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

• A. 731
• B. 64
• C. 8
• D. 16

Hint:

Treat the three links as one pipe; the overall transfer rate is set by the slowest (bottleneck) link, here 500 kbps.

Answer 1

The Correct Option is B

Step 1: Convert the file size into bits.
The file size is 4 million bytes, so using the given convention \(1M=10^6\), the file size is \(4 \times 10^6\) bytes. Since 1 byte equals 8 bits, the file size in bits is
\[ 4 \times 10^6 \times 8 = 32 \times 10^6 \text{ bits} \]

Step 2: Convert every link's bandwidth into bits per second using the given convention.
Link 1: \(2\) Mbps \(= 2 \times 10^6\) bps.
Link 2: \(500\) kbps \(= 500 \times 10^3 = 0.5 \times 10^6\) bps.
Link 3: \(1\) Mbps \(= 1 \times 10^6\) bps.

Step 3: Identify why the slowest link controls the transfer rate.
Since processing delay and propagation delay are both negligible and there is no other traffic on the path, the three links effectively act like one continuous pipe carrying the file from source to destination. A pipe made of segments of different capacities can never move data faster overall than its narrowest segment allows: link 2, at \(0.5 \times 10^6\) bps, cannot accept or pass bits any faster than that rate, no matter how fast link 1 feeds it or how fast link 3 could carry them onward. So the end-to-end effective rate of the whole path is bounded by the minimum bandwidth among the three links, here \(0.5 \times 10^6\) bps (500 kbps).

Step 4: Compute the total transfer time using the bottleneck rate.
\[ \text{Time} = \frac{\text{file size in bits}}{\text{bottleneck bandwidth}} = \frac{32 \times 10^6}{0.5 \times 10^6} = 64 \text{ seconds} \]

Step 5: Rule out the other options.
Using link 1's rate (2 Mbps) alone would wrongly give 16 seconds, and using link 3's rate (1 Mbps) alone would wrongly give 32 seconds; neither is achievable because the file still has to pass through the slower 500 kbps link somewhere on the path. The path behaves as one pipe rate-limited by its slowest link, giving 64 seconds.

Final Answer:
The total time to transfer the file is 64 seconds. \[ \boxed{64} \]