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GATE | GATE-CS-2016 (Set 1) | Question 63

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An IP datagram of size 1000 bytes arrives at a router. The router has to forward this packet on a link whose MTU (maximum transmission unit) is 100 bytes. Assume that the size of the IP header is 20 bytes. The number of fragments that the IP datagram will be divided into for transmission is :
Note : This question was asked as Numerical Answer Type.
(A) 10
(B) 50
(C) 12
(D) 13

Answer: (D)

Explanation: MTU = 100 bytes
Size of IP header = 20 bytes
So, size of data that can be transmitted in one fragment = 100 – 20 = 80 bytes
Size of data to be transmitted = Size of datagram – size of header = 1000 – 20 = 980 bytes
Now, we have a datagram of size 1000 bytes.
So, we need ceil(980/80) = 13 fragments.
Thus, there will be 13 fragments of the datagram.
So, D is the correct choice.

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Last Updated : 28 Jun, 2021
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