Communication channel capacity. Internet connection speed. Network Bandwidth What is Bandwidth

This characteristic depends on several factors. First of all, this is the diameter of the pipe, as well as the type of liquid, and other indicators.

For hydraulic calculation of a pipeline, you can use the hydraulic pipeline calculation calculator.

When calculating any systems based on fluid circulation through pipes, there is a need to accurately determine pipe capacity. This is a metric value that characterizes the amount of liquid flowing through pipes over a certain period of time. This indicator is directly related to the material from which the pipes are made.

If we take, for example, plastic pipes, they differ in almost the same throughput throughout their entire service life. Plastic, unlike metal, is not prone to corrosion, so a gradual increase in deposits is not observed in it.

As for metal pipes, they throughput decreases year after year. Due to the appearance of rust, the material inside the pipes peels off. This leads to surface roughness and the formation of even more plaque. This process occurs especially quickly in hot water pipes.

The following is a table of approximate values, which was created to make it easier to determine the throughput of pipes in apartment wiring. This table does not take into account the reduction in throughput due to the appearance of sedimentary build-ups inside the pipe.

Table of pipe capacity for liquids, gas, water vapor.

Type of liquid	Speed (m/sec)
City water
Water pipeline
Central heating water
Pressure system water in pipeline line
Hydraulic fluid	up to 12m/sec
Oil pipeline line
Oil in the pressure system of the pipeline line
Steam in the heating system
Steam central piping system
Steam in a high temperature heating system
Air and gas in the central piping system

Most often, ordinary water is used as a coolant. The rate of decrease in throughput in pipes depends on its quality. The higher the quality of the coolant, the longer the pipeline made of any material (steel, cast iron, copper or plastic) will last.

Calculation of pipe capacity.

For accurate and professional calculations, you must use the following indicators:

The material from which pipes and other elements of the system are made;
Pipe length
Number of water consumption points (for water supply system)

The most popular calculation methods:

1. Formula. A rather complex formula, which is understandable only to professionals, takes into account several values at once. The main parameters that are taken into account are the material of the pipes (surface roughness) and their slope.

2. Table. This is a simpler way by which anyone can determine the throughput of a pipeline. An example is the engineering table of F. Shevelev, from which you can find out the throughput capacity based on the pipe material.

3. Computer program. One of these programs can be easily found and downloaded on the Internet. It is designed specifically to determine the throughput for pipes of any circuit. In order to find out the value, you need to enter initial data into the program, such as material, pipe length, coolant quality, etc.

It should be said that last method, although the most accurate, is not suitable for calculations of simple household systems. It is quite complex and requires knowledge of the values of a wide variety of indicators. To calculate a simple system in a private house, it is better to use tables.

An example of calculating pipeline capacity.

Pipeline length - important indicator when calculating capacity, the length of the highway has a significant impact on capacity indicators. The greater the distance water travels, the less pressure it creates in the pipes, which means the flow speed decreases.

Here are some examples. Based on tables developed by engineers for these purposes.

Pipe capacity:

0.182 t/h with a diameter of 15 mm
0.65 t/h with pipe diameter 25 mm
4 t/h with a diameter of 50 mm

As can be seen from the examples given, a larger diameter increases the flow rate. If the diameter is doubled, the throughput will also increase. This dependence must be taken into account when installing any liquid system, be it plumbing, drainage or heat supply. This is especially true for heating systems, since in most cases they are closed, and the heat supply in the building depends on the uniform circulation of the liquid.

Ilya Nazarov
System engineer at INTELCOM Line

After assessing the required throughput on each section of the IP network, it is necessary to decide on the choice of OSI network and link layer technologies. In accordance with the selected technologies, the most suitable models of network equipment are determined. This question is also difficult, since throughput directly depends on hardware performance, and performance, in turn, depends on the hardware and software architecture. Let's take a closer look at the criteria and methods for assessing the capacity of channels and equipment in IP networks.

Bandwidth Evaluation Criteria

Since the emergence of teletraffic theory, many methods have been developed for calculating channel capacity. However, unlike calculation methods applied to circuit-switched networks, calculating the required throughput in packet networks is quite complex and is unlikely to provide accurate results. First of all, this is due to a huge number of factors (especially those inherent in modern multiservice networks), which are quite difficult to predict. In IP networks, a common infrastructure is typically used by many applications, each of which may use its own different traffic pattern. Moreover, within one session, traffic transmitted in the forward direction may differ from traffic transmitted in the opposite direction. In addition to this, calculations are complicated by the fact that the speed of traffic between individual network nodes can change. Therefore, in most cases, when building networks, the capacity estimate is actually determined by general recommendations manufacturers, statistical studies and the experience of other organizations.

To more or less accurately determine how much bandwidth is required for the network being designed, you must first know what applications will be used. Next, for each application, you should analyze how data will be transferred during the selected periods of time, and what protocols are used for this.

For simple example Let's consider applications of a small corporate network.

Example of bandwidth calculation

Let's assume there are 300 work computers and the same number of IP phones on the network. It is planned to use the following services: email, IP telephony, video surveillance (Fig. 1). For video surveillance, 20 cameras are used, from which video streams are transmitted to the server. Let's try to estimate what maximum bandwidth is required for all services on the channels between the network core switches and at the junctions with each of the servers.

