Consider a network path P-Q-R between nodes P and R via router Q. Node P sends a file of size 10^6 b

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GATE-CS is a competitive examination conducted in India. This question is from the 2024 GATE-CS Computer Networks paper.

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This GATE-CS 2024 Computer Networks question has a medium difficulty level and is marked as an important question.

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Question

Consider a network path P-Q-R between nodes P and R via router Q. Node P sends a file of size 10^6 bytes to R via this path by splitting the file into chunks of 10^3 bytes each. Node P sends these chunks one after the other without any wait time between successive chunk transmissions. Extra headers are negligible; chunk size is less than MTU. Each link has bandwidth 10^6 bits/sec, negligible propagation latency. Router Q immediately transmits every packet it receives. Q can simultaneously receive on P-Q and transmit on Q-R. P starts at t=0. Which option gives the time (in seconds, rounded to 3 decimal places) at which R receives all the chunks?

Accepted Answer

Correct Answer: (B) 8.008 secondsConcept -- Store-and-Forward Packet Switching (from Kurose and Ross, Computer Networking):File size = 10^6 bytes = 8 x 10^6 bits. Chunk size = 10^3 bytes = 8000 bits. Number of chunks = 10^6 / 10^3 = 1000 chunks.Bandwidth = 10^6 bits/sec. Transmission time per chunk = 8000 / 10^6 = 0.008 seconds.P begins transmitting chunk 1 at t=0. Q starts receiving chunk 1 at t=0 and finishes at t=0.008 s (after complete receipt in store-and-forward). Q immediately starts transmitting chunk 1 to R.However, Q can simultaneously receive and transmit (full duplex). So Q starts forwarding chunk 1 to R at t=0.008 s and finishes at t=0.016 s.Meanwhile P is transmitting chunk 2 from t=0.008 to t=0.016, chunk 3 from t=0.016 to t=0.024, etc.P finishes transmitting all 1000 chunks at t = 1000 x 0.008 = 8.000 s.Q finishes receiving the last chunk at t=8.000 s and immediately starts forwarding it to R. Q finishes forwarding the last chunk to R at t = 8.000 + 0.008 = 8.008 seconds.Since propagation delays are negligible, R receives the last chunk at t = 8.008 s.

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