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 3 ×10⁸ 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 ______.

The correct answer is 7.08.

Concept:

The given data,

Bandwidth B= 100 Mbps.
Distance D= 2100 Km.
Propagation speed S= 3 x 10⁸ m/sec.

Packet size L= 1000 bytes 

Explanation:
Time is taken to receive the packet = Transmission time (T_x) + Propagation time(T_P)

Transmission time T_x= Packet size / Bandwidth = L / B

Propagation time Tp= Distance / Propagation speed = D/S

Analysis:​

Transmission time Tx= 1000 bytes / 100 Mbps

Transmission time = 1000 x 8 bits / 100 x 10⁶ bits/sec

Transmission time = 0.08 Msec.

Propagation time T_p= Distance / Propagation speed = D/S

Propagation time = 2100 Km / 3x 10⁸ m/sec

Propagation time = 2100 x 10³ m / 3x 10⁸ m/sec

Propagation time = 7 x 10^-3 sec

Propagation time =7 msec

Time is taken to receive the packet = 7+0.08

Time is taken to receive the packet =7.08 msec

Hence the correct answer is 7.08 msec.