Question

Consider sending a 1,600-byte datagram into a link that has an mtu of 500 bytes. suppose the original datagram is stamped with the identification number 291. how many fragments are generated? what are the values in the various fields in the ip datagram(s) generated related to fragmentation?

Answer 1

Explanation:

Step one

The maximum size of data field in each fragment = 480

(because there are 20 bytes IP header) Thus the number of required

fragments  $=\frac{1600-20}{480} \\\\= \frac{1580}{480} \\\\=3.29$

thus the number of required fragment is 4

Step two

Each fragment will have identification number 291. each fragment except the last one will be of size 500 bytes (including IP header). the offset of the fragments will be 0, 60, 120, 180. each of the first 3 fragments will have

flag = 1; the last fragment will have flag =0

Answer 2

Using the appropriate formula, the number of fragments which would be present in the datagram to be sent would be 4

The minimum length of IP header = 20 bytes

Maximum transmission unit (mtu) = 500 bytes

Hence, the payload would be calculated thus :

• mtu - header ;
• Payload = 500 - 20 = 480

Hence, the maximum size of data field per Fragment = 480 bytes

The number of fragments required :

• $\frac{datagram \: size - Header \: size}{payload}$

$Number \: of \: fragments = \frac{1600 - 20}{480} = 3.29$

Hence, the number of fragments is 4

• Size per Fragment would be 500 bytes each ; the last Fragment would be about 100 bytes

• Each Fragment would bear the identification number 291.

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