Question

2. Suppose that we run the sliding window algorithm with SWS=5 and RWS=3, and no out-of-order arrivals. Also, assume the sequence numbers start with 0. a) Find the smallest value for MaxSeqNum so that we do not have the problem of being able to distinguish between different incarnations of the same sequence numbers. You may assume that it suffices to find the smallest MaxSeqNum such that if DATA[MaxSeqNum] is in the receive window, then DATA[O] can no longer arrive. b) Given an example showing that MaxSeqNum-1 is not sufficient. c) State a general rule for the minimum MaxSeqNum in terms of SWS and RWS. Justify your answer.

Answer 1

Given that we are running sliding window algo with SWS=5 and RWS=3 and no Out of order arrivals.

a) MaxSeqNum = SWS+RWS-1 = 5+3-1= 7

b) In part a) max seq number = 7. Sender sends first 6 frames 0, 1, 2, 3, 4, 5. The receiver receives all the frames but all the ACKs are lost. Now the receiver is expecting 6, 0 (since seq number wraps around, instead of 7 we have seq number 0). After timeout, the sender sends 0, 1, 2, 3, 4, 5 again. Receiver accepts frame 0 but it was the old incarnation instead of the new frame with seq number 0

c) N = SWS + RWS - 1

Assume that the sequence numbers are 0 based. Suppose a max. seq. number = 3 means sequence numbers 0, 1, 2, and 3 can be used.