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Official PYQ · 2024Set 2 · Q45Data Link Layer: Framing, Error Detection, MAC, Ethernet BridgingNAT2 marks

GATE CS 2024 Set 2 Q45 — Data Link Layer: Framing, Error Detection, MAC, Ethernet Bridging

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Consider an Ethernet segment with a transmission speed of 108bits/sec10^{8} \mathrm{bits} / \mathrm{sec} and a maximum segment length of 500500 meters. If the speed of propagation of the signal in the medium is 2×108 meters/sec2 \times 10^{8} \mathrm{~meters} / \mathrm{sec}, then the minimum frame size (in bits) required for collision detection is ___________.

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Answer: 500

For successful collision detection in a CSMA/CD network (like Ethernet), the transmission delay (TtT_t) of the frame must be at least twice the propagation delay (TpT_p) of the signal between the two farthest nodes.

The condition is given by:

Tt≥2×TpT_t \ge 2 \times T_p

Where:

  • Tt=LBT_t = \frac{L}{B} (Transmission delay)
  • Tp=dvT_p = \frac{d}{v} (Propagation delay)
  • LL = Minimum frame size (in bits)
  • BB = Transmission speed (Bandwidth) = 108 bits/sec10^8 \text{ bits/sec}
  • dd = Maximum segment length = 500 meters500 \text{ meters}
  • vv = Speed of propagation = 2×108 meters/sec2 \times 10^8 \text{ meters/sec}

Substituting the formulas into the condition:

LB≥2×dv\frac{L}{B} \ge 2 \times \frac{d}{v}

Solving for LL:

L≥2×dv×BL \ge 2 \times \frac{d}{v} \times B

Substitute the given values:

L≥2×5002×108×108L \ge 2 \times \frac{500}{2 \times 10^8} \times 10^8L≥2×250×10−8×108L \ge 2 \times 250 \times 10^{-8} \times 10^8L≥500 bitsL \ge 500 \text{ bits}

Thus, the minimum frame size required for collision detection is 500 bits.

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