There is an error on page 102 of the ``Weeks 5 and 6'' Noters.

When an ACK with sequence number seq arrives at the source, the 
source_hacked value can be either (seq - 1) or seq .
(seq - 1) is the ``prefered'' and ``usual'' value.
But seq can also happen.

Excercise: give an example of how  it can occur that on arrival of an 
ACK we find that  source_hacked = seq. 
What should we do when that happens?

I did not make this error on purpose. But now that I made it, it is 
a good example to show that protocols need ``proof of correctness''. 

On page 104 I have it right in my notes, but I was unclear in class.

At the destination, if a frame arrives with 
seq == dest_hacked,  the dest  MUST  send  ACK(seq).

Why?  Excercise.

Hint:  What happens if the dest receives frame(seq)
(for the first time), sends ACK(seq), but the ACK is lost?

Combine these two items:

If the source receives ACK(seq) with seq == source_hacked, 
it  MUST NOT  send  frame(seq + 1).

If the destination receives Frame(seq) with seq == dest_hacked,
it  MUST  send ACK(seq).

Call this the (dest yes, source no) combination.
Show that all other three combinations
(dest yes, source yes) ; (dest no, source no) ; 
(dest no, source yes) eventually will lead to 
``bad things''.
(Anything that CAN happen eventually WILL happen).

This question is likely to be asked on the next quiz.