Question

Consider a CSMA/CD network that transmits data at a rate of 100 Mbps (${10}^8$ bits per second) over a 1 km (kilometer) cable with no repeaters. If the minimum frame size required for this network is 1250 bytes, what is the signal speed (km/sec) in the cable?

Answer 1

Band width $=100\times {10}^6$bps
d = 1000 m
For CSMA/CD : $t_d\ge 2P_d$
$\frac{1250\times 8}{100\times {10}^6}=2\times \frac{1000}{xm/s}$
$x=\frac{2\times 1000\times 100\times {10}^6}{1250\times 8}$
$=2\times {10}^7=20000$(km/sec)