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TOPIC: IP Datagram Fragmentation

IP Datagram Fragmentation 5 years 8 months ago #36013

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Hello Guys,

How are you all? I hope fine.
I have one question related to IP Datagram Fragmentation.
I was reading an article here:
http://penguin.dcs.bbk.ac.uk/academic/networks/network-layer/fragmentation/index.php

Now, if you move over to fig.5 on that page it is saying that that 4 octates are wasted. I am not able to understand how they have been wasted.

Also if you could explain why offset is always multiple of 8?

Thanks

~Ping
The greatest pleasure in life is doing what people say you can not do..!!
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Re: IP Datagram Fragmentation 5 years 8 months ago #36015

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It is by design that the fragment offset is in multiple of 8 octet. In IP addressing 8 is a 'magic number' just like in base 10 arithmetic 10, 100, 1000 etc. In IP addressing, a character is stored in an octect which is 8bits

11111111 = 1 octet = 255 = 1 byte also
00000111 = 1 octet = 7

These days you can take 8 bits = 1byte = 1octet. It is not usually like this in the past.

In the second octet 5 bits are wasted, but by design it will still be used to store 7
The 8 octect multiple emanated from the 8 bit multiple for memory storage space on routers because a router is a store and forward device( it recieve packets in full,examine the header and forward.

Therefore a router that cannot handle 1500 MTU had to fragment it to 20 octet header and 1480 octet data. 20 + 976 = 996. 4 octet will be wasted because from the 1000 octet maximum 20 octet is taken for header remaining 980 octect, 980/8 = 122 remainder 4( the remainder 4 will be wasted because is not up to 8 octet). Remember, by definition " The header length is an integer that specifies the length of the segment header measured in 32 bit multiple" . If you run a network sniffer on a network which gives you the value of say 7 as the header length( this is usually in hexadecimal). It actually means 7x32/8 = 224/8 = 8
bytes. The role of the header length is to allow the receiving computer know where the data portion starts in the packet transmitted. since IP header+TCP header+Data= MTU
also, TCP header + Data = Segment, and Data = Maximum segment in multiple of 8
Finally IP header + sement = IP Datagram


Referances
www.firewall.cx/osi-encap-decap.php
www.firewall.cx/tcp-analysis.php
TCP/IP Illustrated Richard W Stephen
RFC 1122
RFC 1119
RFC 791
sose
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Re: IP Datagram Fragmentation 5 years 8 months ago #36016

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It actually means 7x32/8 = 224/8 = 8

sorry this is actually 224/8 = 122
sose
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Re: IP Datagram Fragmentation 5 years 8 months ago #36017

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It actually means 7x32/8 = 224/8 = 8
sorry this is actually 224/8 = 122

sorry
224/8=28 byte

hehe ..I hate wasting disk space, it cost money
sose
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Re: IP Datagram Fragmentation 5 years 8 months ago #36022

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Thank you sose... it could not have been clearer..

Thanks again..
The greatest pleasure in life is doing what people say you can not do..!!
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Re: IP Datagram Fragmentation 5 years 8 months ago #36031

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I am glad to be helpful
sose
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