Hi all For the given below example, the fragment offset in the second packet should be 185 right? The TCP header has not been considered while calcuating the offset in the given below example.
Snippet from http://www.tech-faq.com/packet-fragmentation.html A Packet Fragmentation Example If a 2,366 byte packet enters an Ethernet network with a default MTU size, it must be fragmented into two packets. The first packet will: - Be 1,500 bytes in length. 20 bytes will be the IP header, 24 bytes will be the TCP header, and 1,456 bytes will be data. - Have the DF bit equal to 0 to mean "May Fragment" and the MF bit equal to 1 to mean "More Fragments." - Have a Fragmentation Offset of 0. The second packet will: - Be 910 bytes in length. 20 bytes will be the IP header, 24 bytes will be the TCP header, and 866 bytes will be data. - Have the DF bit equal to 0 to mean "May Fragment" and the MF bit equal to 0 to mean "Last Fragment." - Have a Fragmentation Offset of 182 (Note: 182 is 1456 divided by 8). With regards Kings
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