GATE 2022 CS Q.59: solution and answer

Official answer: 7.07 to 7.09Source: Official answer key, GATE 2022 CS (IIT Kharagpur)

1. The receiver has the whole packet only when the last bit arrives. The last bit leaves the sender after one transmission time and then travels for one propagation time, so $T=t_{\text{trans}}+t_{\text{prop}}$.

2. Transmission time: the packet is $1000\times 8=8000$ bits and the link rate is $100\text{Mbps}=100\times 10^6$ bits per second, so $t_{\text{trans}}=\frac{8000}{100\times 10^6}=8\times 10^{-5}\text{s}=0.08\text{ms}$.

3. Propagation time: the distance is the altitude, $2100\text{km}=2.1\times 10^6\text{m}$, and the speed is $3\times 10^8\text{m/s}$, so $t_{\text{prop}}=\frac{2.1\times 10^6}{3\times 10^8}=7\times 10^{-3}\text{s}=7\text{ms}$.

4. Total: $T=0.08+7=7.08\text{ms}$.