48 votes 48 votes An $IP$ router with a $\text{Maximum Transmission Unit (MTU)}$ of $1500$ bytes has received an $IP$ packet of size $4404\text{ bytes}$ with an $IP$ header of length $20\text{ bytes}$. The values of the relevant fields in the header of the third $IP$ fragment generated by the router for this packet are: $\text{MF bit}$: $0,$ Datagram Length:$1444;$ Offset$: 370$ $\text{MF bit}$: $1,$ Datagram Length$: 1424;$ Offset$: 185$ $\text{MF bit}$: $1,$ Datagram Length$: 1500;$ Offset$: 370$ $\text{MF bit}$: $0,$ Datagram Length$: 1424;$ Offset$: 2960$ Computer Networks gatecse-2014-set3 computer-networks ip-packet normal + – go_editor asked Sep 28, 2014 • edited Apr 16, 2019 by Pooja Khatri go_editor 21.3k views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply PRANAV M commented Jul 13, 2018 reply Follow Share HOW TO FIND THE OFFSET VALUE?, CAN ANYONE EXPLAIN PLEASE!!!! 1 votes 1 votes Harshada commented Dec 9, 2018 reply Follow Share "Total length of the datagram" field is of 16 bits and "Fragmentation Offset" field is of 13 bits. 2^16-1 cannot be represented using 13 bits therefore 2^16/2^13=8, 8 is the scaling factor and that's why offset is divided by 8. 1st Fragment's Offset :0 2nd Fragment's offset = Previous Fragment's Offset +(Data bytes of prev fragment/8) and so on. 4 votes 4 votes shaktisingh commented Nov 28, 2019 reply Follow Share If they ask what is the length of Internet header length(IHL)? IHL has 4 bit field and we have IP header length = 20bytes answer would be 20/4 = 5 where 4 as scaling factor saw somewhere. please explain this? 0 votes 0 votes Ankit Kabi commented Oct 6, 2020 reply Follow Share Yes, as header length is 4bits and number represented by it is 0-15. As we can represent 20-60 header size so we scale down by 4 to represent 20-60. And 0-3 is not used. And required padding bits should be added to make it divisible by 4. 1 votes 1 votes Please log in or register to add a comment.
0 votes 0 votes IP packet = 4384+20 in which data = 4384 MTU = 1480+20 in which data = 1480 no of packets = ceil(log(4384/1480) = 3 first packet offset = 0*1480/8 second packet offset = 1480/8 third packet offset = 3*1480/8 = 370 MF bit 0 as more fragments = 0 as it is the third and last packet we are talking about. shashankrustagi answered Aug 3, 2020 shashankrustagi comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Given datagram length = 4404B (payload = 4384, header = 20B) Ist IInd IIIrd 4384 ---→ [1480:20], [1480:20], [1424:20] At Third fragment MF= 0 frame offset= (1480+1480)/8=370 datagram length=(1424+20)=1444B ZaheenAfzal answered Sep 29, 2021 ZaheenAfzal comment Share Follow See all 0 reply Please log in or register to add a comment.