| Author |
Message |
Guest
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 Posted:
Fri Dec 02, 2005 11:58 pm Post subject:
Max. no. of stations on a LAN |
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Hi all,
I have a basic question about LAN oerformance.
Given a 1km 10Mbps bus lan using simple TDM with 100bits slots for
Medium access and a propagation speed of 2 x 10^8 m/s, with each
station transmitting at rate R bps.
What would be the max. no. of stations it can accomodate?
Moreover, can anyone point me to any online documents regarding max.
stations and max. throughput in LANs.
Thanks
PL. |
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Walter Roberson
Guest
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| Posted:
Fri Dec 02, 2005 11:58 pm Post subject:
Re: Max. no. of stations on a LAN |
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In article <1133546302.971664.195900@g43g2000cwa.googlegroups.com>,
<parasiteslost@bigfoot.com> wrote:
| Quote: | I have a basic question about LAN oerformance.
Given a 1km 10Mbps bus lan using simple TDM with 100bits slots for
Medium access and a propagation speed of 2 x 10^8 m/s, with each
station transmitting at rate R bps.
What would be the max. no. of stations it can accomodate?
|
You mention TDM. Is it correct that you do not do any bus arbitration
as such? That each station in turn has a time slot to send data in, and
if one station has nothing to send then the slot just doesn't carry any
useful information during that slot? Presumably with there being either
being the ability to send "nothing" during part of a slot, or else with
there being some encoding or protocol signifying "this is idle data"?
Should we also assume that there is a master clock signal controlling
the synchronization, and that that clock signal is carried in such
a way that it does not get skewed beyond tolerance during transmission
or at intermediate switches/hubs/connectors ?
And should we also assume that the stations have the slots pre-assigned,
so we don't have to worry about protocols for automatically
assigning slots to stations (statically or dynamically) ?
Your bus length is 1000 m / (2 x 10^8 m/s) = 5 x 10^-6 seconds long,
multiply by 10 x 10^6 bit/s (10 Mbps) to get a bus latency of 50 bits.
As that is less than your TDM slot, you won't be using the bus as
a storage medium ;-)
So, assuming strict pre-assigned TDM, the problem simplifies quite a bit.
You just need to find the maximum number of stations, N, such that
N * R <= 10 x 10^6. This does, though, presume that when you say
that each station transmits at rate R, that it is always ready to send 100
further bits when its time-slot comes around.
The actual throughput rate achievable would depend upon the details
of the addressing protocols, and of the mechanisms for dealing with
changing destinations mid-slot and so on.
The problem would change considerably if there were mechanisms to
take advantage of unused slots (such as for token-ring), or if
R is an average rate for each station but the data transmission is
not constant speed.
--
If you lie to the compiler, it will get its revenge. -- Henry Spencer |
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Hansang Bae
Guest
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| Posted:
Sat Dec 03, 2005 9:20 am Post subject:
Re: Max. no. of stations on a LAN |
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parasiteslost@bigfoot.com wrote:
| Quote: | Hi all,
I have a basic question about LAN oerformance.
Given a 1km 10Mbps bus lan using simple TDM with 100bits slots for
Medium access and a propagation speed of 2 x 10^8 m/s, with each
station transmitting at rate R bps.
What would be the max. no. of stations it can accomodate?
Moreover, can anyone point me to any online documents regarding max.
stations and max. throughput in LANs.
|
Take home exam?
--
hsb
"Somehow I imagined this experience would be more rewarding" Calvin
**************************ROT13 MY ADDRESS*************************
Due to the volume of email that I receive, I may not not be able to
reply to emails sent to my account. Please post a followup instead.
******************************************************************** |
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PL
Guest
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| Posted:
Sat Dec 03, 2005 5:20 pm Post subject:
Re: Max. no. of stations on a LAN |
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Thanks Walter for your help. I am actually learning on my own and came
across tutorials on this subject on the web.
Here is the original question:
http://www.uwasa.fi/~riku/opetus/arkkitehtuurit/TLT103_2005_E4.pdf
(Question 4)
I guess you are right, the limitation is N * R <= 10 x 10^6
and N * R <= 6..67 x 10^6 if taking into account the max. propagation ? |
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James Knott
Guest
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| Posted:
Sun Dec 04, 2005 3:24 am Post subject:
Re: Max. no. of stations on a LAN |
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parasiteslost@bigfoot.com wrote:
| Quote: | Hi all,
I have a basic question about LAN oerformance.
Given a 1km 10Mbps bus lan using simple TDM with 100bits slots for
Medium access and a propagation speed of 2 x 10^8 m/s, with each
station transmitting at rate R bps.
What would be the max. no. of stations it can accomodate?
|
Do I detect a homework question here? |
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Walter Roberson
Guest
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| Posted:
Tue Dec 06, 2005 6:32 pm Post subject:
Re: Max. no. of stations on a LAN |
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In article <1133610071.066689.128400@g43g2000cwa.googlegroups.com>,
PL <parasiteslost@bigfoot.com> wrote:
| Quote: | I am actually learning on my own and came
across tutorials on this subject on the web.
Here is the original question:
http://www.uwasa.fi/~riku/opetus/arkkitehtuurit/TLT103_2005_E4.pdf
(Question 4)
I guess you are right, the limitation is N * R <= 10 x 10^6
and N * R <= 6..67 x 10^6 if taking into account the max. propagation ?
|
The achievable rate depends upon a number of design decisions about
the bus, such as whether it is simplex or duplex, whether it is a ring
or linear, whether there is a master clock whose timing pulses
for a given slot have to reach the system for that slot before that
system is allowed to transmit, whether there is a master clock which
does a master synchronization periodically but mostly acts to protect
against clock drift with each station being allowed to "predict"
when to send...
For the case you give, 10 Mbit, 100 bit slots, and a bus that works
out to be 50 bits long, then N * R <= 6.67 x 10^6 would correspond
to a simplex ring topology with a master clock signalling the start of
each slot; a simplex linear topology would be N * R < 4.975 x 10^6.
In this case, the transmission layout would be as follows:
Master clock at end A transmits "go-ahead station B" sync pulse at
time T. This pulse takes 50 bit times to reach the other end of the
wire. Station B starts transmitting at the T+51 slot and ends
transmitting with the T+51+100-1 slot. The bits take 50 bit times to
travel back to station A, so the last of the bits arrives in the
T+51+100-1+50 = T+200 slot, so the master clock can signal the next
"go-ahead" starting in the T+201 slot.
Notice this is 1 bit-time later than the T+200 start you might expect,
so the rate becomes 10/2.01 instead of 10/2, unless I have an off-by-1
error somewhere there.
--
Is there any thing whereof it may be said, See, this is new? It hath
been already of old time, which was before us. -- Ecclesiastes |
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