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Suppose TCP operates over a 40-Gbps STS-768 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) 14.32 minutes (b) 116.99 years

Step by step solution

01

Calculate the total number of sequence numbers

TCP sequence numbers are represented by a 32-bit sequence number field. Therefore, the total number of sequence numbers is given by \[2^{32} = 4294967296\]
02

Determine the data transmission rate

The link bandwidth is 40 Gbps. This means that the transmission rate is: \[40 \text{ Gbps} = 40 \times 10^9 \text{ bits per second}\]
03

Convert the transmission rate to bytes per second

Since 1 byte = 8 bits, we convert the transmission rate: \[\frac{40 \times 10^9 \text{ bits per second}}{8 \text{ bits per byte}} = 5 \times 10^9 \text{ bytes per second}\]
04

Calculate the time to exhaust sequence numbers

Using the number of sequence numbers and the transmission rate, we find the wraparound time: \[\text{Time} = \frac{\text{Total sequence numbers}}{\text{Transmission rate}} = \frac{4294967296 \text{ bytes}}{5 \times 10^9 \text{ bytes per second}} = 858.9934592 \text{ seconds}\]
05

Convert the wraparound time to a meaningful unit

Convert seconds to a more comprehensible unit: \[\text{858.9934592 seconds} = \frac{858.9934592}{60} \text{ minutes} = 14.31655765 \text{ minutes} \approx 14.32 \text{ minutes}\]
06

Define the increment rate for the timestamp field

The 32-bit timestamp field increments 1000 times during the wraparound time: \[\text{Increment Time} = \frac{\text{Wraparound Time}}{1000} = \frac{858.9934592 \text{ seconds}}{1000} = 0.8589934592 \text{ seconds}\]
07

Calculate the time to wrap around the timestamp

Since the timestamp is a 32-bit field, the total number of timestamps is: \[2^{32} = 4294967296\] The time to wrap around is: \[\text{Total Time} = 4294967296 \text{ timestamps} \times 0.8589934592 \text{ seconds per increment} = 3689348814.376258 \text{ seconds}\]
08

Convert the timestamp wraparound time to meaningful units

Convert the result to years: \[\text{3689348814.376258 seconds} = \frac{3689348814.376258}{60 \times 60 \times 24 \times 365.25} \text{ years} \approx 116.99 \text{ years}\]

Key Concepts

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

TCP/IP protocol
TCP/IP (Transmission Control Protocol/Internet Protocol) is essential for data transmission across networks. TCP ensures reliable data transfer through features like connection establishment, data integrity checks, and retransmission of lost packets. It uses sequence numbers to manage the order of packets. The 32-bit sequence number field enables tracking of data packets, ensuring that the information arrives correctly and in order. Understanding TCP sequence numbers is crucial for solving network-related exercises because it directly affects how data is managed and verified.
network bandwidth
Network bandwidth refers to the data transfer capacity of a network link, measured in bits per second (bps). In this exercise, the given link has a bandwidth of 40 Gbps. High bandwidth allows more data to be transmitted in a given time, enhancing network performance. To utilize the bandwidth effectively in TCP, data packets are transmitted continuously at the maximum rate. Calculating the transmission rate in bytes per second helps in understanding how quickly sequence numbers can be exhausted, providing insights into network efficiency and performance.
wraparound time
Wraparound time is the duration it takes for TCP sequence numbers to reach their maximum and reset back to zero. The 32-bit sequence number field allows up to \(2^{32}\) unique numbers. Given the 40 Gbps link, we can calculate the wraparound time by dividing the total sequence numbers by the transmission rate (in bytes per second). Here, it takes approximately 14.32 minutes for the sequence numbers to wrap around. This concept is vital in network management as it impacts data flow and tracking, ensuring continuous transmission without overlapping sequence numbers.
timestamp field
The timestamp field in TCP is critical for managing data packets' timing and ordering. In this exercise, an added 32-bit timestamp field increments 1000 times during the sequence number wraparound time. By dividing the wraparound time by 1000, we determine the increment period. Over approximately 14.32 minutes, the timestamp wraps around after a vastly longer period (about 116.99 years). This calculation demonstrates the longevity and reliability of the TCP timestamp field in managing data packet sequencing over extended periods, ensuring precision in data transfer tracking.

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Most popular questions from this chapter

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.

Suppose, in TCP's adaptive retransmission mechanism, that EstimatedRTT is \(4.0\) at some point and subsequent measured RTTs all are \(1.0\). How long does it take before the TimeOut value, as calculated by the Jacobson/Karels algorithm, falls below \(4.0\) ? Assume a plausible initial value of Deviation; how sensitive is your answer to this choice? Use \(\delta=1 / 8\).

Suppose a host wants to establish the reliability of a link by sending packets and measuring the percentage that are received; routers, for example, do this. Explain the difficulty of doing this over a TCP connection.

One of the purposes of TIME_WAIT is to handle the case of a data packet from a first incarnation of a connection arriving very late and being accepted as data for the second incarnation. (a) Explain why, for this to happen (in the absence of TIME_WAIT), the hosts involved would have to exchange several packets in sequence after the delayed packet was sent but before it was delivered. (b) Propose a network scenario that might account for such a late delivery.

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.

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