0 votes 0 votes Consider Noisy station that detects transmissions and disrupts them by beginning a competing transmission as soon as it hears the beginning of the transmitted frame, thereby causing a collision. Assume Detector machine, which is on this Ethernet with bandwidth 10 Mbps, detects collision during the transmission of its 12th bit on the wire (including any preamble). If the speed of the signal in the wire is 2 × 108 meter/second, then the distance (in meters) of the Noisy station from Detector machine is _______ Somoshree Datta 5 asked Nov 2, 2018 Somoshree Datta 5 1.3k views answer comment Share Follow See all 13 Comments See all 13 13 Comments reply Utkarsh Joshi commented Nov 2, 2018 reply Follow Share 24? 0 votes 0 votes Somoshree Datta 5 commented Nov 2, 2018 reply Follow Share Utkarsh Joshi answer is given as 120 0 votes 0 votes Utkarsh Joshi commented Nov 2, 2018 reply Follow Share okay! are you getting the same answer? 0 votes 0 votes Somoshree Datta 5 commented Nov 2, 2018 reply Follow Share actually i did it this way..dont know whether the approach is correct or not.. Let the distance between the noisy station and detector be d metres. As the collision is detected during the transmission of the 12th bit, so 12 bits have already been transmitted into the network. So, as Tt>=2*Tp, so 12/(10*106) = (2*d)/(2*108) SO d=120m. But dont know whether this approach is correct..Is it correct? 1 votes 1 votes Magma commented Nov 2, 2018 reply Follow Share nice approach It seems to be right 0 votes 0 votes Utkarsh Joshi commented Nov 2, 2018 reply Follow Share why are you considering Tt >= 2Tp?Somoshree Datta 5 0 votes 0 votes kumar.dilip commented Nov 2, 2018 reply Follow Share @Utkarsh Joshi This is Collision Detection Condition Tt >= 2Tp One Important thing if The packet size if less than this then we can't detect the collision L >= 2*Tp*B 0 votes 0 votes Somoshree Datta 5 commented Nov 2, 2018 reply Follow Share Utkarsh Joshi because this is the condition for detecting the collision which needs to be fulfilled..this is the reason why detector m/c is being able to detect the collision 0 votes 0 votes Utkarsh Joshi commented Nov 2, 2018 reply Follow Share Somoshree Datta 5 By enforcing this condition the transmitting station can detect the collision on its own. But in this question, we have a detector machine which can sense the collision. So I don't see any reason for satisfying this condition. 1 votes 1 votes kumar.dilip commented Nov 2, 2018 reply Follow Share Bro, Ethernet uses CSMA/CD as access control method. So, here Detector machine is CSMA/CD. So, we have to Apply that condition. 0 votes 0 votes Utkarsh Joshi commented Nov 2, 2018 reply Follow Share okay, i agree it can be 120. 0 votes 0 votes kumar.dilip commented Nov 2, 2018 reply Follow Share This is the Question from Lan Technology (Ethernet). 0 votes 0 votes jatin khachane 1 commented Nov 26, 2018 reply Follow Share @Somoshree Datta 5 @Utkarsh Joshi please check this..correct me if i am wrong In LAN technology Why Tt >= 2Tp ?.. because in worst case for collision signal to come back to sender can take 2Tp time and during this time we want sender to be in sending state hence we equate Tt >= 2Tp to help Sender understand that collision happened because of him so Stop transmission..But this is for Sender only what about other how can they understand collision ..Hence JAM signal is send when collision signal reached to sender or receiver first. If we can consider Collision detector as Simple Host in Ethernet which detects collision during its transmission Hence Tt >= 2TP 0 votes 0 votes Please log in or register to add a comment.
Best answer 1 votes 1 votes Using the collision Detection Formula L >= 2*Tp*B We can derive the distance as d <= (L/2)*(V/B ) Here d is the maximum distance up to which we can detect the collision So, Here L = 12 , B = 107 , v = 2*108 then distance d = 120 kumar.dilip answered Nov 2, 2018 • selected Nov 2, 2018 by Somoshree Datta 5 kumar.dilip comment Share Follow See 1 comment See all 1 1 comment reply Magma commented Nov 2, 2018 reply Follow Share this same approach is also followed my Somoshree Datta 5 what's the difference ?? 1 votes 1 votes Please log in or register to add a comment.