7 votes 7 votes A IP packet has arrived in which the fragmentation offset value is $100,$ the value of $\textsf{HLEN}$ is $5$ and the value of total length field is $200.$ What is the number of the last byte? $194$ $394$ $979$ $1179$ Computer Networks computer-networks ip-packet isro2014 + – ajit asked Sep 23, 2015 edited Dec 4, 2022 by Lakshman Bhaiya ajit 9.5k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 16 votes 16 votes Data length $ = 200 – 4 \times 5=180$ Starting number of the first byte of fragment $ = 100\times8=800$ Number of the last byte $ = 800+179 = 979$ $\mathbf{Ans}\;\textbf C$ Pooja Palod answered Sep 23, 2015 edited Nov 1, 2021 by `JEET Pooja Palod comment Share Follow See all 11 Comments See all 11 11 Comments reply abhishek tripathi commented Jun 29, 2016 reply Follow Share why 179 is added could you please explain it? 0 votes 0 votes Danny commented Jul 1, 2016 reply Follow Share coz there are 180 bytes so + 0-179 @abhishek 0 votes 0 votes abhishek tripathi commented Jul 1, 2016 reply Follow Share @ danny thanx dude..I got it 0 votes 0 votes Vishal Goyal commented Jun 18, 2017 reply Follow Share 4*5 means 1 votes 1 votes Divya Krishnan commented Mar 25, 2018 reply Follow Share Someone please answer this 0 votes 0 votes ayush palak commented Apr 1, 2018 reply Follow Share hlen is of 4 bits and it can represent a maximum value of 15. while the header length can g upto 60. so we use a scale factor of 4. so now 4 bits can represent upto 15*4 = 60. so if the value of hlen is 5 then it means it is actually representing 5*4 = 20. hope that clears! 4 votes 4 votes pankajbelwal commented Apr 7, 2018 reply Follow Share why 100 is multiply by 8. Does it convert into bits? 1 votes 1 votes Raju Kalagoni commented Apr 18, 2018 reply Follow Share @ pankajbelwal , scaling factor of fragment offset is 8. 3 votes 3 votes shraddha priya commented Mar 6, 2019 reply Follow Share @ayush palak @Raju Kalagoni But shouldn't data be a multiple of 8? Here data size is 180, which is not a multiple of 8. 1 votes 1 votes Raju Kalagoni commented Aug 8, 2019 reply Follow Share @shraddha priya, data need not be a multiple of 8. it doesn't make sense to have data always in multiple of 8. We should always be in a position to send data across network either in multiple of 8 or not. 1 votes 1 votes commenter commenter commented Jan 3, 2020 reply Follow Share @shraddha priya Data should be a multiple of 8 in cases where there are more than 1 fragmented packets and all the packets except the last packet must be a multiple of 8. If there is only one packet or the packet is last fragmented packet then it need not be a multiple of 8. 1 votes 1 votes Please log in or register to add a comment.
4 votes 4 votes $\mathbf{Option \;C}$ Total Length=$200$ Data field excluding header=$ 200-20 = 180(0 \dots 179)$ No of byte ahead from this packet is $ = 100\times8 = 800$ so last byte = $800+179=979$ Paras Nath answered Sep 10, 2016 edited Nov 1, 2021 by `JEET Paras Nath comment Share Follow See all 0 reply Please log in or register to add a comment.
3 votes 3 votes Header Payload An IP packet looks like this. Header Length = $5*4=20$ So, Payload = $200-20=180$ Fragmentation offset = $100*8=800$ The last byte would be $180 + 800$th = $980th$ but since we number everything from 0 in computer science, it'll be $979th$ JashanArora answered Dec 8, 2019 JashanArora comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Here in this case answer should be 979. eshita1997 answered May 8, 2021 eshita1997 comment Share Follow See all 0 reply Please log in or register to add a comment.