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Re: [HSSG] BER Objective

Pat & Marcus,

I think you will find it easier if we refer to "error ratios" not "error rates" when discussing BER :-)


Pat Thaler wrote:
That equation isn't right. If I've got 10 parallel links, the number of bit errors on the accumulation is the sum of the bit errors on the individual links.
Send a seconds worth of data on 10 10 Gb/s links with BER10G and one will get 10^10 * BER10G bit errors on each one. Therefore one will get a total of
10 * 10^10 * BER10G = 10^11 * BER10G errors during the transmission of 10^11 bits. The BER of the accumulated link is therefore
10^11 * BER10G / 10^11 = BER10G. Your equation would yield BER100G = BER10G^10/10 which isn't the same as BER10G.
It is true that one gets the same bit error rate for the 10 Gig links and the accumulated 100 Gig link but I don't understand what you mean by:
"which means that the error accumulation time is given by the 10G speed, not the 100G speed."
Both error rates are the same so for a given number of bits transmitted one gets the same error rate from sending on 10 parallel 10 Gig links with a given BER as one would get on one 100 Gig link with the same BER.

From: Marcus Duelk [mailto:duelk@xxxxxxxxxx]
Sent: Tuesday, August 29, 2006 10:53 AM
To: STDS-802-3-HSSG@xxxxxxxxxxxxxxxxx
Subject: Re: [HSSG] BER Objective


I guess you have to look at it also from the "packet world". A transmission error
means that TCP does not get an acknowledgment which means that the packet
gets resend. If we target 100 Gb/s (1E11 b/s) and you have a BER of 1E-12
then you have an error every 10 seconds. This is a little too high and will cause
too many packets being resent. I don't know exactly what the right or good
numbers are but I would guess that BER goals in the range of 1E-14 to 1E-15
will be practical. BER 1E-15 will mean an error every 1E4 (10,000) seconds,
that should be good enough for Carrier Ethernet equipment.

The interesting thing about the time it takes to accumulate an error is
that it depends a lot on the PHY. If you use 100G serial PHY then the
numbers I have given above are correct. If you use 10x10G PHY then
the BER accumulate rather at 10G rate, i.e.

BER100G = [BER10G(1)*BER10G(2)*....BER10G(10)]/10

where BER10G(i) is the BER on each of ten 10G lanes. If they
are all the same then BER10G = BER100G, which means that
the error accumulation time is given by the 10G speed, not the 100G