Consider a selective repeat sliding window protocol that uses a frame size of…
2014
Consider a selective repeat sliding window protocol that uses a frame size of 1 KB to send data on a 1.5 Mbps link with a one-way latency of 50 msec. To achieve a link utilization of 60%, the minimum number of bits required to represent the sequence number field is ________.
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Correct answer: 5
Key data: frame size = 1 KB, bandwidth = 1.5 Mbps, one-way latency = 50 ms, target utilization = 60%.
Frame size in bits = 1 KB = 1024 bytes = 1024 × 8 = 8192 bits.
Round-trip time (RTT) = 2 × one-way latency = 2 × 50 ms = 100 ms = 0.1 s.
Bits that must be in flight to get 60% utilization = utilization × bandwidth × RTT = 0.6 × 1.5×10^6 × 0.1 = 90,000 bits.
Minimum window size W (in frames) = 90,000 / 8192 ≈ 10.99 → must send at least 11 frames.
For selective repeat the sequence-number space must be at least 2W to avoid ambiguity, so need at least 2 × 11 = 22 distinct sequence numbers.
Smallest k with 2^k ≥ 22 is k = 5 (since 2^5 = 32).
Answer: 5 bits
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