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

@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