ME-FT-8

Question

Assume that A and B are only active stations on an ethernet.both has a steady queue of frames to send .to get the control on the channel they uses binary exponential algorithm.both are attempting to transmit the frame.the probability that both are successfully allowed to send the frame on 4th round and in earlier attempts both A and B collides is _

answer is given as  0.11

Comments

according to me it should be ,

p=p(collision in 1st attempt )*p(collision in  2​​​nd attempt )*p(collision in 3rd attempt )*p(no collision in 4th attempt )

=1/2*1/4*1/8*(1-1/16)

=1/64*15/16=.0146

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I think in ethernet, only 1 station can trasmit data at a given time. Here how can both stations simultaneously send data? Someone Explain

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Its 1*1/2*1/4*(1-1/8)

= 0.1093

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@Shamim Ahmed

how $1$ st collision takes probability of collision $1$ ?

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Check this https://gateoverflow.in/30257/csma-cd

What I think is, rounds are being counted from the first time of transmission only. Its that transmission which finally leads to collision and starts the back off algo..

Here according to @Prateek Raghuvanshi ,I think your round started with the backoff algo i.e. after the 1st collision. That is why it had 2 contention slots per station and probability of collision was $\frac{1}{2}$.

But in the 1st time, when both transmit at the same time, there will be only 1 contention slot (backoff hasn't started yeti.e. before any collision) and they will transmit in that one slot only which is surely lead to collision. That is why probability of collision in 1st round=1.

I would have also done like you, but then see the link which I gave. The answer there was selected by Bikram Sir.

Correct me here..

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thanks alot @MiNiPanda 

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@MiNiPanda @Prateek Raghuvanshi

Plz chk the comment of shivam chauhan

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@srestha

http://omikron.eit.lth.se/ETSN01/ETSN01/tutorials/Tutorial8.pdf

Pg 5 problem 4

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@srestha

Ma'am, Both A and B would send frames and there is a collision. Both would wait for Backoff algorithm.

 

So both A and B can take value from single contention slot. Thats why the probability is 1.

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@MiNiPanda

what is 1 meaning in 1st slot u told?

chk this

After c collisions, a random number of slot times between 0 and 2c − 1 is chosen. After the first collision, each sender will wait 0 or 1 slot times. After the second collision, the senders will wait anywhere from 0 to 3 slot times (inclusive). After the third collision, the senders will wait anywhere from 0 to 7 slot times (inclusive), and so forth. As the number of retransmission attempts increases, the number of possibilities for delay increases exponentially.

https://en.wikipedia.org/wiki/Exponential_backoff 

Answer 1

1*0.5*0.25*(1-0.125)=0.11