Consider a network connecting two systems located 8000 kilometers apart. The…
2015
Consider a network connecting two systems located 8000 kilometers apart. The bandwidth of the network is 500×10^6 bits per second. The propagation speed of the media is 4×10^6 meters per second. It is needed to design a Go-Back-𝑁 sliding window protocol for this network. The average packet size is 10^7 bits. The network is to be used to its full capacity. Assume that processing delays at nodes are negligible. Then, the minimum size in bits of the sequence number field has to be ___________.
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Correct answer: 8
Key insight: the window must accommodate all packets in flight (bandwidth × RTT divided by packet size), and for Go-Back-N the window W must satisfy W ≤ 2^k − 1 where k is the number of sequence number bits.
Compute delays: distance = 8000 km = 8,000,000 m; propagation speed = 4×10^6 m/s, so one-way propagation delay = 8,000,000 / 4,000,000 = 2 s. RTT ≈ 2 × 2 s = 4 s.
Bandwidth-delay product = 500×10^6 b/s × 4 s = 2×10^9 bits (this is the number of bits that can be in the link at once).
Number of packets in flight = (2×10^9 bits) / (10^7 bits/packet) = 200 packets. So the window must be at least 200.
Go-Back-N requirement: W ≤ 2^k − 1. We need 2^k − 1 ≥ 200 ⇒ 2^k ≥ 201 ⇒ k = ceil(log2 201) = 8.
Therefore, the minimum size of the sequence number field is 8 bits.
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