A 1Mbps satellite link connects two ground stations. The altitude of the…
2008
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.
Answer: A. 120 bytes — Link rate: R=1 Mbps=106 bpsR = 1Satellite altitude = 36,504 km → distance ground–satellite–ground=2×36504 km=73008 km=7.3008×107 m Propagation speed: v=3×108…
- A.
120 bytes
- B.
60 bytes
- C.
240 bytes
- D.
90 bytes
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Correct answer: A
Link rate: R=1 Mbps=106 bpsR = 1Satellite altitude = 36,504 km → distance ground–satellite–ground=2×36504 km=73008 km=7.3008×107 m
Propagation speed: v=3×108 m/s
Tp=d/v= 7.3008×107/3×108≈0.24336 s
Go-back-127 ⇒ window size W=127W = 127W=127.
Let Tf be frame transmission time, a=Tp/Tf
For Go-Back-N with finite window (no errors) and utilization U:
U=W/(1+2a)
Given U=0.25:
on solvignwe get 0.00096 s
frame size L=RTf =106×0.00096=960 bits=120 bytes
Hand written note
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