Consider a file of size 4 million bytes being transferred between two hosts connected via a path con

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GATE-CS is a competitive examination conducted in India. This question is from the 2026 GATE-CS Computer Networks paper.

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This GATE-CS 2026 Computer Networks question has a medium difficulty level and is marked as an important question.

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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

Accepted Answer

Correct Answer: (B) 64 secondsConcept -- Network File Transfer and Bottleneck Bandwidth: File size = 4 × 10^6 bytes = 32 × 10^6 bits. Bandwidths: 2 Mbps = 2×10^6 bps, 500 kbps = 5×10^5 bps, 1 Mbps = 10^6 bps. In a store-and-forward network with negligible propagation delay, the total transfer time is determined by the bottleneck (slowest) link because the file must be transmitted over each link, and the slowest link limits the throughput. Bottleneck = 500 kbps = 5×10^5 bps. Time = File size / bottleneck bandwidth = (32×10^6) / (5×10^5) = 64 seconds. The other links are faster, so they do not add significant extra time after the pipeline is filled.Why other options are incorrect: • A (731) – Might come from using 1M=2^20 and miscalculating.• C (8) – Possibly using the fastest link (2 Mbps) only.• D (16) – Using 1 Mbps as bottleneck gives 32 seconds, not 16; 16 could be half of that.Note: In a path with multiple links, the maximum achievable throughput is the minimum of the link bandwidths (bottleneck). The transfer time = file size / bottleneck bandwidth, assuming no other delays and the sender can saturate the bottleneck.

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