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GATE CS 2012 Q44 — Concept of Layering: OSI and TCP/IP Protocol Stacks

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Consider a source computer (S)(S) transmitting a file of size 10610^{6} bits to a destination computer (D)(D) over a network of two routers (R1 and R2)(R_{1}\text{ and }R_{2}) and three links (L1,L2, and L3)(L_{1},L_{2},\text{ and } L_{3}). L1L_{1} connects SS to R1R_{1}; L2L_{2} connects R1R_{1} to R2R_{2}; and L3L_{3} connects R2R_{2} to DD. Let each link be of length 100 km\text{100 km}. Assume signals travel over each link at a speed of 10810^{8} meters per second. Assume that the link bandwidth on each link is 1 Mbps\text{1 Mbps}. Let the file be broken down into 10001000 packets each of size 10001000 bits. Find the total sum of transmission and propagation delays in transmitting the file from SS to DD?

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Answer: A

We can find the total time taken to transmit the file using the store-and-forward packet switching model with NN packets and kk links.

1. Calculate individual delays:

  • Propagation delay per link (TpT_p): Tp=DistanceSpeed=100×103 m108 m/s=10−3 s=1 msT_p = \frac{\text{Distance}}{\text{Speed}} = \frac{100 \times 10^3 \text{ m}}{10^8 \text{ m/s}} = 10^{-3} \text{ s} = 1 \text{ ms}
  • Transmission delay per packet (TtT_t): Tt=Packet SizeBandwidth=1000 bits106 bps=10−3 s=1 msT_t = \frac{\text{Packet Size}}{\text{Bandwidth}} = \frac{1000 \text{ bits}}{10^6 \text{ bps}} = 10^{-3} \text{ s} = 1 \text{ ms}

2. Calculate total time:

For N=1000N = 1000 packets and k=3k = 3 links (since there are 2 routers between SS and DD), the total time is given by the formula:

Total Time=(N+k−1)Tt+kTp\text{Total Time} = (N + k - 1) T_t + k T_p Total Time=(1000+3−1)×1 ms+3×1 ms\text{Total Time} = (1000 + 3 - 1) \times 1 \text{ ms} + 3 \times 1 \text{ ms} Total Time=1002 ms+3 ms=1005 ms\text{Total Time} = 1002 \text{ ms} + 3 \text{ ms} = 1005 \text{ ms}

Thus, the total sum of transmission and propagation delays is 1005 ms.

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