Consider a network path $\text{P—Q—R}$ between nodes $\text{P}$ and $\text{R}$ via router $\text{Q}$. Node $\text{P}$ sends a file of size $10^6$ bytes to $\text{R}$ via this path by splitting the file into chunks of $10^3$ bytes each. Node $\text{P}$ sends these chunks one after the other without any wait time between the successive chunk transmissions. Assume that the size of extra headers added to these chunks is negligible, and that the chunk size is less than the MTU. Each of the links $\text{P—Q}$ and $\text{Q—R}$ has a bandwidth of $10^6$ bits/sec, and negligible propagation latency. Router $\text{Q}$ immediately transmits every packet it receives from $\text{P}$ to $\text{R}$, with negligible processing and queueing delays. Router $\text{Q}$ can simultaneously receive on link $\text{P—Q}$ and transmit on link $\text{Q—R}$. Assume $\text{P}$ starts transmitting the chunks at time $t = 0$. Which one of the following options gives the time ($\textit{in seconds}$, rounded off to 3 decimal places) at which $\text{R}$ receives all the chunks of the file?

To find the total time taken for node $\text{R}$ to receive all chunks of the file:

1. Parameters given:

   • Total file size: $F = 10^6\text{ bytes}$

   • Chunk size (packet size): $L = 10^3\text{ bytes} = 10^3 \times 8\text{ bits} = 8000\text{ bits}$

   • Total number of chunks: $N = \frac{10^6}{10^3} = 1000\text{ chunks}$

   • Bandwidth of each link: $B = 10^6\text{ bits/sec}$

   • Number of links (hops): $k = 2$

   • Propagation, processing, and queuing delays are negligible.

2. Transmission time for a single chunk over one link:

   $$T_t = \frac{L}{B} = \frac{8000\text{ bits}}{10^6\text{ bits/sec}} = 0.008\text{ seconds}$$

3. Pipelined transmission across 2 hops:

   • Node $\text{P}$ finishes transmitting all $1000$ chunks over link $\text{P—Q}$ at time:

     $$t_1 = N \times T_t = 1000 \times 0.008\text{ s} = 8.000\text{ seconds}$$

   • At $t = 8.000\text{ s}$, router $\text{Q}$ receives the final ($1000^{\text{th}}$) chunk completely.

   • Router $\text{Q}$ then transmits this final chunk over link $\text{Q—R}$, which takes one additional transmission delay of $0.008\text{ seconds}$.

4. Total time to receive all chunks at $\text{R}$:

   $$\text{Total Time} = (N + k - 1) \times T_t = (1000 + 2 - 1) \times 0.008\text{ s} = 1001 \times 0.008\text{ s} = 8.008\text{ seconds}$$

Thus, the correct option is (B).