Consider a file of size 4 million bytes being transferred between two hosts connected via a path consisting of three consecutive links of bandwidth 2 Mbps, 500 kbps, and 1 Mbps, respectively. All processing delays and propagation delays are negligible. Assume that there is no other background traffic over the path and no other additional overhead to transfer the file. Which one of the following is the total time (in seconds) to transfer the file? Note: 1M=$10^6$, 1k=$10^3$

Given:

• File size $F = 4 \times 10^6\text{ bytes} = 4 \times 10^6 \times 8\text{ bits} = 32 \times 10^6\text{ bits} = 32\text{ Mb}$

• Link bandwidths along the path:

  • $R_1 = 2\text{ Mbps} = 2 \times 10^6\text{ bps}$

  • $R_2 = 500\text{ kbps} = 0.5 \times 10^6\text{ bps}$

  • $R_3 = 1\text{ Mbps} = 1 \times 10^6\text{ bps}$

The end-to-end throughput across a multi-hop path with no interfering traffic is constrained by the bottleneck link bandwidth $R_{\text{bottleneck}}$:

With negligible propagation and processing delays and a continuous bit stream transmission, the total transfer time is dominated by the time taken to push the entire file across the bottleneck link:

Hence, the correct option is B.