The teacher went over the following problem in class. However, I do not understand how he explains the work. He skips too many steps and jumps around. I have the final answer but I need a clear and complete explanation on how this problem is solved:
Let A and B be two stations attempting to transmit on an Ethernet. Each has a steady queue of frames ready to send; A's frames will be numbered A1, A2, so on, and B's similarly. Let T = 51.2 microseconds be exponential backoff base unit. Suppose A and B simultaneously attempt to send frame 1, collide and happen to chose backoff times of 0xT and 1xT, respectively, meaning A wins the race and transmits A1 while B waits. At the end of this transmission, B will attempt to restransmit B1 while A will attempt to transmit A2. These first attempts will collide, but now A backs off for either 0xT or 1xT, while B backs off for time equal to one of 0xT,...3xT.
a) Give the probability that A wins this second backoff race immediately after this first collision; that is, A's first choice of backoff time k x 51.2 is less than B's.
b) Suppose A wins this second backoff race. A transmits A3, and when it is finished, A and B collide again as A tries to transmit A4 and B tries once more to transmit B1. Give the probability that A wins this third backoff race immediately after the first collision.
c) Give a reasonable lower bound for the probability that A wins all the remaining backoff races.
d) What then happens to the frame B1?
(a) A can choose kA = 0 or 1; B can choose kB = 0, 1, 2, 3.
A wins outright if (kA, kB) is (0,1), (0,2), (0,3), (1,2) or (1,3), which can occur 5 out of 8 times. Therefore p = 5/8 = 0.625
(b) A can choose kA = 0 or 1; B can choose kB = 0, 1, ...