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The sequence number field in the TCP header is 32 bits long, which is big enough to cover over 4 billion bytes of data. Even if this many bytes were never transferred over a single connection, why might the sequence number still wrap around from \(2^{32}-1\) to 0 ?

Short Answer

Expert verified
If data is transmitted at a high rate, the sequence number can quickly reach its maximum value of (2^{32}-1) and wrap around to 0.

Step by step solution

01

Understand the Sequence Number

In the TCP header, the sequence number is used to keep track of the bytes of data transmitted. It is a 32-bit field, meaning it can represent values from 0 to (2^{32}-1)
02

Calculate Maximum Value

The maximum value a 32-bit number can hold is (2^{32}-1), which is 4,294,967,295 bytes.
03

Examine Data Transfer Speed

Consider a high-speed network where data transfer rates can be extremely high, potentially reaching gigabits per second (Gbps).
04

Analyze Wrap Around

If data is transmitted at a very high rate, the sequence number can reach (2^{32}-1) quickly and then wrap around to 0. This occurs because the 32-bit field is not sufficient to hold numbers larger than (2^{32}-1).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

TCP header
The Transmission Control Protocol (TCP) header is an essential part of how data is effectively transmitted across networks. It contains several important fields, including the source and destination ports, sequence numbers, acknowledgment numbers, and more. Each field in the TCP header helps manage the data packets and ensures they are sent, received, and organized correctly. One key field is the sequence number, which we will delve into next.
32-bit sequence number
The sequence number in the TCP header is a 32-bit field. This means it can record up to 4,294,967,295 distinct numbers, from 0 to (2^{32}-1). The sequence number plays a critical role in tracking the order of data bytes between sender and receiver, ensuring that data is reassembled in the correct sequence, even if packets arrive out of order.
data transmission rates
Data transmission rates refer to how much data can be transmitted over a network in a given period. These rates can vary based on network conditions and technology used. In modern networks, data rates can be incredibly high, reaching up to gigabits per second (Gbps). This ability to transfer large amounts of data rapidly can influence how quickly sequence numbers are used up and why understanding TCP sequence numbers is crucial.
gigabits per second (Gbps)
Gigabits per second (Gbps) is a measure of data transfer rate. 1 Gbps equals 1,000,000,000 bits per second. At such high speeds, a large number of bytes can be transmitted in a very short time. For example, in a 10 Gbps network, hundreds of millions of bytes can be transmitted per second. This rapid transfer is why the 32-bit sequence numbers in the TCP header can quickly reach their maximum value in high-speed networks.
wrap around
Wrap around occurs when a sequence number reaches its highest value (4,294,967,295) and resets to zero. This can happen when data is transmitted at high speeds over extended periods. Since the sequence number is limited to 32 bits, it will reset after hitting the maximum value, continuing from zero again. This concept ensures continuity in data transmission, although it demands careful management to avoid confusion and data overlap.

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Most popular questions from this chapter

Find out the generic format for TCP header options from Request for Comments \(793 .\) (a) Outline a strategy that would expand the space available for options beyond the current limit of 44 bytes. (b) Suggest an extension to TCP allowing the sender of an option a way of specifying what the receiver should do if the option is not understood. List several such receiver actions that might be useful, and try to give an example application of each.

When TCP sends a \(\langle\) SYN, SequenceNum \(=x\rangle\) or \(\langle\) FIN, SequenceNum \(=x\rangle\), the consequent ACK has Acknowledgment \(=x+1\); that is, SYNs and FINs each take up one unit in sequence number space. Is this necessary? If so, give an example of an ambiguity that would arise if the corresponding Acknowledgment were \(x\) instead of \(x+1 ;\) if not, explain why.

Suppose party A connects to the Internet via a dial-up IP server (e.g., using SLIP or PPP), has several open Telnet connections (using TCP), and is cut off. Party B then dials in and is assigned the same IP address that A had. Assuming B was able to guess to what host(s) A had been connected, describe a sequence of probes that could enable B to obtain sufficient state information to continue with A's connections.

The Nagle algorithm, built into most TCP implementations, requires the sender to hold a partial segment's worth of data (even if PUSHed) until either a full segment accumulates or the most recent outstanding ACK arrives. (a) Suppose the letters abcdefghi are sent, one per second, over a TCP connection with an RTT of \(4.1\) seconds. Draw a timeline indicating when each packet is sent and what it contains. (b) If the above were typed over a full-duplex Telnet connection, what would the user see? (c) Suppose that mouse position changes are being sent over the connection. Assuming that multiple position changes are sent each RTT, how would a user perceive the mouse motion with and without the Nagle algorithm?

Suppose, in TCP's adaptive retransmission mechanism, that EstimatedRTT is 90 at some point and subsequent measured RTTs all are 200 . How long does it take before the TimeOut value, as calculated by the Jacobson/Karels algorithm, falls below 300 ? Assume initial Deviation value of 25 ; use \(\delta=1 / 8\).

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