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An $IP$ router with a $\text{Maximum Transmission Unit (MTU)}$ of $1500$ bytes has received an $IP$ packet of size $4404\text{ bytes}$ with an $IP$ header of length $20\text{ bytes}$. The values of the relevant fields in the header of the third $IP$ fragment generated by the router for this packet are:

  1. $\text{MF bit}$: $0,$ Datagram Length:$1444;$ Offset$: 370$
  2. $\text{MF bit}$: $1,$ Datagram Length$: 1424;$ Offset$: 185$
  3. $\text{MF bit}$: $1,$ Datagram Length$: 1500;$ Offset$: 370$
  4. $\text{MF bit}$: $0,$ Datagram Length$: 1424;$ Offset$: 2960$
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ans a)
4 votes
4 votes
Number of packet fragments = ⌈ (total size of packet)/(MTU) ⌉
                           = ⌈ 4404/1500 ⌉
                           = ⌈ 2.936 ⌉ 
                           = 3

So Datagram with data 4404 byte fragmented into 3 fragments. 

The first frame carries bytes 0 to 1479 (because MTU is 1500 bytes and HLEN is 20 byte so the total bytes in fragments is maximum 1500-20=1480).

the offset for this datagram is 0/8 = 0.

The second fragment carries byte 1480 to 2959.

The offset for this datagram is 1480/8 = 185.

finally the third fragment carries byte 2960 to 4404.

the offset is 370.and for all fragments except last one the M bit is 1.

so in the third bit M is 0.

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Since we have to add ip header and then ultimately remove it , it is better to not take it in consideration while solving such questions and taking MTU by excluding IP header 

Data : 4404 B

F1 : 1480 offset : 0 MF:1

F2 : 2960 offset : 185 MF:1

F3 : 1444 offset : 370 MF: 0

Fragmentation is done on data not on header

1 votes
1 votes

Number of packet fragments = ⌈ (total size of packet)/(MTU) ⌉ 
So Datagram with data 4404 byte fragmented into 3 fragments. 

souce --> https://pyq.ravindrababuravula.com/subjects/

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