It should be noted right away that all calculations must be carried out for the time of greatest network activity of users (in teletraffic theory - peak hours), since usually during such periods network performance is most important and delays and failures in application operation associated with a lack of bandwidth occur. , are unacceptable. In organizations, the greatest load on the network may occur, for example, at the end of the reporting period or during a seasonal influx of customers, when the largest number of telephone calls are made and the majority of email messages are sent.

Email
Returning to our example, consider an email service. It uses protocols that run on top of TCP, meaning the data transfer rate is constantly adjusted to take up all the available bandwidth. Thus, we will start from the maximum delay value for sending a message - let’s say 1 second will be enough to make the user comfortable. Next, you need to estimate the average size of the message sent. Let us assume that during peaks of activity mail messages will often contain various attachments (copies of invoices, reports, etc.), so for our example we will take the average message size to be 500 KB. And finally, last parameter, which we need to select is the maximum number of employees who simultaneously send messages. Suppose, during emergency times, half of the employees simultaneously press the "Send" button in mail client. The required maximum throughput for email traffic would then be (500 kB x 150 hosts)/1 s = 75,000 kB/s or 600 Mbps. From here we can immediately conclude that to connect the mail server to the network it is necessary to use a Gigabit Ethernet channel. At the core of the network, this value will be one of the terms that makes up the total required throughput.

Telephony and video surveillance
Other applications - telephony and video surveillance - are similar in their stream transmission structure: both types of traffic are transmitted using the UDP protocol and have a more or less fixed transmission rate. The main differences are that in telephony the streams are bidirectional and limited by the time of the call, while in video surveillance the streams are transmitted in one direction and, as a rule, are continuous.

To estimate the required throughput for telephony traffic, assume that during peak activity the number of simultaneous connections passing through the gateway can reach 100. When using the G.711 codec in Ethernet networks the speed of one stream, taking into account headers and service packets, is approximately 100 kbit/s. Thus, during periods of greatest user activity, the required bandwidth in the network core will be 10 Mbit/s.

Video surveillance traffic is calculated quite simply and accurately. Let’s say that in our case, video cameras transmit streams of 4 Mbit/s each. The required bandwidth will be equal to the sum of the speeds of all video streams: 4 Mbit/s x 20 cameras = 80 Mbit/s.

All that remains is to add up the resulting peak values for each of network services: 600 + 10 + 80 = 690 Mbps. This will be the required bandwidth in the network core. The design should also include the possibility of scaling so that communication channels can serve the traffic of a growing network for as long as possible. In our example, it will be enough to use Gigabit Ethernet to meet the requirements of the services and at the same time be able to seamlessly develop the network by connecting more nodes

Of course, the example given is far from being a standard one - each case must be considered separately. In reality, the network topology can be much more complex (Fig. 2), and capacity assessment must be made for each section of the network.

It should be taken into account that VoIP traffic (IP telephony) is distributed not only from phones to the server, but also between phones directly. In addition, network activity may vary in different departments of the organization: the technical support service makes more phone calls, the project department uses e-mail more actively than others, the engineering department consumes more Internet traffic than others, etc. As a result, some parts of the network may require more bandwidth than others.

Usable and full throughput

In our example, when calculating the IP telephony flow rate, we took into account the codec used and the size of the packet header. This is an important detail to keep in mind. Depending on the encoding method (codecs used), the amount of data transmitted in each packet, and the link-layer protocols used, the total throughput of the stream is formed. It is the total throughput that must be taken into account when estimating the required network throughput. This is most relevant for IP telephony and other applications that use real-time transmission of low-speed streams, in which the size of the packet headers is a significant part of the size of the entire packet. For clarity, let's compare two VoIP streams (see table). These streams use the same compression, but different payload sizes (actually, the digital audio stream) and different link layer protocols.

The data transfer rate in its pure form, without taking into account network protocol headers (in our case, a digital audio stream), is useful bandwidth. As you can see from the table, with the same useful throughput of streams, their total throughput can vary greatly. Thus, when calculating the required network capacity for telephone calls during peak loads, especially for telecom operators, the choice of channel protocols and flow parameters plays a significant role.

Equipment selection

The choice of link-layer protocols is usually not a problem (today the question more often arises of how much bandwidth an Ethernet channel should have), but choosing the right equipment can cause difficulties even for an experienced engineer.

The development of network technologies, along with the growing demands of applications for network bandwidth, forces network equipment manufacturers to develop ever new software and hardware architectures. Often, from a single manufacturer there are seemingly similar equipment models, but designed to solve different network problems. Take, for example, Ethernet switches: most manufacturers, along with conventional switches used in enterprises, have switches for building data storage networks, organizing operator services, etc. Models of one price category differ in their architecture, “tailored” for specific tasks.

In addition to overall performance, the choice of equipment should also be based on supported technologies. Depending on the type of hardware, a certain set of functions and types of traffic can be processed at the hardware level without using CPU and memory resources. At the same time, traffic from other applications will be processed at the software level, which greatly reduces overall performance and, as a result, maximum throughput. For example, multi-layer switches, thanks to their complex hardware architecture, are capable of transmitting IP packets without reducing performance when all ports are at maximum load. Moreover, if we want to use more complex encapsulation (GRE, MPLS), then such switches (at least inexpensive models) are unlikely to be suitable for us, since their architecture does not support the corresponding protocols, and in the best case, such encapsulation will occur at the expense of a low-performance central processor. Therefore, to solve such problems, we can consider, for example, routers whose architecture is based on a high-performance central processor and depends to a greater extent on software rather than hardware implementation. In this case, at the expense of maximum throughput, we get a huge range of supported protocols and technologies that are not supported by switches in the same price category.

