question about 100mbits/sec

[Chris]

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

Hi,

Assume that several packets with a total amount of 100 mbits (not in bytes
to simplify) are send together . Assume that the theoretical speed of 100
Mbits/sec is a fact.

Does it mean that they are sent all together (in a kind of parallel
channels) and that the total broadband is busy for a second or are they sent
one after one but the sum of the busy time of all the packets will be one
second?

Thanks
Chris

 

[Guest]

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

"Chris" <sdqfsd@qdvqsd.cv> wrote:
>Assume that several packets with a total amount of 100 mbits (not in bytes
>to simplify) are send together . Assume that the theoretical speed of 100
>Mbits/sec is a fact.

They are not 'sent together', they are queued up for transmission.
100 megabits per second is indeed a fact, it's the wire speed.

>Does it mean that they are sent all together (in a kind of parallel
>channels) and that the total broadband is busy for a second or are they sent
>one after one but the sum of the busy time of all the packets will be one
>second?

The latter is true. Either way (plus or minus overhead, collisions,
etc) your LAN will be busy for one second.

 

[Guest]

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

In article <TrWdnbT-38I7lCDfRVnygQ@scarlet.biz>,
Chris <sdqfsd@qdvqsd.cv> wrote:
:Assume that several packets with a total amount of 100 mbits (not in bytes
:to simplify) are send together . Assume that the theoretical speed of 100
:Mbits/sec is a fact.

oes it mean that they are sent all together (in a kind of parallel
:channels) and that the total broadband is busy for a second or are they sent
ne after one but the sum of the busy time of all the packets will be one
:second?

Neither -- the time required to send 100 Mbits of packets over
a 100 Mbit/s link is more than 1 second, due to the mandatory
IFG (intra-frame gap) of 96 bit-times. Even if you include the size of
the preamble and CRC in your 100 Mbit count, during the IFG nothing
is being sent. Including preamble and CRC in the bit counts, you
can get slightly over 99 Mbits through in one seconds of 100 Mbit/s.


Different signalling methods are used over different media, some
involving discrete bits and some involving phase analysis of
particular sample points of analog waveforms. In the analog case,
considering the error correction methods and other fine details,
bits are not transmitted independantly of each other but rather in
small groups. (The interconnection is noticably stronger for gigabit.)
Thus the question cannot be resolved without reference to a particular
media -- and to a particular definition of what it means to
transit bits in parallel.


There are numerous good sources of information on the 'net. One
with a useful (but not up-to-date) overview is
http://ckp.made-it.com/ieee8023.html
--
"I want to make sure [a user] can't get through ... an online
experience without hitting a Microsoft ad"
-- Steve Ballmer [Microsoft Chief Executive]

 

[Chris]

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

thanks both



"Walter Roberson" <roberson@ibd.nrc-cnrc.gc.ca> wrote in message
news:d9k2im$efn$1@canopus.cc.umanitoba.ca...
> In article <TrWdnbT-38I7lCDfRVnygQ@scarlet.biz>,
> Chris <sdqfsd@qdvqsd.cv> wrote:
> :Assume that several packets with a total amount of 100 mbits (not in
bytes
> :to simplify) are send together . Assume that the theoretical speed of 100
> :Mbits/sec is a fact.
>
> oes it mean that they are sent all together (in a kind of parallel
> :channels) and that the total broadband is busy for a second or are they
sent
> ne after one but the sum of the busy time of all the packets will be
one
> :second?
>
> Neither -- the time required to send 100 Mbits of packets over
> a 100 Mbit/s link is more than 1 second, due to the mandatory
> IFG (intra-frame gap) of 96 bit-times. Even if you include the size of
> the preamble and CRC in your 100 Mbit count, during the IFG nothing
> is being sent. Including preamble and CRC in the bit counts, you
> can get slightly over 99 Mbits through in one seconds of 100 Mbit/s.
>
>
> Different signalling methods are used over different media, some
> involving discrete bits and some involving phase analysis of
> particular sample points of analog waveforms. In the analog case,
> considering the error correction methods and other fine details,
> bits are not transmitted independantly of each other but rather in
> small groups. (The interconnection is noticably stronger for gigabit.)
> Thus the question cannot be resolved without reference to a particular
> media -- and to a particular definition of what it means to
> transit bits in parallel.
>
>
> There are numerous good sources of information on the 'net. One
> with a useful (but not up-to-date) overview is
> http://ckp.made-it.com/ieee8023.html
> --
> "I want to make sure [a user] can't get through ... an online
> experience without hitting a Microsoft ad"
> -- Steve Ballmer [Microsoft Chief Executive]