In a computer network using CSMA/CD, the minimum frame size for reliable…

In CSMA/CD, a sending station must still be transmitting when a collision signal from the far end of the link returns, otherwise it cannot detect the collision. This is only guaranteed when the frame transmission time is at least twice the one-way propagation delay of the link (the round-trip time for a collision signal), giving the governing relation: Bandwidth = Frame size (in bits) / (2 x Propagation delay).

1. Convert the minimum frame size to bits: 1500 bytes x 8 = 12,000 bits.

2. Find the one-way propagation delay for the 1 Km cable at a signal speed of 250000 Km/sec: Propagation delay = distance / speed = 1 Km / 250000 Km/sec = 4 microseconds.

3. Double the propagation delay to get the round-trip (collision-detection) time: 2 x 4 microseconds = 8 microseconds.

4. Since 1500 bytes is stated as the minimum frame size, apply the relation at equality: Bandwidth = Frame size / round-trip time = 12,000 bits / 8 microseconds = 1.5 x 109 bps = 1.5 Gbps.

Check: at 1.5 Gbps, transmitting a 12,000-bit frame takes 12,000 / (1.5 x 109) = 8 microseconds, exactly matching the round-trip time above -- confirming that 1500 bytes is indeed the minimum frame size for reliable collision detection on this 1 Km link, at a data rate of 1.5 Gbps.