Overall Equipment Performance

In the documentation for their equipment, manufacturers often indicate two maximum throughput values: one expressed in packets per second, the other in bits per second. This is due to the fact that most of the performance of network equipment is spent, as a rule, on processing packet headers. Roughly speaking, the equipment must receive the packet, find a suitable switching path for it, generate a new header (if necessary) and transmit it further. Obviously, in this case it is not the volume of data transmitted per unit of time that plays a role, but the number of packets.

If we compare two streams transmitted at the same speed, but with different sizes packets, then transmitting a stream with a smaller packet size will require more performance. This fact should be taken into account if, for example, a large number of IP telephony streams are supposed to be used on the network - the maximum throughput in bits per second here will be much less than declared.

It is clear that with mixed traffic, and even taking into account additional services (NAT, VPN), as happens in the vast majority of cases, it is very difficult to calculate the load on equipment resources. Often equipment manufacturers or their partners conduct Stress Testing different models under different conditions and the results are published on the Internet in the form of comparative tables. Familiarization with these results greatly simplifies the task of choosing suitable model.

Pitfalls of modular equipment

If selected network hardware is modular, then, in addition to the flexible configuration and scalability promised by the manufacturer, you can get many pitfalls.

When choosing modules, you should carefully read their description or consult the manufacturer. It is not enough to be guided only by the type of interfaces and their number - you also need to become familiar with the architecture of the module itself. For similar modules, it is not uncommon that when transmitting traffic, some are able to process packets autonomously, while others simply forward packets to the central processing module for further processing (accordingly, for externally identical modules, the price for them can differ several times). In the first case, the overall performance of the equipment and, as a consequence, its maximum throughput are higher than in the second, since part of its work CPU transfers to module processors.

In addition, modular equipment often has a blocking architecture (when the maximum throughput is lower than the total speed of all ports). This is due to the limited capacity of the internal bus through which the modules exchange traffic with each other. For example, if a modular switch has a 20 Gbps internal bus, its 48-port Gigabit Ethernet line card can only use 20 ports when fully loaded. You should also keep such details in mind and carefully read the documentation when choosing equipment.

When designing IP networks, bandwidth is a key parameter that will determine the architecture of the network as a whole. For a more accurate assessment of throughput, you can follow the following recommendations:

Study the applications that you plan to use on the network, the technologies they use, and the volume of transmitted traffic. Use the advice of developers and the experience of colleagues to take into account all the nuances of these applications when building networks.
Dive deep into the network protocols and technologies used by these applications.
Read the documentation carefully when choosing equipment. To have some reserve ready-made solutions, check out the product lines of different manufacturers.

As a result, when making the right choice technologies and equipment, you can be sure that the network will fully satisfy the requirements of all applications and, being sufficiently flexible and scalable, will last for a long time.

Throughput is a universal characteristic that describes the maximum number of units of objects passing through a channel, node, section. The characteristic is widely used by signalmen, transport workers, hydraulics, optics, acoustics, and mechanical engineering. Everyone gives their own definition. Usually they draw the line by using units of time, clearly linking the physical meaning to the speed of the process. The communication channel transmits information. Therefore, the characteristic of throughput is the bitrate (bit/s, baud).

Unit

Standard bit/s is often supplemented with prefixes:

Kilo: kbps = 1000 bps.
Mega: Mbps = 1,000,000 bps.
Giga: Gbit/s = 1 billion bit/s.
Tera: Tbit/s = 1 trillion. bps
Peta: Pbit/s = 1 quadrillion bit/s.

Byte dimensions are used less frequently (1B = 8 bits). The value usually refers to the physical layer of the OSI hierarchy. Part of the channel capacity is taken away by protocol conventions: headers, start bits... Bauds are used to measure modulated speed, indicating the number of symbols per unit time. For the binary system (0, 1) both concepts are equivalent. Coding levels, for example, with pseudo-noise sequences changes the balance of power. The baud rate becomes smaller at the same bitrate; the difference is determined by the base of the superimposed signal. The theoretically achievable upper limit of the modulated rate is related to the channel spectrum width by the Nyquist law:

baud ≤ 2 x width (Hz).

In practice, the threshold is achieved by the simultaneous fulfillment of two conditions:

Single sideband modulation.
Linear (physical) coding.

Commercial channels show throughput at half the rate. The real network also transmits frame bits, redundant error correction information. The latter applies doubly to wireless protocols and ultra-high-speed copper lines. The headers of each subsequent OSI layer successively reduce the actual channel throughput.

Separately, experts stipulate peak values - numbers obtained using ideal conditions. Real speed connections are established with specialized equipment, less often software. Online meters often show unrealistic values describing the state of a single branch of the World Wide Web. Lack of standardization adds to the confusion. Sometimes bitrate implies physical speed, less often – network speed (subtracting the amount of service information). The values are related as follows:

network speed = physical speed x code speed.

