We looked at three protocols for providing reliable transport: Alternating Bit, Go-Back-N and Selective-Repeat. Assume we have a delay-bandwidth product of 1MByte and a maximum segment size (MSS) of 1KByte. Assume the local memory available for buffering is not a limiting factor. Assuming the send window size has been set to the delay-bandwidth product (with the exception of alternating bit which has been set to the MSS), what would be size of the receiver buffer for each of the protocols?
Try an answer before revealing the guidance below.
Key Concepts
- Receive window
- In-order delivery
- Out-of-order buffering
- MSS
- Delay-bandwidth product
Answer Approach
- For each protocol, ask how many segments the receiver ever needs to hold.
- Alternating Bit and Go-Back-N accept only the next in-order segment.
- Selective Repeat buffers out-of-order segments up to the send window.
- Convert segment counts to bytes using MSS = 1 KB.
Full Answer
Answer status: Draft answer (unofficial). Revision notes, not an official marking scheme.
Alternating Bit: 1 segment = 1 KByte. Go-Back-N: 1 segment = 1 KByte, because the receiver discards out-of-order segments and only accepts the next expected one. Selective Repeat: the full window, 1 MByte (1024 segments of 1 KByte), because it buffers every out-of-order segment within the window.
Shortcuts: K concepts · A approach · F answer · R reviewed · B bookmark · ← / → previous / next