Two systems are 8000 km apart. Bandwidth is $500\times10^6$ bits/s; propagation speed is $4\times10^6$ metres/s; average packet size is $10^7$ bits. Neglect processing delay. Find the minimum sequence-number field length for a Go-Back-N protocol that uses full link capacity.
One-way propagation is 8×10⁶/(4×10⁶)=2 s. Packet transmission is 10⁷/(500×10⁶)=0.02 s, so a=100. Full utilization needs window W≥1+2a=201. Go-Back-N requires W≤2ᵐ−1. Seven bits allow only 127; eight allow 255. Therefore eight bits are necessary.
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