Routing
September 12, 2024Computer-Networks
September 12, 2024Computer-Networks
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Question 88
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A 1Mbps satellite link connects two ground stations. The altitude of the satellite is 36,504 km and speed of the signal is 3 × 108 m/s. What should be the packet size for a channel utilization of 25% for a satellite link using go-back-127 sliding window protocol? Assume that the acknowledgment packets are negligible in size and that there are no errors during communication.
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120 bytes
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60 bytes
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240 bytes
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90 bytes
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Question 88 Explanation:
Time to reach satellite = 36504000/3×108 = 0.121685
RTT = 4×Time to reach satellite (S1→S, S→S2, S2→S, S→S1)
∴ RTT = 0.48
Efficiency = N×Tt/Tt+2Tp
= N×Tt/Tt+RTT
0.25 = 127×Tt/Tt+0.48
0.25Tt + 0.25 × 0.48 = 127Tt
0.25 × 0.48 = 126.5Tt
0.25 × 0.48 × 106/126.5 = L
L = 952 bit ≈ 120 byte
RTT = 4×Time to reach satellite (S1→S, S→S2, S2→S, S→S1)
∴ RTT = 0.48
Efficiency = N×Tt/Tt+2Tp
= N×Tt/Tt+RTT
0.25 = 127×Tt/Tt+0.48
0.25Tt + 0.25 × 0.48 = 127Tt
0.25 × 0.48 = 126.5Tt
0.25 × 0.48 × 106/126.5 = L
L = 952 bit ≈ 120 byte
Correct Answer: A
Question 88 Explanation:
Time to reach satellite = 36504000/3×108 = 0.121685
RTT = 4×Time to reach satellite (S1→S, S→S2, S2→S, S→S1)
∴ RTT = 0.48
Efficiency = N×Tt/Tt+2Tp
= N×Tt/Tt+RTT
0.25 = 127×Tt/Tt+0.48
0.25Tt + 0.25 × 0.48 = 127Tt
0.25 × 0.48 = 126.5Tt
0.25 × 0.48 × 106/126.5 = L
L = 952 bit ≈ 120 byte
RTT = 4×Time to reach satellite (S1→S, S→S2, S2→S, S→S1)
∴ RTT = 0.48
Efficiency = N×Tt/Tt+2Tp
= N×Tt/Tt+RTT
0.25 = 127×Tt/Tt+0.48
0.25Tt + 0.25 × 0.48 = 127Tt
0.25 × 0.48 = 126.5Tt
0.25 × 0.48 × 106/126.5 = L
L = 952 bit ≈ 120 byte
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