Suppose a 1 Gbps CSMA/CD local area network uses a 1 km cable. If the signal…

Concept

In CSMA/CD, a transmitting station must still be sending when a worst-case collision signal returns.

Therefore, the minimum frame transmission time equals the round-trip propagation delay: Tmin = 2d/v, and the minimum frame size is Lmin = R × Tmin.

Application

1. Let d = 1 km, v = 2 × 105 km/s, and R = 1 Gbps = 109 bit/s.

2. The one-way propagation delay is d/v = 1/(2 × 105) s = 5 × 10−6 s = 5 microseconds.

3. The round-trip propagation delay is 2 × 5 microseconds = 10 microseconds = 10 × 10−6 s.

4. Hence, Lmin = (109 bit/s)(10 × 10−6 s) = 104 bits = 10,000 bits.

Cross-check

At 1 Gbps, transmitting 10,000 bits takes 10,000/109 s = 10 microseconds, exactly the round-trip propagation delay.

Result

The minimum frame size is 10,000 bits.