Ahmad although 1480 + 1480 + 1040 = 4000....the math doesn’t add up as to why the last fragment isn’t 1020 instead...but obviously doing that it would make it 3980...yeah. I’m also confused about the last part
The original datagram also has headers, if you remove that from it, you get the right answer. 4000(Total Length) - 20(Header) = 3980 1480+1480+1020 = 3980
doesn't have to, you wanna focus on data length which in this case is 1480*2 + 1020 which adds up to 3980... remember to substract the header for each fragment
The initial fragment has 4,000 total length, which includes the header of 20 bytes, so only 3,980 bytes of actual data. The individual fragments each have 20 byte headers, so: (1,500 - 20) + (1,500 - 20) + (1,040 - 20) = 3,980 (or 4,000 - 20)
He messed up the last part of the offset. You get 370 by dividing last position i.e. (1480+1480)/8= 370
why did we divide those first 2? what does last position mean>
the first 1480 is the data transfered by fragment 1 and the secend 1480 the data transfered by frag 2
ya i was like wtf is he saying lol
1500 + 1500 + 1040 doesnt add up to 4000
Ahmad although 1480 + 1480 + 1040 = 4000....the math doesn’t add up as to why the last fragment isn’t 1020 instead...but obviously doing that it would make it 3980...yeah. I’m also confused about the last part
Thanks for saving my semester
Thanks, my actual teacher was absolutely useless at explaining this.
If i heard correctly you said tcp 6 and udp 20 upper layer which is incorrect tcp 6 and udp 17
Hello! Can u upload sections 4.5 and 4.6 4.7 please?
if the first 2 have 1480 bytes of actual data and the last one has 1040 - 20(header) doesnt that add up to 3980 bytes?
The original datagram also has headers, if you remove that from it, you get the right answer.
4000(Total Length) - 20(Header) = 3980
1480+1480+1020 = 3980
Thanks so much for clearing that for me! much appreciated
raheem abdul it is divisible by 8
A life saver, thanks!
@@waleed_tq Do you basically mean that during reassembly our datagram will be 3980 and isn't that not correct?
Thank You So much. really thankful for the content!
Perceived sole insecurity in his voice
big box through a small hole lol
How was the last offset calculated?
(1480+1480)/8= 370
You're such a great teacher I regret I'd not subscribe to your channel earlier. Thanks for sharing this awesome stuff.
Nice vids! Thanks for the info.
1500 + 1500 + 1040 doesnt add up to 4000
doesn't have to, you wanna focus on data length which in this case is 1480*2 + 1020 which adds up to 3980... remember to substract the header for each fragment
The initial fragment has 4,000 total length, which includes the header of 20 bytes, so only 3,980 bytes of actual data.
The individual fragments each have 20 byte headers, so:
(1,500 - 20) + (1,500 - 20) + (1,040 - 20) = 3,980 (or 4,000 - 20)