Maximum maximum packets per second on 10 megabit ethernet ..

Archived from groups: comp.dcom.lans.ethernet (More info?)

Hi,

I am trying to figure out how many maximum sized packets can be
sent/received over 10 megabit ethernet per second in theory (or ideal
circumstances :D)

They say 10 megabit ethernet runs at exactly 10.000.000 bits per second. (?)

The maximum payload size is 1500 bytes.
The overhead is: 12 gap + 8 preamble + 14 header + 4 trailer = 38 bytes.

So the maximum size of an ethernet packet is 1538 bytes.

10.000.000 bits / 8 bits = 1.250.000 bytes

1.250.000 bytes / 1538 bytes = 812,74382314694408322496749024707

I am kinda surprised that it's not a whole number. I was excepting a whole
number for maximum efficiency :D

Am I missing something ? maybe extra overhead bits ?

Bye,
Skybuck.
7 answers Last reply
More about maximum maximum packets megabit ethernet
  1. Archived from groups: comp.dcom.lans.ethernet (More info?)

    "Skybuck Flying" <nospam@hotmail.com> writes:

    >The maximum payload size is 1500 bytes.
    >The overhead is: 12 gap + 8 preamble + 14 header + 4 trailer = 38 bytes.

    Maybe add in 4 byte for a VLAN tag header.

    >1.250.000 bytes / 1538 bytes = 812,74382314694408322496749024707

    >I am kinda surprised that it's not a whole number. I was excepting a whole
    >number for maximum efficiency :D

    Huh? Why would you gain efficiency by having an integral number of
    packets fit into one second of realtime? What does Ethernet know
    about real time seconds? I just don't see it.

    best regards
    Patrick
  2. Archived from groups: comp.dcom.lans.ethernet (More info?)

    The difference in efficiency is because the per-bit overhead
    is higher for the 813th frame. Your analysis accounts for the
    full 38-byte overhead even though only a total of 1144 bytes were
    sent in that frame. The analysis of the website assumes the
    per-bit overhead is always that encountered for a 1538-byte frame.

    To get the same result as the website, you would need to reduce
    your overhead for the last frame to 38/1538*1144 bytes.
    Alternatively, you could go through the analysis for a larger
    interval than 1 second, say 1000000 seconds, where you have a
    steady stream of 1538 byte packets until the very end of the
    measurement interval; i.e. a frame smaller than 1538 bytes is
    sent only at the end of the 1000000th second.

    Nice puzzle :-)

    Anoop

    Skybuck Flying wrote:
    > "Patrick Schaaf" <mailer-daemon@bof.de> wrote in message
    > news:4216f9d5$0$4615$9b622d9e@news.freenet.de...
    > > "Skybuck Flying" <nospam@hotmail.com> writes:
    > >
    > > >The maximum payload size is 1500 bytes.
    > > >The overhead is: 12 gap + 8 preamble + 14 header + 4 trailer = 38
    bytes.
    > >
    > > Maybe add in 4 byte for a VLAN tag header.
    >
    > Nope that gives:
    >
    > 1.250.000 bytes / 1542 bytes = 810.6355382619974059662775616083
    >
    > >
    > > >1.250.000 bytes / 1538 bytes = 812,74382314694408322496749024707
    > >
    > > >I am kinda surprised that it's not a whole number. I was excepting
    a
    > whole
    > > >number for maximum efficiency :D
    > >
    > > Huh? Why would you gain efficiency by having an integral number of
    > > packets fit into one second of realtime? What does Ethernet know
    > > about real time seconds? I just don't see it.
    >
    > The maximum ammount of bits that can be sent per second is 10.000.000
    bits.
    >
    > To minimize overhead it would make sense to send maximum sized
    packets only.
    >
    > In this case 812 packets * 1538 bytes = 1248856 bytes.
    >
    > Or 1248856 * 8 bits = 9990848 bits.
    >
    > The last (813th) packet would have to be:
    >
    > 1.250.000 - 1.248.856 = 1144 bytes.
    >
    > The interval between the first 812 packets would have to be:
    >
    > 1538 bytes * 8 bits = 12304 bits.
    >
    > 1 bit takes 1 second / 10.000.000 bits = 0.0000001 seconds
    >
    > So 12304 bits * 0.0000001 = 0.0012304 seconds
    >
    > or 0.0012304 * 1000 = 1.2304 milliseconds per bit
    >
    > So after (812 packets * 1.2304 milliseconds =) 999.0848 milliseconds
    the
    > 813th packet would have to be sent ;)
    >
    > 1144 * 8 bits = 9152 bits for the 813 th packet.
    >
    > 9152 bits * 0.0000001 = 0.0009152 seconds
    >
    > or (0.0009152 * 1000 =) 0.9152 milliseconds to send it.
    >
    > check: 999.0848 + 0.9152 = 1000 milliseconds ;)
    >
    > Exactly one second ;)
    >
    > Let's see how much real ethernet payload/data can be sent per second
    ;)
    >
    > (812 * 1500) + (1144 - 38) = 1218000 + 1106 = 1219106 bytes
    >
    > 1219106 bytes * 8 bits = 9.752.848 bits of payload ;)
    >
    > (9.752.848 / 10.000.000) * 100% = 97.52848%
    >
    > 97.52848% efficieny ;)
    > 2.47152% overhead :D
    >
    > 1.219.106 / 1024 = 1190.533203125 KByte
    >
    > 1.219.106 / (1024*1024) = 1.1626300811767578125 Mbyte / sec :D
    >
    > The funny thing is that the figures on this website are therefore not
    > correct ;)
    >
    > http://sd.wareonearth.com/~phil/net/overhead/
    >
    > The website reads:
    > "
    > Ethernet Payload data rates are thus:
    > (1500/(38+1500)) * 100 = 97.529258777633289986996098829649 without
    802.1q
    > tags.
    > "
    >
    > Websites 97.5293 % versus my 97.52848% ;)
    >
    > De website is assuming 97.5293 efficiency.
    >
    > Bogus answer: maximum payload bits per second:
    > 0.97529258777633289986996098829649 * 10.000.000 bits =
    > 9752925.8777633289986996098829649 (half a bit? yeah right ;) :D )
    >
    > Real answer: maximum payload bits per second:
    > 9752848 bits
    >
    > Difference:
    > 9752925.8777633289986996098829649 - 9752848 =
    > 77.877763328998699609882964889467 bits.
    >
    > The amazing thing is ;) de website is off by about 77.8 bits ;D
    >
    > Bye,
    > Skybuck.
  3. Archived from groups: comp.dcom.lans.ethernet (More info?)

