A TCP segment of 3000 bytes is to be transmitted over a network with MTU of 262

Question

A TCP segment of 3000 bytes is to be transmitted over a network with MTU of 262 bytes. Assuming the header in each IP datagram requires 20 bytes, would fragmentation take place? Explain why or why not? If fragmentation takes place, derive the number of datagrams (fragments) required. Also, show how many bytes are in each fragment, and how many of those bytes correspond to headers and data (payload) fields. [Hint: in IPv4, the fragmentation-offset field is expressed in multiple of 8 bytes (see Section 4.3.2 on IP Datagram Fragmentation in textbook), i.e. the amount of original payload data from the original datagram that each fragment carries (except the last fragment) must be multiple of 8 bytes.]

Explanation / Answer

The MTU or maximum transmissable unit of the network is 262 bytes. This includes the 20 byte IP header as well. Hence, only 262-20=242 bytes of data can be sent in each datagram.

Given that the size of the segment is 3000 bytes, fragmentation will take place, as the network can only send 242 bytes of data in each datagram and the entire data cannot be sent in one datagram.

The number of datagrams required are - 3000/242 = 12.39. Hence 13 fragments are required with 12 fragments of 242 bytes of data or 262 byte datagrams and one datagram with 96 bytes of data. But since the payload must be in multiples of 8 bytes, the size of the datagrams will be 12 datagrams of 260 bytes (240 bytes of data and 20 bytes of header as 240 is the nearest multiple of 8 from 242) and one datagram with 140 bytes (120 bytes of data and 20 byte header).

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.