GATE CS 2024 Set 2 | Question 55
Consider an Ethernet segment with a transmission speed of \(10^{8}\) bits/sec and a maximum segment length of 500 meters. If the speed of propagation of the signal in the medium is \(2 \times 10^{8}\) meters/sec, then the minimum frame size (in bits) required for collision detection is _________
Explanation
Answer: 500
With CSMA/CD a station detects a collision only while it is still transmitting. In the worst case the collision happens at the far end of the segment, and the noise needs another one-way delay to come back, so a frame must last at least one round-trip propagation time.
- One-way delay: \(500 / (2 \times 10^{8}) = 2.5\ \mu\text{s}\).
- Round trip: \(5\ \mu\text{s}\).
- Minimum frame size: \(10^{8} \times 5 \times 10^{-6} = 500\) bits.
Using the one-way delay alone would give 250 bits, too short to guarantee detection.
Answer: 500