    "Patrick Schaaf" <mailer-daemon@bof.de> wrote in message
    news:4216f9d5$0$4615$9b622d9e@news.freenet.de...
    > "Skybuck Flying" <nospam@hotmail.com> writes:
    >
    > >The maximum payload size is 1500 bytes.
    > >The overhead is: 12 gap + 8 preamble + 14 header + 4 trailer = 38 bytes.
    >
    > Maybe add in 4 byte for a VLAN tag header.

    Nope that gives:

    1.250.000 bytes / 1542 bytes = 810.6355382619974059662775616083

    >
    > >1.250.000 bytes / 1538 bytes = 812,74382314694408322496749024707
    >
    > >I am kinda surprised that it's not a whole number. I was excepting a
    whole
    > >number for maximum efficiency :D
    >
    > Huh? Why would you gain efficiency by having an integral number of
    > packets fit into one second of realtime? What does Ethernet know
    > about real time seconds? I just don't see it.

    The maximum ammount of bits that can be sent per second is 10.000.000 bits.

    To minimize overhead it would make sense to send maximum sized packets only.

    In this case 812 packets * 1538 bytes = 1248856 bytes.

    Or 1248856 * 8 bits = 9990848 bits.

    The last (813th) packet would have to be:

    1.250.000 - 1.248.856 = 1144 bytes.

    The interval between the first 812 packets would have to be:

    1538 bytes * 8 bits = 12304 bits.

    1 bit takes 1 second / 10.000.000 bits = 0.0000001 seconds

    So 12304 bits * 0.0000001 = 0.0012304 seconds

    or 0.0012304 * 1000 = 1.2304 milliseconds per bit

    So after (812 packets * 1.2304 milliseconds =) 999.0848 milliseconds the
    813th packet would have to be sent ;)

    1144 * 8 bits = 9152 bits for the 813 th packet.

    9152 bits * 0.0000001 = 0.0009152 seconds

    or (0.0009152 * 1000 =) 0.9152 milliseconds to send it.

    check: 999.0848 + 0.9152 = 1000 milliseconds ;)

    Exactly one second ;)

    Let's see how much real ethernet payload/data can be sent per second ;)

    (812 * 1500) + (1144 - 38) = 1218000 + 1106 = 1219106 bytes

    1219106 bytes * 8 bits = 9.752.848 bits of payload ;)

    (9.752.848 / 10.000.000) * 100% = 97.52848%

    97.52848% efficieny ;)
    2.47152% overhead :D

    1.219.106 / 1024 = 1190.533203125 KByte

    1.219.106 / (1024*1024) = 1.1626300811767578125 Mbyte / sec :D

    The funny thing is that the figures on this website are therefore not
    correct ;)

    http://sd.wareonearth.com/~phil/net/overhead/

    The website reads:
    "
    Ethernet Payload data rates are thus:
    (1500/(38+1500)) * 100 = 97.529258777633289986996098829649 without 802.1q
    tags.
    "

    Websites 97.5293 % versus my 97.52848% ;)

    De website is assuming 97.5293 efficiency.

