An IPv6 packet has a 40 byte base header, a 20 byte destination options extension header (which is not used for routing) and 2000 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.

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

go to edit profile, preferences, then choose level and select the desired choice.

only #2 represents an object at rest on a table.

in #1, there is no normal force due to the table; the object is accelerating downward.

in #3, the vertical forces are not balanced; the object is accelerating downward.

in #4, neither the vertical nor the horizontal forces are balanced. the object is accelerating upwards and to the right.