GATE | Gate IT 2008 | Question 62

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.
(A) 120 bytes
(B) 60 bytes
(C) 240 bytes
(D) 90 bytes


Answer: (A)

Explanation: According to given data,

Transmission time (Tx) = distance / speed,
Propagation time (Tp) = size of packet / bandwidth,
a = Tp / Tx

Now, efficiency is,

efficiency = window size / (1 + 2a)
→ 0.25 = 127 / (1 + 2((36504*103 / (3*108)) / size of packet / 1*106))
→ 0.25 = (127*size of packet) / (L + 0.48*106)
→ 0.25*(size of packet) + 0.25*0.48*106 = 127*(size of packet)
→ size of packet = 960 bits
→ size of packet = 120 bytes

So, option (A) is correct.



This explanation is contributed by Deep Shah.

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