[e2e] on local ethernet throughput?
vjs at calcite.rhyolite.com
Thu Oct 18 06:41:09 PDT 2001
> From: Dan HE <he at enseirb.fr>
> I was confused by one on the questions of the available bandwidth in a
> local segment.
> This question could be very fundemantal but it would be confused.
> Here is a graph that addresses a segment of ethernet topology.
> |A| |B|
> | |
> --------------------------------- 10Mb
> | | |
> |A1| |C| |B1|
> There are five stations sharing a segment of ethernet. every stations
> 10Mb/s interface card. A and A1 are communicating, B and B1 are
> C is an observer.
> (1) how much maximum throughput can be reached between the communicating
> pairs(A-A1, B-B1) at the same time? is it about 10Mb/s for both (A-A1)
> and (B-B1)? or when communicating, both pairs only can share average
> half of 10Mb/s bandwidth?
> is it possible to say that adding both the average throughput of (A-A1)
> and the average throughput of (B-B1) is lager than 10Mb/s!?
If we are talking about an 802.3 10MHz CSMA/CD network instead of
something involving multi-port bridges ("switches" in marketoon talk),
then the total available TCP/IP payload bandwidth is less than 10
Mbit/sec. If all of the stations use classic BEB with fully compliant
802.3 state machines and no BLAM and no "polite" MACs like the venerable
AMD 7990 LANCE and all use subtantial TCP/IP windows, then the total
throughput will be about 6.5 Mbit/sec because of the Ethernet Capture
Effect. If all of the MACs use BLAM instead of BEB, all are "polite"
(delay when they ought to defer), or if all of the TCP windows are
tiny, then the total throughput will approach 9.3 Mbit/sec.
> (2) for the observer, how much bandwidth has been used in this segment
> when the others
> are communicating? because in some adaptive applications based on the
> local segment adaptation, they usually addressed that they are measuring
> the local available bandwidth.
> but I don't know how a C can infer the available bandwidth on the link?
I don't understand that question. If it is intended to ask about the
fairness of CSMA/CD with or without BLAM, then the answer is that
CSMA/CD is very fair. Except for the Capture Effect, which is in some
sense fair, the available bandwidth is fairl distributed among stations
that want to transmit. If there are 2 stations transmitting, each
will have access to about 50% of the available 10 MHz (which is less
than 10 Mbit/sec because of TCP/IP headers, IFG's, and collisions).
If there are 5 stations, as would be the case with a heavy TCP stream
from A to A1 (don't forget TCP ACKs), a heavy TCP stream from B to
B1, and some from C, then each will have access to about 20%.
The most extensive collection of measurements and simulations on
that issue is probably Mart Molle's. The last I heard it was at
That collection also includes discussions of BLAM, the capture
effect, and related issues.
Finally, before giving credence to the old 802.5 and FDDI token ring
salescritter and trade rag exspurt claim that CSMA/CD collisions
consume lots of bandwidth, use a 'scope or a calculator. (This is
only a pre-emptive note. It's nice that so far in this thread that
old silly lie has not been revived.)
Vernon Schryver vjs at rhyolite.com
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