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28 June, 07:57

Assume that network MTU limitations necessitate that an IP datagram be split into two fragments of different sizes. In the resulting IP datagrams, indicate which of the following header fields is guaranteed to be the same and which could be different. You should be comparing the headers of the two fragments to each other, not to the header of the original. Briefly justify each answer.

Header fields:

IHL

Total length

Identification

D Flag

M flag

Fragment Offset

Header checksum

+1
Answers (1)
  1. 28 June, 08:22
    0
    Consider a packet of size 756 bytes. Enters a network having mtu=500. Now packet will be fragmented in two fragment.

    Fragment 1 : 480 data + 20 byte header.

    Fragment 2 : 276 data + 20 byte header.

    Let's compare header of the two fragments.

    Internet header length (IHL) ⇒ Due to same size of both fragment header, IHL value will be same. If header length is 20 IHL, then will contain 0101.

    Total length ⇒ Total length can be different. Looking at our case, fragment 1 contain 480 byte data while fragment 2 contains 276 byte so total length will be different.

    Identification ⇒ Identification is same for all fragment belonging to same packet

    D flag (don't fragment) ⇒ This is used to indicate weather packet is fragmented or not. D=1 not a fragment. D=0 fragment. So this D flag will be same.

    M flag (more fragment) ⇒ This is used to indicate more fragment present or not. M=1 more fragment are present. M=0 last fragment.

    Fragment offset ⇒This is used to indicate position of fragment among fragments. This value will be different for two fragment.

    Header checksum : As many other fields of two fragment is different thus checksum of two fragment will also be different.

    Explanation:

    See all the explanation in the answer.
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