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Consider a scnario of CSMA/CD network. Suppose A & B attempt to transmit a frame simultaneously and collide. Using exponential backoff algorithm A chooses K=0 and B chooses K=1. Again they collide and after the 2nd collision both A & B will choose K with equal probability from set:

- For A {0,1} and for B {0,1}
- For A {0,1,2,3} and for B {0,1}
- For A {0,1} and for B {0,1,2,3}
- For A {0,1,2,3} and for B {0,1,2,3}

0

tell me one thing that after ist collision

A chooses K = 0 and B chooses =1

then there's no collision right ??

then they send 2nd packet

again they collide (ist collision)

after the 2nd collision means 2nd packet collide 2 times

so both chooses the number between (0,1,2,3)

therefore D right ??

My concept is right or wrong brother ??

+1

First both A and B send at time slot = 0

and collide for ist time

Now they send 2nd time in which

A = 0 B = 1 which means A need not wait at all while B has to wait for 1 time slot before re-transmit the packet

Now A send the packet , after 1 time slot when B is try to send his ist packet at that same time slot A is also send his 2nd packet

and there's a collision

so there's ist collision for packet 2nd which is send by A can take value from (0 to $2^{n} -1$ ) [where n is no of collision] = ( 0 to = 1 )

and 2nd collision for packet ist which is send by B , so B can take value from (0 to $2^{n} -1$ ) = (0 to 2^2 - 1) = (0 to 3)

and collide for ist time

Now they send 2nd time in which

A = 0 B = 1 which means A need not wait at all while B has to wait for 1 time slot before re-transmit the packet

Now A send the packet , after 1 time slot when B is try to send his ist packet at that same time slot A is also send his 2nd packet

and there's a collision

so there's ist collision for packet 2nd which is send by A can take value from (0 to $2^{n} -1$ ) [where n is no of collision] = ( 0 to = 1 )

and 2nd collision for packet ist which is send by B , so B can take value from (0 to $2^{n} -1$ ) = (0 to 2^2 - 1) = (0 to 3)

0

I think it would be $(C)$

What would happen according to me,

After first collision occurs, $A$ chooses from $\left \{ 0,1\right \}$ and same for $B$

Now, as $A$ won the first backoff race as it chooses $0$ and $B$ chooses $1$, so after $2^{nd}$ collision it still chooses from $\left \{0,1 \right\}$ but $B$ has to choose from $\left \{ 0,1,2,3\right \}$

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