fragmentation

Question

Assume a datagram of length 5*10³ bytes needs to pass through five networks to reach its destination. The MTU’s of each network is 1000, 820, 850, 950 and 900 respectively. Then at destination how many datagrams has to reached and what is the offset value of 3rd fragment after fragmentation?

1.   7, 200
2.   6, 400
3.  6,  300

Comments

generally these are connection oriented. So the overall flow and packet size is decided by considering minimum mtu. This is all done during connection establishment.

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so,you are saying that from starting only,we should divide our packet acc to minimum MTU??

https://gateoverflow.in/75993/networks-fragmentation

in this question,if the mtu of first network would have been 800 instead of 1024..then we would have divided according 800 first and then acc to 520 OR from starting we should have divided according to the minimum 512 bytes MTU ??

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see the minimum mtu in the question given in the link is 512

now amount of data can be transfer = 512 - 20 - max mtu header

max mtu header is 12

amount of data can be transfer = 512 - 20 - 12 = 480

Now why I am saying like this. In the question it is already given that TCP connection is used. That means connection oriented. So this mtu information will be known

Answer 1

even if we fragment a/c to 1000 bytes, again we have to do fragmenation for 820 byte

so its better we pick minimum MTU which is 820 bytes

size of data = 800

size of datagram = 5000 bytes (data=4080byte and header =20 byte)

now 4080 is fragmented into 6 fragments which are 800,800,800,800,800 and last is 80 byte

so 6 fragment will reach to destination and

offset of 3rd fragment = total data ahead of it/8

                                =(800 + 800) /8

                                = 200

Comments

that what my doubt was @target2017..i am also confused about this thing that which way to go

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@cse23

"size of datagram = 5000 bytes (data=4080byte and header =20 byte"

It should've been:  data=4980B and header=20B right?

So 4980 can be fragmented as: 800*6 + 180*1

Therefore totally 6+1 = 7 fragments.

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This approch: Choosing minimum MTU out of all the given MTUs, is wrong.