GATE 2024 CS2 Q.55: solution and answer

GATE 2024 · CS2Q.55 · 2 marks · NAT

Consider an Ethernet segment with a transmission speed of $1 0^{8}$ bits/sec and a maximum segment length of 500 meters. If the speed of propagation of the signal in the medium is $2 \times 1 0^{8}$ meters/sec, then the minimum frame size (in bits) required for collision detection is ______

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Answer and solution

ANSWER

The minimum frame size required for collision detection is 500 bits.

Official answer: 500Source: Official answer key, GATE 2024 CS2 (IISc Bengaluru)

For CSMA/CD the sender must still be transmitting when a collision signal returns from the farthest station. So the transmission time of the frame must be at least the round-trip propagation delay: $T_{t} \geq 2 T_{p}$ .
One-way propagation delay: $T_{p} = \frac{500 m}{2 \times 1 0 ^{8} m/s} = 2.5 \times 1 0^{- 6} s = 2.5 μ s$ .
Round-trip time: $2 T_{p} = 5 μ s$ .
Minimum frame size: $L_{m i n} = bandwidth \times 2 T_{p} = 1 0^{8} bits/s \times 5 \times 1 0^{- 6} s$ .
$L_{m i n} = 500 bits$

KEY IDEA

Collision detection needs the frame's transmission time to be at least the round-trip propagation time, so $L_{m i n} = B \times 2 T_{p}$ .

WATCH OUT FOR

Using the one-way delay $T_{p}$ instead of $2 T_{p}$ . That gives 250 bits, which is half the correct value.

From the GATE 2024 CS2 paper. Every question of this paper · Download the full paper · All 3 Data link & MAC questions, with notes