3 votes 3 votes header size not given. tiger asked Jan 4, 2016 edited Jul 7, 2017 by Prashant. tiger 1.9k views answer comment Share Follow See all 7 Comments See all 7 7 Comments reply Show 4 previous comments tiger commented Jan 4, 2016 reply Follow Share ceil(1600/480) = 4 fragments each fragment take 20B header ie extra 80bytes 1600+80 = 1680 0 votes 0 votes resuscitate commented Jan 4, 2016 reply Follow Share yeah 1680 is answer,ur approach is wrong ..i made mstk previously..i fragmented the packet is A also,that why 1700 is coming.. –1 votes –1 votes Gaurab Ghosh commented Oct 20, 2016 reply Follow Share @resuscitate I don't know why you changed your approach but you were right the first time. Answer must be 1700 bytes only. And number of fragments must be 5. At A we fragment the main bigger datagram into 2 fragments where one fragment contains 1480 bytes of data (divisible by 8) and the second one contains 120 bytes. Now at the first router R the first fragment is further fragmented into 4 fragments each of 456 data bytes (divisible by 8). So at B we received total 5 packets each with 20B of header. Therefore extra bytes received is (5x20)=100B. answer =1600+100=1700. 1 votes 1 votes Please log in or register to add a comment.
Best answer 5 votes 5 votes There are 5 fragments. After first network 1480+20 120+20 After second network 1480 packet will be divided into 456+20 456+20 456+20 112+20 And second fragment is 120+20 So total will be (120+20)+(456+20)+(456+20)(456+20)+(112+20)=1700 bytes... so 1700 Bytes transmitted. The almost same question also asked in gate 2016 set 1 http://quiz.geeksforgeeks.org/gate-gate-cs-2016-set-1-question-63/ also same type of problem you can found from Kuroe & ross book , chapter 4 , No-19p https://www.chegg.com/homework-help/consider-sending-2400-byte-datagram-link-mtu-700-bytes-suppo-chapter-4-problem-19p-solution-9780133128093-exc Bikram answered Jan 11, 2017 Bikram comment Share Follow See all 2 Comments See all 2 2 Comments reply MohammedSunasra commented Jan 11, 2017 reply Follow Share How did you get 5 fragments? 0 votes 0 votes Bikram commented Jan 11, 2017 reply Follow Share At A we fragment the main bigger datagram into 2 fragments where one fragment contains 1480 bytes and the second one contains 120 bytes. At the first router R the first fragment is further fragmented into 4 fragments each of 456 data bytes. so in total 4 + 1 = 5 fragments. 1 votes 1 votes Please log in or register to add a comment.
1 votes 1 votes mtu is 480. payload is 460 B. A sends payload 1600 B to B so total data A sends to B = (460+20,460+20,460+20 ,220+20) = 1680 B. resuscitate answered Jan 4, 2016 resuscitate comment Share Follow See all 3 Comments See all 3 3 Comments reply tiger commented Jan 4, 2016 reply Follow Share ohh thanx –1 votes –1 votes Bikram commented Oct 22, 2016 reply Follow Share 5 fragments. After first network 1480+20 120+20 After second network 1480 packet will be divided into 456+20 456+20 456+20 112+20 And second fragment is 120+20 So total will be (120+20)+(456+20)+(456+20)(456+20)+(112+20)=1700 bytes... so 1680 is not the answer. The almost same question also asked in gate 2016 set 1 http://quiz.geeksforgeeks.org/gate-gate-cs-2016-set-1-question-63/ also same type of problem you can found from Kuroe & ross book , chapter 4 , No-19p https://www.chegg.com/homework-help/consider-sending-2400-byte-datagram-link-mtu-700-bytes-suppo-chapter-4-problem-19p-solution-9780133128093-exc 2 votes 2 votes kaival commented Aug 23, 2020 reply Follow Share Here we took 456 (not 460) as it is closest divisible by 8 that is required in Offset field, thats the reason, So if you calculate through 460 and has 4 fragments than the answer is 1680 which is wrong. Thanks 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes 1700 is the actual size of the data sh2mohit111 answered Oct 9, 2017 sh2mohit111 comment Share Follow See all 0 reply Please log in or register to add a comment.