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- Subject
- Posted on
- Ethernet utilization formula per RFC2819
- 01-19-2006
- Thomas Bahls
January 19, 2006, 10:19 pm
Hi folks,
Usually not being so clumsy about maths, I don't get the clue in this
case though it MUST be simple, I guess... RFC2819 presents a formula
that can be used to approximate the utilization of an 10 Mbps Ethernet
link as follows:
Pkts * (9.6 + 6.4) + (Octets * .8)
Utilization = -------------------------------------
Interval * 10,000
What I don't get now seems to be two-fold:
(1) Why do we use both pkts and octets in the formula? Won't octets,
bandwidth, and an interval be enough? What do the pkts help here?
(2) What about the 'magic numbers' 9.6, 6.4, and 0.8? What might be the
source of these figures and what's their purpose? There must be some
obvious one I guess because otherwise one would have summed up 9.6+6.4
in the equation, at least. However, I don't get it anyway.
Is there anybody who would point me to what I am missing apparently, or
just explain the entire thing I am struggeling with? Any idea will be
appreciated!
Thanks,
/Thomas.
--
Thomas Bahls | "Wege entstehen dadurch, daß man sie geht."
Greifswald, Germany | -- Franz Kafka
ICQ #119230485 +++ PGP Key 0x6E70B6AE +++ http://solitaryhiker.net /
Re: Ethernet utilization formula per RFC2819
Thomas Bahls wrote:
Ugrgh. Someone cancelled out a factor of ten.
rewrite as:
Pkts * (96 + 64) + (Octets * 8)
Utilization = -------------------------------------
Interval * 100,000
96 is the number of bit times (minimum) for the interframe gap. 64 is
the number of bits in the preamble + SFD. (The octet count doesn't
include these, so we have to add them in for each packet.)
8 is the number of bits in each octet.
Utilization will be in %, so rewrite as:
Pkts * (96 + 64) + (Octets * 8)
Utilization = ------------------------------------- * 100%
Interval * 10,000,000
The 10000000 in the denominator is the raw bit rate, in bits per
second.
In other words, Utilization (sic) is the number of bits transmitted,
divided by the number of bits that could have been transmitted,
multiplied by 100%.
RFC2819 lists its author as Steve Waldbusser. Perhaps you'd like to
contact him and ask him why he chose to make this hard to understand.
There could be a valid reason!
Regards,
Allan
Re: Ethernet utilization formula per RFC2819
Hi Allan,
On 20 Jan 2006 02:31:02 -0800, allanherriman@hotmail.com wrote:
This makes it quite clear, so clear that one wonders why I didn't get it
myself. I apparently overlooked the fact that we transfer more than the
raw octets starting with dstMAC. The point was nicely disguised by one of
those tricks mathematicians seems to enjoy: They almost 'randomly' divide
and expand entire equations by 'magic' factors. We loved that at
university, kind of. ;-)
I think I'd better not bore Steve with this now I see what the parts of
the equation represent. Instead, I will carefully keep a copy of your
explanation and enjoy the understanding.
Thanks for your prompt f'up (also to Robert),
/Thomas.
--
Thomas Bahls | "Wege entstehen dadurch, daß man sie geht."
Greifswald, Germany | -- Franz Kafka
ICQ #119230485 +++ PGP Key 0x6E70B6AE +++ http://solitaryhiker.net /
Re: Ethernet utilization formula per RFC2819
Because there's a certain amount of overhead per packet (src
& dest MAC, inter frame gap, etc). A network might be very
heavily loaded if all the data is in very small packets.
As the other response, someone has divided top & bottom by
ten. Octets are 8 bits. 96 + 64 would correspond to packet
overheads, IIRC 96 for preamble & MACs, 64 for IFG.
-- Robert
Re: Ethernet utilization formula per RFC2819
Noah Davids wrote:
No, it is that simple. 10Gbps Ethernet also follows the same rule.
The line encodings are different (and have different overhead) but that
is made up for by changing the symbol rate on the line to make the
effective usable bit rate 10M, 100M, 1G, 10G, etc.
The IFGs and preambles also stay the same (nominal) number of bits.
Regards,
Allan
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