Here's the question bank on all the computer networks topics.
Station A uses 32 byte packets to transmit messages to Station B using a sliding window protocol. The round trip delay between A and B is 80 ms and the bottleneck bandwidth on the path between A and B is 128 kbps. What is the optimal window size that A should use?
Data:window size = NRound trip time = 80 mspropagation delay = Tp ? 2 Tp = 80 mspacket size = L = 32 byte = 32 × 8 bitsBand width = BW = 128 kbps = 128 × 103 bpsCalculation:\({T_t} = \frac{L}{{BW}} = \frac{{32 \times 8}}{{128 \times {{10}^3}}} = 2\;ms\)\(Efficiecny\left( \eta \right) = \frac{{N \times {T_t}}}{{{T_t}\; + \;2 \times {T_p}}}\)To get optimal window size: ? = 1\(N = \;\frac{{{T_t}\; + 2 \times {T_p}}}{{{T_t}\;}}\)\(N = 1 + \frac{{2 \times {T_p}}}{{{T_t}}}\)\(N = 1 + \frac{{80}}{2}\)\(N = 41\)Sometimes 1 is ignored. If 1 is ignored, then\(N = \frac{{2 \times {T_p}}}{{{T_t}}}\)\(N = \frac{{80}}{2}\)\(N = 40\)The optimal window size that A should use is 40
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