Fragmentation at Network Level | Computer Networks - Computer Science Engineering (CSE) PDF Download

Introduction

Fragmentation is done by the network layer when the maximum size of datagram is greater than maximum size of data that can be held in a frame i.e., its Maximum Transmission Unit (MTU). The network layer divides the datagram received from the transport layer into fragments so that data flow is not disrupted. 

  • Since there are 16 bits for total length in IP header so, the maximum size of IP datagram = 216 – 1 = 65, 535 bytes.
  • It is done by the network layer at the destination side and is usually done at routers.
  • Source side does not require fragmentation due to wise (good) segmentation by transport layer i.e. instead of doing segmentation at the transport layer and fragmentation at the network layer, the transport layer looks at datagram data limit and frame data limit and does segmentation in such a way that resulting data can easily fit in a frame without the need of fragmentation.
  • Receiver identifies the frame with the identification (16 bits) field in the IP header. Each fragment of a frame has the same identification number.
  • Receiver identifies the sequence of frames using the fragment offset(13 bits) field in the IP header
  • Overhead at the network layer is present due to the extra header introduced due to fragmentation.

Fields in IP header for fragmentation

  • Identification (16 bits): use to identify fragments of the same frame.
  • Fragment offset (13 bits): use to identify the sequence of fragments in the frame. It generally indicates a number of data bytes preceding or ahead of the fragment.
    Maximum fragment offset possible = (65535 – 20)  = 65515
    {where 65535 is the maximum size of datagram and 20 is the minimum size of IP header}
    So, we need ceil(log265515) = 16 bits for a fragment offset but the fragment offset field has only 13 bits. So, to represent efficiently we need to scale down the fragment offset field by 216/213 = 8 which acts as a scaling factor. Hence, all fragments except the last fragment should have data in multiples of 8 so that fragment offset ∈ N.
  • More fragments (MF = 1 bit): tells if more fragments are ahead of this fragment i.e. if MF = 1, more fragments are ahead of this fragment and if MF = 0, it is the last fragment.
  • Don’t fragment (DF = 1 bit): if we don’t want the packet to be fragmented then DF is set i.e. DF = 1.

Reassembly of Fragments

It takes place only at the destination and not at routers since packets take an independent path(datagram packet switching), so all may not meet at a router and hence a need of fragmentation may arise again. The fragments may arrive out of order also. 

The document Fragmentation at Network Level | Computer Networks - Computer Science Engineering (CSE) is a part of the Computer Science Engineering (CSE) Course Computer Networks.
All you need of Computer Science Engineering (CSE) at this link: Computer Science Engineering (CSE)
21 videos|113 docs|66 tests

Top Courses for Computer Science Engineering (CSE)

21 videos|113 docs|66 tests
Download as PDF
Explore Courses for Computer Science Engineering (CSE) exam

Top Courses for Computer Science Engineering (CSE)

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

Objective type Questions

,

MCQs

,

Exam

,

Viva Questions

,

Fragmentation at Network Level | Computer Networks - Computer Science Engineering (CSE)

,

Summary

,

Fragmentation at Network Level | Computer Networks - Computer Science Engineering (CSE)

,

shortcuts and tricks

,

video lectures

,

Extra Questions

,

practice quizzes

,

Fragmentation at Network Level | Computer Networks - Computer Science Engineering (CSE)

,

Previous Year Questions with Solutions

,

Sample Paper

,

Semester Notes

,

Important questions

,

mock tests for examination

,

ppt

,

pdf

,

study material

,

Free

,

past year papers

;