CN - Fragmentation Offset [Modified Gate Question]

Question

A host X send a message to host Y having two intermediate  networks in between them.A TCP message consisting of 2100 bytes is passed to IP for delivery across two hosts. In first network has MTU of  1200 bytes  second network has MTU of 400 bytes excluding network overhead. Assume that IP overhead per packet is 20 bytes. What is the fragmentation offset for the Last fragment that reaches destination

Answer 1

File size = 2100 when it goes to network layer 20 bytes are added so it becomes 2120B but this is just for understanding as we can solve it without considering IP overhead because fragmentation is done on data not headers.

At sender S : 2100 B to intermediate router P(1200 bytes without header) to intermediate router Q(400 B w/o header)  to destination D.

S to P :

fragement 1 : 1200 fragment 2 : 900 to P

P to Q :

for 1200B fragment 

F1 : 400(data length in bytes) offset : 0 

F2 : 400 offset : 400/8=50

F3 : 400 offset : 100

for 900B fragment 

F4: 400 offset : 150

F5: 400 offset : 200

F6: 100 offset : 250 last fragment 

In other way 250*8 = 2000 bytes ahead of this last fragment (from 0 to 1999) and sequence number of last fragment is 2000.

So 250 shoud be answer.

Comments

offdet is divide by 8 field ?

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yes because fragmentation offset field is of 13 bit and total length is of 16  bit so we need to do scaling of 8.

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 excluding network overhead

This line indirectly says that MTU is 1200 B and 400 B without IP header 20 B.

network overhead is 20 B that is IP header length , as header length is considered as network overhead .

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And why Offset is divide by 8 ?

fragmentation offset field is of 13 bit and total length is of 16  bit.

totallength = header size + payload = 16  bits

fragment offset field is 13 bits

so flag is  16 - 13 = 3 bit 

and 2³ is 8 so that's why IP fragments must be  multiples of 8 bytes .

Reference :

https://cs.nyu.edu/courses/fall98/G22.2262-001/class11.txt

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here fragmentation will be based on minimum MTU i.e 400.

1st fragment: 20|376

2nd fragemnt:20|376

3rd fragment:20|376

4th fragment: 20|376

5th fragment: 20|376

6th fragment:20|220

fragment offset of last fragment is 235.

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You r not right bcoz if fragment will be according to 400 mtu then full utilization of first channel 1200 mtu cant be used . And there is concept of fragmentation at each router if possible.

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Many Thanks @Ravi_1511 sir

two doubt cleared nice explanation

Answer 2

let P,Q are intermediate host....