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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 (59.4k points) | 2.5k views
+9


----->>>>i think it helps... 

+1

3 Answers

+37 votes
Best answer

Find this solution. 

answered by (171 points)
selected by
+1
nice explanation
0
So if a station wins the back-off, its K resets to 0?
0
k is the attempt number, if a packet is sent for the first time k=1, second time k=2 and so on.............
0
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
0
why did you add the two probabilities? , he is justing asking for the second time.
0
Yes this is wrong way,i dont know what i did, please find the correct one:::::::

at K=1 waiting time of station A=(0,1)*p.t.

 at K=1 waiting time of station B=(0,1)*p.t.

For the second time,  Since st.A wins the race(means A has sent data safely) so it will transfer new frame and   for the second time station B will continue with old frame(it has not transferred old data)

So for A, waiting time=(0 to (2^1)-1)*p.t.

Station A transferring again after successful transferring at k=1, So we will start backoff algorithm for this frame with fresh parameter,so  K=1 because its transferring new frame

That means waiting time of station of A ={0,1*p.t.}

For station B waiting time would be (0 to (2^2)-1)*p.t. here k=2 because count for THIS frame(old frame) become 2 since old backoff algo is not ended.

That means waiting time of station B  {0, 1*p.t, 2*p.t, 3*p.t.}

You can see there are 8 possible pair  out which 5 would be favourable case for A to win the race

Pair in which A wins i.e (A,B)={.   (0,1*p.t),(0.,2*p.t,),   (0,3*p.t ),(1*p.t,2*p.t),(1*p.t.,3*p.t)    }

So 5/8=.625

Hope you Will get my approach.
+6 votes

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

answered by Active (2.2k points)
edited by
–2 votes
ans b)
answered by Active (5.2k points)
+6
can you explain how did you calculate?


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