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 × 10^8 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_________.

GATE CSE 2022 Q59: Time to receive a packet over an earth–satellite link

Computer Networks
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NAT
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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 $3 \times 10^8$ 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_________.

Try it: numerical answer: type the value

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Numerical answer. No negative marking.

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Answer: 7.07 to 7.09 (official GATE 2022 final answer key)

Concept used: Total time = transmission delay + propagation delay.

First step: Transmission: $\frac{8000 \text{ bits}}{10^8 \text{ bps}} = 0.08$ ms; propagation: $\frac{2.1 \times 10^6}{3 \times 10^8}$ s.

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Source: official GATE 2022 CS question paper and final answer key, organised by IIT Kharagpur. Question paper · Answer key

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GATE CSE 2022 Q59: Time to receive a packet over an earth–satellite link

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