IP Fragmentation-

 

• IP Fragmentation is a process of dividing the datagram into fragments during its transmission.
• It is done by intermediary devices such as routers at the destination host at network layer.

 

 

Need-

 

• Each network has its maximum transmission unit (MTU).
• It dictates the maximum size of the packet that can be transmitted through it.
• Data packets of size greater than MTU can not be transmitted through the network.
• So, datagrams are divided into fragments of size less than or equal to MTU.

 

Datagram Fragmentation-

 

When router receives a datagram to transmit further, it examines the following-

• Size of the datagram
• MTU of the destination network
• DF bit value in the IP header

 

Then, following cases are possible-

 

Case-01:

 

• Size of the datagram is found to be smaller than or equal to MTU.
• In this case, router transmits the datagram without any fragmentation.

 

Case-02:

 

• Size of the datagram is found to be greater than MTU and DF bit set to 1.
• In this case, router discards the datagram.

 

Case-03:

 

• Size of the datagram is found to be greater than MTU and DF bit set to 0.
• In this case, router divides the datagram into fragments of size less than or equal to MTU.
• Router attaches an IP header with each fragment making the following changes in it.
• Then, router transmits all the fragments of the datagram.

 

Changes Made By Router-

 

Router makes the following changes in IP header of each fragment-

• It changes the value of total length field to the size of fragment.
• It sets the MF bit to 1 for all the fragments except the last one.
• For the last fragment, it sets the MF bit to 0.
• It sets the fragment offset field value.
• It recalculates the header checksum.

 

Also Read- IPv4 Header

 

IP Fragmentation Examples-

 

Now, lets us discuss some examples of IP fragmentation to understand how the fragmentation is actually carried out.

 

Example-01:

 

Consider-

• There is a host A present in network X having MTU = 520 bytes.
• There is a host B present in network Y having MTU = 200 bytes.
• Host A wants to send a message to host B.

 

 

Consider router receives a datagram from host A having-

• Header length = 20 bytes
• Payload length = 500 bytes
• Total length = 520 bytes
• DF bit set to 0

 

Now, router works in the following steps-

 

Step-01:

 

Router examines the datagram and finds-

• Size of the datagram = 520 bytes
• Destination is network Y having MTU = 200 bytes
• DF bit is set to 0

 

Router concludes-

• Size of the datagram is greater than MTU.
• So, it will have to divide the datagram into fragments.
• DF bit is set to 0.
• So, it is allowed to create fragments of the datagram.

 

Step-02:

 

Router decides the amount of data that it should transmit in each fragment.

 

Router knows-

• MTU of the destination network = 200 bytes.
• So, maximum total length of any fragment can be only 200 bytes.
• Out of 200 bytes, 20 bytes will be taken by the header.
• So, maximum amount of data that can be sent in any fragment = 180 bytes.

 

Router uses the following rule to choose the amount of data that will be transmitted in one fragment-

 

RULE   The amount of data sent in one fragment is chosen such that- It is as large as possible but less than or equal to MTU. It is a multiple of 8 so that pure decimal value can be obtained for the fragment offset field.   NOTE   It is not compulsory for the last fragment to contain the amount of data that is a multiple of 8. This is because it does not have to decide the fragment offset value for any other fragment.

 

Following the above rule,

• Router decides to send maximum 176 bytes of data in one fragment.
• This is because it is the greatest value that is a multiple of 8 and less than MTU.

 

Step-03:

 

Router creates three fragments of the original datagram where-

• First fragment contains the data = 176 bytes
• Second fragment contains the data = 176 byes
• Third fragment contains the data = 148 bytes

 

 

The information contained in the IP header of each fragment is-

 

Header Information Of 1st Fragment-

 

• Header length field value = 20 / 4 = 5
• Total length field value = 176 + 20 = 196
• MF bit = 1
• Fragment offset field value = 0
• Header checksum is recalculated.
• Identification number is same as that of original datagram.

 

Header Information Of 2nd Fragment-

 

• Header length field value = 20 / 4 = 5
• Total length field value = 176 + 20 = 196
• MF bit = 1
• Fragment offset field value = 176 / 8 = 22
• Header checksum is recalculated.
• Identification number is same as that of original datagram.

 

Header Information Of 3rd Fragment-

 

• Header length field value = 20 / 4 = 5
• Total length field value = 148 + 20 = 168
• MF bit = 0
• Fragment offset field value = (176 + 176) / 8 = 44
• Header checksum is recalculated.
• Identification number is same as that of original datagram.

 

Router transmits all the fragments.

 

Step-04:

 

At destination side,

• Receiver receives 3 fragments of the datagram.
• Reassembly algorithm is applied to combine all the fragments to obtain the original datagram.

 

Example-02:

 

Consider-

• There is a host A present in network X having MTU = 520 bytes.
• There is a host B present in network Y having MTU = 200 bytes.
• There exists a network Z having MTU = 110 bytes.
• Host A wants to send a message to host B.

 

 

Consider Router-1 receives a datagram from host A having-

• Header length = 20 bytes
• Payload length = 500 bytes
• Total length = 520 bytes
• DF bit set to 0

 

Consider Router-1 divides the datagram into 3 fragments as discussed in Example-01.

 

Then,

• First fragment contains the data = 176 bytes
• Second fragment contains the data = 176 byes
• Third fragment contains the data = 148 bytes

 

Now, consider-

• First and third fragment reaches the destination directly.
• However,  second fragment takes its way through network Z and reach the destination through Router-3.

 

Journey Of Second Fragment-

 

Now, let us discuss the journey of fragment-2 and how it finally reaches the destination.

