Since the range of sequence numbers is so large relative to the window size, whe
ID: 672473 • Letter: S
Question
Since the range of sequence numbers is so large relative to the window size, when answering the questions, you don’t need to worry about the necessity of wrapping the sequence numbers from the higher end (1023) back to the lower end (0).
1.Consider the go-back-n protocol with a sender window size N=4 and a sequence number range of 0-1023. (Apparently, for some reason, this sender is not using the maximum possible value 1023 as its window size, but that is ok). Suppose at time t, the next in-order packet that the receiver is expecting has a sequence number of k, i.e. = k. Assume the medium does not reorder messages. Answer the following questions. Justify your answer.
a.What are the possible values of on the sender side?
b.For each of the send_base above, what are the possible values of on the sender side?
Explanation / Answer
Go-Back-N ARQ is a automatic retransmission request(ARQ) protocol, in which the sending process continues to send a number of frames specified by a window size even without receiving an acknowledgement(ACK) packet from the receiver..
The receiver process keeps track of the sequence number of the next frame it expects to receive, and sends that number with every ACK it sends. The receiver will discard any frame that does not have the exact sequence number it expects and will resend an ACK for the last correct in-order frame. Once the sender has sent all of the frames in its window, it will detect that all of the frames since the first lost frame are outstanding, and will go back to the sequence number of the last ACK it received from the receiver process and fill its window starting with that frame and continue the process over again.
a)
given window size =4
which means for first frame 4 packet will send to the reciever
sequence no range 0-1023
which means total 1024 packets
the possible values or packets on sender side=1 to 1024
1,2,3,4,5,6,7,8,9,10,11…………………………………………..1024packets
Since window size=4 For first cycle 1,2,3,4 packets will sent to the receiver
For second cycle 5,6,7,8 packets will sent to the receiver and so on
For 1st packet the sequence no is 0,
For 2nd packet the sequence no is 1
For 1023rd packet sequence no is 1024
b)
given window size =4
sequence no range 0-1023
if sequence no=0 then 1s packet value=1
if sequence no=1 then 2 packet value=2
and so on
time=t
if sequence no =k
1,2,3,4,5,6,7,8,9,10,11…………………………………………..(k-1)packets
the possible values or packets on sender side is 1 to k-1 values or packets