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Consider sending a 2400-byte datagram into a link that has an mtu of 700 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(s) generated related to fragmentation?

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Limosa

Explanation:

Let, DG is the datagram so, DG= 2400.

Let, FV is the Value of Fragment and F is the Flag and FO is the Fragmentation Offset.

Let, M is the MTU so, M=700.

Let, IP is the IP header so, IP= 20.

Let, id is the identification number so, id=422

Required numbers of the fragment = [tex][\frac{DG-IP}{M-IP} ][/tex]

Insert values in the formula = [tex][\frac{2400-20}{700-20} ][/tex]

Then,        = [tex][\frac{2380}{680} ][/tex] = [tex][3.5][/tex]

The generated numbers of the fragment is 4

  • If FV = 1 then, bytes in data field of DG= [tex]720-20 = 680[/tex] and id=422 and FO=0 and F=1.
  • If FV = 2 then, bytes in data field of DG= [tex]720-20 = 680[/tex] and id=422 and FO=85[tex](85*8=680 bytes)[/tex] and F=1.
  • If FV = 3 then, bytes in data field of DG= [tex]720-20 = 680[/tex] and id=422 and FO=170[tex](170*8=1360 bytes)[/tex] and F=1.
  • If FV = 4 then, bytes in data field of DG= [tex]2380-3(680) = 340[/tex] and id=422 and FO=255[tex](255*8=2040 bytes)[/tex] and F=0.

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