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dimulka [17.4K]
3 years ago
5

an IPv6 packet has a 40 byte base header, a 20 byte destination options extension header (which is not used for routing) and 200

0 bytes of payload data (total packet length is 2060 bytes). This is to be fragmented to travel over a network with a MTU of 1500 bytes. Considering the fragmentable and unfragmentable parts of this packet, create a sketch showing the headers and data in each fragment, including the number of bytes. Don’t forget the fragmentation header, which is 8 bytes.
Computers and Technology
1 answer:
NISA [10]3 years ago
7 0

Answer:

Explanation:

Given, total payload is: 2000 bytes

 MTU is: 1500 bytes

 Header is: 20 bytes

 Base header is: 40 bytes

Total number of fragments = Total Payload / MTU

                                            = 2500 / 1500

                                            = 2 fragments

Fragment     Base header    Fragmentation header     Authentication  

                                                                                                     header + Data

1                  40 bytes                  8 bytes                              20 bytes +772bytes

2                 40 bytes                   8 bytes                                 792 bytes

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