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 ????1, ????2 and so on, and B’s similarly. Let T = 51.2 ????????????c be the exponential backoff base unit. Suppose A and B simultaneously attempt to send frame 1, collide, and happen to choose backoff times of 0 × T and 1 × T, respectively. As a result, Station A transmits ????1 while Station B waits. At the end of this transmission, B will attempt to retransmit ????1 while A will attempt to transmit ????2. These first attempts will collide, but now A backs off for either 0 × T or 1 × T (with equal probability), while B backs off for time equal to one of 0 × T, 1 × T, 2 × T and 3× T (with equal probability).(a) Give the probability that A wins this second backoff race immediately after his first collision.(b) Suppose A wins this second backoff race. A transmits ????2 and when it is finished, A and B collide again as A tries to transmit ????3 and B tries once more to transmit ????1. Give the probability that Awins this third backoff race immediately after the first collision.(c) What is the probability that A wins all the ???? backoff races. (???? is a given constant)(d) Assume that there are 3 stations sharing the Ethernet. Will the chance for A to win all the backoffraces decrease or increase? Why?

[tex]= P[kA(2) = 0] \times P[kB(2) > 0] + P[kA(2) = 1] \times P[kB(2) > 1]\\\\= \frac{1}{2} \times \frac{3}{4} + \frac{1}{2} \times \frac{2}{4} \\\\=\frac{3}{8} +\frac{2}{8} \\\\= \frac{3+2}{8}\\\\= \frac{5}{8}[/tex]

For Point b:

Throughout this example, [tex]kA(3)[/tex] also selects to be either 0 or 1 with such a [tex]\frac{1}{2}[/tex] probability. So, although B chooses [tex]kB(3)[/tex] from [tex](0, 1, 2, 3, 4, 5, 6, 7),[/tex] the probabilities each are [tex]\frac{1}{8}[/tex]:

[tex]\to P[A \ wins] = P[kA(3) < kB(3)][/tex]

[tex]= P[kA(3) = 0] \times P[kB(3) > 0] + P[kA(3) = 1] \times P[kB(3) > 1]\\\\= \frac{1}{2} \times \frac{7}{8} + \frac{1}{2} \times \frac{6}{8}\\\\= \frac{7}{16} + \frac{6}{16}\\\\= \frac{7+6}{16} \\\\= \frac{13}{16}\\\\[/tex]

For point c:

Assume that B tries again 16 times (typical value), and it destroys. In addition, throughout the exponential background n is obtained at 10 when choosing k between 0 to 2n−1. The probability of A winning all 13 backoff events is: [tex]P[A \text{wins remaining races}] = 16\pi i =4P[A \ wins \ i |A \ wins \ i -1 ][/tex]

Let the k value kA(i) be A for the backoff race I select. Because A retains the breed

[tex]=(kA(i)] \cdot P[kA(i+ 1)< kB(i+ 1)] \geq P[kA(i) + 1<kB(i)] \cdot P[kA(i+ 1)< kB(i+1)]+P[kA(i) + 1 \geq kB(i)] \cdot P[kA(i+ 1)< kB(i+ 1)] \\\\= (P[kA(i) + 1< kB(i)] +P[kA(i) + 1 \geq kB(i)]) \times P[kA(i+ 1) < kB(i+ 1)]\\\\=P[kA(i+ 1)< kB(i+ 1)]\\\\[/tex]

For point d:

Two stations A and B are supposed. They assume that B will try 16 times afterward. Even so, for A, 16 races were likely to also be won at a rate of 0.82 For Just higher expectations of three A, B, and C stations. For Station A, possibility to win all backoffs