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anastassius [24]
3 years ago
10

Consider sending a 2400-byte datagram into a link that has an mtu of 700 bytes. suppose the original datagram is stamped with th

e 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?
Computers and Technology
1 answer:
Feliz [49]3 years ago
3 0

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 = [\frac{DG-IP}{M-IP} ]

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

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

The generated numbers of the fragment is 4

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

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