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A and B are the only two stations on an Ethernet. Each has a steady queue of frames to send. Both A and B attempt to transmit a frame, collide, and A wins the first backoff race. At the end of this successful transmission by A, both A and B attempt to transmit and collide. The probability that A wins the second backoff race is

  1. 0.5
  2. 0.625
  3. 0.75
  4. 1.0
asked in Computer Networks by Veteran (69k points) | 2.2k views


----->>>>i think it helps... 

3 Answers

+36 votes
Best answer

Find this solution. 

answered by (171 points)
selected by
nice explanation
So if a station wins the back-off, its K resets to 0?
k is the attempt number, if a packet is sent for the first time k=1, second time k=2 and so on.............
Exponential backoff algo says in the event of collision all stations will wait till (0 to 2^k-1)*propagation time.

Where k=no of times stations applied back off algorithm.

So in first time k=1 when collision occurs

Both stations applies back off algo .

For A : waiting time=0,1*pt

For B : waiting time=0,1*pt

In first case A wins the race  so probability ={favorable cases in which A can wins}/{all possible cases in which they can applies back off algo}

=1/4=.250

Out of 4 only in 1 case (when wt of A=0 and wt of B=pt)

 

For second time(k=2) A wins =6/16=.375

6? ( Bcz possible case with 0 ,1*pt,2*pt,3*pt) in which A wins race

.{0,pt},{0,2*pt},{0,3*pt}=3

{1*pt,2*pt},{1pt,3*pt}=2

Last case in which A wins race is {2*pt,3*pt}=1

So total =6

Hence .250+.375=.625
+5 votes

watch the first 12 mins .. it's explained very well . 

answered by Loyal (4k points)
edited by
–2 votes
ans b)
answered by Boss (5.1k points)
can you explain how did you calculate?


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