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 the processing delay is negligible, the $\text{minimum}$ number of bits for the sequence number field in a frame for maximum utilization, when the $\text{sliding window protocol}$ is used, is:

  1. $\lceil \log_2 \frac{2LtR +2K}{K} \rceil$
  2. $\lceil \log_2 \frac{2LtR}{K} \rceil$
  3. $\lceil \log_2 \frac{2LtR +K}{K} \rceil$
  4. $\lceil \log_2 \frac{2LtR +2K}{2K} \rceil$
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B Correct answer
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