ISRO CS 2023 Q.13: solution and answer

Official answer: DSource: ISRO's revised and final answer key of 16 February 2024

1. For CSMA/CD the sender must still be transmitting when a collision signal can return from the far end. So the minimum frame transmission time must be at least the round-trip propagation time: $T_{\text{tx}}\ge 2T_p$, that is $\frac{L}{B}\ge 2T_p$.

2. One-way propagation delay: $T_p=\frac{1\text{km}}{250000\text{km/s}}=4\mu \text{s}$.

3. Round-trip time: $2T_p=8\mu \text{s}$.

4. Minimum frame size: $L=1500\times 8=12000$ bits.

5. Rearranging $\frac{L}{B}\ge 2T_p$ gives $B\le \frac{L}{2T_p}$, so $\frac{L}{2T_p}$ is the highest rate at which a frame of this minimum size still lasts for the whole round trip. The rate at which the minimum frame lasts exactly $2T_p$ is $$B=\frac{12000}{8\times 10^{-6}\text{s}}=1.5\times 10^9\text{bps}=1.5\text{Gbps}.$$

6. Check with the options: at $1.5\text{Gbps}$, $12000/(1.5\times 10^9)=8\mu \text{s}=2T_p$ exactly. At $1\text{Gbps}$, $100\text{Mbps}$ and $500\text{Mbps}$ the transmission times are $12\mu \text{s}$, $120\mu \text{s}$ and $24\mu \text{s}$, all longer than $8\mu \text{s}$. Those rates are slower than the limit, so the frame is longer than it needs to be. The network speed that matches the stated minimum frame size and cable is $1.5\text{Gbps}$, option (d).