Given that in a CSMA/CD protocol using binary exponential back off, the adapter waits K × 512 bit times after a collision, where K is drawn randomly. What is the maximum length of time the adapter will wait until the next transmission attempt given 7 collision has occured so far. Assume that you have a 10 Mbps broadcast channel?
Try an answer before revealing the guidance below.
Key Concepts
- K ∈ {0, …, 2^m − 1}
- m = 7 collisions
- Bit time at 10 Mbps
- Maximum backoff
Answer Approach
- Find the range of K after 7 collisions.
- Multiply the largest K by 512 bit times.
- Convert bit times to seconds at 10 Mbps.
Full Answer
Answer status: Draft answer (unofficial). Revision notes, not an official marking scheme.
After 7 collisions K is chosen from {0, 1, …, 2^7 − 1} = {0, …, 127}. The maximum wait is 127 × 512 = 65,024 bit times. At 10 Mbps one bit time is 0.1 µs, so the wait is 65,024 × 0.1 µs = 6,502.4 µs ≈ 6.5 ms.
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