Question

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?

Answer 1

Following are the solution to the given question:

Step-by-step explanation:

Please find the complete and correct question in the attachment file.

For Point a:

For the second round,

A is selects kA(2) either 0 or 1, so for each of them, that is
`(1)/(2)`.

B selects
`kB(2)\ from\ (0, 1, 2, 3)` for each choice with the probability of
`(1)/(4)`.

If
`kA(2) < kB(2)` wins the second rear race.

`\to P[A \ wins] = P[kA(2) < kB(2)]`

`= P[kA(2) = 0] * P[kB(2) > 0] + P[kA(2) = 1] * P[kB(2) > 1]\\\\= (1)/(2) * (3)/(4) + (1)/(2) * (2)/(4) \\\\=(3)/(8) +(2)/(8) \\\\= (3+2)/(8)\\\\= (5)/(8)`

For Point b:

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

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

`= P[kA(3) = 0] * P[kB(3) > 0] + P[kA(3) = 1] * P[kB(3) > 1]\\\\= (1)/(2) * (7)/(8) + (1)/(2) * (6)/(8)\\\\= (7)/(16) + (6)/(16)\\\\= (7+6)/(16) \\\\= (13)/(16)\\\\`

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:
`P[A \text{wins remaining races}] = 16\pi i =4P[A \ wins \ i |A \ wins \ i -1 ]`

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

`=(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)]) * P[kA(i+ 1) < kB(i+ 1)]\\\\=P[kA(i+ 1)< kB(i+ 1)]\\\\`

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