[NTLUG:Discuss] RE: Switch Question -- timing and Fast Ethernet

[Bryan J. Smith]

On Sun, 2004-11-14 at 16:02, Paul Ingendorf wrote:
> I understand the electrical specs.  I'm not one of those people who thinks
> manchester encoding has something to do with secret societies.

Understood.  You're a very knowledgeable and proper technologist, unlike
many others.

> I just don't understand why there is 5-4-3 for 10MB and with 100Mb there
> only appears to be a single hub and they cannot be chained.

First off, you _can_ chain exactly _one_ 100 hub to another.  But only
if three is no transceiver anywhere (no media conversion), be it in
either hub or at a station.  This is where the Class I / Class II
terminology comes in (see below).

Secondly, again, it has to do with rates.  This is a good exercise to
refresh myself with.

The speed of light is too slow for established Ethernet timing
considerations.  Because of how Carrier Sense Multiple Access /
Collision Detect (CSMA/CD) aka "Ethernet" works (especially the
collision detect which is really more of a "collision assume" ;-), if
signals are not received in an alloted time, these assumptions are made.

Ideally the speed of light is roughly 300,000,000m/s = 0.3m/ns.
So figure it takes 5ns/meter to go through UTP or fiber.
So given 100m, this is 0.5us or 500ns per segment of UTP or fiber for a
full sequence of 1500 byte frames.
Also remember that the timing my accommodate a full _round_trip_ to the
end node and back (especially if there is a collision), so it's 2x
total.

Given _modern_ clocked boolean logic (CBL) and PHY analog circuitry,
figure 0.5um or 500ns for a repeat.  Half that for a common end-node (or
media transceiver), 0.25um or 250ns.  Again, remember the trip is
_round_trip_, so 2x total also.

10Base is 10Mbps at 1500 bytes frames ~ 14,400 frames/second
100Base is 100Mbps at 1500 byte frames ~ 144,000 frames/second

That results in 70um or 70,000ns allotted timing for each 10Base frame.
That results in 7um or 7,000ns allotted timing for for 100Base frame.

4 hops means up to 6 segments with 5 repeats between 2 nodes.
  6 * 500 + 5 * 500 + 2 * 250 = 6,500ns
Now figure the round-trip:  2 * 6,500ns = 13,000ns

With modern combinational boolean logic, that's well under 70,000ns. 
Much of the old 5-4-3 rule is because older switches were much slower at
repeating.  But still there was a lot of play in there, hence why you
could typically use even more.

1 hop means up to 3 segments with 2 repeats between 2 nodes.
  3 * 500 + 2 * 500 + 3 * 250 = 3,250ns
Now figure the round-trip:  2 * 3,250ns = 6,500ns

That's very close to the 7,000ns timing allotted for each 100Base frame.


-- 
Bryan J. Smith                                    b.j.smith at ieee.org 
-------------------------------------------------------------------- 
Subtotal Cost of Ownership (SCO) for Windows being less than Linux
Total Cost of Ownership (TCO) assumes experts for the former, costly
retraining for the latter, omitted "software assurance" costs in 
compatible desktop OS/apps for the former, no free/legacy reuse for
latter, and no basic security, patch or downtime comparison at all.