Question

Consider sending a 2400 byte diagram into a link that has an MTU of 700 bytes. Only the second fragment of the datagram has to further go through another link with MTU of 400 bytes. Suppose the original datagram is stamped with the identification number 422. How many fragments are generated? What are the values in the various fields in the IP datagram generated related to fragmentation?

Answer 1

datagram size is 2400 bytes including ip header size.so only payload size is 2380bytes.

MTU 700B also includes ip header.allowable size 700-20=680B.

1)first fragment is of size 680B.offset is (0-84)

2)second fragment is also size 680B.it again fragmented because it is greater than MTU size.MTU ip header,allowable datasize is 380B.

i)first fragment size in second one is 376B.offset is (85-131)

ii)second fragment size in second one is 304B.offset is(132-169)

3)third fragment is also size 680B. offset is (170-254)

4)fourth fragment is of size 344B.offset is (255-297).it includes 4 padding bits.

so total 5 fragments .

Comments

packet size is in the form of bytes,so it should be divisible by 8.

in the first fragment 680mod 8=0(so it is divisible by 8)we can send entire 680B of data.

offset is nothing but address of the packet.680=85*8.(so 0-84).

coming to 2nd MTU(380)

same as above it is not exactly divisible by 8.376 is divisible by 8 so we can send only 376B of data at a time through network.