Backoff Algorithm

After some transmission on Ethernet, current frame of A's is collided 2 times and B's is collided 4 times. The probability that A wins the present back off race is____.

Answer

Number of times A has been involved in collision = 2
Number of times B has been involved in collision = 4
 A can choose any number between (0 to 22-1 ) , i.e, (0,1,2,3)
B can choose any number between (0 to 24-1 ) , i.e, (0,1,2,3,......,15)
It is given that A wins the race, so choices for (A,B) will be:
(0,1) , (0,2) ,(0,3)................ , (0,15)
(1,2), (1,3),...........................,(1,15)
(2,3),(2,4),............................,(2,15)
(3,4),(3,5),.............................,(3,15)
Total choices = 15 + 14 + 13 + 12 = 54 out of 64 choices.

So probability = 54 / 64 = 0.84

7Comments
Anurag Lahon @anuraglahona
28 Dec 2016 09:43 pm

wins the race means ,A takes less time?

Shraddha @shraddhagami
30 Dec 2016 12:28 am

Yeah if it takes less time then it's waiting time is less than B,therefore A wins.

Busy Lover @asdfasdf
28 Dec 2016 09:45 pm

Because that's what <b>it</b> means!

Parth Gadoya @parthgadoya
31 Dec 2016 10:44 am

What is probability of collisons? is it 4/64.

Jagmeet @jagmt
5 Jan 2017 04:27 pm

Yes!
1 Collision: (0,0)(1,1)(2,2)(3,3) {4/64}.
2 B wins the race: (1,0)(2,0)(3,0)(2,1)(3,1)(3,2) {6/64}

Anmol Verma @avdominic
26 Jan 2017 03:52 pm

I think the ans given in this question is of B's winning the race which is 54/64

For A's winning the race it is 6/64......

and for collision it is 4/64....

Correct me If I'm wrong......:-)

Govindarajan @govindj
4 Feb 2017 11:04 am

For collision both A and B's waiting time must be equal  (0,0) (1,1) (2,2) (3,3)      - only 4 possibilities out of 64              for collision

For  B to win it has to transmit before A (1,0) (2,0) (2,1) (3,0) (3,1) (3,2)   -    only 6 possibilities out of 64                           for B to win the race. 

In remaining all cases, A will have lesser waiting time than B  so A will win the race  in  -  54 possibilities out of 64 .       for A to win the race.