GATE CS 2026 Set 2 Q.21: OSI/TCP-IP Layering, Physical Layer and Switching

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: $1\text{M}=10^6$ , $1\text{k}=10^3$

$731$

$64$

$8$

$16$

Official answer and solution

Official answer: B
From: Official GATE 2026 answer key

Solution

Find the end-to-end bottleneck bandwidth across the three links and divide the total file size in bits by the bottleneck bandwidth.

The file size in bits is:

L = 4 \times 10^6 \text{ bytes} \times 8 \text{ bits/byte} = 32 \times 10^6 \text{ bits}

The three links in series have bandwidths $R_1 = 2\text{ Mbps}$ , $R_2 = 500\text{ kbps} = 0.5\text{ Mbps}$ , and $R_3 = 1\text{ Mbps}$ . For a large fluid/packetized file transfer with negligible propagation and processing delays, the end-to-end throughput is determined by the bottleneck link:

R_{\text{bottleneck}} = \min(2\text{ Mbps}, 0.5\text{ Mbps}, 1\text{ Mbps}) = 0.5 \times 10^6 \text{ bps}

Therefore, the total transfer time is:

T = \frac{32 \times 10^6 \text{ bits}}{0.5 \times 10^6 \text{ bps}} = 64 \text{ seconds}

Final answer

Option (B)

Common trap

Forgetting to convert 4 million bytes into 32 million bits (which gives 8 seconds).

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2026 Computer Science and Information Technology - Session 2 (CS2)
Question: 21 of 65
Section: Computer Science and Information Technology

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