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Re: Modem rates (was: Re: Rumors)



On Fri, 01 Oct 1999 14:47:36 +0200 (EET), guy keren <choo@actcom.co.il> wrote:

>
>On Thu, 30 Sep 1999, Stanislav Malyshev a.k.a Frodo wrote:
>
>> YE>> I meant 25 Kilobytes per second,
>
>if you mean what you say - then say what you mean (not you, frodo.. but
>rather yoni). bps = bits per second. Bps = bytes per second. there was
>alredy a thread about these conventions few month back. lets try to stick
>with them. b = bit, B = byte.
>
>> In fact, if you count data bytes, it isn't. I don't remember exactly how
>> much bitstakes to pass one byte, but more than 8. 
>
>normally, it'll be 10 bits per byte (1 start bit, one stop bit, and 8 data
>bits in between). but to that you need to add:
>
>1. compression made by the modem.
>2. header compression done by the PPP and CSLIP protocols.
>3. protocol overhead (be that IP's and TCPs headers, etc.).

You forgot another important one:

All modems in the last 10 years use either MNP5 and up, or LAPB.42 (if I
hadn't misspelled that). Anyway, MNP5 use hardware frames, which means that
every 16 characters are packed as one long asyn character with 128 bits (i.e.
1 start bit, 128 data bits, 1 stop bits). Since this is done only at the
modem-to-modem level over the phone line, this is transparent to the user,
which keeps using his normal UART, albeit at higher rates.

For this reason, even if you are sending totally random, non-compressible
data, you must use a DTE speed which is at least 20% faster than your DCE.
i.e. even if you use a DTE speed of 38.4Kb/s on your 33.6Kb/s modem, you are
NOT getting 100% utilization on your modem. i.e.:

(33.6K / (16*8+2) * (16 * (8+2)) = 41.3K

which means you need at least 41.3K/s to saturate your 33.6Kb/s modem.

Ofcourse, this is all true with true external modems. I don't know if internal
modems (old ISA ones, not WinModems) actually respect the UART baud rate
registers or simply ignores them.

Udi

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