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the predominant protocol where CSMA/CD is used, the random wait time is generated using a rule called frame, a random number K is uniformly drawn
the predominant protocol where CSMA/CD is used, the random wait time is generated using a rule called frame, a random number K is uniformly drawn from {0,1,,2i1}, which is the number of time slots that the sender will wait before transmitting the frame again. Now assume, in an Ethernet CSMACD network, two nodes N and M attempt to send frames and both have many frames ready to send. Let's denote node N 's frames as N1,N2,N3, etc., and node M 's 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 0T or 1T. Assume that M picked 0T while N picked 1T, and hence M wins and gets to transmit first while N has to wait. Now we are at the end of M 's 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. (5) What is the probability that N wins this second backoff race after the collision, i.e., N's backoff time is less than M's backoff time? (3) Assume the link that they are competing on is 10Mbps. 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.) (6) 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=)
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