self-doubt backoff algorithm

Question

As we know according to backoff algorithm, Suppose there are two stations A and B the range of slots that can be choosen is 0 to 2^n -1 , where n is collision number.
My doubt is :-
What is the initial value of n(collision number) of each stations A and B? 1 or 0??
What happens to collision number, if  A(suppose) wins, is it's n is reinitialized to 1 or 0?

Answer 1

I think you would better understand Back off algorithm after I provide some more clarity.

The range from which a number can be selected lies between $0$ to $2^{n-1}$.

When the collision number is 1, then the range is between $0$ to $2^{1-1}$ = $0$ to $2^{0}$ = $0$ to $1$. So now A and B can choose a number between this range.

Subsequently if collision occurs again, say A and B both chose $0$ then this time collision number would be $1$. So now number would be between $0$ to $2^{2-1}$ = $0$ to $2^{1}$ = $0$ to $2$

and process goes so on..

So, as you can see there is nothing as re-initialization of particular number, only range is re-initialized.

Comments

Thanks for you reply, But i think the range should be  2^n -1 not 2^(n-1).

Answer 2

The initial value of collision number is 1 for each of the stations  because station's data packet will come into collision for the first  time so 0 to 2^n -1 =0 to 1 =>{0 ,1} for each station

when A wins so value of collision number will be initialised to 1 because A has sent the data packet and when A want to send second data packet ,so second data packet will come into collision for the first time  and for B collision number will be 2 mean 0 to 3 =>{0,1,2,3} because B lost for the first time and that first packet will come into collision for the second time that's why collision number for B is 2 .