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Suppose TCP operates over a 1-Gbps link. (a) Assuming TCP could utilize the full bandwidth continuously, how long would it take the sequence numbers to wrap around completely? (b) Suppose an added 32-bit timestamp field increments 1000 times during the wraparound time you found above. How long would it take for the timestamp to wrap around?

Short Answer

Expert verified
a) 34.36 seconds, b) 147,323.84 seconds

Step by step solution

01

- Calculate the Total Number of Sequence Numbers

TCP sequence numbers are 32-bit numbers. Calculate the total number of unique sequence numbers available by using the formula: \( 2^{32} \).
02

- Convert the Total Number of Sequence Numbers to Bits

Each sequence number represents a byte, so the total number of sequence numbers corresponds to \( 2^{32} \) bytes. Since there are 8 bits in a byte, convert this to bits: \( 2^{32} \times 8 \).
03

- Calculate the Time to Wrap Around (Part a)

To determine how long it takes to send \( 2^{32} \) bytes over a 1-Gbps link, use the formula: \( \text{Time} = \frac{\text{Total bits}}{\text{Bandwidth}} = \frac{2^{32} \times 8}{10^9} \) seconds.
04

- Simplify the Time Calculation

Calculate the simplified value: \( \frac{2^{32} \times 8}{10^9} = \frac{34,359,738,368}{1,000,000,000} = 34.36 \) seconds.
05

- Calculate Timestamp Wraparound (Part b)

The timestamp field increments 1000 times during the sequence number wraparound time of 34.36 seconds. Thus, the time taken for one increment is \( \frac{34.36}{1000} \) seconds. Calculate the total wraparound time for the timestamp field using 32 bits: \( 2^{32} \times \frac{34.36}{1000} \) seconds.
06

- Simplify the Timestamp Wraparound Time

Calculate the simplified value: \( 2^{32} \times \frac{34.36}{1000} = 34.36 \times 4,294,967.296 = 147,323.84 \) seconds.

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Key Concepts

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

32-bit Sequence Numbers
TCP (Transmission Control Protocol) uses 32-bit sequence numbers for managing data packets. A 32-bit sequence number means that you have a total of \(2^{32}\) unique sequence numbers. This is over 4 billion distinct values. When a TCP connection starts, the sequence number begins at a randomly selected value and increments with each byte of data sent.

Due to the 32-bit limit, eventually the sequence numbers will 'wrap around.' This means the sequence number will go back to zero after reaching the maximum value. This will happen every \(2^{32}\) bytes of data. For a high-speed Gigabit connection (1 Gbps), the time to wrap around can be calculated as follows:
  • Convert bytes to bits: \(2^{32}\) bytes \(\times 8\) bits/byte.
  • Calculate the wraparound time: \(\frac{2^{32} \times 8}{10^9} = 34.36\) seconds.
So, at a speed of 1 Gbps, the sequence numbers will wrap around approximately every 34.36 seconds.
Timestamp Wraparound
To address sequence number wraparound issues, TCP can use 32-bit timestamps, which help manage data packets more effectively over long connections. Normally, a timestamp field might increment several times during a sequence number cycle.

Given that the timestamp increments 1000 times during a sequence number wraparound, we can calculate this more precisely:
  • Time per increment: \(\frac{34.36}{1000} = 0.03436\) seconds.
  • Total wraparound time for the timestamp: \(2^{32} \times 0.03436 = 147,323.84\) seconds.
Therefore, with 32-bit timestamps, the wraparound period is significantly longer, making it easier to manage and track data packets over extended periods.
Data Transmission Speed
The speed at which data is transmitted over a network link significantly impacts sequence number wraparound and data management. Here, we consider a 1-Gbps connection:

  • High-speed data transmission allows for large amounts of data to be sent rapidly.
  • This results in quicker sequence number wraparound, needing efficient tracking mechanisms like timestamps.
The data transmission speed (in Gbps) determines how quickly sequence numbers and potentially timestamps wrap around. For instance, on a 1-Gbps link, \(2^{32}\) bytes translate to 34.36 seconds for a full wraparound. Higher speeds would reduce this time, exacerbating wraparound management challenges. In essence, faster data transmission requires sophisticated methods to ensure seamless data flow and tracking.

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Most popular questions from this chapter

Design a simple UDP-based protocol for retrieving files from a server. No authentication is to be provided. Stop-and-wait transmission of the data may be used. Your protocol should address the following issues: (a) Duplication of the first packet should not duplicate the "connection." (b) Loss of the final ACK should not necessarily leave the server in doubt as to whether the transfer succeeded. (c) A late-arriving packet from a past connection shouldn't be interpretable as part of a current connection.

Propose an extension to TCP by which one end of a connection can hand off its end to a third host; that is, if \(\mathrm{A}\) were connected to \(\mathrm{B}\), and \(\mathrm{A}\) handed off its connection to \(\mathrm{C}\), then afterwards \(\mathrm{C}\) would be connected to \(\mathrm{B}\) and \(\mathrm{A}\) would not. Specify the new states and transitions needed in the TCP state transition diagram, and any new packet types involved. You may assume all parties will understand this new option. What state should A go into immediately after the handoff?

Explain why TIME_WAIT is a somewhat more serious problem if the server initiates the close than if the client does. Describe a situation in which this might reasonably happen.

The Jacobson/Karels algorithm sets TimeOut to be 4 mean deviations above the mean. Assume that individual packet round-trip times follow a statistical normal distribution, for which 4 mean deviations are \(\pi\) standard deviations. Using statistical tables, for example, what is the probability that a packet will take more than TimeOut time to arrive?

A sender on a TCP connection that receives a 0 advertised window periodically probes the receiver to discover when the window becomes nonzero. Why would the receiver need an extra timer if it were responsible for reporting that its advertised window had become nonzero (i.e., if the sender did not probe)?

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