cse123A  Probem set 4: More problems from Chapter3 Datalink Layer
			For Discussion and posting on the web


               
       Problem 12.  The propagation time for a 300 km long trunk is 18 msec.  T1
       speed is 1.536  Mbps and a 64 byte  frame  takes  0.3  msec.  Round  trip       
       (frame  +  ack)  takes   18+0.3+18=   36.3  msec  (ack  is  small  size).
       Trans,itter  must have  window  size to keep going for 36.3 mesc and with
       0.3 msec per frame, it takes 36.3/0.3=121 frames to fill the pipe.  seven
       bit sequence numbers would do (Protocol 5 - go back n).



       Problem 14.  The protocol will be wrong  (fails).  Suppose 3 bit sequence
       numbers are in use.  Consider the following scenario.

       A sent 7; B gets it and sends  piggybacked  Ack; A gets the Ack and sends
       frames 0 thru 6 all of which  get lost; B times out and  retransmits  its
       current frame with the Ack 7; when A receives this, it has:

       r.ack=7,  AckExpected=0,  NextFrameTosend=7  The  between as given in the
       problem  would return  TRUE,  causing A to take that the lost frames were
       acknowledged by B.  QED


       Problem 16.  This is the only place where incoming  acknowledgements  are
       processed.  Without  this, the sender  would keep timing out for ever and
       make no progress - the protocol fails.

       
       Problem17.  This would  defeat the purpose of having  NAKs and  selective
       repeat/retransmission will be on timeouts.  The performance would degrade
       though  correctness  of the  protocol  is  not  affected.  NAKs  are  not
       essential.