    Bogus answer: maximum payload bits per second:
    0.97529258777633289986996098829649 * 10.000.000 bits =
    9752925.8777633289986996098829649 (half a bit? yeah right ;) :D )

    Real answer: maximum payload bits per second:
    9752848 bits

    Difference:
    9752925.8777633289986996098829649 - 9752848 =
    77.877763328998699609882964889467 bits.

    The amazing thing is ;) de website is off by about 77.8 bits ;D

    Bye,
    Skybuck.
  4. Archived from groups: comp.dcom.lans.ethernet (More info?)

    "Skybuck Flying" <nospam@hotmail.com> writes:

    >> Huh? Why would you gain efficiency by having an integral number of
    >> packets fit into one second of realtime? What does Ethernet know
    >> about real time seconds? I just don't see it.

    >The maximum ammount of bits that can be sent per second is 10.000.000 bits.

    >To minimize overhead it would make sense to send maximum sized packets only.

    >In this case 812 packets * 1538 bytes = 1248856 bytes.

    >Or 1248856 * 8 bits = 9990848 bits.

    >The last (813th) packet would have to be:

    >1.250.000 - 1.248.856 = 1144 bytes.

    Again, what makes one second so special, that would cut the length
    of the 813th packet so it falls on a second-boundary in realtime?

    In reality, the 813th packet will also be full-size, and sending
    all 813 packets will take slightly more than a single second.
    So what? who cares?

    best regards
    Patrick
  5. Archived from groups: comp.dcom.lans.ethernet (More info?)

    Skybuck Flying wrote:

    > I am kinda surprised that it's not a whole number. I was excepting a whole
    > number for maximum efficiency :D
    >
    > Am I missing something ? maybe extra overhead bits ?

    Why would it be a whole number? There was no intended relationship between
    packet size and bit rate. Try using your argument on the size of cars that
    can travel on a highway, to see if it stands up.
  6. Archived from groups: comp.dcom.lans.ethernet (More info?)

    In article <cv6ost$lpt$1@news2.zwoll1.ov.home.nl>,
    Skybuck Flying <nospam@hotmail.com> wrote:
    :I am trying to figure out how many maximum sized packets can be
    :sent/received over 10 megabit ethernet per second in theory (or ideal
    :circumstances :D)

    :They say 10 megabit ethernet runs at exactly 10.000.000 bits per second. (?)

    That's the first mistake. Any ethernet implimentation you are likely
    to encounter runs asynchronous, no synchronous. 1E7 is the -maximum-
    data rate, which will never be achieved in practice. Especially at
    10 megabit ethernet, if no station has anything to send, then the
    line is quiet. When a station has something to send and the line
    has been quiet for at least the inter-frame gap, then the station
    just starts sending, rather than waiting to synchronize with a
    clock.

    Synchronous ethernet is a lot harder, because you have to synchronize
    the bit edges of all the stations on the segment, even though there
    are different propogation delays due to distance, and even though
    there may be repeaters along the way.

    :1.250.000 bytes / 1538 bytes = 812,74382314694408322496749024707

    :I am kinda surprised that it's not a whole number.

    Then you will be even more surprised when you look at 100 megabit
    per second or gigabit speeds, which send multiple bits per symbol
    and which have error correction built in. 100 Mbps for example sends
    in chunks of 4 bits of data, encoded as 5 bits.

    :I was excepting a whole
    :number for maximum efficiency :D

    That statement presumes that all of the systems are running with
    the same clock (to within a fraction of a bit time) and that there
    is something special about 1 second boundaries. Once you give up
    on sychronizing the systems together then you can see that the
    data keeps running, and that expecting a whole number of packets per
    second is about as useful as expecting that on a car that is driving
    at 100 kilometers per hour, that each wheel will make a whole number
    of revolutions per second.
    --
    Live it up, rip it up, why so lazy?
    Give it out, dish it out, let's go crazy, yeah!
    -- Supertramp (The USENET Song)
  7. Archived from groups: comp.dcom.lans.ethernet (More info?)

    Walter Roberson wrote:

    > That's the first mistake. Any ethernet implimentation you are likely
    > to encounter runs asynchronous, no synchronous.

    The packets are asynchronous, but the data within a packet is synchronous.
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