The latter value takes into account the ability to correct errors, always less than one. Network speed is definitely lower than physical speed. Example:

The network speed of the IEEE 802.11a protocol is 6..54 Mbit/s. Pure bitrate – 12..72 Mbit/s.
The actual transmission speed of 100Base-TX Ethernet is 125 Mbps, thanks to the adopted 4B5B encoding system. However, the applied NRZI linear modulation technique allows a symbol rate of 125 Mbaud to be specified.
Ethernet 10Base-T is devoid of error correction code, the network speed is equal to physical speed (10 Mbit/s). However, the Manchester code used determines the assignment of the final symbolic value to 20 Mbaud.
The asymmetry of the speed of the upstream (48 kbit/s) and downstream (56 kbit/s) channels of a V.92 voice modem is well known. Networks of many generations of cellular communications work similarly.

The channel capacity was named Shannon - the theoretical upper limit of the network bitrate in the absence of errors.

Capacity Enhancement Theory

Information theory was developed by Claude Shannon, observing the horrors of World War II, introduced the concept of channel capacity, and developed mathematical models. Simulation of a connected line includes three blocks:

Transmitter.
Noisy channel (presence of an interference source).
Receiver.

Transmitted and received information are presented conditional functions distributions. Shannon's capacitive model is described by graphs. The Wikipedia example gives an overview of an environment characterized by five discrete levels of wanted signal. The noise is selected from the interval (-1..+1). Then the channel capacity is equal to the sum of the useful signal and interference modulo 5. The resulting value is often fractional. Therefore, it is difficult to determine the size of the initially transmitted information (round up or down).

Values that are further apart (for example, 1; 3) cannot be confused. Each set formed by three or more distinguishable messages is supplemented by one fuzzy one. Although the nominal capacity of the channel allows 5 values to be transmitted simultaneously, a pair that allows messages to be encoded without errors is effective. To increase the volume, use the following combinations: 11, 23, 54, 42. The code distance of the sequences is always greater than two. Therefore, interference is powerless to prevent the correct recognition of the combination. Multiplexing becomes possible, significantly increasing the throughput of the communication channel.

Five discrete values are also combined by an equilateral graph. The ends of the edges indicate pairs of values that the receiver may confuse due to the presence of noise. Then the number of combinations is represented by an independent set of the composed graph. Graphically, the set is assembled by combinations that exclude the presence of both points of one edge. Shannon's model for a five-level signal is composed exclusively of pairs of values (see above). Attention, question!

What do complex theoretical calculations have to do with the discussed topic of channel capacity?

The most direct thing. The first digital system for transmitting encoded information Green Bumblebee (Second World War) used a 6-level signal. The theoretical calculations of scientists provided the allies with reliable encrypted communications, allowing them to hold over 3,000 conferences. The computational complexity of Shannon graphs remains unknown. They tried to obtain meaning in a roundabout way, continuing the series as the case became more complex. We consider the Lovas number to be a colorful example of what has been said.

Bitrate

The capacity of a real channel is calculated according to theory. A noise model is constructed, for example, additive Gaussian, and the expression of the Shannon-Hartley theorem is obtained:

C = B log2 (1 + S/N),

B – bandwidth (Hz); S/N – signal/noise ratio. The base 2 logarithm allows you to calculate the bitrate (bit/s). The magnitude of the signal and noise are written in square volts or watts. Substituting decibels gives the wrong result. The formula for peer-to-peer wireless networks is slightly different. Take the spectral density of the noise multiplied by the bandwidth. Separate expressions for fast and slow fading channels are derived.

Multimedia files

In relation to entertainment applications, bitrate shows the amount of information stored and played back every second:

Data sampling rates vary.
Samples of different sizes (bits).
Sometimes encryption is carried out.
Specialized algorithms compress information.

A golden mean is selected that helps minimize the bitrate and ensures acceptable quality. Sometimes compression irreversibly distorts the source material with compression noise. Often the speed shows the number of bits per unit of time of audio or video being played (displayed by the player). Sometimes the value is calculated by dividing the file size by the total duration. Since the dimension is specified in bytes, a multiplier of 8 is entered. Often the multimedia bitrate fluctuates. The entropy rate is called the minimum rate that ensures complete preservation of the original material.

CDs

The audio CD standard requires the stream to be transmitted at a sampling rate of 44.1 kHz (16-bit depth). Typical stereo music is composed of two channels (left, right speakers). Bitrate doubles to mono. The throughput of the pulse code modulation channel is determined by the expression:

bitrate = sampling rate x depth x number of channels.

The audio CD standard gives a final figure of 1.4112 Mbit/s. A simple calculation shows: 80 minutes of recording takes up 847 MB, excluding headers. The large file size determines the need to compress the content. Here are the MP3 format numbers:

32 kbit/s – acceptable for articulate speech.
96 kbps – low quality recording.
.160 kbit/s is a weak level.
192 kbps is something in between.
256 kbps is typical for most tracks.
320 kbps – premium quality.

The effect is obvious. Reducing speed while increasing playback quality. The simplest telephone codecs take 8 kbit/s, Opus - 6 kbit/s. Video is more demanding. A 10-bit uncompressed Full HD stream (24 frames) takes up 1.4 Gbps. The need for providers to continually exceed previously set records becomes clear. Basic family Sunday viewing is measured by the overall experience of the audience. It is difficult to explain to loved ones what an error in image digitization is.

Real channels are being built, providing a solid supply. Similar reasons are behind the progress of digital media standards. Dolby Digital(1994) explicitly provided for information loss. The first showing of Batman Returns (1992) was played from 35mm film carrying compressed sound(320 kbps). The video frames were transferred by a CCD scanner, and the equipment unpacked the audio along the way. Equipped with a 5.1 Digital Surround system, the hall required further digital processing flow.

