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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

Suppose an RPC request is of the form "Increment the value of field X of disk block \(\mathrm{N}\) by \(10 \%\)." Specify a mechanism to be used by the executing server to guarantee that an arriving request is executed exactly once, even if the server crashes while in the middle of the operation. Assume that individual disk block writes are either complete or else the block is unchanged. You may also assume that some designated "undo log" blocks are available. Your mechanism should include how the RPC server is to behave at restart.

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.

The sequence number field in the TCP header is 32 bits long, which is big enough to cover over 4 billion bytes of data. Even if this many bytes were never transferred over a single connection, why might the sequence number still wrap around from \(2^{32}-1\) to 0 ?

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.

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.

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