GATE CS 2024 Set 2 — Question 55

NAT+2 / -0MediumCSMA/CDMedium Access & LANsComputer NetworksEthernet & MAC

Computer Networks → Medium Access & LANs → CSMA/CD

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Question

Consider an Ethernet segment with a transmission speed of 10810^8 bits/sec and a maximum segment length of 500 meters. If the speed of propagation of the signal in the medium is 2×1082 \times 10^8 meters/sec, then the minimum frame size (in bits) required for collision detection is _________

Correct answer

500 to 500

Solution

In an Ethernet network using CSMA/CD, to ensure that a collision is detected by the transmitting station before it finishes transmitting the frame, the transmission time (TtT_t) must be at least twice the propagation delay (TpT_p) between the two farthest points in the segment.
Given:
  • Transmission speed (Bandwidth, BB) = 10810^8 bits/sec
  • Maximum segment length (dd) = 500 meters
  • Speed of propagation (vv) = 2×1082 \times 10^8 meters/sec
Condition for collision detection:Tt2×TpT_t \geq 2 \times T_pWhere:
  • Tt=Minimum Frame Size (L)BT_t = \frac{\text{Minimum Frame Size (L)}}{B}
  • Tp=dvT_p = \frac{d}{v}
Substituting the expressions into the condition:LB2×dv\frac{L}{B} \geq 2 \times \frac{d}{v}Solving for LL:L2×dv×BL \geq 2 \times \frac{d}{v} \times BSubstituting the given values:L2×5002×108×108L \geq 2 \times \frac{500}{2 \times 10^8} \times 10^8L2×5002L \geq 2 \times \frac{500}{2}L500 bitsL \geq 500 \text{ bits}Thus, the minimum frame size required for collision detection is 500 bits.

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