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

• MHB
• shivajikobardan
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