Tag: Fragmentation in Computer Networks

IP Header | IP Fragmentation | Problems

IP Fragmentation-

 

Before you go through this article, make sure that you have gone through the previous article on IP Fragmentation.

 

We have discussed-

  • IP Fragmentation is a process of dividing the datagram into fragments during its transmission.
  • It is performed by intermediary devices at destination side at network layer.

 

Also Read- IP Header

 

In this article, we will discuss practice problems based on IP Header and IP Fragmentation.

 

PRACTICE PROBLEMS BASED ON IP HEADER AND IP FRAGMENTATION-

 

Problem-01:

 

The intermediate routers between source and destination need the following information in IP header-

  1. Version
  2. Protocol
  3. Identification Number
  4. Source IP Address

 

Solution-

 

Option-A:

 

  • Version field indicates the version of IP used.
  • This information is required to process the packet appropriately based on its version.

 

Option-B:

 

  • Protocol field indicates the next level protocol.
  • This information is required by the router to accept or discard the packet if its buffer is full.
  • Based on the priority, router takes its decision.

 

Option-C:

 

  • Identification number field identifies the fragments of the same datagram.
  • This information is required while re-assembling the datagram fragments.

 

Option-D:

 

  • Source IP Address field indicates the IP Address of the source.
  • This information is required by the router to send ICMP packet to the source.
  • ICMP packet informs the source that its packet has been discarded.

 

Thus, All these fields are required in the IP Header.

 

Problem-02:

 

Fragmentation of a datagram is needed in-

  1. Datagram circuit only
  2. Virtual circuit only
  3. Both (A) and (B)
  4. None

 

Solution-

 

  • Each network has its Maximum Transmission Unit (MTU).
  • If the size of data packet is greater than MTU, then it will have divided into fragments to transmit it through the network.
  • So, fragmentation may be required in datagram circuits as well as virtual circuits.
  • Thus, Option (C) is correct.

 

Problem-03:

 

What are all the fields required from IP header to allow the destination to perform reassembly of fragments?

  1. Identification, MF, Offset, Header length and Total length
  2. MF, Offset and Destination IP
  3. MF, Datagram length, Source IP
  4. MF, Options and Offset

 

Solution-

 

Clearly, Option (A) is correct.

 

Problem-04:

 

The checksum in IP must be recomputed at every router because of change in ____ fields.

  1. TTL, Options, Identification Number, Offset
  2. TTL, Options, Datagram Length, Offset
  3. TTL, Options, Data, Offset
  4. TTL, Header Length, Offset, ToS

 

Solution-

 

Clearly, Option (B) is correct.

 

Problem-05:

 

If the value available in “fragment offset” field of IP header is 100, then the number of bytes ahead of this fragment is ___ ?

  1. 100 B
  2. 400 B
  3. 800 B
  4. 200 B

 

Solution-

 

  • Fragment offset field use a scaling factor of 8.
  • If Fragment offset field value = 100, then fragment offset = 8 x 100 = 800.
  • It suggests 800 bytes of data is ahead of this fragment.
  • Thus, Option (C) is correct.

 

Problem-06:

 

When the source does not trust the routers to route properly or source wishes to make sure that the packet does not stray from specified path, what options can be used?

  1. Loose source routing
  2. Trace route
  3. Strict source routing
  4. Internet Time Stamp

 

Solution-

 

Clearly, Option (C) is correct.

 

Problem-07:

 

The checksum computation in IP header includes-

  1. IP header only
  2. IP header and data
  3. IP header and Pseudo header
  4. None

 

Solution-

 

  • Checksum computation in IP header includes IP header only.
  • Errors in the data field are handled by the encapsulated protocol.
  • Thus, Option (A) is correct.

 

Problem-08:

 

Suppose a router receives an IP packet containing 600 data bytes and has to forward the packet to a network with maximum transmission unit of 200 bytes. Assume that IP header is 20 bytes long. What are fragment offset values for divided packets?

  1. 22, 44, 66, 88
  2. 0, 22, 44
  3. 0, 22, 44, 66
  4. 22, 44, 66

 

Solution-

 

Given-

  • MTU size of the destination network = 200 bytes
  • IP header length = 20

 

Now,

  • Maximum amount of data that can be sent in one fragment = 200 – 20 = 180 bytes.
  • Amount of data sent in a fragment must be a multiple of 8.
  • So, maximum data sent that can be in one fragment = 176 bytes.

 

Thus, 4 fragments are created-

  • 1st fragment contains 176 bytes of data.
  • 2nd fragment contains 176 bytes of data.
  • 3rd fragment contains 176 bytes of data.
  • 4th fragment contains 72 bytes of data

 

So,

  • Fragment offset value for 1st fragment = 0
  • Fragment offset value for 2nd fragment = 176 / 8 = 22
  • Fragment offset value for 3rd fragment = (176+176) / 8 = 44
  • Fragment offset value for 4th fragment = (176 + 176 + 176) / 8 = 66

 

Thus, Option (C) is correct.

 

To watch video solution and practice other problems,

Watch this Video Lecture

 

Next Article- Transmission Control Protocol | TCP

 

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IP Fragmentation | Fragmentation in Networking

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 bitOffset valueRepresents
101st Fragment
1!=0Intermediate Fragment
0!=0Last Fragment
00No 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

 

Get more notes and other study material of Computer Networks.

Watch video lectures by visiting our YouTube channel LearnVidFun.