GATE 2026 CS2 Q.21: solution and answer

Official answer: BSource: Official answer key, GATE 2026 CS2 (IIT Guwahati)

1. File size in bits: $4\times 10^6\text{bytes}\times 8=32\times 10^6\text{bits}$.

2. Link rates (with $1\text{M}=10^6$, $1\text{k}=10^3$): $2\text{Mbps}=2\times 10^6$, $500\text{kbps}=5\times 10^5$ and $1\text{Mbps}=10^6$ bits per second.

3. Processing and propagation delays are negligible, and there is no other traffic or overhead. The file therefore flows through the three links as one continuous stream, and the slowest link sets the end-to-end throughput.

4. The slowest link is the $500\text{kbps}$ link, so the end-to-end throughput is $5\times 10^5\text{bits/s}$.

5. $$T=\frac{32\times 10^6}{5\times 10^5}=64\text{s}$$

6. Check: the same data would take $\frac{32\times 10^6}{2\times 10^6}=16\text{s}$ on the first link and $\frac{32\times 10^6}{10^6}=32\text{s}$ on the third. Both links are faster than the bottleneck and work at the same time as it, so they add nothing. The answer is option (B).