Tanenbaum

Question

In selective repeat SR protocol, MAX_SEQ = 2^n - 1. While this condition is obviously desirable to make efficient use of header bits, we have not demonstrated that it is essential. Does the protocol work correctly for MAX_SEQ = 4, for example?

Comments

i have not understood your question properly but what i can say is total available sequence numbers should be >= sender window size + reciever windwo size

in SR,sender window= reciever window

so,2ⁿ=2*senderwindow 

so,window size should be 2^n-1

now,why is this condition required is to identify duplicate frames

suppose sender window size =4 and reciever obviously also 4 ,then total sequence numbers required are >=8

suppose we send 0,1,2,3 frames in one go adn all are delivered successfully to the reciever.but the acknowledgment of each one is lost,

now the scene is reciver is waiting for frame with sequence number 4,but sender on the other hand sends 0,1,2,3 again(because after time out timer on sender's side,it will retranmit again).so when reciever will recieve frame 0 again,then it will understand that it is a duplicate packet and hence will discard.

if we would not have taken these many sequence numbers then reciver on reciveing the retransmitted 0th frame again would have considered it as new frame and accepted it.