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GATE CS 2005 Q74 — Data Link Layer: Framing, Error Detection, MAC, Ethernet Bridging

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Suppose the round trip propagation delay for a 10 Mbps10\text{ Mbps} Ethernet having 48-bit48\text{-bit} jamming signal is 46.4 μs46.4\ \mu s. The minimum frame size is:

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Answer: D

To ensure reliable collision detection in a CSMA/CD network (like Ethernet), a transmitting station must continue transmitting long enough for the collision signal to travel back to it, even in the worst-case scenario. This requires the frame transmission time (TxT_x) to be at least the round-trip propagation delay (RTTRTT) plus the time required to transmit the jamming signal (TjamT_{jam}):

Tx≥RTT+TjamT_x \ge RTT + T_{jam}

We are given:

  • Bandwidth (BB) = 10 Mbps=10×106 bits/sec10\text{ Mbps} = 10 \times 10^6\text{ bits/sec}
  • Round-trip propagation delay (RTTRTT) = 46.4 μs=46.4×10−6 s46.4\ \mu\text{s} = 46.4 \times 10^{-6}\text{ s}
  • Jamming signal size = 48 bits48\text{ bits}

The minimum frame size (in bits) can be calculated as:

Minimum Frame Size=B×(RTT+Tjam)=(B×RTT)+Jamming Signal Size=(10×106 bps×46.4×10−6 s)+48 bits=464 bits+48 bits=512 bits\begin{aligned} \text{Minimum Frame Size} &= B \times (RTT + T_{jam}) \\ &= (B \times RTT) + \text{Jamming Signal Size} \\ &= (10 \times 10^6 \text{ bps} \times 46.4 \times 10^{-6} \text{ s}) + 48 \text{ bits} \\ &= 464 \text{ bits} + 48 \text{ bits} \\ &= 512 \text{ bits} \end{aligned}

Thus, the minimum frame size is 512 bits.

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