CSMA CD in Ethernet

In CSMA/CD, minimum length of packet to detect collision, L >= 2 * Tp * Bandwidth

example:Determine the maximum length of the cable (in km) for transmitting data at a rate of 500 Mbps in an Ethernet LAN with frames of size 10,000 bits. Assume the signal speed in the cable to be 2,00,000 km/s.

Tt >=2*Tp
L(length of the cable)<=(Data size*Signal speed)/(2*Bandwidth)
L<=(10,000 bits * 2 *108m/s)/(2*500*106b/s)
L<=2000m
Hence L<=2Km.

Backoff Algorithm in CSMA/CD

Example:After some transmission on Ethernet, current frame of A's is collided 2 times and B's is collided 4 times. The probability that A wins the present back off race is____.

Number of times A has been involved in collision = 2
Number of times B has been involved in collision = 4
A can choose any number between (0 to 22-1 ) , i.e, (0,1,2,3)
B can choose any number between (0 to 24-1 ) , i.e, (0,1,2,3,......,15)
It is given that A wins the race, so choices for (A,B) will be:
(0,1) , (0,2) ,(0,3)................ , (0,15)
(1,2), (1,3),...........................,(1,15)
(2,3),(2,4),............................,(2,15)
(3,4),(3,5),.............................,(3,15)
Total choices = 15 + 14 + 13 + 12 = 54 out of 64 choices.

So probability = 54 / 64 = 0.84