Question

suppose two devices using CSMA/CD and the binary exponential backoff algorithm have just sent transmission that have each experienced collision Twice. what is tye probability that both devices will transmit successfully during the bext two time slots where n=3.

Answer 1

Exponential Backoff algorithm:

It is basically a collision resolution mechanism. It is mainly used in ethernet.

It is used schedule schedule retransmission after the collision.

e.g.,

A and B are two transmitting stations as shown in below figure:

А B

Both start transmitting at collide, and returned back to their respective stations because of collision.

Here, n = 1 i.e it is the collision number.

Therefore, waiting time will be to $2^n-1$ i.e., 0 to 21-1 = [0,1]

Therefore, in next transmission can either be at T=0 or T=1

A	B	Collision ( Yes/ No )
0	0	Yes
0	1	No
1	0	No
1	1	Yes

Probability of success tranmission after n=1 $= \frac{ total \ transmission \ in \ which \ there \ is \ no \ collision}{Total \ possible \ collision}$

2 4 1 2

Now,

Here total collision,

Waiting time can be $0 \ to \ 2^3 -1 = \left [ 0,7 \right ]$

So,

• Station A can transmit at T = 0 to T = 7
• Station B can transmit at T = 0 to T = 7

So, Total possible combinations = 8*8 = 64

but (0,0),(1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (7,7) , Total 8 transmissions will lead to collision,

Probability for successful transmission = $\frac{Total \ transmission \ in \ which \ there \ is \ no \ collision}{Total \ number \ of \ transmission}$

$=\frac{64-8}{64} = \frac{56}{64}$

$= \frac{7}{8}$

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