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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.

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

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 that, when a TCP segment is sent more than once, we take SampleRTT to be the time between the original transmission and the ACK, as in Figure \(5.10(\mathrm{a}) .\) Show that if a connection with a 1-packet window loses every other packet (i.e., each packet is transmitted twice), then EstimatedRTT increases to infinity. Assume TimeOut = EstimatedRTT; both algorithms presented in the text always set TimeOut even larger. Hint: EstimatedRTT \(=\) EstimatedRTT \(+\beta \times(\) SampleRT \(-\) EstimatedRTT).

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.

Diagnostic programs are commonly available that record the first 100 bytes, say, of every TCP connection to a certain (host, port). Outline what must be done with each received TCP packet, P, in order to determine if it contains data that belongs to the first 100 bytes of a connection to host HOST, port PORT. Assume the IP header is P.IPHEAD, the TCP header is P.TCPHEAD, and header fields are as named in Figures \(4.3\) and 5.4. Hint: To get initial sequence numbers (ISNs) you will have to examine every packet with the SYN bit set. Ignore the fact that sequence numbers will eventually be reused.

When closing a TCP connection, why is the two-segment-lifetime timeout not necessary on the transition from LAST_ACK to CLOSED?

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