Consider a file of size 4 million bytes being transferred between two hosts…
2026
Consider a file of size 4 million bytes being transferred between two hosts connected via 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.
What is the total time (in seconds) to transfer the file?
Note: 1M=106 , 1k=103
Answer: B. 64 — To find the total transfer time, first convert the file size from bytes to bits. Then, identify the bottleneck bandwidth among the three links. Finally,…
- A.
731
- B.
64
- C.
8
- D.
16
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Show answer & explanation
Correct answer: B
To find the total transfer time, first convert the file size from bytes to bits. Then, identify the bottleneck bandwidth among the three links. Finally, divide the total bits by the bottleneck bandwidth.
File Size = 4 million bytes = 4 * 10^6 * 8 bits = 32 * 10^6 bits.
Bandwidths: 2 Mbps, 500 kbps, 1 Mbps. Bottleneck = 500 kbps = 0.5 * 10^6 bps.
Time = 32 * 10^6 / 0.5 * 10^6 = 64 seconds.
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