Discussion
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=106, 1k=103
Answer: B
The transfer takes 64 seconds.
With no per-packet overhead and negligible delays, the data streams through the path, and the throughput is limited by the slowest link, the bottleneck of 500 kbps:
\[ \begin{aligned} \text{Size} &= 4 \times 10^{6} \times 8 = 32 \times 10^{6}\ \text{bits} \\ t &= \frac{32 \times 10^{6}}{500 \times 10^{3}} = 64\ \text{s} \end{aligned} \]
8 s divides the size in bytes, not bits, by the bottleneck rate, and 16 s is the time over the 2 Mbps link alone. 731 matches no consistent calculation. The faster links do not add to the time, because the stream flows through them while the bottleneck is still sending.
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