Question

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: \(1\mathrm{M}=10^6\), \(1\mathrm{k}=10^3\)

• A. 731
• B. 64
• C. 8
• D. 16

Hint:

The total transfer time over a series of links with negligible propagation and processing delay is governed by the bottleneck, lowest-bandwidth link: time = file size in bits divided by minimum bandwidth.

Answer 1

The Correct Option is B

Think of the three links as pipes of different widths joined end to end: 2 Mbps, 500 kbps and 1 Mbps. Data can flow through the whole path only as fast as the narrowest section allows, so the 500 kbps link decides the overall speed of the transfer. First express the file in bits: \(4 \times 10^6\) bytes times \(8 = 32 \times 10^6\) bits. Since there is no propagation or processing delay and no other traffic, the time to push all the bits through is simply the file size divided by this narrowest, bottleneck rate: \(\frac{32 \times 10^6 \text{ bits}}{5 \times 10^5 \text{ bps}} = 64\) seconds. So the transfer takes 64 seconds, matching option B.