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GATE CS 2022 · Computer Networks

Question 59

NAT2 marksNo negative markingEasy (estimate)

Question

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/s3 \times 10^8\text{ 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_________.

Official answer and solution
Accepted answers
7.07 to 7.09
From
Official GATE 2022 answer key

Solution

Compute total transmission time plus propagation delay from the earth station to the satellite.

Given parameters:

  • Bandwidth B=100 Mbps=100×106 bits/s=108 bpsB = 100\text{ Mbps} = 100 \times 10^6\text{ bits/s} = 10^8\text{ bps}
  • Packet size L=1000 bytes=8000 bitsL = 1000\text{ bytes} = 8000\text{ bits}
  • Distance d=2100 km=2.1×106 md = 2100\text{ km} = 2.1 \times 10^6\text{ m}
  • Propagation speed v=3×108 m/sv = 3 \times 10^8\text{ m/s}

  1. Transmission time (TtxT_{\text{tx}}):

    Ttx=LB=8000 bits108 bits/s=8×10−5 s=0.08 msT_{\text{tx}} = \frac{L}{B} = \frac{8000\text{ bits}}{10^8\text{ bits/s}} = 8 \times 10^{-5}\text{ s} = 0.08\text{ ms}

  2. Propagation delay (TpropT_{\text{prop}}):

    Tprop=dv=2.1×106 m3×108 m/s=0.7×10−2 s=7×10−3 s=7.00 msT_{\text{prop}} = \frac{d}{v} = \frac{2.1 \times 10^6\text{ m}}{3 \times 10^8\text{ m/s}} = 0.7 \times 10^{-2}\text{ s} = 7 \times 10^{-3}\text{ s} = 7.00\text{ ms}

  3. Total time for the receiver to completely receive the packet:

    Ttotal=Ttx+Tprop=0.08 ms+7.00 ms=7.08 msT_{\text{total}} = T_{\text{tx}} + T_{\text{prop}} = 0.08\text{ ms} + 7.00\text{ ms} = 7.08\text{ ms}
Final answer

7.08

Common trap

Computing only the propagation delay (7.00 ms) and forgetting to include the packet transmission time (0.08 ms).

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2022 Computer Science and Information Technology (CS)
Question
59 of 65
Section
Computer Science and Information Technology

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