Solve Number of Bytes to be Transferred Not Being in Multiples of Eight in IPv4

[shivajikobardan]

Theory that I am trying to understand-:

https://lh5.googleusercontent.com/Cb2IV2YzbeqrasCd6ACQrrPo8rQVUutCMwB6eF09qc9XB0KdDbMo42ssv2S6-mjNNvbnEVeuOJA5eUrogtZacaUFjQMhoxP_s6zSUxOoKSLCjZx18cPGSRYDSSUKZKh-JOq0RYAa-TTmDb_vfQ
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

 

[Future Bruno]

THINGS.The picture you attached is showing the format of an IP packet header. It has a total of 20 bytes of header information. This includes fields like source, destination, protocol and flags. The Flag field is used to indicate whether the packet is fragmented or not (1 indicates fragmented and 0 indicates not fragmented). The Fragmented Offset field indicates the offset of the fragment from the start of the original packet. The Total Length field indicates how many bytes the packet contains.In your case, you have a total of 1440 bytes that is routed through an interface with MTU of 576 bytes. This means that the packet needs to be divided into 3 fragments in order to be sent. The first fragment will have a total length of 576 (20 bytes of header + 556 bytes of data). The flag will be set to 1 indicating that it is a fragmented packet. The offset will be 0. The second fragment will also have a total length of 576 (20 bytes of header + 556 bytes of data). The flag will still be set to 1 indicating that it is a fragmented packet. The offset will be 576. The third fragment will have a total length of 288 (20 bytes of header + 308 bytes of data). The flag will still be set to 1 indicating that it is a fragmented packet. The offset will be 1152. So all the packets need to have a total length that is divisible by 8. If the total length of any packet is not divisible by 8, the data part of the packet can be reduced so that the total length is divisible by 8.