I previously posted this question and received a very solid answer to part one.
ID: 3528926 • Letter: I
Question
I previously posted this question and received a very solid answer to part one. I forgot to post part two of the question and I am looking for help on that part. Here is the full question so the context of part two is clear: From Computer Networking: A Top-down Approach 6th Edition:
Consider the rdt2.2 receiver in Figure 3.14, and the creation of a new packet in the self-transition (i.e., the transition from the state back to itself) in the Wait-for-0-from-below and the Wait-for-1-from-below states: sndpkt=make_pkt(ACK,1,checksum) and sndpkt=make_pkt(ACK,0,checksum). PART ONE (ANSWERED): Would the protocol work correctly if this action were removed from the self-transition in the Wait-for-1-from-below state? Justify your answer.
PART TWO (NEEDS ANSWERED): What if this event were removed from the self-transition in the Wait-for-0-from-below state? [Hint: In this latter case, consider what would happen if the first sender-to-receiver packet were corrupted.]
Explanation / Answer
The answer is simple...
The sender will stuck in Wait-for -ACK-0-from below and never move forward to wait-for-ACK-1 -state;!
If the first sender to receiver packet was corrupted and sender has no feedback from receiver(sender will be in the state of Wait-For-ACK-0),the sender will keep on sending packet with sequence number 0 and receiver will keep on rejecting the received packet....;