Real systems are often formed by a set of channels. Today, the former chic is being replaced by Dolby Surround 7.1, and Atmos is growing in popularity. Identical technologies can be implemented in almost original ways. Here are examples of eight-channel (7.1) audio:

Dolby Digital Plus (3/1.7 Mbps).
Dolby TrueHD (18 Mbit/s).

The specified bandwidth varies.

Channel Capacity Examples

Let's consider the evolution of digital information transmission technologies.

Modems

Acoustic pair (1972) – 300 baud.
Modem Vadik&Bell 212A (1977) – 1200 baud.
ISDN channel (1986) – 2 channels 64 kbit/s (total speed – 144 kbit/s).
32bis (1990) – up to 19.2 kbit/s.
34 (1994) – 28.8 kbps.
90 (1995) – 56 kbit/s downstream, 33.6 kbit/s upstream.
92 (1999) – 56/48 kbps downstream/upstream.
ADSL (1998) – up to 10 Mbit/s.
ADSL2 (2003) – up to 12 Mbit/s.
ADSL2+ (2005) – up to 26 Mbit/s.
VDSL2 (2005) – 200 Mbit/s.
fast (2014) – 1 Gbit/s.

Ethernet LAN

Experimental version (1975) – 2.94 Mbit/s.
10BASES (1981, coaxial cable) – 10 Mbit/s.
10BASE-T (1990, twisted pair) – 10 Mbit/s.
Fast Ethernet (1995) – 100 Mbit/s.
Gigabit Ethernet (1999) – 1 Gbit/s.
10 Gigabit Ethernet (2003) – 10 Gbit/s.
100 Gigabit Ethernet (2010) – 100 Gbit/s.

WiFi

IEEE 802.11 (1997) – 2 Mbit/s.
IEEE 802.11b (1999) – 11 Mbit/s.
IEEE 802.11a (1999) – 54 Mbit/s.
IEEE 802.11g (2003) – 54 Mbit/s.
IEEE 802.11n (2007) – 600 Mbit/s.
IEEE 802.11ac (2012) – 1000 Mbps.

cellular

First generation:
1. NMT (1981) – 1.2 kbit/s.
2G:
1. GSM CSD, D-AMPS (1991) – 14.4 kbit/s.
2. EDGE (2003) – 296/118.4 kbps.
3G:
1. UMTS-FDD (2001) – 384 kbit/s.
2. UMTS HSDPA (2007) – 14.4 Mbit/s.
3. UMTS HSPA (2008) – 14.4/5.76 Mbit/s.
4. HSPA+ (2009) – 28/22 Mbit/s.
5. CDMA2000 EV-DO Rev. B (2010) – 14.7 Mbit/s.
6. HSPA+ MIMO (2011) – 42 Mbit/s.
3G+:
1. IEEE 802.16e (2007) – 144/35 Mbit/s.
2. LTE (2009) – 100/50 Mbit/s.
4G:
1. LTE-A (2012) – 115 Mbit/s.
2. WiMAX 2 (2011-2013, IEEE 802.16m) – 1 Gbit/s (maximum provided by fixed objects).

Japan is introducing the fifth generation today mobile communications, increasing the capabilities of transmitting digital packets.

In today's IP networks, with the emergence of many new network applications, it becomes increasingly difficult to estimate the required bandwidth: typically, you need to know what applications you plan to use, what data protocols they use, and how they will communicate

Ilya Nazarov
System engineer at INTELCOM Line

Bandwidth Evaluation Criteria

Since the emergence of teletraffic theory, many methods have been developed for calculating channel capacity. However, unlike calculation methods applied to circuit-switched networks, calculating the required throughput in packet networks is quite complex and is unlikely to provide accurate results. First of all, this is due to a huge number of factors (especially those inherent in modern multiservice networks), which are quite difficult to predict. In IP networks, a common infrastructure is typically used by many applications, each of which may use its own different traffic pattern. Moreover, within one session, traffic transmitted in the forward direction may differ from traffic transmitted in the opposite direction. In addition to this, calculations are complicated by the fact that the speed of traffic between individual network nodes can change. Therefore, in most cases when building networks, the capacity assessment is actually determined by the general recommendations of manufacturers, statistical studies and the experience of other organizations.

For a simple example, consider applications on a small corporate network.

Example of bandwidth calculation

Email
Returning to our example, consider an email service. It uses protocols that run on top of TCP, meaning the data transfer rate is constantly adjusted to take up all the available bandwidth. Thus, we will start from the maximum delay value for sending a message - let’s say 1 second will be enough to make the user comfortable. Next, you need to estimate the average size of the message sent. Let's assume that during peak activity, email messages will often contain various attachments (copies of invoices, reports, etc.), so for our example we'll take the average message size to be 500 KB. Finally, the last parameter we need to select is the maximum number of employees who can simultaneously send messages. Let's say that during emergency times, half of the employees simultaneously press the "Send" button in the email client. The required maximum throughput for email traffic would then be (500 kB x 150 hosts)/1 s = 75,000 kB/s or 600 Mbps. From here we can immediately conclude that to connect the mail server to the network it is necessary to use a Gigabit Ethernet channel. At the core of the network, this value will be one of the terms that makes up the total required throughput.

