Multiple choice

The distance between two stations M and N is L kilometers. All frames are K bits long. The propagation delay per kilometer is t seconds. Let R bits/second be the channel capacity. Assuming that processing delay is negligible, the minimum number of bits for the sequence number field in a frame for maximum utilization, when the sliding window protocol is used, is:

  1. $\left[ log_2 \dfrac{2LtR +2K}{K} \right]$
  2. $\left[ log_2 \dfrac{2LtR +2K}{K} \right]$
  3. $\left[ log_2 \dfrac{2LtR +2K}{K} \right]$
  4. $\left[ log_2 \dfrac{2LtR +2K}{K} \right]$
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B Correct answer
Explanation

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