In a sliding window ARQ scheme, the transmitter's window size is N and the…

Answer: M + N

Explanation: The minimum number of distinct sequence numbers S must be large enough that sequence numbers in use by the sender and those within the receiver's acceptance window cannot be confused.

• The sender can have up to N outstanding frames (sent but not yet acknowledged).

• The receiver can accept up to M frames ahead within its window; these sequence numbers must not be mistaken for new transmissions.

• To avoid reuse that could cause ambiguity, the sequence-number space must cover both ranges simultaneously, so S >= N + M.

Example: If the transmitter window N = 4 and the receiver window M = 3, you need at least 7 distinct sequence numbers (for example 0..6). Reusing a sequence number sooner could make the receiver confuse a new frame with an outstanding one, or an acknowledgement with a different frame.

Therefore, the minimum required number of distinct sequence numbers is M + N.