To estimate the required throughput for telephony traffic, assume that during peak activity the number of simultaneous connections passing through the gateway can reach 100. When using the G.711 codec on Ethernet networks, the speed of one stream, taking into account headers and service packets, is approximately 100 kbit/s. With. Thus, during periods of greatest user activity, the required bandwidth in the network core will be 10 Mbit/s.

All that remains is to add up the resulting peak values for each of the network services: 600 + 10 + 80 = 690 Mbit/s. This will be the required bandwidth in the network core. The design should also include the possibility of scaling so that communication channels can serve the traffic of a growing network for as long as possible. In our example, it will be enough to use Gigabit Ethernet to meet the requirements of the services and at the same time be able to seamlessly develop the network by connecting more nodes

Usable and full throughput

Equipment selection

The development of network technologies, along with the growing demands of applications for network bandwidth, forces network equipment manufacturers to develop ever new software and hardware architectures. Often, from a single manufacturer there are seemingly similar equipment models, but designed to solve different network problems. Take, for example, Ethernet switches: most manufacturers, along with conventional switches used in enterprises, have switches for building data storage networks, organizing operator services, etc. Models of the same price category differ in their architecture, “tailored” for specific tasks.

In addition to overall performance, the choice of equipment should also be based on supported technologies. Depending on the type of hardware, a certain set of functions and types of traffic can be processed at the hardware level without using CPU and memory resources. At the same time, traffic from other applications will be processed at the software level, which greatly reduces overall performance and, as a result, maximum throughput. For example, multi-layer switches, thanks to their complex hardware architecture, are capable of transmitting IP packets without reducing performance when all ports are at maximum load. Moreover, if we want to use more complex encapsulation (GRE, MPLS), then such switches (at least inexpensive models) are unlikely to suit us, since their architecture does not support the corresponding protocols, and at best such encapsulation will occur at the expense of the central processor low productivity. Therefore, to solve such problems, we can consider, for example, routers whose architecture is based on a high-performance central processor and depends to a greater extent on software rather than hardware implementation. In this case, at the expense of maximum throughput, we get a huge range of supported protocols and technologies that are not supported by switches in the same price category.

Overall Equipment Performance

If you compare two streams transmitted at the same speed but with different packet sizes, then the stream with a smaller packet size will require more performance to transmit. This fact should be taken into account if, for example, a large number of IP telephony streams are supposed to be used on the network - the maximum throughput in bits per second here will be much less than declared.

It is clear that with mixed traffic, and even taking into account additional services (NAT, VPN), as happens in the vast majority of cases, it is very difficult to calculate the load on equipment resources. Often, equipment manufacturers or their partners load test different models under different conditions and publish the results on the Internet in the form of comparison tables. Familiarization with these results greatly simplifies the task of choosing the appropriate model.

Pitfalls of modular equipment

If the selected network equipment is modular, then in addition to the flexible configuration and scalability promised by the manufacturer, you can get many pitfalls.

When choosing modules, you should carefully read their description or consult the manufacturer. It is not enough to be guided only by the type of interfaces and their number - you also need to become familiar with the architecture of the module itself. For similar modules, it is not uncommon that when transmitting traffic, some are able to process packets autonomously, while others simply forward packets to the central processing module for further processing (accordingly, for externally identical modules, the price for them can differ several times). In the first case, the overall performance of the equipment and, as a consequence, its maximum throughput are higher than in the second, since the central processor shifts part of its work to the processors of the modules.

Study the applications that you plan to use on the network, the technologies they use, and the volume of transmitted traffic. Use the advice of developers and the experience of colleagues to take into account all the nuances of these applications when building networks.
Dive deep into the network protocols and technologies used by these applications.
Read the documentation carefully when choosing equipment. To have some stock of ready-made solutions, check out the product lines of different manufacturers.

As a result, with the right choice of technologies and equipment, you can be sure that the network will fully satisfy the requirements of all applications and, being sufficiently flexible and scalable, will last for a long time.

1.What is the process of information transfer?

Transfer of information- a physical process through which information is moved in space. We recorded the information on a disk and moved it to another room. This process characterized by the presence of the following components:

A source of information. Information receiver. Information carrier. Transmission medium.

Information transmission scheme:

Source of information – information channel – receiver of information.

Information is presented and transmitted in the form of a sequence of signals and symbols. From the source to the receiver, the message is transmitted through some material medium. If technical means of communication are used in the transmission process, they are called information transmission channels (information channels). These include telephone, radio, TV. Human sense organs play the role of biological information channels.

The process of transmitting information through technical communication channels follows the following scheme (according to Shannon):

The term “noise” refers to various types of interference that distort the transmitted signal and lead to loss of information. Such interference, first of all, arises for technical reasons: poor quality of communication lines, insecurity of different streams of information transmitted over the same channels from each other. To protect against noise, various methods are used, for example, the use of various types of filters that separate the useful signal from the noise.

Claude Shannon developed a special coding theory that provides methods for dealing with noise. One of the important ideas of this theory is that the code transmitted over the communication line must be redundant. Due to this, the loss of some part of the information during transmission can be compensated. However, the redundancy should not be too large. This will lead to delays and increased communication costs.

2. General scheme of information transfer

3.List the communication channels you know

Communication channel (English channel, data line) - system technical means and a signal propagation medium for transmitting messages (not just data) from source to destination (and vice versa). A communication channel, understood in a narrow sense (communication path), represents only the physical medium of signal propagation, for example, a physical communication line.

