Logout Open In App
Open In App
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

Chapter 2: Problem 52

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 \(N\).

Short Answer

Expert verified
\(\frac{5}{\frac{N}{2} + 5}\)

Step by step solution

01

Understanding the problem

Determine the time required for sorting out which station transmits next and the time taken for the average packet transmission.
02

Calculate sorting time

Given that it takes \(N/2\) slot times to sort out who transmits next.
03

Determine packet transmission time

Given that the average packet size equates to 5 slot times.
04

Compute total time per transmission

The total time for one complete cycle (sorting + transmission) is \(\frac{N}{2} + 5\) slot times.
05

Express the available bandwidth

The available bandwidth in terms of the number of stations \(N\) and the number of successful transmissions per slot time is given by \(\frac{5}{\frac{N}{2} + 5}\). This represents the fraction of time utilized for transmitting useful data.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

network stations
Network stations are the computers or devices connected to a network. In an Ethernet network, these stations communicate with each other by transmitting data packets. Each station must wait for its turn to send data to avoid collisions, which can happen if two stations try to send data at the same time. The number of network stations, represented as \(N\), impacts the overall network performance and bandwidth.
slot time
Slot time is a critical concept in Ethernet networks. It represents the time window during which the network is checked for collisions. The length of a slot time is determined by the maximum network propagation delay. In this problem, sorting out which station will transmit next takes \( \frac{N}{2} \) slot times, where \(N\) is the number of stations.
packet transmission
Packet transmission refers to the sending of data packets from one network station to another. Each packet has a size, and in this problem, the average packet size is given as 5 slot times. This means it takes 5 slot times to send one packet completely without any interruptions.
available bandwidth
Available bandwidth is the measure of the network's ability to transmit data. It's affected by the time spent sorting out which station will send next and the actual transmission time of packets. It's expressed as the fraction of time used effectively for transmission. For \(N\) stations, the available bandwidth is represented as \( \frac{5}{\frac{N}{2} + 5} \). This formula shows that as the number of stations \(N\) increases, the time needed to sort out transmissions also increases, reducing the available bandwidth.
network efficiency
Network efficiency is about how well a network uses its available resources. In the context of Ethernet, it's about maximizing the amount of data successfully transmitted versus the time spent on managing transmissions and collisions. The available bandwidth formula \( \frac{5}{\frac{N}{2} + 5} \) indicates network efficiency. With fewer stations, sorting time is less, and efficiency is higher. As the number of network stations increases, more time is spent deciding who transmits next, lowering efficiency.

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

Assume that a SONET receiver resynchronizes its clock whenever a 1 bit appears; otherwise, the receiver samples the signal in the middle of what it believes is the bit's time slot. (a) What relative accuracy of the sender's and receiver's clocks is required in order to receive correctly 480 bytes (one ATM AAL. 5 cell's worth) in a row? (b) Consider a forwarding station A on a SONET STS-1 line, receiving frames from the downstream end \(B\) and retransmitting them upstream. What relative accuracy of A's and B's clocks is required to keep A from accumulating more than one extra frame per minute?

Suppose a \(100-\mathrm{Mbps}\) delayed-release token ring has 10 stations, a ring latency of \(30 \mu \mathrm{s}\), and an agreed-upon TTRT of \(350 \mu \mathrm{s}\). (a) How many synchronous frame bytes could each station send, assuming all are allocated the same amount? (b) Assume stations \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) are in increasing order on the ring. Due to uniform synchronous traffic, the TRT without asynchronous data is \(300 \mu \mathrm{s}\). B sends a \(200-\mu \mathrm{s}(2.5-\mathrm{Kb})\) asynchronous frame. What TRT will \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) then see on their next measurement? Who may transmit such a frame next?

Coaxial cable Ethernet was limited to a maximum of \(500 \mathrm{~m}\) between repeaters, which regenerate the signal to \(100 \%\) of its original amplitude. Along one \(500-\mathrm{m}\) segment, the signal could decay to no less than \(14 \%\) of its original value \((8.5 \mathrm{~dB})\). Along \(1500 \mathrm{~m}\), then, the decay might be \((0.14)^{3}=0.3 \%\). Such a signal, even along \(2500 \mathrm{~m}\), is still strong enough to be read; why then are repeaters required every \(500 \mathrm{~m}\) ? $$ \begin{array}{l|l} \hline \text { Item } & \text { Delay } \\ \hline \text { Coaxial cable } & \text { propagation speed } .77 c \\ \text { Link/drop cable } & \text { propagation speed .65c } \\ \text { Repeaters } & \text { approximately 0.6 } \mu \text { s each } \\ \text { Transceivers } & \text { approximately 0.2 } \mu \text { s each } \\ \hline \end{array} $$

An IEEE \(802.5\) token ring has five stations and a total wire length of \(230 \mathrm{~m}\). How many bits of delay must the monitor insert into the ring? Do this for both \(4 \mathrm{Mbps}\) and \(16 \mathrm{Mbps}\); use a propagation rate of \(2.3 \times 10^{8} \mathrm{~m} / \mathrm{s}\).

Suppose \(A, B\), and \(C\) all make their first carrier sense, as part of an attempt to transmit, while a fourth station D is transmitting. Draw a timeline showing one possible sequence of transmissions, attempts, collisions, and exponential backoff choices. Your timeline should also meet the following criteria: (i) initial transmission attempts should be in the order \(A, B, C\), but successful transmissions should be in the order \(\mathrm{C}, \mathrm{B}, \mathrm{A}\), and (ii) there should be at least four collisions.
See all solutions

Recommended explanations on Computer Science Textbooks

Blockchain Technology

Read Explanation

Data Representation in Computer Science

Read Explanation

Computer Organisation and Architecture

Read Explanation

Algorithms in Computer Science

Read Explanation

Game Design in Computer Science

Read Explanation

Databases

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.

Sign-up for free