1 votes 1 votes The round trip propagation delay for a 100Mbps Ethernet having 48-bit jamming signal is 64 ms. What is the minimum frame size (in bytes)? Computer Networks computer-networks mac-layer ethernet csma-cd + – Sumit Singh Chauhan asked Aug 7, 2018 Sumit Singh Chauhan 1.6k views answer comment Share Follow See all 12 Comments See all 12 12 Comments reply Sumit Singh Chauhan commented Aug 7, 2018 reply Follow Share I am getting the answer as 6.49Mb. Please confirm is it correct or not. 0 votes 0 votes Shaik Masthan commented Aug 7, 2018 reply Follow Share what the answer they given ? i am getting 0.8 M 0 votes 0 votes Sumit Singh Chauhan commented Aug 7, 2018 reply Follow Share I don't have the answer of this. I think you did like this, Minimum frame size = RTT * Bandwidth = 64 * 10-3 * 100 * 106 = 0.8 Mbps My Query is that the 48 bit has to be included in Frame if we are considering TT = RTT and if this is the case we have to delete that right? Please let me know your view. 0 votes 0 votes Shaik Masthan commented Aug 7, 2018 reply Follow Share 48 bit has to be included in Frame why? 0 votes 0 votes Sumit Singh Chauhan commented Aug 7, 2018 reply Follow Share As we are taking TT = 2PT that is we are talking about csma/cd and there the jamming bit has to be included in the frame right. 0 votes 0 votes Shaik Masthan commented Aug 7, 2018 reply Follow Share according to me, frame means either data or jamming signal but not both at a time, moreover, when collision identified by the station, then it transmit jam signal 1 votes 1 votes Sumit Singh Chauhan commented Aug 7, 2018 reply Follow Share Might be i am misunderstanding the concept. Please let me know one thing that what is the use of mentioning the Jam signal size. They can simply mention that the jamming signal without size. Isn't it? 0 votes 0 votes Shaik Masthan commented Aug 7, 2018 reply Follow Share i hope there is no need of even talking JAM signal 0 votes 0 votes Sumit Singh Chauhan commented Aug 7, 2018 reply Follow Share But if the question doesn't talks about the Jamming signal, how do we got to know that we have to consider TT = 2PT? 0 votes 0 votes Shaik Masthan commented Aug 7, 2018 reply Follow Share how do we got to know that for detect collision, we have L ≥ 2 * Tp 0 votes 0 votes Sumit Singh Chauhan commented Aug 7, 2018 reply Follow Share Please share the way you solved it. 0 votes 0 votes Sumit Singh Chauhan commented Aug 7, 2018 reply Follow Share Can you please help me with theses as well? https://gateoverflow.in/231193/made-easy-workbook https://gateoverflow.in/231326/made-easy-workbook-computer-networks 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes WHENVER THERE IS COLLISION THEN JAMMING SIGNAL IS USED UPTO 2PT. FOR CALCULATING MIN FRAME SIZE WE USE TT = 2*PT GIVEN RTT = 2*PT = 64 MILLISEC LET FRAME SIZE = X BITS -> X/100*10^6 = 64*10^-3 -> X = 64 * 10^5 BITS = 8*10^5 BYTE = 800000 BYTES sunnykg answered Aug 12, 2018 sunnykg comment Share Follow See 1 comment See all 1 1 comment reply Ashish Goyal commented Dec 12, 2018 reply Follow Share It would be 8000006 bytes. The jamming-signal bits would be added, not because of the sender sending jamming signal in the frame but because of the increase in the Tt for sender. Check out my solution on https://gateoverflow.in/270488/geeksforgeeks?show=277512#a277512 and Arjun Sir's response here- https://gateoverflow.in/1397/gate2005-74 1 votes 1 votes Please log in or register to add a comment.