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Maximum number of stations on IEEE 802.3 networks

IEEE 802.3-2005 refers in sections 9.1 (Repeater unit for 10 Mb/s baseband networks), 27.1.1.1 (100 Mb/s) and 41.1.1.1 (1000 Mb/s) to an "overall limit of 1024 stations on a network".

This value's origin, however, is not explained at any place in the standard; the only rationale I can think of would be the backoff timing described in section 4.2.3.2.5, with a random delay between 0 and 2^k (with k=min(retry;10)).

On the other hand, collisions between the backoff timers may already occur with fewer stations online; and many more stations might be present on a network unless more than 1024 of them try to simultaneously start a transmission.

Thanks for any hints,

Dominik Neubauer

PS: Don't worry, the question is purely academic, I will NOT set up networks with that many stations within a single segment ;-)

Reply to
Dominik Neubauer
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This is discussed periodically on this NG. I think you're correct. The limit is not a very credible number, it relates only to the backoff timer of half duplex Ethernets, and even then it doesn't make a lot of sense.

Even if there are a total of 1024 different values for backoff timers, there is no guarantee whatever that two or more nodes in a 1024-node network will not pick exactly the same value. And more importantly, if there are far fewer than 1024 nodes in a single Ethernet, it is still possible that two or more nodes will select the same backoff value anyway. But I suppose you can say that if there are more than 1024 stations, and they all try to get access at the same time, you are guaranteed to always have two or more colliding.

It seems clear that max number of nodes in a single network would be a function of traffic level and traffic patterns, and could be far less than 1024.

I suppose if you have to pick an arbitrary number which looks large,

1024 is as good as any.

Bert

Reply to
Albert Manfredi

Yes, the original limit of 1024 stations was based on the backoff algorithm, as you both note.

Agreed, although we didn't fully recognize this at the time the number was selected (1979-80).

Of course that is true, but that problem is dealt with by having multiple backoffs. The problem we were addressing was where *every station* (not just two) was trying to transmit simultaneously. With a backoff selection range of 0-1023, if there are more than 1024 stations sharing the LAN, it becomes unlikely for the collision to ever resolve.

That was our thinking at the time.

Today, the issue is moot. Remember, in all full-duplex (and most switched, half-duplex) LANs, there are only two stations on any Ethernet, regardless of the number of ports on the switch or the total number of stations in the catenet.

-- Rich Seifert Networks and Communications Consulting 21885 Bear Creek Way (408) 395-5700 Los Gatos, CA 95033 (408) 228-0803 FAX

Send replies to: usenet at richseifert dot com

Reply to
Rich Seifert

(snip)

Consider the case where there have already been enough collisions such that 1024 stations all have their counters on 1024. Each chooses a number between 0 and 1023, how many will choose 0?

On the average, 1024/1024 or 1 will choose zero, sometimes more, sometimes less.

I believe it is Poisson, with y=1024/1024=1, so the number of stations, n, that choose zero is

P(n)=y**n exp(-y)/n! = exp(-1)/n!

0 1 2 3 4 5 1/e 1/e 1/2e 1/6e 1/24e 1/120e ...

The probability that one station chooses zero is 1/e, a little less than 1/2. If no stations choose 0, then how many choose 1? 1/e.

The sum, then, should be the sum of exp(-n), 1/(e-1), or about 0.58

With N stations all trying to transmit with 1024 slots, the probability that only one chooses zero,

P(1)=(N/1024) exp(-N/1024)

I believe the maximum probability of only one station choosing 0 out of 1024 is with 1024 stations trying to transmit.

-- glen

Reply to
glen herrmannsfeldt
  • Rich Seifert:

Thank you, Rich and Albert, for the good answers. That's pretty much what I expected (and what a co-worker of mine wouldn't believe without 'proper sources').

Kind regards,

Dominik Neubauer

Reply to
Dominik Neubauer

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