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In <NEWTNews.820626346.32539.jrhodes@jrhodespc.nwest.mccaw.com>, jrhodes@jrhodespc.nwest.mccaw.com writes:
>
>In article <4c1q90$n37@mercury.near.net>, <fgoldstein@bbn.com> writes:
>> 
>Jeffrey Rhodes at jcr@creator.nwest.attws.com
>
>ps. with 64kbps DOSBS using V.120 error correction, would the 
>infrequent "robbed" bits in b) be corrected anyway? Half the 
>robbed bits will match the DOSBS data stream, so a V.120 packet
>would usually get 11 out of 12 octets thru unharmed. How long 
>is a V.120 packet and is there a Hamming code included to 
>correct transmission errors?
>
Jeffrey,

Fred and Jim missed part of your question. V.120 packets are usually 
256 bytes max; sometimes shorter, e.g., 128 bytes.
If every sixth byte is robbed, and the robbery is no-corrupting half 
the time, you still don't have much likeliehood of getting more than
12-byte packets through.  The probablility for 128 or 256 is dismal.

Since V.120 is just another HDLC derivative, when a frame fails, it 
will be re-transmitted.  Maybe the same bits will be robbed--maybe 
not--because byte alignment might be different eith respect to the 
superframe (every sixth frame, where robbery occurs) and bit alignment
might also be different (because you did some zero-bit insertion the 
first time).  But the frame length is never shortened (no 'learning' 
here), so re-transmits aren't especially likely to succeed when the 
original has failed.

Laurence V. Marks
IBM Corp. - Research Triangle Park, NC