1 votes 1 votes 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} Computer Networks gateforum-test-series computer-networks lan-technologies + – Gupta731 asked Jan 13, 2019 edited Mar 12, 2019 by ajaysoni1924 Gupta731 585 views answer comment Share Follow See all 7 Comments See all 7 7 Comments reply Magma commented Jan 13, 2019 reply Follow Share A ) ?? 0 votes 0 votes Gupta731 commented Jan 13, 2019 reply Follow Share No 0 votes 0 votes Magma commented Jan 13, 2019 reply Follow Share @Shobhit Joshi 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 ?? 0 votes 0 votes Gupta731 commented Jan 13, 2019 reply Follow Share C is provided as answer. 0 votes 0 votes Magma commented Jan 13, 2019 reply Follow Share Ouu I got it brother :3 wait I explain you 0 votes 0 votes Magma commented Jan 13, 2019 reply Follow Share 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) 3 votes 3 votes Shobhit Joshi commented Jan 13, 2019 reply Follow Share 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 $\left \{ 0,1,2,3\right \}$ See here 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes The answer is D because after collision, n=2 for both A and B therefore, k={0,1,2,3} for both A and B anwesha-me answered Oct 20, 2020 anwesha-me comment Share Follow See all 0 reply Please log in or register to add a comment.