Reliable transport protocols must have at least a unique sequence number for each segment in the window. However, the protocol used can have additional requirements on the minimum unique sequence numbers beyond this requirement. What is the consequence if these minimum requirements are not met? Give an explanation of precisely why or how they will fail. You may use an example of a failure case or an explanation, but must provide either a general explanation that covers both selective repeat and go back n or a failure case for each of the protocols.
Try an answer before revealing the guidance below.
Key Concepts
- Lost ACKs
- Retransmission
- Duplicate detection
- Sequence number reuse
Answer Approach
- Pick a small window and too-small sequence space.
- Let the receiver accept a full window, then lose every ACK.
- Compare the number the receiver now expects with the number on the sender’s retransmission.
Full Answer
Answer status: Draft answer (unofficial). Revision notes, not an official marking scheme.
If the sequence space is too small, the receiver cannot tell a retransmitted old packet from a new packet with the same number, so it accepts duplicate data as new and delivers corrupted data to the application. Example (GBN, W = 3, only 3 numbers 0–2): the sender sends 0, 1, 2; all arrive, so the receiver now expects 0 again, but all ACKs are lost. The sender times out and resends the old packet 0, which the receiver wrongly accepts as new. Example (SR, W = 3, only 4 numbers 0–3): the receiver accepts 0, 1, 2 and its window moves to 3, 0, 1; if the ACKs are lost the sender resends old 0, which falls inside the receiver’s window and is buffered as new data.
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