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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

Consult Request for Comments 793 to find out how TCP is supposed to respond if a FIN or an RST arrives with a sequence number other than NextByteExpected. Consider both when the sequence number is within the receive window and when it is not.

Suppose a client \(C\) repeatedly connects via TCP to a given port on a server \(S\), and that each time it is \(\mathrm{C}\) that initiates the close. (a) How many TCP connections a second can C make here before it ties up all its available ports in TIME_WAIT state? Assume client ephemeral ports are in the range 1024-5119, and that TIME_WAIT lasts 60 seconds. (b) Berkeley-derived TCP implementations typically allow a socket in TIME WAIT state to be reopened before TIME_WAIT expires, if the highest sequence number used by the old incarnation of the connection is less than the ISN used by the new incarnation. This solves the problem of old data accepted as new; however, TIME_WAIT also serves the purpose of handling late final FINs. What would such an implementation have to do to address this and still achieve strict compliance with the TCP requirement that a FIN sent anytime before or during a connection's TIME_WAIT receive the same response?

Suppose party A connects to the Internet via a dial-up IP server (e.g., using SLIP or PPP), has several open Telnet connections (using TCP), and is cut off. Party B then dials in and is assigned the same IP address that A had. Assuming B was able to guess to what host(s) A had been connected, describe a sequence of probes that could enable B to obtain sufficient state information to continue with A's connections.

A sender on a TCP connection that receives a 0 advertised window periodically probes the receiver to discover when the window becomes nonzero. Why would the receiver need an extra timer if it were responsible for reporting that its advertised window had become nonzero (i.e., if the sender did not probe)?

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.

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