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GATE CSE 2024 Session 1 Q36: Time for a file split into chunks to cross a store-and-forward router

Question

Consider a network path P—Q—R between nodes P and R via router Q. Node P sends a file of size 10610^6 bytes to R via this path by splitting the file into chunks of 10310^3 bytes each. Node P sends these chunks one after the other without any wait time between the successive chunk transmissions. Assume that the size of extra headers added to these chunks is negligible, and that the chunk size is less than the MTU.

Each of the links P—Q and Q—R has a bandwidth of 10610^6 bits/sec, and negligible propagation latency. Router Q immediately transmits every packet it receives from P to R, with negligible processing and queueing delays. Router Q can simultaneously receive on link P—Q and transmit on link Q—R.

Assume P starts transmitting the chunks at time t=0t = 0.

Which one of the following options gives the time (in seconds, rounded off to 3 decimal places) at which R receives all the chunks of the file?

Try it: multiple choice: exactly one correct option

Select one option
Wrong answer costs 2/3 mark in GATE.
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Answer: B (official GATE 2024 final answer key)

Concept used: Pipelined store-and-forward: total time = all chunks on the first link + one chunk on each later link.

First step: One chunk is 8,000 bits, so it takes 8 ms per link; P finishes sending 1,000 chunks at 8 s.

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Source: official GATE 2024 Session 1 CS question paper and final answer key, organised by IISc Bangalore. Question paper · Answer key