Project ForgeComputer Networks
Transport Protocols & IP Addressing
GATE 2014
Expert
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
5
Explanation
Transmission time Tt = 8192 bits / 1.5×10⁶ bps ≈ 5.461 ms. RTT = 2×50 = 100 ms. For utilization U=0.6: window size W ≥ U×(Tt+RTT)/Tt = 0.6×105.461/5.461 ≈ 11.59, round up to 12. Selective repeat needs window size ≤ 2^(n−1), so 2^(n−1) ≥ 12 requires n−1≥4 (2^4=16≥12), so n≥5.
Source: GATE CS 2014 (Session 1) (2014)
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