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GATE 2026 CS (CS2) – Question 21

Computer Networks · Principles of Layering · 1 mark · Multiple choice

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 additional overhead to transfer the file. Which one of the following is the total time, in seconds, to transfer the file?

Note: $1\text{ M}=10^6$ and $1\text{ k}=10^3$.

  1. 731
  2. 64
  3. 8
  4. 16

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Correct answer: (B) 64

Explanation

The file size is 4 million bytes, or $32\times 10^6$ bits. Since the three links are in series and propagation and processing delays are negligible, the bottleneck bandwidth determines the transfer time. The minimum link bandwidth is 500 kbps, i.e. $0.5\times 10^6$ bps. Thus the transfer time is $$\frac{32\times 10^6}{0.5\times 10^6}=64\text{ s}.$$ Therefore, option (B) is correct.