Comp Sci Ipv4 fragmentation number of bytes not in multiple of 8

[shivajikobardan]

Homework Statement

What should we do if the number of bytes to be transferred are not in multiples of eight in ipv4?

Relevant Equations

ipv4 header

Theory that I am trying to understand-:

So i found a question that is relevant to this-:
->a total of 1440 bytes that is routed through an interface with MTU of 576 bytes. Calculate flag, fragmented offset, total length and data transmitted in each packet after fragmentation. Assume IP header to be 20 bytes.

Solution-:

1440=20+1420

MTU is 576.
Number of fragments=1420/576=3

So let’s call 3 packets P1,P2,P3.

P1=>20+556
P2=>20+556
P3=>20+308

So I am trying to understand what the above picture is trying to say.

There are 2 cases-:
-> Is it trying to say that total length of P1 should be divisible by 8?

-> Is it trying to say that “only data” part should be divisible by 8?

I have even further questions about it.

->Say, the total length of P1 should be divisible by 8. What will we do if it is not?

->(I believe) Say the “only data” part should be divisible by 8, then what should we do as neither 556 nor 308 is divided by 8.

So say I reiterate and do this arrangement(I believe this is correct way)-:
P1->20+552
P2>20+552
P3->20+316

Still 316 isn’t divisible by 8, what should I do now?

i don't understand the solution that is written in the picture that i attested above. how can we use that solution to our case?

IRRESPECTIVE OF WHATEVER I WROTE,if you want,YOU CAN EXPLAIN LIKE I AM BEGINNER TO ALL THESE

 

[Tom.G]

A rather detailed explanation/example can be found at:
https://en.wikipedia.org/wiki/IPv4#Fragmentation_and_reassembly
See especially REASSEMBLY.

 

[shivajikobardan]

A rather detailed explanation/example can be found at:
https://en.wikipedia.org/wiki/IPv4#Fragmentation_and_reassembly
See especially REASSEMBLY.

wikipedia feels slightly above my level of understanding. i don't find it that simple. anyway i will try reading that. it would be nice if you would help me with this case.

 

[Tom.G]

Well, slightly above is good, it means your view is enlarging and you are learning something!

See also the sentence immediately above the REASSEMBLY heading that reads:
The last offset and last data size are used to calculate the total data size

That implies that any leftover room in the last octet is to be considered irrelevant and should be ignored.

Cheers,
Tom