Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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).

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose a client \(C\) repeatedly connects via TCP to a given port on a server \(S\), and that each time it is \(\mathrm{C}\) that initiates the close. (a) How many TCP connections a second can C make here before it ties up all its available ports in TIME_WAIT state? Assume client ephemeral ports are in the range 1024-5119, and that TIME_WAIT lasts 60 seconds. (b) Berkeley-derived TCP implementations typically allow a socket in TIME WAIT state to be reopened before TIME_WAIT expires, if the highest sequence number used by the old incarnation of the connection is less than the ISN used by the new incarnation. This solves the problem of old data accepted as new; however, TIME_WAIT also serves the purpose of handling late final FINs. What would such an implementation have to do to address this and still achieve strict compliance with the TCP requirement that a FIN sent anytime before or during a connection's TIME_WAIT receive the same response?

Suppose that, when a TCP segment is sent more than once, we take SampleRTT to be the time between the most recent transmission and the ACK, as in Figure \(5.10\) (b). Assume, for definiteness, that TimeOut \(=2 \times\) EstimatedRTT. Sketch a scenario in which no packets are lost but EstimatedRTT converges to a third of the true RTT, and give a diagram illustrating the final steady state. Hint: Begin with a sudden jump in the true RTT to just over the established TimeOut.

Consider a simple UDP-based protocol for requesting files (based somewhat loosely on the Trivial File Transport Protocol, TFTP). The client sends an initial file request, and the server answers (if the file can be sent) with the first data packet. Client and server then continue with a stop-and-wait transmission mechanism. (a) Describe a scenario by which a client might request one file but get another; you may allow the client application to exit abruptly and be restarted with the same port. (b) Propose a change in the protocol that will make this situation much less likely.

The RPC-based "NFS" remote file system is sometimes considered to have slower than expected write performance. In NFS, a server's RPC reply to a client write request means that the data is physically written to the server's disk, not just placed in a queue. (a) Explain the bottleneck we might expect, even with infinite bandwidth, if the client sends all its write requests through a single logical CHAN channel, and explain why using a pool of channels could help. Hint: You will need to know a little about disk controllers. (b) Suppose the server's reply means only that the data has been placed in the disk queue. Explain how this could lead to data loss that wouldn't occur with a local disk. Note that a system crash immediately after data was enqueued doesn't count because that would cause data loss on a local disk as well. (c) An alternative would be for the server to respond immediately to acknowledge the write request, and to send its own separate CHAN request later to confirm the physical write. Propose different CHAN RPC semantics to achieve the same effect, but with a single logical request/reply.

Suppose an idle TCP connection exists between sockets A and B. A third party has eavesdropped and knows the current sequence number at both ends. (a) Suppose the third party sends A a forged packet ostensibly from B and with 100 bytes of new data. What happens? Hint: Look up in Request for Comments 793 what TCP does when it receives an ACK that is not an "acceptable ACK." (b) Suppose the third party sends each end such a forged 100-byte data packet ostensibly from the other end. What happens now? What would happen if \(\mathrm{A}\) later sent 200 bytes of data to B?

See all solutions

Recommended explanations on Computer Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free