Let Ash and Beech be two stations attempting to transmit on an Ethernet. Each has a steady queue of frames ready to send: Ashs frames will be numbered A1, A2, and so on, and Beechs similarly. Let T = 51.2 us be the exponential backoff base unit. Suppose Ash and Beech simultaneously attempt to send frame 1, collide, and happen to choose backoff times of 0xT and 1xT, respectively, meaning Ash wins the race and transmits A1 while Beech waits. At the end of this transmission, Beech will attempt to retransmit B1 while A will attempt to transmit A2. These first attempts will collide, but now A backs off for either 0xT or 1xT, while Beech backs off for time equal to one of 0xT, 1xT, 2xT or 3xT. Note that you should not consider any inter-frame delay for this problem.

(a) Give the probability that Ash wins this second backoff race immediately after this first collision; that is, Ashs first choice of backoff time k x51.2 us is less than Beechs.

(b) Suppose Ash wins this second backoff race. Ash transmits A3, and when it is finished, Ash and Beech collide again as Ash tries to transmit A4 and Beech tries once more to transmit B1. Give the probability that Ash wins this third backoff race immediately after the first collision.

(c) For this part, assume that Beech does not limit the range of values it uses for backoff after its 10th collision. Also, assume that Ash and Beech continue this process indefinitely (i.e., Ash continues to win each race and Beech keeps trying). Calculate the probability that Ash wins the i-th backoff race given than it has won all the previous races. Note you have already calculated the probability that Ash wins the second race in part (a) and the third race in part (b).

(d) Using the result from part (c) give the probability that Ash wins all the remaining races given that it has won the 1st, 2nd, and 3rd. It is not necessary to find a closed form solution. Also, using a spread sheet or simple program give an approximate numerical answer.