Question
Download Solution PDFConsider a selective repeat sliding window protocol that uses a frame size of 1 KB to send data on a 1.5 Mbps link with a one-way latency of 50 msec. To achieve a link utilization of 60%, the minimum number of bits required to represent the sequence number field is ________.
Answer (Detailed Solution Below) 5
Detailed Solution
Download Solution PDFData:
Frame Size = 1 KB = 8 Kb = 8 × 103 b
Bandwidth = 1.5 Mbps = 1.5 × 106 bps
Propagation delay = 50 ms
Link Utilization = 60% = 0.6
Formula:
2a = (2 × propagation delay) ÷ Transmission delay
Calculation:
Window size = N = 11.85
Minimum sequence number = 2 × N = 23.7
Bits required r = ⌈ log2(23.7) ⌉ = 5
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