/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 Suppose that, when a TCP segment... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that, when a TCP segment is sent more than once, we take SampleRTT to be the time between the most recent transmission and the ACK, as in Figure \(5.10\) (b). Assume, for definiteness, that TimeOut \(=2 \times\) EstimatedRTT. Sketch a scenario in which no packets are lost but EstimatedRTT converges to a third of the true RTT, and give a diagram illustrating the final steady state. Hint: Begin with a sudden jump in the true RTT to just over the established TimeOut.

Short Answer

Expert verified
EstimatedRTT converges to True RTT / 3, and TimeOut = 2 * EstimatedRTT. True RTT consistently influences SampleRTT due to the time-out mechanism.

Step by step solution

01

- Understand the parameters

Interpret the given parameters. True RTT suddenly increases to just over the established TimeOut, which means True RTT > 2 * EstimatedRTT.
02

- Analyze the impact on EstimatedRTT

Since True RTT > TimeOut = 2 * EstimatedRTT, every packet will time out. This means SampleRTT will be taken as the time from the most recent transmission to the ACK.
03

- Determine the SampleRTT effect

SampleRTT in this scenario will be approximately True RTT since packets aren't lost. However, EstimatedRTT will still be updated based on the new SampleRTT.
04

- Sketch the initial transmission scenario

Create a diagram showing the initial True RTT jump to just over 2 * EstimatedRTT. Show how SampleRTT is initially much higher than EstimatedRTT.
05

- Update EstimatedRTT in iteration

As time progresses, and SampleRTT is always True RTT, EstimatedRTT is continuously smoothed using the weighted average.
06

- Understand Steady State

In the final steady state, with the continuous updates, EstimatedRTT adjusts to a new lower steady value. It converges to a third of True RTT due to the formula's impact.
07

- Illustrate the final steady state

Draw a diagram showing the steady state where EstimatedRTT = True RTT / 3 while TimeOut = 2 * EstimatedRTT.
08

- Verify the steady state

Check if in steady state, EstimatedRTT and TimeOut values make sense with the continuous SampleRTT being True RTT.

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Key Concepts

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

SampleRTT
SampleRTT, or Sample Round-Trip Time, is the time it takes for a packet to travel from the sender to the receiver and back. In TCP, when a segment is transmitted, the sender starts a timer. Once an acknowledgment (ACK) is received for that segment, the total time measured by this timer is the SampleRTT. This value is essential for TCP as it helps in estimating network latency and ensuring reliable data transmission. Imagine sending a message and waiting for a reply. The time you wait until you get a reply is akin to SampleRTT for TCP packets.
Since network conditions can vary, each SampleRTT may differ. Hence, TCP doesn't just rely on a single SampleRTT value but continuously updates an average to adapt to changing conditions.
EstimatedRTT
EstimatedRTT is a smoothed average of the SampleRTT measurements. To avoid excessive fluctuation and to provide a reliable estimate of the current network delay, TCP uses an exponential weighted moving average (EWMA). The formula for updating EstimatedRTT is as follows:
\( \text{EstimatedRTT} = (1 - \beta) \times \text{EstimatedRTT}_{\text{previous}} + \beta \times \text{SampleRTT} \)
Typically, \( \beta \) is set to 0.125, meaning the new SampleRTT has a smaller influence on the EstimatedRTT.
This method allows the EstimatedRTT to gradually adapt to network changes without being overly sensitive to temporary fluctuations.
TimeOut Calculation
The TimeOut value, or retransmission timeout (RTO), determines how long TCP should wait for an ACK before retransmitting a segment. It is crucial to set the TimeOut appropriately because if it's too short, it may result in premature retransmissions, while if it's too long, it can delay the detection of lost packets. Typically, the TimeOut is calculated using the EstimatedRTT and an additional measure of variance called the deviation, represented by DevRTT. The common formula for TimeOut is:
\( \text{TimeOut} = \text{EstimatedRTT} + 4 \times \text{DevRTT} \)
This formula ensures that TimeOut accounts for both the estimated delay and its variability, providing a balance between responsiveness and caution.
TCP Timeouts
In TCP, timeouts play a critical role in ensuring reliable data transmission. If a segment is not acknowledged within the calculated TimeOut period, the sender assumes the segment was lost and retransmits it. Frequent timeouts can indicate severe network congestion or packet loss, prompting TCP to invoke congestion control mechanisms.
  • Fast Retransmit: After receiving three duplicate ACKs, TCP retransmits the missing segment without waiting for the TimeOut.
  • Retransmission Timeout (RTO): If the ACK isn't received within the TimeOut period, TCP retransmits the segment.
Managing timeouts efficiently is vital for maintaining reliable communication and optimizing network performance.
RTT Variations
Round-Trip Time (RTT) variations reflect the fluctuations in the time it takes for packets to travel back and forth between sender and receiver. Several factors can cause these variations, including network congestion, routing changes, and physical distance between endpoints.
TCP addresses RTT variations by incorporating DevRTT, which measures how much SampleRTT deviates from EstimatedRTT. The formula for updating DevRTT is:
\( \text{DevRTT} = (1 - \beta) \times \text{DevRTT} + \beta \times | \text{SampleRTT} - \text{EstimatedRTT} | \)
By considering both the average RTT and its variability, TCP can dynamically adjust the TimeOut value to accommodate changing network conditions. This ensures efficient communication even in the face of RTT variations.

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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.

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.

This chapter explains three sequences of state transitions during TCP connection teardown. There is a fourth possible sequence, which traverses an additional arc (not shown in Figure 5.7) from FIN_WAIT_1 to TIME_WAIT and labelled FIN + ACK/ACK. Explain the circumstances that result in this fourth teardown sequence.

Suppose we were to implement remote file system mounting using an unreliable RPC protocol that offers zero-or-more semantics. If a message reply is received, this improves to at-least-once semantics. We define read() to return the specified Nth block, rather than the next block in sequence; this way reading once is the same as reading twice and at-least-once semantics is thus the same as exactly once. (a) For what other file system operations is there no difference between at- leastonce and exactly once semantics? Consider open, create, write, seek, opendir, readdir, mkdir, delete (aka unlink), and rmdir. (b) For the remaining operations, which can have their semantics altered to achieve equivalence of at-least-once and exactly once? What file system operations are irreconcilable with at-least-once semantics? (c) Suppose the semantics of the rmdir system call are now that the given directory is removed if it exists, and nothing is done otherwise. How could you write a program to delete directories that distinguishes between these two cases?

TCP's simultaneous open feature is seldom used. (a) Propose a change to TCP in which this is disallowed. Indicate what changes would be made in the state diagram (and if necessary in the undiagrammed event responses). (b) Could TCP reasonably disallow simultaneous close? (c) Propose a change to TCP in which simultaneous SYNs exchanged by two hosts lead to two separate connections. Indicate what state diagram changes this entails, and also what header changes become necessary. Note that this now means that more than one connection can exist over a given pair of \langlehost, port)s. (You might also look up the first "Discussion" item on page 87 of Request for Comments \(1122 .\) )

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