 

Router-2 receives a datagram (second fragment of original datagram) where-

• Header length = 20 bytes
• Payload length = 176 bytes
• Total length = 196 bytes
• DF bit set to 0

 

Now, Router-2 works in the following steps-

 

Step-01:

 

Router-2 examines the datagram and finds-

• Size of the datagram = 196 bytes
• Destination is network Z having MTU = 110 bytes
• DF bit is set to 0

 

Router-2 concludes-

• Size of the datagram is greater than MTU.
• So, it will have to divide the datagram into fragments.
• DF bit is set to 0.
• So, it is allowed to create fragments of the datagram.

 

Step-02:

 

Router-2 decides the amount of data that it should transmit in each fragment.

 

Router-2 knows-

• MTU of the destination network = 110 bytes.
• So, maximum total length of any fragment can be only 110 bytes.
• Out of 110 bytes, 20 bytes will be taken by the header.
• So, maximum amount of data that can be sent in any fragment = 90 bytes.

 

Following the rule,

• Router-2 decides to send maximum 88 bytes of data in one fragment.
• This is because it is the greatest value that is a multiple of 8 and less than MTU.

 

Step-03:

 

Router-2 creates two fragments of the received datagram where-

• First fragment contains the data = 88 bytes
• Second fragment contains the data = 88 byes

 

 

The information contained in the IP header of each fragment is-

 

Header Information Of 1st Fragment-

 

• Header length field value = 20 / 4 = 5
• Total length field value = 88 + 20 = 108
• MF bit = 1
• Fragment offset field value = 176 / 8 = 22
• Header checksum is recalculated.
• Identification number is same as that of original datagram.

 

NOTE-

 

• This fragment is NOT the first fragment of the original datagram.
• It is the first fragment of the datagram received by Router-2.
• The datagram received by Router-2 is the second fragment of the original datagram.
• This datagram will serve as the second fragment of the original datagram.
• Therefore, fragment offset field is set according to the first fragment of the original datagram.

 

Header Information Of 2nd Fragment-

 

• Header length field value = 20 / 4 = 5
• Total length field value = 88 + 20 = 108
• MF bit = 1
• Fragment offset field value = (176 + 88) / 8 = 33
• Header checksum is recalculated.
• Identification number is same as that of original datagram.

 

NOTE-

 

• This fragment is NOT the last fragment of the original datagram.
• It is the last fragment of the datagram received by Router-2.
• The datagram received by Router-2 is the second fragment of the original datagram.
• This datagram will serve as the third fragment of the original datagram.
• There is another fragment of the original datagram that follows it.
• That is why, here MF bit is not set to 0.

 

Router-2 transmits both the fragments which reaches the destination through Router-3.

Router-3 performs no fragmentation.

 

Step-04:

 

At destination side,

• Receiver receives 4 fragments of the datagram.
• Reassembly algorithm is applied to combine all the fragments to obtain the original datagram.

 

 

Reassembly Algorithm-

 

Receiver applies the following steps for reassembly of all the fragments-

1. It identifies whether datagram is fragmented or not using MF bit and Fragment offset field.
2. It identifies all the fragments belonging to the same datagram using identification field.
3. It identifies the first fragment. Fragment with offset field value = 0 is the first fragment.
4. It identifies the subsequent fragments using total length, header length and fragment offset.
5. It repeats step-04 until MF bit = 0.

 

Fragment Offset field value for the next subsequent fragment = ( Payload length of the current fragment / 8 ) + Offset field value of the current fragment = ( Total length – Header length / 8 ) + Offset field value of the current fragment

 

Fragmentation Overhead-

 

• Fragmentation of datagram increases the overhead.
• This is because after fragmentation, IP header has to be attached with each fragment.

 

Total Overhead = (Total number of fragmented datagrams – 1) x size of IP header

 

Efficiency = Useful bytes transferred / Total bytes transferred OR Efficiency = Data without header / Data with header

 

Bandwidth Utilization or Throughput = Efficiency x Bandwidth

 

Important Notes-

 

Note-01:

 

• Source side does not require fragmentation due to wise segmentation by transport layer.
• The transport layer looks at the datagram data limit and frame data limit.
• Then, it performs segmentation in such a way that the resulting data can easily fit in a frame.
• Thus, there is no need of fragmentation at the source side.

 

Note-02:

 

• Datagrams from the same source to the same destination may take different routes in the network.

 

Note-03:

 

• Fragment offset field value is set to 0 for the first fragmented datagram.
• MF bit is set to 1 for all the fragmented datagrams except the last one.

 

Note-04:

 

• Unique combinations of MF bit value and fragment offset value.

 

MF bit	Offset value	Represents
1	0	1st Fragment
1	!=0	Intermediate Fragment
0	!=0	Last Fragment
0	0	No Fragmentation

 

Note-05:

 

• Identification number for all the fragments is same as that of the original datagram.
• This is to identify all the fragments of the same datagram while re-assembling them.

 

Note-06:

 

• Consider datagram goes through a path where different intermediaries having different bandwidths.
• Then, while calculating the throughput, consider the minimum bandwidth since it act as a bottleneck.

 

Note-07:

 

• Fragmentation is done by intermediary devices such as routers.
• The reassembly of fragmented datagrams is done only after reaching the destination.

 

Note-08:

 

Reassembly is not done at the routers because-

• All the fragments may not meet at the router.
• Fragmented datagrams may reach the destination through independent paths.
• There may be a need for further fragmentation.

 

Note-09:

 

If a fragment (say parent) is re fragmented into multiple datagrams then-

1. The fragment offset value for the first re-fragment is always same as its parent.
2. The MF bit bit value for the last re-fragment is always same as its parent.

 

To gain better understanding about IP Fragmentation,

Watch this Video Lecture

 

Next Article- Practice Problems On IP Fragmentation

 

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