Why would token ring protocol be inefficient if a LAN had a very large perimeter? Explain your answer in terms of the average length of frame length, L (bytes), and the transmission rate of the links, R (bytes per second) and propagation delay.
Try an answer before revealing the guidance below.
Key Concepts
- Token passing
- Ring latency
- Transmission time L/R
- Propagation delay
- Idle time
Answer Approach
- Compare the useful time spent sending one frame (L/R) with the time the token spends travelling the ring.
- Ask how the ring propagation delay grows with perimeter.
- Express efficiency as a ratio of useful time to total time.
Full Answer
Answer status: Draft answer (unofficial). Revision notes, not an official marking scheme.
A station can only send when it holds the token, and the token (and each frame) must travel around the ring. Useful time per frame is the transmission time L/R, but the channel also spends the ring propagation delay (perimeter / propagation speed) carrying the token and frame around with no new data being sent. Efficiency is roughly (L/R) / (L/R + d_prop). With a very large perimeter d_prop becomes large compared with L/R, so most of the time is spent waiting for the token to arrive and efficiency falls.
Efficiency ≈ (L/R) / (L/R + d_prop)Shortcuts: K concepts · A approach · F answer · R reviewed · B bookmark · ← / → previous / next