Consider a CSMA/CD network that transmits data at the rate of 100 Mbps over a 1 kilometre cable with no repeater. If the minimum frame size required for this network is 1250 bytes, what is the signal speed (km/sec) in the cable ?

Minimum frame size:

$1250 \text{ bytes} = 1250 \times 8 = 10000 \text{ bits}$

Transmission rate:

$100 \text{ Mbps} = 100 \times 10^6 \text{ bits/sec}$

Transmission time:

$T = \frac{10000}{100 \times 10^6}$

$T = 10^{-4} \text{ sec}$

In CSMA/CD:

$T = 2 \times$ propagation delay

So propagation delay:

$\frac{T}{2} = 5 \times 10^{-5}$ sec

Distance = 1 km

Signal speed:

$v = \frac{distance}{time}$

$v = \frac{1}{5 \times 10^{-5}}$

$v = 20000 \text{ km/sec}$