Based on the type of distribution medium, communication channels are divided into:

wired; acoustic; optical; infrared; radio channels.

4. What are telecommunications and computer telecommunications?

Telecommunications(Greek tele - into the distance, far away and lat. communicatio - communication) is the transmission and reception of any information (sound, image, data, text) over a distance via various electromagnetic systems (cable and fiber optic channels, radio channels and other wired and wireless channels communications).

Telecommunications network is a system of technical means through which telecommunications are carried out.

Telecommunication networks include:

1. Computer networks (for data transmission)

2. Telephone networks (transmission of voice information)

3. Radio networks (transmission of voice information - broadcast services)

4. Television networks (voice and video - broadcast services)

Computer telecommunications are telecommunications whose terminal devices are computers.

The transfer of information from computer to computer is called synchronous communication, and through an intermediate computer, which allows messages to be accumulated and transmitted to personal computers as requested by the user - asynchronous.

Computer telecommunications are beginning to be introduced into education. In higher education they are used to coordinate scientific research, prompt exchange of information between project participants, distance learning, and consultations. In the school education system - to increase the efficiency of students’ independent activities related to various types of creative work, including educational activities, based on the widespread use of research methods, free access to databases, and exchange of information with partners both within the country and abroad.

5. What is the bandwidth of an information transmission channel?

Bandwidth- metric characteristic, showing the ratio maximum number of passing units ( information, objects, volume ) per unit of time through a channel, system, node.

In computer science, the definition of bandwidth is usually applied to a communication channel and is determined by the maximum amount of information transmitted/received per unit of time.

Bandwidth is one of the most important factors from a user's point of view. It is estimated by the amount of data that the network can, in the limit, transfer per unit of time from one device connected to it to another.

The speed of information transfer depends largely on the speed of its creation (source performance), encoding and decoding methods. The highest possible information transmission speed in a given channel is called its throughput. The channel capacity, by definition, is

the information transmission rate when using the “best” (optimal) source, encoder and decoder for a given channel, so it characterizes only the channel.

5. In what units is the capacity of information transmission channels measured?

Can be measured in various, sometimes very specialized, units - pieces, bits/sec, tons, Cubic Meters etc.

6. Classification of computer communication channels (by coding method, by communication method, by signal transmission method)

broadcast networks; networks with transmission from node to node.

7. Characteristics of cable channels for information transmission (coaxial cable, twisted pair, telephone cable, fiber optic cable)

wired – telephone, telegraph (air) communication lines; cable – copper twisted pairs, coaxial, fiber optic;

and also based on electromagnetic radiation:

radio channels terrestrial and satellite communications; based on infrared rays.

cables based on twisted (twisted) pairs of copper wires; coaxial cables (central core and copper braid); fiber optic cables.

Twisted pair cables

Cables based on twisted pairs are used to transmit digital data and are widely used in computer networks. It is also possible to use them to transmit analog signals. Twisting the wires reduces the influence of external interference on useful signals and reduces the radiated electromagnetic vibrations into the external space. Shielding increases the cost of the cable, complicates installation and requires high-quality grounding. In Fig. A typical UTP design based on two twisted pairs is presented.

Rice. Cable design with unprotected twisted pair.

Depending on the presence of protection - an electrically grounded copper braid or aluminum foil around twisted pairs, the types of cables based on twisted pairs are determined:

unprotected twisted pair UTP (Unshielded twisted pair) – there is no protective shield around an individual pair;

foil twisted pair FTP (Foiled twisted pair) – there is one common external shield in the form of foil;

protected twisted pair STP (Shielded twisted pair) – there is a protective screen for each pair and a common external screen in the form of a mesh;

foil shielded twisted pair S/FTP (Screened Foiled twisted pair) – there is a protective screen for each pair in foil braid and an outer screen made of copper braid;

unprotected shielded twisted pair SF/UTP (Screened Foiled Unshielded twisted pair) – double external shield made of copper braid and foil, each twisted pair without protection.

1.5.2.2. Coaxial cable

Purpose coaxial cable– signal transmission in various fields of technology: communication systems; broadcast networks; computer networks; antenna-feeder systems of communication equipment, etc. This type of cable has an asymmetrical design and consists of an internal copper core and braid, separated from the core by an insulation layer.

A typical coaxial cable design is shown in Fig. 1.22.

Rice. 1.22. Typical coaxial cable design

Thanks to the metal shielding braid, it has high noise immunity. The main advantage of coax over twisted pair is its wide bandwidth, which provides potentially higher data transfer rates of up to 500 Mbps compared to twisted pair cables. In addition, coaxial provides significantly greater permissible signal transmission distances (up to a kilometer), it is more difficult to mechanically connect to it for unauthorized eavesdropping of the network, and it is also noticeably less polluting environment electromagnetic radiation. However, installation and repair of coaxial cable is more difficult than twisted pair cable, and the cost is higher.

It uses conventional LED transceivers, which reduces cost and increases service life compared to single-mode cable. In Fig. 1.24. The characteristic of signal attenuation in optical fiber is given. Compared to other types of cables used for communication lines, this type of cable has significantly lower signal attenuation values, which usually range from 0.2 to 5 dB per 1000 m of length. Multimode optical fiber is characterized by attenuation transparency windows in the wavelength ranges 380-850, 850-1310 (nm), and single-mode fiber, respectively, 850-1310, 1310-1550 (nm).

