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Re: Cable modem is cooomminggg !!!



On Wed, 30 Sep 1998, Yarin Benado wrote:

> I think you all forget somthing very important.
> ALL networking K's are K bit and not K byte
> when you have a 64k you acctually have up to 8K BYTE
> and a 33.6k has somthing about 4.5K BYTE.
> So ALL what you wrote here is incorrect.. there is no THAT difference between
> 64k to 33.6k and ofcourse not 8 times faster from 33.6k BIT to 4k BYTE.
> (every 8-10K BYTE are 1k BIT. 8K of buffer and 1start bit and 1 end bit)

For modem pools etc there is a 1/10 rule which says that there should be 1
modem for every 10 customers who may use a PoP, assuming normal ISP use. 
The number is more accurate as the number of users is larger. This means
that only at most 10% of the users are online at any one time, or else
someone gets a busy tone. Since most users go online between certain hours
(the lemming effect) this is not always true, and some ISPs were forced by
user complaints to put in much more modems than 1:10 on certain PoPs. User
patterns and populations can be made out 'on the run' and corrected for.

For networks connected all the time, similar rules apply. F.ex. a 10 MBps
ethernet hub with 16 workstations on it in a business environment is
loaded with less tham 2 MBps on a busy day. This is only 0.125 MBps
(equivalent to a 2 channel ISDN connection). During this time, the users
never notice the load, they see a full 10 MBps connection most of the time
(assuming a 10 MBps back end). This is due to the usage factor of the
channel for each user. Each user can f.ex. query a database w. 1 packet
and receive an answer of 5 packets. That's 6 packets every 2 seconds, or 6
x 1500 / 2 = 1500 x 3 = 4500 bytes/sec over 10 MBps == 1E6 bytes/sec. So
the channel load for each user is 4E3/1E6 = 0.004 = 0.4%. Say, 1%. 6 users
doing this add up to 16% (worst case), with the average load being MUCH
lower (say, 4%). CSMA/CD makes this more complicated but the figures
remain like this.

For 64k divided by 8 it's more or less like this: they do not mostly see
64/8 = 8 kbit/sec ( 1 kbyte/sec), they MOSTLY see about 64k == 8
kbytes/sec. In fact, a surfing person loads a page about 1x per 20
seconds, with a total of <50 kBytes of contents. That's 50 kbytes / 20 sec
= 2.5 kbytes/sec. BUT the load factor he generates on a 64k connection is
2.5 kBytes / 64 kBits = 2.5 / 8 ~= 0.31 = 31%. So a 64k connection
relates to the modem pool rule as a 3+ modems per 10 users solution,
except there is no busy sound and you can always download heavy stuff at 3
a.m. using cron ;). Methinks that 3:10 at 32k and 24/7/365 for about
$50/month is rather good as deals like this go in this country, EXCEPT for
the shopping mall and personal server limits. 

This is not magic, my numbers may be a little bit off because they do not
rely on real statistics and do not take the 1kbit = 1024 or 1000 bits
controversy into account. However, there are books and manuals that deal
with this. Read them before you complain about how 'bad' this is. Packet
switched data network statistics are very complicated in the real world
(and I know very little about the subject).

sorry for the long post,

Peter