Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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).

Short Answer

Expert verified
EstimatedRTT increases exponentially due to retransmissions, and ultimately becomes infinite.

Step by step solution

01

Understand the given formula

The formula for EstimatedRTT is given by \( \text{EstimatedRTT} = \text{EstimatedRTT} + \beta \cdot (\text{SampleRTT} - \text{EstimatedRTT}) \). We need to understand how this formula behaves when packets are retransmitted due to loss.
02

Define initial conditions

Assume the initial EstimatedRTT is a finite value, say \( \text{EstimatedRTT}_0 \). Also, \( \beta \) is a constant value between 0 and 1 (typically \( \beta = 0.875 \)).
03

Analyze the time between retransmissions

When a packet is lost and retransmitted, SampleRTT is measured as the time interval from the original transmission to the acknowledgment (ACK) of the retransmitted packet. This interval includes the timeout period plus the time to send the retransmitted packet and receive its ACK. Hence, SampleRTT increases more than in the case of no loss.
04

Calculate SampleRTT for retransmitted packets

Given that each packet is retransmitted, SampleRTT becomes larger with each loss because it waits for an extra timeout period before the ACK. If the timeout period equals the EstimatedRTT, SampleRTT for a retransmitted packet is approximately \(2 \times \text{EstimatedRTT}_n\) where \( \text{EstimatedRTT}_n \) is the EstimatedRTT before the retransmission.
05

Show the update for EstimatedRTT

After receiving the ACK for the retransmitted packet, EstimatedRTT is updated using the formula: \( \text{EstimatedRTT}_{n+1} = \text{EstimatedRTT}_n + \beta \cdot (2 \times \text{EstimatedRTT}_n - \text{EstimatedRTT}_n) \). Simplifying, this becomes\( \text{EstimatedRTT}_{n+1} = \text{EstimatedRTT}_n + \beta \cdot \text{EstimatedRTT}_n \).
06

Observe the growth of EstimatedRTT

Notice that each update makes EstimatedRTT larger: \( \text{EstimatedRTT}_{n+1} = \text{EstimatedRTT}_n (1 + \beta) \).Given that \( \beta \) is a constant value between 0 and 1, this reveals exponential growth. Thus, as the number of retransmissions increases, EstimatedRTT will grow without bound, eventually increasing to infinity.

Key Concepts

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

EstimatedRTT calculation
Estimated Round-Trip Time (EstimatedRTT) is critical in TCP to predict the time it takes for a packet to travel to the receiver and back. The formula to update EstimatedRTT after each new measurement is:\[\text{EstimatedRTT} = \text{EstimatedRTT}_{previous} + \beta \cdot (\text{SampleRTT} - \text{EstimatedRTT}_{previous})\]Here, \(\beta\) is a smoothing factor (typically 0.875). The goal is to gradually adjust EstimatedRTT based on new measurements (SampleRTT), maintaining a balance between past knowledge and new data. This helps anticipate network delays accurately.

Each time a new SampleRTT is recorded, EstimatedRTT gets updated, incorporating this new information while retaining the majority of the old value due to the factor \(\beta\). This weighted updating mechanism helps keep EstimatedRTT close to the true average RTT but smooths out any abrupt spikes.
TCP packet loss
TCP packet loss occurs when packets of data traveling across a network fail to reach their destination. In such cases, TCP must retransmit the lost packets. Packet loss can be caused by network congestion, hardware failures, or software issues.

When a packet is lost, TCP presumes the packet has not been received and retransmits it. However, this retransmission affects the dynamics of RTT calculations.

If every other packet is lost, each packet is sent twice. This causes delays and increases the SampleRTT because the lost packet waits for a timeout period before being resent. The retransmission ensures reliability but complicates the RTT estimation process since it introduces additional delay.
SampleRTT measurement
SampleRTT measures the time elapsed from sending a TCP packet until its acknowledgment (ACK) is received. It provides real-time data about current network conditions.

Ideally, SampleRTT is taken when packets are delivered successfully without retransmissions. However, in scenarios with packet loss, SampleRTT includes the timeout period and the time for retransmitted packets.

If a packet is lost and retransmitted, SampleRTT essentially becomes the initial sending time plus the timeout duration plus the transmission time of the retransmitted packet. This extended time increases the SampleRTT value, which can significantly affect the EstimatedRTT when continuous losses occur.
timeout period
The timeout period in TCP is the duration the sender waits for an acknowledgment before assuming the packet is lost and retransmitting it. This period is generally set based on the EstimatedRTT with an added safety margin.

.Timeout is crucial to TCP's reliability, ensuring lost packets are retransmitted. However, it complicates RTT calculations in the presence of packet loss. A longer timeout can mean more delay before retransmission, making SampleRTT measurements significantly larger.

When packets are consistently lost and retransmitted, the timeout period causes a cascading effect. Each SampleRTT after a lost packet will include this extended waiting time, thus disrupting the EstimatedRTT calculation by inflating it with each loss occurrence.
exponential growth in RTT
Due to the nature of the retransmission process and the way EstimatedRTT is calculated, consistent packet loss can lead to exponential growth in RTT.

Let's illustrate with the given formula update: \[\text{EstimatedRTT}_{n+1} = \text{EstimatedRTT}_n + \beta \cdot (2 \times \text{EstimatedRTT}_n - \text{EstimatedRTT}_n)\]This simplifies to: \[\text{EstimatedRTT}_{n+1} = \text{EstimatedRTT}_n (1 + \beta)\]Given a \(\beta\) of 0.875, each retransmission cycle multiplies the EstimatedRTT by 1.875. This multiplication causes the EstimatedRTT to grow exponentially, moving towards infinity as more packets are lost and retransmitted. The larger the EstimatedRTT, the more distorted the timeout period and SampleRTT values become, causing further cyclic amplification until it can no longer reflect meaningful RTT measurements.This exponential growth underscores TCP's vulnerability to packet loss scenarios.

Understanding these underlying mechanisms is crucial for improving TCP algorithms and developing resilient network systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

TCP is a very symmetric protocol, but the client/server model is not. Consider an asymmetric TCP-like protocol in which only the server side is assigned a port number visible to the application layers. Client-side sockets would simply be abstractions that can be connected to server ports. (a) Propose header data and connection semantics to support this. What will you use to replace the client port number? (b) What form does TIME_WAIT now take? How would this be seen through the programming interface? Assume that a client socket could now be reconnected arbitrarily many times to a given server port, resources permitting. (c) Look up the \(\mathrm{rsh} / \mathrm{rlogin}\) protocol. How would the above break this?

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

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.

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 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.

See all solutions

Recommended explanations on Computer Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free