GATE 2024 CS1 Q.36: solution and answer

Official answer: BSource: Official answer key, GATE 2024 CS1 (IISc Bengaluru)

1. The file has $10^6$ bytes and each chunk has $10^3$ bytes, so there are $10^6/10^3=1000$ chunks. Each chunk is $10^3\times 8=8000$ bits.

2. Each link runs at $10^6$ bits/s, so one chunk takes $\frac{8000}{10^6}=0.008\text{s}$ to transmit on one link. Propagation, processing and queueing delays are negligible.

3. P sends the chunks back to back. Chunk $k$ is fully received at Q at $0.008k$ seconds, so the last chunk (k = 1000) is fully received at Q at $8.000\text{s}$.

4. Q receives a whole chunk before forwarding it (store and forward), and it can receive and transmit at the same time. Chunk $k$ reaches Q at $0.008k$, and the previous chunk finished leaving Q at $0.008(k-1)+0.008=0.008k$. Link Q—R is therefore free exactly when each chunk arrives, and no chunk waits in a queue.

5. So Q starts sending the last chunk to R at $8.000\text{s}$. This takes $0.008\text{s}$ more.

6. R has the whole file at $8.000+0.008=8.008\text{s}$. In general, for $n$ chunks over 2 equal-rate links the time is $(n+1)\frac{L}{B}=1001\times 0.008=8.008\text{s}$. The answer is option (B).