0 votes 0 votes closed as a duplicate of: Made Easy Mock In the given network system, station A needs to send a payload of 1600 B from its network layer to station B. If fragmentation is done, then the actual data size to be transmitted is ______________ Computer Networks made-easy-test-series + – MiNiPanda asked Jan 18, 2019 • closed Jan 18, 2019 by MiNiPanda MiNiPanda 839 views comment Share Follow See all 8 Comments See all 8 8 Comments reply kumar.dilip commented Jan 18, 2019 reply Follow Share At network one we have to divide the 1600 byte data. At network 1:==> divide the packet into two part ( 1480 + 120 ) and add the 20 bytes of the head to each. At network 2 :==> MTU = 480, here we have to divide 1480 into three parts + 20 byte of header to each. So, total packets we have ==== 4. At network 3:==> No need to divide. the packes. So, we got 4 packets. 1680 === 1600 (data) + 80 ( header for four packets). 1680. 2 votes 2 votes balchandar reddy san commented Jan 18, 2019 reply Follow Share how are the packets sent? is it 460B+20B, since 460B is not divisible by 8,456B+30B is sent? 0 votes 0 votes MiNiPanda commented Jan 18, 2019 reply Follow Share @kumar.dilip Thank you :) I did a mistake which I realized from your solution :D 0 votes 0 votes Somoshree Datta 5 commented Jan 18, 2019 reply Follow Share What about the fragment containing 120B data and 20B header?why aren't u adding that? 2 votes 2 votes MiNiPanda commented Jan 18, 2019 reply Follow Share @Somoshree Datta 5 You are right.. @kumar.dilip then 20 B has to be added more right? 1700 B should be the ans then 0 votes 0 votes Somoshree Datta 5 commented Jan 18, 2019 reply Follow Share https://gateoverflow.in/296435/made-easy-mock#c296509 0 votes 0 votes kumar.dilip commented Jan 18, 2019 reply Follow Share Somoshree Datta 5 Yes, it should be 1700. 0 votes 0 votes OneZero commented Jan 18, 2019 reply Follow Share I am getting as follows network 1 : (1480B payload + 20B Header) + (120B payload + 20B Header) network 2 : (460B payload + 20 Header)*3 + ((120B payload + 20B Header)) + (120B payload + 20B Header) Hence total of 100B overhead. 0 votes 0 votes Please log in or register to add a comment.