/*! 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 23 Consider an ARQ algorithm runnin... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 23

Consider an ARQ algorithm running over a \(20-\mathrm{km}\) point-to-point fiber link. (a) Compute the propagation delay for this link, assuming that the speed of light is \(2 \times 10^{8} \mathrm{~m} / \mathrm{s}\) in the fiber. (b) Suggest a suitable timeout value for the ARQ algorithm to use. (c) Why might it still be possible for the ARQ algorithm to time out and retransmit a frame, given this timeout value?

Short Answer

Expert verified
Propagation delay is 100 ms. Suitable timeout value is 250 ms. Timeouts can occur due to congestion and processing delays.

Step by step solution

01

Calculate the propagation delay

To compute the propagation delay, use the formula: \[ \text{Propagation Delay} = \frac{\text{Distance}}{\text{Speed}} \] Given the distance is 20 km (or 20,000 meters) and the speed of light in the fiber is \( 2 \times 10^8 \text{ m/s} \), we can plug these values into the formula: \[ \text{Propagation Delay} = \frac{20{,}000 \text{ meters}}{2 \times 10^8 \text{ m/s}} = 0.0001 \text{ s} = 100 \text{ ms}\]
02

Suggest a suitable timeout value

A suitable timeout value for the ARQ algorithm should be slightly more than the round-trip time to account for processing and potential delays. First, calculate the round-trip time:\[ \text{RTT} = 2 \times \text{Propagation Delay} = 2 \times 100 \text{ ms} = 200 \text{ ms}\] To account for additional delays and processing time, it's good to add a margin to the timeout value. For example, adding 50 ms would make the timeout value:\[ \text{Timeout Value} = 200 \text{ ms} + 50 \text{ ms} = 250 \text{ ms} \]
03

Explain why the ARQ algorithm might time out

Even with the calculated timeout value, the ARQ algorithm might still time out and retransmit a frame. This can happen due to various unforeseen factors such as network congestion, errors in the data link causing retransmissions, or additional delays in the processing time at either end of the communication.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Propagation Delay
Propagation delay refers to the time it takes for a signal to travel from the sender to the receiver. It's crucial in networking because it affects the overall latency. In our example, we use the formula: \( \text{Propagation Delay} = \frac{\text{Distance}}{\text{Speed}} \) Given a 20 km (20,000 m) fiber link and the speed of light in fiber at \( 2 \times 10^8 \text{ m/s} \), the propagation delay is calculated as 100 milliseconds. This delay directly influences network performance.
Timeout Value
The timeout value in an ARQ algorithm determines how long the sender waits for an acknowledgment before retransmitting the frame. It's essential to set this value correctly to avoid unnecessary retransmissions. Based on our computed propagation delay, we calculate the round-trip time (RTT) and suggest a slightly higher timeout value to account for processing and potential delays. If the RTT is 200 milliseconds, adding 50 milliseconds as a safety margin results in a 250 milliseconds timeout value. This helps ensure that the network can handle minor delays without unnecessary retransmissions.
Round-Trip Time (RTT)
Round-trip time, or RTT, is the time it takes for a signal to go from the sender to the receiver and back again. It's a critical factor in determining the efficiency of the ARQ algorithm. To find the RTT, we multiply the one-way propagation delay by two: \( \text{RTT} = 2 \times \text{Propagation Delay} \) For a propagation delay of 100 milliseconds, the RTT is 200 milliseconds. Adding a buffer to account for additional delays means that our final timeout value should be a bit more than the RTT, like 250 milliseconds.
Network Congestion
Network congestion occurs when the network is overloaded with more data than it can handle efficiently. This congestion can cause increased delays, packet loss, and retransmissions. Even with an optimized timeout value, congestion can lead to timeouts because packets or acknowledgments are delayed beyond their expected arrival times. This is why it's essential to consider typical network traffic patterns and potential congestion when setting timeout values for ARQ algorithms.
Data Link Errors
Data link errors happen when there are problems in transmitting data over a network link, causing frames to be corrupted or lost. These errors can trigger retransmissions in ARQ algorithms, leading to increased delays. Factors like physical defects in the transmission medium, interference, or weak signals can cause such errors. Even with a well-calculated timeout value, these errors can lead to timeouts if the frame doesn't arrive correctly or if the acknowledgment from the receiver doesn't get back in time. Therefore, robust error detection and correction mechanisms are vital to minimize the impact of data link errors.

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

Suppose you want to send some data using the BISYNC framing protocol, and the last 2 bytes of your data are DLE and ETX. What sequence of bytes would be transmitted immediately prior to the CRC?

Assuming a framing protocol that uses bit stuffing, show the bit sequence transmitted over the link when the frame contains the following bit sequence: 110101111101011111101011111110 Mark the stuffed bits.

Consider a token ring with a ring latency of \(200 \mu\). Assuming that the delayed token release strategy is used, what is the effective throughput rate that can be achieved if the ring has a bandwidth of \(4 \mathrm{Mbps}\) ? What is the effective throughput rate that can be achieved if the ring has a bandwidth of \(100 \mathrm{Mbps}\) ? Answer for both a single active host and for "many" hosts; for the latter, assume there are sufficiently many hosts transmitting that the time spent advancing the token can be ignored. Assume a packet size of \(1 \mathrm{~KB}\).

Suppose that \(N\) Ethernet stations, all trying to send at the same time, require \(N / 2\) slot times to sort out who transmits next. Assuming the average packet size is 5 slot times, express the available bandwidth as a function of \(N\).

For a 100-Mbps token ring network with a token rotation time of \(200 \mu\) s that allows each station to transmit one 1-KB packet each time it possesses the token, calculate the maximum effective throughput rate that any one host can achieve. Do this assuming (a) immediate release and (b) delayed release.
See all solutions

Recommended explanations on Computer Science Textbooks

Computer Programming

Read Explanation

Algorithms in Computer Science

Read Explanation

Cybersecurity in Computer Science

Read Explanation

Issues in Computer Science

Read Explanation

Problem Solving Techniques

Read Explanation

Computer Organisation and Architecture

Read Explanation
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.