GATE CS 2024 Set 2 Q.55: Medium Access Control (ALOHA, CSMA/CD) and Ethernet/Switching

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 _________

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Official answer and solution

Official answer: 500
From: Official GATE 2024 answer key

Solution

Use the CSMA/CD condition $T_t \ge 2 T_p$ , so $L_{\min} = 2 \times \frac{d}{v} \times B$ .

For CSMA/CD collision detection on an Ethernet segment, the frame transmission time $T_t = \frac{L}{B}$ must be at least the round-trip propagation delay $2T_p = 2\frac{d}{v}$ :

L_{\min} = 2 \times \frac{d}{v} \times B = 2 \times \frac{500\text{ m}}{2 \times 10^8\text{ m/s}} \times 10^8\text{ bits/s} = 500\text{ bits}

Final answer

500

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2024 Computer Science and Information Technology - Set 2 (CS2)
Question: 55 of 65
Section: Computer Science and Information Technology

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