Directions (Q36–40): A 3000 km long trunk operates at 1.536 Mbps and is used to transmit 64-byte frames using a sliding window protocol. The propagation delay is 6 μs/km.
The minimum number of bits required in the sequence number field of the packet is
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Correct answer: B. 7 bits
Explanation
a = Tp/Tt = 54, so the window for full utilisation is 1 + 2a = 109 frames. ⌈log2 109⌉ = 7 bits.More Data Communication and Computer Networks questions
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