[43186] in North American Network Operators' Group
RE: Long haul latency calculation?
daemon@ATHENA.MIT.EDU (Martin, Christian)
Mon Oct 1 21:14:47 2001
Message-ID: <94B9091E1149D411A45C00508BACEB359CDF22@entmail.gnilink.com>
From: "Martin, Christian" <cmartin@gnilink.net>
To: "'chris@bblabs.com'" <chris@bblabs.com>, nanog@merit.edu
Date: Mon, 1 Oct 2001 21:13:56 -0400
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Christopher,
The factors influencing latency are propagation delay (Pd), transmission
delay (Td), and queueing or processing delay (Qd). For simplicity, assume
Qd is negligible, then L = Pd + Td. Pd, as suggested by others, is:
distance/.66c
Transmission delay is the time it takes to transmit X bits over a link of
bandwidth Y, without acknowledgement. For the single packet case, this
reduces to a worst case scenario of the MTU/Bandwidth. For TCP
applications, assuming a fully realized window WIN = (min{cwnd, rwnd}), this
reduces to WIN/BW.
Since it is simpler to dicuss the MTU case, you get the folowwing formula:
L = ((MTU/BW) + (dist/(.66c)))
So, given a T1 that traverses 4000 meters and an MTU of 1500 bytes (8000
bits), you get a one-way Latency:
L = ((8000/1.536E6) + (4e6/(.66*3E8)) = 25.4 msec.
For an OC-48
L = ((8000/2.448E9) + (4e6/(.66*3E8)) = 20.2 msec
For 2900 mile circuit, you get
L = ((8000/1.536E6) + (2900*1.6e3/(.66*3E8)) = 28.2 msec
Note that distance begins to dominate the delay as distance increases. That
is, a short OC-48 transmits 1500 bytes much faster than a T1, but an OC-48
to Jupiter transmits the data at nearly the same rate as a T1 (both take a
really long time!)
chris
> -----Original Message-----
> From: Christopher Wolff [mailto:chris@bblabs.com]
> Sent: Sunday, September 30, 2001 10:09 PM
> To: nanog@merit.edu
> Subject: Long haul latency calculation?
>
>
>
> Dear Nanog:
>
> I was wondering if there is a benchmark for long-haul circuit
> latency... For example if I had a T1 circuit with 2900 miles
> between the two end-points (and assuming the provider is best
> case scenario) can I do something like (miles*latencyfactor)
> = 5 ms for 2900 miles?
>
> Thanks,
> Christopher
>
