Synchronous transmission, multi-rate bit-stuffing
 with non-integer-multiple rates.

 Example:

  2 data streams:

  stream A: 0.75Mbps
  stream B: 0.33Mbps
  
  1 Mbps = 1 bit / 1 micro second  (\mu s) ->

  stream A: 3 bits / 4 \mu s
  stream B: 1 bit  / 3 \mu s

  In the same time interval:

  stream A: 9 bits / 12 \mu s
  stream B: 4 bits / 12 \mu s

  When using frames with 1-byte-size slots (or cells):
  8 bits per slot -> use time to send 1 byte in stream B as common time.

  stream A: 18 bits / 24 \mu s
  stream B:  8 bits / 24 \mu s

  Upto now, we have only re-written
  stream A: 0.75Mbps
  stream B: 0.33Mbps
  in a different form.

  Stream A is 18/8 = 2.25 times faster than stream B.
  If stream A were a bit faster, it would be 3 times (an integer number)
  as fast as stream B.
  So, let's MAKE stream A 3 times as fast as stream B by creating
  a new stream A' from stream A.
  If stream A' is to be 3 times (24/8) as fast as stream B, it needs to 
  provide 24 bits per 24 \mu s. This can be achieved by "stuffing" the
  bit stream with dummy stuff bits. The stuff bits will be removed 
  on the receiving side to re-constitute the stream A.
  Every 8 \mu s, stream A provides 6 bits. If we stuff 2 bits every
  8 \mu s, we will get 8 bits (1 byte) every 8 \mu s in the stream A'.
  During a period of 24 \mu s we can thus produce 3 bytes (which only 
  contain 18 meaningful bits however).
  So, during 24 \mu s, stream A' provides 3 bytes and stream B provides
  1 byte. 
  Together, this makes for 4 bytes (32 bits) / 24 \mu s or a multiplexed
  bitrate of 1.33 Mbps.