csma/cd

Question

Two csma/cd stations are trying to send frames..After each frame is sent they contend for channel using backoff exponential algorithm?What is probability that contention ends on round k?

Comments

This is tanenbaum problem

Chapter 4 Problem 16P , edition- 5th

http://www.chegg.com/homework-help/two-csma-cd-stations-trying-transmit-long-multiframe-files-f-chapter-4-problem-16p-solution-9780133072624-exc

Answer 1

Collision always occurs in round $1$. So probability that successful transmission happens in round 1 = 0.

P(one of station transmits data in kth round) $= 1*(1/2)*(1/2^2)*......*(1/2^{k-2})*(1 - 1/2^{k-1})$

Comments

@harsh 

i am confirm now, this answer is correct and the formula you can see clearly here http://cesg.tamu.edu/~sshakkot/courses/ECEN424/Solution-3.pdf  in page 5 check 2nd line 

it says 

P( k-1 failures followed by success ) =1∗(1/2)∗(1/2²)∗......∗(1/2^k−2)∗(1−1/2^k−1)

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@Habibkhan 
The link sir has given here: http://omikron.eit.lth.se/ETSN01/ETSN01/tutorials/Tutorial8.pdf

I still don't get how they obtained Pk here.

why are we multiplying the first term here?
If you know any other method you may tell that too. Thanks in advance.

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@shraddha priya maam, I am unable to understand too.

Have you got any idea how it is working?

Answer 2

1-(1/2  + 1/4 + 1/8 + ....1/2^k)

Answer 3

Contention ends on round k means that earlier in k-1 trial it didn't succeed and finally it succeeded on kth Contention.

Probability of no success in ith Contention = 1/2^i

probability that contention ends on round k =  (1/2)(1/2^2)(1/2^3)...(1/2^k)*(1-1/2^k)

Comments

Can u give any resource reference for your explanation ? From where did you get -

Probability of no success in ith Contention = 1/2^i ?

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probability that contention ends on round k =  $\frac{1}{2}*\frac{1}{2^2}*\frac{1}{2^3}*..........*\frac{1}{2^{k-1}}*(1-\frac{1}{2^k})$