Based on the following passage, answer the Questions: A 3000 km long trunk operates at 1.536 mbps and is used to transmit 64 bytes frames and uses sliding window protocol. The propagation speed is 6 μ sec/km.
For the link in the passage, use a data rate of 1.536 Mbps and interpret 6 μs/km as propagation delay per kilometre. Assume error-free Go-Back-N transmission with continuously available data. ACKs are sent immediately after complete frame reception on a separate return channel with the same propagation delay; ACK transmission and processing times are negligible. What is the minimum number of sequence-number bits needed to allow a sender window large enough for full forward-link utilization?
7 bits
Step 1 — Frame and propagation times:
\(T_t=\frac{64\times8}{1.536\times10^6}=\frac{1}{3000}\ \mathrm{s},\qquad T_p=3000\times6\times10^{-6}=0.018\ \mathrm{s}.\)
Step 2 — Window needed for continuous transmission: An ACK for the first frame returns after its transmission time plus the round-trip propagation delay.
\(a=\frac{T_p}{T_t}=54,\qquad W_{\mathrm{full}}=\left\lceil1+2a\right\rceil=109.\)
Step 3 — Sequence-number field: For Go-Back-N, the sender window must be smaller than the sequence-number space.
\(2^m-1\ge109.\)
\(2^6-1=63<109,\qquad2^7-1=127\ge109.\)
Thus the minimum is 7 bits.
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