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An IP datagram of size $1000$ $\text{bytes }$arrives at a router. The router has to forward this packet on a link whose MTU (maximum transmission unit) is $100$ $\text{bytes }$. Assume that the size of the IP header is $20$ $\text{bytes }.$

The number of fragments that the IP datagram will be divided into for transmission is________.
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Best answer
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67 votes
IP Datagram size $=1000B$
MTU $=100B$
IP header size $=20B$
So, each packet will have $20B$ header + $80B$ payload.
Therefore, $80 \times 12 = 960$
now remaining $20B$ data could be sent in next fragment.
So, total $12 + 1 = 13$ fragments.
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48 votes

MTU (M)  is $80 + 20 \text{ bytes}$

Datagram size (DS) is $980 + 20$

No. of fragments are $\dfrac{\text{DS}}{\text{M}}=\dfrac{980}{80}=12.25$

So Answer is 13.

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