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48 votes
48 votes

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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10 Answers

Best answer
67 votes
67 votes
IP packet length is given $4404$ which includes ip header of length $20$
So, data is $4384$.

Now, router divide this data in $3$ parts
$1480\quad 1480\quad 1424$

After adding ip header in last packet size is: $1444$ and since its the last packet therefore $MF =0$

And offset is $\dfrac{2960}{8}=370$

Correct Answer: $A$
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9 votes
9 votes
Ip packet size excluding the header=1404-20=4384

MTU of router/Network/Gateway=1500

No. Of fragments=ceil(4384/1500)=~3

These three fragments will be multiple of 8 but max but less 1500 like

1480,1480 and 1424

Offset of last fragments is:1480+1480=2960/8=370
5 votes
5 votes

Number of fragments =ceil(4404-20/1500-2)=3(approx) Take integer

First Fragments.

1480 20

Offset  of first fragments  :: 0 and MF =1

Second Fragments:

1480 20

Offset  of IInd fragments  :: 185 and MF =1

Third Fragments::

1424 20

Offset  of IIIrd fragments  :: 370 and MF =1

And datagram length is nothing but total length=1424+20=1444 Byte

Therefore option A will be right.

Answer:

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