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GATE CS 2024 Set 1 · Computer Networks

Question 36

MCQ2 marksWrong answer: −2/3Easy (estimate)

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?

Choose one option
Official answer and solution
Official answer
B
From
Official GATE 2024 answer key

Solution

Calculate the number of chunks NN, the single-hop transmission time TcT_c of one chunk, and the pipelined two-hop completion time (N+1)Tc(N + 1) T_c.

  1. Number of chunks:
    N=106 bytes103 bytes/chunk=1000 chunksN = \frac{10^6\text{ bytes}}{10^3\text{ bytes/chunk}} = 1000\text{ chunks}
  2. Transmission time of one chunk on each link:
    Each chunk has size 103 bytes=8×103 bits10^3\text{ bytes} = 8 \times 10^3\text{ bits}, and link bandwidth is 106 bits/s10^6\text{ bits/s}:
    Tc=8×103 bits106 bits/s=0.008 sT_c = \frac{8 \times 10^3\text{ bits}}{10^6\text{ bits/s}} = 0.008\text{ s}
  3. Total time with store-and-forward pipelining across 2 equal-bandwidth links:
    • Node P finishes transmitting the 1000th1000\text{th} chunk to Q at t=1000×0.008=8.000 st = 1000 \times 0.008 = 8.000\text{ s}.
    • Router Q immediately transmits the 1000th1000\text{th} chunk over link Q-R, which takes one additional chunk transmission time Tc=0.008 sT_c = 0.008\text{ s}.
    Ttotal=(N+1)×Tc=1001×0.008 s=8.008 sT_{\text{total}} = (N + 1) \times T_c = 1001 \times 0.008\text{ s} = 8.008\text{ s}
Final answer

Option (B)

Common trap

Forgetting the store-and-forward delay of the last chunk on the second link (giving 8.000 s) or ignoring pipelining across the two links (giving 16.000 s).

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

Source

Paper
GATE 2024 Computer Science and Information Technology - Set 1 (CS1)
Question
36 of 65
Section
Computer Science and Information Technology

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