# Why is Fragmentation Offset Multiple of 8? Understanding IPv4 Fragmentation

• MHB
• shivajikobardan
In summary: So, if the fragment offset is 1, the fragment will start at byte offset 0x8000 in the original datagram, and will end at byte offset 0x807f in the new datagram. If the fragment offset is 0, the fragment will start at byte offset 0x8000 in the original datagram, and will end at byte offset 0x7fff in the new datagram.In summary, the fragment offset field in the IPv4 header is 8 bits wide, and can store a value between 0 and 7FFFFFFF.
shivajikobardan
I am studying about ipv4 fragmentation.
This is ipv4 header. The fields of concern here are-:
-> identification(16 bit)

-> Flags(3 bit)

-> fragment offset(13 bit)

-> Total length(16 bit)

In order to understand these concepts of fragmentation, I have solved multiple problems. One of them is below.
Here initially we have 5000 byte datagram of which 20 byte is header.
The MTU is 1500 bytes.

But I am unable to realize this simple concept that I believe even a 5th grader can understand even though I am college level undergraduate student. It is being pretty shameful for me.I am leaving my shame aside and asking this instead of giving up. I have asked this everywhere and everyone says this is basic math but it is not clicking in my head. Imma memorize this lol. I can solve any numerical problem related to fragmentation but this curiosity isn’t letting me to learn further and go to ipv6.
It would be immensely helpful to me if anyone can help me via that above example to understand this simple concept. I am unable to relate with this concept.
I agree there are 2^13 offset values ie 0,1,2,3,4,5,6,7,8,9…..8191. So that is the total no. of fragments possible 8192 fragments. But datagram can be only as big as 2^16-1 bytes. But why are we dividing them? It doesn’t make any sense. Maybe it is trying to say that 8192 fragments need to span 65535 bytes which gives 7.99987793 as the answer. But I have calculated it later, and fragmentation doesn’t work like that.

Opinion 1-:

Max “true” fragment offset possible=sum of all previous DATA ONLY excluding header
=65535-n*20 bytes.

Where 65535 is the max total length possible.n is the number of fragments and 20 bytes is the header length
The max value of offset possible is 8191. 65535-n*20 bytes should be represented by 8191 (both max values). But this opinion takes nowhere.Opinion 2-:
Max. 13 bit number=8191
Max 16 bit number=65535
So 65535/8191=8.0000sthsth (not exactly 8).

So the problem is like how can max offset value represent the max data size. But as i said in opinion 1, max data size can never be 65535 as header also takes up some space.

Opinion 3–:
Say i want to understand this example by a dummy example. Say fragmentation offset field is of 1 bit. And total length field is of 4 bit. What would happen?

Here are some answers that I have read which gave me further confusions-:
https://networkengineering.stackexc...26/why-ip-fragmentation-is-on-8-byte-boundary

https://learningnetwork.cisco.com/s/question/0D53i00000Kt7dxCAB/fragment-offset-concept

@shivajikobardan, you posted this on mathhelpboard, but I don't remember whether you also posted this here at PF.

From the above, "The fragment offset ... is measured in unit (sic) of 8 bytes." It's not in units of single bytes. So for example, if the fragment offset is 9, what this means is that the fragment offset is really 72 (= 9 * 8). In binary, 9 is 1001 (base-2). Multiplying by 8 gives us 72, which in binary is 100 1000 (base 2). In hex, this is 0x48, or decimal 72.

Another example: Suppose the first fragment has a length of 32,768 bytes, at byte offsets 0 through 32,767. The next available storage location would be at offset 32,768. In hex, this is 0x8000, and in binary it would be 1000 0000 0000 0000 (base 2). Since the fragment offset is in units of 8 bytes, we need to divide 32,768 by 8, which is 4096. In hex, this is 0x1000, or in binary it would be 0001 0000 0000 0000 (base 2), which will fit into 13 bits.

shivajikobardan

## 1. Why is the fragmentation offset multiple of 8?

The fragmentation offset in IPv4 is always a multiple of 8 because each IPv4 datagram is divided into smaller fragments, each containing a header with a fixed size of 20 bytes. This means that the offset, which indicates the starting position of a fragment within the original datagram, must also be a multiple of 8 in order for the header to align properly with the rest of the datagram.

## 2. What is fragmentation in IPv4?

Fragmentation in IPv4 is the process of dividing a large datagram into smaller fragments in order to fit within the maximum transmission unit (MTU) of a network. This is necessary because different networks may have different MTUs, and a datagram that is too large may not be able to be transmitted without being fragmented.

## 3. How does fragmentation affect network performance?

Fragmentation can have a negative impact on network performance because it increases the amount of processing and overhead required by routers and hosts. This can lead to slower data transmission and higher network latency. Additionally, excessive fragmentation can result in packet loss and retransmissions, further reducing performance.

## 4. Can IPv4 fragments be reassembled?

Yes, IPv4 fragments can be reassembled at the destination host by using the fragment offset and identification fields in the IP header to properly order and combine the fragments. However, if any of the fragments are lost or damaged during transmission, the entire datagram will not be able to be reassembled and will need to be transmitted again.

## 5. Why is IPv4 fragmentation necessary?

IPv4 fragmentation is necessary to ensure that data can be transmitted across networks with different MTUs. Without fragmentation, large datagrams would not be able to be transmitted and communication between networks would be limited. However, excessive fragmentation can lead to performance issues, so it is important for network engineers to properly configure MTUs to minimize the need for fragmentation.

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