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Official PYQ · 2022Q49Transport Layer: Flow Control, Congestion Control, UDP, TCP, SocketsNAT2 marks

GATE CS 2022 Q49 — Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets

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Consider a 100  Mbps100 \; \text{Mbps} link between an earth station (sender) and a satellite (receiver) at an altitude of 2100  km.2100 \; \text{km}. The signal propagates at a speed of 3×108  m/s.3 \times 10^{8} \; \text{m/s.} The time taken (in milliseconds,  rounded off to two decimal places)\text{(in milliseconds,} \; \textit{rounded off to two decimal places}) for the receiver to completely receive a packet of 1000  bytes1000 \; \text{bytes} transmitted by the sender is _______________.
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Answer: 7.07:7.09

To find the total time taken for the receiver to completely receive the packet, we need to calculate the sum of the transmission delay (TtT_t) and the propagation delay (TpT_p).

1. Transmission Delay (TtT_t):

The transmission delay is the time required to push all the packet's bits into the link.

Tt=Packet SizeBandwidth=1000 bytes100 Mbps=1000×8 bits100×106 bits/second=8×10−5 seconds=0.08 milliseconds (ms)\begin{aligned}T_t &= \frac{\text{Packet Size}}{\text{Bandwidth}} \\&= \frac{1000 \text{ bytes}}{100 \text{ Mbps}} \\&= \frac{1000 \times 8 \text{ bits}}{100 \times 10^6 \text{ bits/second}} \\&= 8 \times 10^{-5} \text{ seconds} \\&= 0.08 \text{ milliseconds (ms)}\end{aligned}

2. Propagation Delay (TpT_p):

The propagation delay is the time taken for the first bit of the packet to travel from the sender to the receiver.

Tp=DistancePropagation Speed=2100 km3×108 m/s=2100×103 m3×108 m/s=0.007 seconds=7 milliseconds (ms)\begin{aligned}T_p &= \frac{\text{Distance}}{\text{Propagation Speed}} \\&= \frac{2100 \text{ km}}{3 \times 10^8 \text{ m/s}} \\&= \frac{2100 \times 10^3 \text{ m}}{3 \times 10^8 \text{ m/s}} \\&= 0.007 \text{ seconds} \\&= 7 \text{ milliseconds (ms)}\end{aligned}

3. Total Time Taken (TtotalT_{\text{total}}):

The total time for the receiver to completely receive the packet is:

Ttotal=Tp+Tt=7 ms+0.08 ms=7.08 ms\begin{aligned}T_{\text{total}} &= T_p + T_t \\&= 7 \text{ ms} + 0.08 \text{ ms} \\&= 7.08 \text{ ms}\end{aligned}

Thus, the time taken for the receiver to completely receive the packet is 7.08 ms.

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