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

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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

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?

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.

Suppose a TCP connection, with window size 1, loses every other packet. Those that do arrive have RTT \(=1\) second. What happens? What happens to TimeOut? Do this for two cases: (a) After a packet is eventually received, we pick up where we left off, resuming with EstimatedRTT initialized to its pretimeout value and TimeOut double that. (b) After a packet is eventually received, we resume with TimeOut initialized to the last exponentially backed-off value used for the timeout interval. In the following four exercises, the calculations involved are straightforward with a spreadsheet.

You are hired to design a reliable byte-stream protocol that uses a sliding window (like TCP). This protocol will run over a 1-Gbps network. The RTT of the network is \(140 \mathrm{~ms}\), and the maximum segment lifetime is 60 seconds. How many bits would you include in the AdvertisedWindow and SequenceNum fields of your protocol header?

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.

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