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

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.

Short Answer

Expert verified
Yes, it is necessary to increment the acknowledgment by 1 to avoid confusion between control segments and data segments.

Step by step solution

01

- Understand the basics of sequence and acknowledgment numbers

In TCP, each byte of data has a sequence number. When a segment is sent, it includes the Sequence Number field (indicating the byte number of the first byte in the segment) and the Acknowledgment Number field (indicating the next byte number the sender expects to receive).
02

- Understand SYN and FIN segments

A SYN segment is used to initiate a connection, and a FIN segment is used to terminate a connection. Each of these segments consumes one sequence number space because they need to be uniquely identified and acknowledged.
03

- Analyze the acknowledgment process

When a SYN segment with SequenceNum = x is sent, the receiving end needs to acknowledge it by sending an ACK with Acknowledgment = x + 1. Similarly, when a FIN segment is sent with SequenceNum = x, the receiving end acknowledges it with Acknowledgment = x + 1.
04

- Identify the necessity of incrementing sequence numbers

If the acknowledgment for a SYN or FIN segment were simply x instead of x + 1, it would cause confusion. The sender would not be able to differentiate between an acknowledgment for the control segment (SYN/FIN) and an acknowledgment for data segment x.
05

- Provide an example of ambiguity

Consider a scenario where a SYN is sent with SequenceNum = x, and the receiver sends an acknowledgment without incrementing the sequence number (thus ACK = x). The sender may think the acknowledgment is for some earlier data segment, leading to confusion and potential retransmissions.
06

- Conclusion

To prevent such ambiguities and to clearly distinguish control segments (SYN/FIN) from data segments, it is necessary for the acknowledgment to increment by 1. This practice ensures clarity in communication and proper establishment/termination of connections.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

SYN Segments
When initiating a TCP connection, a special packet called a SYN segment is sent. SYN stands for 'synchronize'. This segment is crucial because it helps set the initial sequence number, which is used to keep track of the bytes transmitted in the session.
  • A SYN segment indicates the start of a connection.
  • It includes a sequence number.
  • The receiver acknowledges it by sending back an ACK with an acknowledgment number equal to the sequence number + 1.
Here’s a simple example:
If a SYN segment has SequenceNum = 1000, the acknowledgment from the receiver will be Acknowledgment = 1001.
FIN Segments
A FIN segment (short for 'finish') is used to terminate a TCP connection. It signifies that the sender has finished sending data.
  • A FIN segment indicates the end of a connection.
  • Similar to SYN, it includes a sequence number.
  • The receiver acknowledges this with an acknowledgment number equal to the sequence number + 1.
For instance:
If a FIN segment has SequenceNum = 3000, the acknowledgment will be Acknowledgment = 3001.
Sequence Number Space
The sequence number space in TCP is a range of numbers used to keep track of the position of each byte in a stream of data.
  • Every byte transmitted in a TCP connection has a unique sequence number.
  • SYN and FIN segments also occupy space in the sequence number space.
  • This is essential for ensuring no data is lost or duplicated.
Imagine sequence numbers as labels on individual packets. Each byte's sequence number ensures data arrives in the correct order and can be reassembled properly.
TCP Acknowledgment Process
The acknowledgment process in TCP helps ensure data is received correctly. The receiver sends an acknowledgment number, which represents the next byte it expects to receive.
  • SYN segments trigger acknowledgments (ACK) from the receiver.
  • FIN segments also expect acknowledgments from the receiver.
  • Regular data segments include sequence numbers, and the receiver acknowledges them by sending an ACK for the next byte expected.
For example, if a segment with SequenceNum = 1500 is sent and received correctly, the receiver sends back an ACK with Acknowledgment = 1501. If SYN or FIN segments are involved, the acknowledgment increments by 1 to indicate the control segment has been received.
Ambiguity in TCP Acknowledgment
Ambiguity in TCP acknowledgment can occur if the acknowledgment number for SYN or FIN segments is not incremented.
  • Without incrementing, it’s unclear whether the acknowledgment is for a control segment or for regular data.
  • This can cause confusion and retransmission of data.
Consider an example:
If a SYN segment with SequenceNum = 500 is acknowledged with ACK = 500 (instead of 501), the sender might think the acknowledgment is for some previous data segment. This leads to unnecessary retransmissions and potential connection issues. Hence, to avoid ambiguity, the acknowledgment for SYN and FIN must increment the sequence number by 1. This ensures clarity in communication, making sure the control segments (SYN/FIN) are distinctly acknowledged apart from regular data.

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

If host \(\mathrm{A}\) receives two SYN packets from the same port from remote host \(\mathrm{B}\), the second may be either a retransmission of the original or else, if B has crashed and rebooted, an entirely new connection request. (a) Describe the difference as seen by host A between these two cases. (b) Give an algorithmic description of what the TCP layer needs to do upon receiving a SYN packet. Consider the duplicate/new cases above, and the possibility that nothing is listening to the destination port.

Read the man page (or Windows equivalent) for the Unix/Windows utility netstat. Use netstat to see the state of the local TCP connections. Find out how long closing connections spend in TIME_WAIT.

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?

Request for Comments 1122 states (of TCP): A host MAY implement a "half-duplex" TCP close sequence, so that an application that has called CLOSE cannot continue to read data from the connection. If such a host issues a CLOSE call while received data is still pending in TCP, or if new data is received after CLOSE is called, its TCP SHOULD send an RST to show that data was lost. Sketch a scenario involving the above in which data sent by (not to!) the closing host is lost. You may assume that the remote host, upon receiving an RST, discards all received data still unread in buffers.

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 \(\delta=1 / 8\).

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