rate of packet arrival at the node (u do not mention that, and that is a
very important factor in queueing theory)
For example, mean time to wait for other queued packets=(mean time of server
to become free from present job)*l/(m-l)
BUT mean time of server to become free from present job=l*(mean squared
value of service duration t)/2  #formula derived from queueing theory
IF the distribution of packet length IS EXPONENTIAL then service duration t
is exponential, so from  what I told in the beginning we conclude:
mean squared value of service duration E(t^2)=var(t)=2E(t)^2
( E(t^2)=E(t)^2 is the case when we have fixed packet length; this could
lead to mistakes)
PLEASE DOUBLE CHECK what i have writen as it is early in the morning and I
have to sleep (eventually - tomorrow i am traveling :)

PS I consider an infinite queue. If the queue is limited (in fact, if you
have dropped packets) the approach should be different. (Look at the State
Machine method to have results as well) Have also in mind that I didnot have
time to look at Lloyd's script....
Good luck,
---------------------------------------------------
Sarafidis Iordanis
University of Surrey
---------------------------------------------------
----- Original Message -----
From: Manimaran Selvaraj <ms8@ece.msstate.edu>
To: <ns-users@ISI.EDU>
Sent: Wednesday, August 01, 2001 10:47 PM
Subject: [ns] average end-to-end delay per packet


>
> Hi
> I am using L.Wood's awk script to calculate end to end delay of packet.
> awk script given at:
> http://www.isi.edu/nsnam/archive/ns-users/webarch/1999/msg02093.html
>
> So can i assume the average end-to-end delay of a particular flow as =
> (sum of delay experienced by each pkt making up the flow)/number of pkts.
>
> Or if i am wrong, how can average end-to-end delay be calculated from
> trace files and how can this be verified by hand? Say if i have a 1Mb
> link and prop delay is 10ms and the average packet size is 500bytes.
> Then delay = (500*8)/1Mb + 10ms = 0.014 sec or 14ms.
>
> But using awk script and my assumption stated above i get different
> results for average ene-to-end delay compared to above calculation.
>
> Could someone tell me as to where i am committing a mistake?
>
> -selvaraj
>