Question

10. Consider sending a 3000-byte IPv4 datagram into a link that has an MTU of 700 bytes. Suppose the original datagram is stamped with the identification number 1001. Assume that all IP datagram headers have a size of 20 bytes. How many fragmented datagrams will be generated? For *each* of the fragmented datagrams, please give: the identification number of the fragmented datagram, the value in the offset field of the datagram header, the length of the fragmented datagram, and the length of the payload in the fragmented datagram.

Answer 1

Answer:

Number of fragmented datagrams: 5

Explanation:

Since the MTU is 700 bytes,the payload will be 680 bytes and Up header will be 20 bytes.

The datagram is of size 3000 bytes. Therefore this datagram is fragmented to send it in 700 bytes MTU.

The first four fragments has the size 680 bytes of payload(4*680= 2720 bytes). The remaining 280 bytes will be sent in the 5th fragment.

The offset value gives how much was sent before the current fragment in terms of 8-byte blocks.

For first fragment it will be 0, as no data was sent before this first fragment.

For second fragment, 680 bytes(680/8=85 8-bytes block) were sent in first fragment before this current fragment. Hence the value will be 0+85=85.

For third fragment it is : 0+85+85=170

Number of fragmented datagrams: 5