Figure 1.24. Fiber optic transparency windows.

Advantages of fiber optic communication:

Wide bandwidth.

Extremely conditioned high frequency carrier vibration. When using the technology of spectral multiplexing of communication channels using the wave

Multiplexing in 2009, signals from 155 communication channels with a transmission speed of 100 Gbit/s each were transmitted over a distance of 7,000 kilometers. Thus, the total data transfer rate over optical fiber was 15.5 Tbit/s. (Tera = 1000 Giga);

Low attenuation of the light signal in the fiber.

Allows you to build long-length fiber-optic communication lines without intermediate signal amplification;

Low noise level in fiber optic cable.

Allows you to increase bandwidth by transmitting various modulations of signals with low code redundancy;

High noise immunity and protection from unauthorized access.

It ensures absolute protection of the optical fiber from electrical interference, interference and a complete absence of radiation into the external environment. This is explained by the nature of light vibration, which does not interact with electromagnetic fields of other frequency ranges, like the optical fiber itself, which is a dielectric. By exploiting a number of light propagation properties in fiber optics, optical link integrity monitoring systems can instantly shut down a compromised link and sound an alarm. Such systems are especially necessary when creating communication lines in government, banking and some other special services that have increased requirements for data protection;

No need for galvanic isolation of network nodes.

Fiber optic networks fundamentally cannot have electrical ground loops, which occur when two network devices have grounding connections at different points of the building;

 High explosion and fire safety, resistance to aggressive environments.

Due to the absence of the possibility of sparking, optical fiber increases network security in chemical plants, oil refineries, and maintenance technological processes increased risk;

 Light weight, volume, cost-effectiveness of fiber optic cable.

The fiber is based on quartz (silicon dioxide), which is a widely available inexpensive material. Currently, the cost of fiber relative to a copper pair is 2:5. The cost of the fiber optic cable itself is constantly decreasing, but the use of special optical receivers and transmitters (fiber optic modems) that convert light signals into electrical signals and vice versa significantly increases the cost of the network as a whole;

 Long service life.

The service life of optical fiber is at least 25 years. Fiber optic cable also has some disadvantages. The main one is the high complexity of installation. When connecting the ends of the cable, it is necessary to ensure high accuracy of the cross-section of the fiberglass, subsequent polishing of the cut and alignment of the fiberglass when installed in the connector. Installation of connectors is carried out by welding the joint or by gluing using a special gel that has the same refractive index of light as fiberglass. In any case, this requires highly qualified personnel and special tools. In addition, fiber optic cable is less durable and less flexible than electrical cable, and is sensitive to mechanical stress. It is also sensitive to ionizing radiation, due to which the transparency of the glass fiber decreases, that is, the signal attenuation in the cable increases. Sudden temperature changes can cause fiberglass to crack. To reduce the influence of these factors, various design solutions are used, which affects the cost of the cable.

Taking into account the unique properties of optical fiber, telecommunications based on it are increasingly being used in all areas of technology. These are computer networks, city, regional, federal, as well as intercontinental underwater primary communication networks, and much more. Using fiber optic communication channels, the following is carried out: cable TV, remote video surveillance, video conferences and video broadcasts, telemetry and other information systems.

8. Characteristics of wireless information transmission channels (satellite,

radio channels, Wi-Fi, Bluetooth)

Wireless technologies- subclass information technologies, serve to transmit information over a distance between two or more points, without requiring them to be connected by wires. Can be used to transmit informationinfrared radiation, radio waves, optical or laser radiation.

Currently there are many wireless technologies, most commonly known to users by their marketing names such as Wi-Fi, WiMAX, Bluetooth. Each technology has certain characteristics that determine its scope of application.

There are different approaches to classifying wireless technologies.

By range:

o Wireless Personal Area Networks ( WPAN - Wireless Personal Area Networks). Examples of technologies are Bluetooth.

o Wireless local networks ( WLAN - Wireless Local Area Networks).

Examples of technologies are Wi-Fi.

o City-scale wireless networks ( WMAN - Wireless Metropolitan Area Networks). Examples of technologies are WiMAX.

o Wireless Wide Area Networks ( WWAN - Wireless Wide Area Network).

Examples of technologies are CSD, GPRS, EDGE, EV-DO, HSPA.

By topology:

o "Point-to-point".

o Point-to-multipoint.

By area of application:

o Corporate (departmental) wireless network- created by companies for their own needs.

o Operator wireless networks - created by telecom operators to provide services for a fee.

A short but concise way of classification can be to simultaneously display the two most significant characteristics of wireless technologies on two axes: maximum information transfer speed and maximum distance.

Tasks Task 1. In 10 s, 500 bytes of information are transmitted over the communication channel. What is it equal to

channel capacity? (500/10=50 bytes/s=400bit/s)

Task 2. How much information can be transmitted over a channel with a bandwidth of 10 kbit/s in 1 minute? (10 kbit/s*60 s = 600 kbit)

Problem 3. The average data transfer speed using a modem is 36864 bps. How many seconds will it take for the modem to transmit 4 pages of text in KOI-8 encoding, assuming that each page has an average of 2304 characters.

Solution: Number of characters in the text: 2304*4 = 9216 characters.

In KOI-8 encoding, each character is encoded by one byte, then the information volume of the text is 9216 * 8 = 73,728 bits.

Time = volume / speed. 73728: 36864 = 2 s

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