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Official PYQ · 2026Set 2 · Q11Routing Protocols: Shortest Path, Flooding, Distance Vector, Link StateMCQ1 mark

GATE CS 2026 Set 2 Q11 — Routing Protocols: Shortest Path, Flooding, Distance Vector, Link State

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Consider a file of size 44 million bytes being transferred between two hosts connected via a path consisting of three consecutive links of bandwidth 22 Mbps , 500500 kbps , and 11 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=106,1k=1031 \mathrm{M}=10^{6}, 1 \mathrm{k}=10^{3}

Choose one option
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Answer: B

To find the total time to transfer the file, we analyze the transmission over the three consecutive links.

1. Convert File Size to Bits:

Given file size, F=4 million bytes=4×106 bytesF = 4 \text{ million bytes} = 4 \times 10^6 \text{ bytes}.

Since 1 byte=8 bits1 \text{ byte} = 8 \text{ bits}:

F=4×106×8=32×106 bitsF = 4 \times 10^6 \times 8 = 32 \times 10^6 \text{ bits}

2. Identify the Bottleneck Link:

The bandwidths of the three consecutive links are:

  • Link 1: R1=2 Mbps=2×106 bpsR_1 = 2 \text{ Mbps} = 2 \times 10^6 \text{ bps}
  • Link 2: R2=500 kbps=0.5×106 bpsR_2 = 500 \text{ kbps} = 0.5 \times 10^6 \text{ bps}
  • Link 3: R3=1 Mbps=1×106 bpsR_3 = 1 \text{ Mbps} = 1 \times 10^6 \text{ bps}

The bottleneck bandwidth (RbottleneckR_{\text{bottleneck}}) is the minimum of these rates:

Rbottleneck=min⁡(2 Mbps,0.5 Mbps,1 Mbps)=0.5 Mbps=0.5×106 bpsR_{\text{bottleneck}} = \min(2 \text{ Mbps}, 0.5 \text{ Mbps}, 1 \text{ Mbps}) = 0.5 \text{ Mbps} = 0.5 \times 10^6 \text{ bps}

3. Calculate the Transfer Time:

For a large file transfer with negligible propagation and processing delays, the total transfer time is dominated by the transmission time through the bottleneck link (fluid flow model / pipelined transmission):

Total Time=File SizeBottleneck Bandwidth=32×106 bits0.5×106 bps=64 seconds\text{Total Time} = \frac{\text{File Size}}{\text{Bottleneck Bandwidth}} = \frac{32 \times 10^6 \text{ bits}}{0.5 \times 10^6 \text{ bps}} = 64 \text{ seconds}

Thus, the total time to transfer the file is 6464 seconds.

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