On Dec 15,  7:56pm, Andrew Gillham wrote:
} Subject: Re: SUP failing, my experience
} > 
} > Sup has been failing for me too.  The symptoms are quite similar to
} > what others have reported; stderr is
} > 
} > SUP: Read error on network: Connection reset by peer
} > SUP: Premature EOF on network input
} > SUP: Network write timed out
} > SUP: Network write timed out
} 
} Doesn't the ftp motd on ftp.netbsd.org say that the T3s are 'burping'
} packets, or something to that effect?  Perhaps that is affecting the
} SUP connection also.  

The comment that someone made about files with lots of NUL bytes having
more trouble than most reminded me of this comp.dcom.telecom article ...

) From: rfranken@cs.umr.edu
) Newsgroups: comp.dcom.telecom
) Subject: Re: Confused About T1 Bandwidths
) Message-ID: <telecom12.763.3@eecs.nwu.edu>
) Date: 5 Oct 92 21:17:30 GMT
) Sender: Telecom@eecs.nwu.edu
) Organization: TELECOM Digest
) Lines: 71
) Approved: Telecom@eecs.nwu.edu
) X-Submissions-To: telecom@eecs.nwu.edu
) X-Administrivia-To: telecom-request@eecs.nwu.edu
) X-Telecom-Digest: Volume 12, Issue 763, Message 3 of 6
) 
) > Data Comm Folks,
) 
) > I am confused about the T1 usable rates.  My understanding is that the
) > T1 lines can come in two speeds:
) 
) > 1. 1,344 Kbps i.e. 24 channels @ 56 Kbps each;  OR
) > 2. 1,536 Kbps i.e. 24 channels @ 64 Kbps each.
)  
) > Now, what do I need to do in order to get the higher 1,536 speed: Is
) > this the function of the terminating CSUs, the phone company circuit
) > or both? The difference is 192 Kbps which is quite a lot and we would
) > like to take advantage of this extra bandwith if possible.
)  
) > Do I need to tell the phone company when I order the circuit that I
) > want the 1,536 Kbps line?
)  
) > My understanding also is that if the two CSUs are connected back to
) > back via an in-house circuit less then 6,000 long they can indeed get
) > the 1,544 Kbps usable bandwith rates since no framing bits are needed.
) 
) A T1 line represents 1's by alternating between high and low voltage
) (called Alternate Mark Inversion) -- that is, 0's are represented by 0
) volts, and 1's are represented by +voltage (if the previous 1 was
) represented by a negative voltage) or by -voltage (if the previous 1
) was represented by positive voltage).  This allows repeaters, which
) must get their timing from the line, to remain in sync over long
) streams of 1 bits (since there will be a voltage change between two
) consecutive 1's).  However, any more than 14 straight zeros, and the
) repeater may lose sync (since there is no voltage change between
) consecutive zeros).  In a T1 carrying voice, 1's can be inserted where
) needed to maintain 1's density (since you won't likely notice the
) difference), but that is obviously not acceptable in a digital
) high-speed data circuit. There are two ways around this:
)  
)  1) Use only 56KBps of each 64KBps DS0 channel.  This means seven out
) of every eight bits are used.  The eighth bit is set to 1, ensuring
) 1's density.
)  
)  2) Use B8ZS coding.  What this does is replace any 00000000 byte with
) 10011010 (I don't think thats the correct bit pattern, but it will
) suffice for this explanation).  Since no 00000000 streams occur,
) sufficient 1's density will be maintained.  But, when the receiver
) received 10011010, how does it know if the original byte really was
) 10011010 or 00000000? When 10011010 is sent in place of 00000000 the
) sender deliberately causes a bi-polar violation on the fifth bit.
) Normally, 10011010 should be sent (voltage wise) as +00-+0-0 or
) -00+-0+0.  If a substitution for 00000000 is occurring, then +00--0+0
) or -00++0-0 is sent.  The receiver then detects this bipolar violation
) and substitutes 00000000.
) 
)  (A third method is to guarantee that your source will never send two
) many zeros consecutively.  This is generally not possible with
) computer data, which could be anything.)
)  
) So, you need to get a circuit from the telephone company that is set
) up for B8ZS.  (If the circuit is a direct line connected by repeaters,
) then you can probably use B8ZS, as most repeaters don't care about
) anything but a voltage transition at least once every 15 bits.  But,
) this is rarely the case.  Usually the telco will have some other
) equipment in the line that will show an error on the line (becasue of
) all the bipolar violations) if it is not configured for B8ZS.
)  
) For a fast serial line, framing bits are never needed (the computers
) take care of figuring out where a byte starts).  If nothing inbetween
) the sender and receiver cares about framing, and you had a CSU that
) would support it, then you could use the full 1.544 (most repeaters
) don't care about framing, but, for example, a DACS probably would).
) 
) 
) Brett    (rfranken@cs.umr.edu)