Gate CS 2022 | Question - 59 | Computer Network

Consider a 100 Mbps link between an earth station (sender) and a satellite (receiver) at an altitude of 2100 km. The signal propagates at a speed of 3x10⁸ m/s. The time taken (in milliseconds, rounded off to two decimal places) for the receiver to
completely receive a packet of 1000 bytes transmitted by the sender is_________.

Correct Answer:

7.08

Solution:

Given,
Data Size = 1000 Bytes = 8000 Bits
Bandwidth = 100 Mbps = 10⁸ bps

The time required for the receiver to receive the packet is transmission time (Tt)+ propagation time(Tp). $T t = \frac{D a t a S i z e}{B a n d w i d t h} = \frac{8000 b i t s}{10^{8} b p s} = 0.08 m s$

$T p = \frac{L e n g t h}{S p e e d} = \frac{2100 * 1000 m}{3 * 10^{8} m / s} = 7 m s$

Total time = 0.08 + 7 = 7.08ms

Gate CS 2022 | Question - 59 | Computer Network

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