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In a sliding window ARQ scheme, the transmitter's window size is N and the receiver's window size is M. The minimum number of distinct sequence numbers required to ensure correct operation of the ARQ scheme is

  1. min(M,N)
  2. max(M,N)
  3. M+N
  4. MN Answer given is M+N but my understanding is

According to the given question the communication between sender and receiver is unidirectional or bi directional?

If it is unidirectional then answer should be max (N,M) , as sender will send N frames and each frame should have unique sequence number and if it is bidirectional then it should be M+N as both will have their Frames In transition with unique sequence number.

Is my interpretation is correct? Please correct me if I am wrong!! If not why M+N is correct?

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Your interpretation of bidirectional is correct.

First of all, it should be bi-directional. Here is the reason. The receiver window size M is not equal to sender window size N so it must be bidirectional. If it is unidirectional, then the receive window should be 1 for Go-Back-N protocol and N for Selective-Repeat protocol.

Then it is very straightforward: the minimum sequence number should be N+M. For example, the sender could choose [0, N-1] as sequence numbers while the receiver could choose [10000, 10000+M-1] as sequence numbers.

One thing to note, this problem is not strict. If you use Selective-Repeat protocol, then the minimum sequence number should be 2*(N+M) to handle both new packet and a retransmission.

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