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: 1M=106, 1k=103

• 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

Step 1: The file size is 4 x 10^6 bytes. Convert to bits: \(4 \times 10^6 \times 8 = 32 \times 10^6\) bits.
Step 2: The path has three links in series with bandwidths 2 Mbps, 500 kbps and 1 Mbps. Since propagation delay and processing delay are negligible and the file is sent as a continuous stream, the effective rate at which the file crosses the entire path is governed by the slowest link, i.e. the bottleneck bandwidth, which is 500 kbps \(= 5 \times 10^5\) bps.
Step 3: Total transfer time \(= \dfrac{\text{file size in bits}}{\text{bottleneck bandwidth}} = \dfrac{32 \times 10^6}{5 \times 10^5} = 64\) seconds.
Final Answer: 64 seconds (Option B)