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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.

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

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?

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.

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.

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.

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?

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