Question

Consider an Ethernet segment with a transmission speed of \(10^8 \, \text{bits/sec}\) and a maximum segment length of \(500 \, \text{meters}\). If the speed of propagation of the signal in the medium is \(2 \times 10^8 \, \text{meters/sec}\), then the minimum frame size (in bits) required for collision detection is ............

Hint:

Collision detection in Ethernet requires the frame size to be large enough to cover the round-trip propagation delay.

Answer 1

The round-trip time for a signal to propagate across the segment is: \[ \text{Time} = 2 \times \frac{\text{Distance}}{\text{Propagation speed}} = 2 \times \frac{500}{2 \times 10^8} = 5 \, \mu\text{s}. \]
To detect collisions, the frame transmission time must be at least equal to the round-trip time. The frame transmission time is: \[ \text{Frame transmission time} = \frac{\text{Frame size}}{\text{Transmission speed}}. \]
Equating the two: \[ \frac{\text{Frame size}}{10^8} = 5 \, \mu\text{s} \implies \text{Frame size} = 10^8 \times 5 \times 10^{-6} = 500 \, \text{bits}. \] Final Answer: \[ \boxed{500 \, \text{bits}} \]