The random wait time after collision in CSMA/CD is generated following certain rules. In Ethernet, which is the predominant protocol where CSMA/CD is used, the random wait time is generated using a rule called binary exponential backoff. Each time slot in CSMA/CD is set as 512 bit times. After the i-th collision of a frame, a random number K is uniformly drawn from {0, 1, , 2i-1}, which is the number of time slots that the sender will wait before transmitting the frame again.

Now assume, in an Ethernet CSMA/CD network, two nodes N and M attempt to send frames and both have many frames ready to send. Lets denote node Ns frames as N1, N2, N3, etc., and node Ms frames as M1, M2, M3, etc.

Assume now N and M each attempt to transmit at the same time. They will collide, and have to pick a random backoff time of either 0*T or 1*T. Assume that M picked 0*T while N picked 1*T, and hence M wins and gets to transmit first while N has to wait. Now we are at the end of Ms transmission of frame M1, and M and N are up for another competition to send out M2 and N1 respectively. And their frames again collide.

A. What is the probability that N wins this second backoff race after the collision, i.e., Ns backoff time is less than Ms backoff time?

B. Assume the link that they are competing on is 10 Mbps. Assume there is no upper bound on K (Kmax = ). For an arbitrary K value chosen, what is the backoff time the node has to wait before starting to transmit again? (Give a formula with K.)

C. If a node is already at its P-th try (P > 1) of sending out the same packet, and yet still failed due to collision. What is the expected wait time of this node after this failure? Still assume there is no upper bound on K (Kmax = ).