Network fundamentals

 


This blog post covers the essential concepts of network fundamentals, offering a comprehensive guide for beginners and a solid refresher for professionals. We explore the basics of networking, including key topics such as network topologies, protocols, devices, and the OSI model. Whether you're new to networking or looking to strengthen your foundational knowledge, this post provides clear explanations and valuable insights to help you understand the core principles of how networks operate.


NetworkFundamentals #NetworkingBasics #OSIModel #NetworkProtocols #ITTraining #CCNA #NetworkEngineering #TechEducation #NetworkingTips #ITSkills




•        Routers: Routers are devices that connect multiple computer networks together and route network traffic between them. They operate at the network layer (Layer 3) of the OSI model and use routing tables to determine the best path for data packets to reach their destination. Routers also provide functions such as network address translation (NAT), which allows multiple devices on a local network to share a single public IP address.


 




 

Switch

•        Switches are devices that connect multiple devices within a local area network (LAN) and forward data packets to their intended destination based on the MAC address of the devices. Unlike hubs, which broadcast data to all devices on a network, switches create dedicated connections between devices, leading to more efficient and secure data transmission within the LAN.

 


 


 

Firewall:

•        A firewall is a network security device or software that monitors and controls incoming and outgoing network traffic based on predetermined security rules. Firewalls can be implemented at the network level (e.g., hardware firewall) or the host level (e.g., software firewall) and help protect networks from unauthorized access, malicious attacks, and other security threats.

 


 

Wireless LAN Controller (WLC):

•        A WLC is a network device that manages multiple wireless access points (APs) in a wireless LAN (WLAN). It centralizes the configuration, security, and management of APs, allowing administrators to easily deploy and maintain wireless networks. WLCs provide features such as roaming support, radio frequency (RF) management, and authentication for wireless clients.

 


 

Access Points (APs):

•        APs are devices that allow wireless devices to connect to a wired network. They transmit and receive wireless signals, providing access to the network for Wi-Fi-enabled devices such as laptops, smartphones, and tablets. APs are typically connected to a wired network infrastructure and can be standalone devices or managed by a WLC.

 


 

Endpoints

•        Endpoints are the devices connected to a network, such as computers, printers, smartphones, tablets, and IoT devices. They initiate and consume network services and communicate with other devices on the network. Endpoints can be both sources and destinations of data packets in a network.

 

Server:

•        A server is a computer or software application that provides services or resources to other devices on a network. Servers can fulfill various roles, including file storage, email hosting, web hosting, database management, and application hosting. They typically have more processing power, memory, and storage capacity than client devices and are designed to handle multiple client requests simultaneously.

 



OSI model vs. TCP/IP model

•        The OSI reference model describes the functions of a telecommunication or networking system, while TCP/IP is a suite of communication protocols used to interconnect network devices on the internet. TCP/IP and OSI are the most broadly used networking models for communication.

•         The main similarity is in their construction, as both use layers, although the OSI model consists of seven layers, while TCP/IP consists of just four layers.

•        Another similarity is that the upper layer for each model is the application layer, which performs the same tasks in each model but may vary according to the information each receives.

•        The functions performed in each model are also similar because each uses a network and transport layer to operate. The OSI and TCP/IP model are mostly used to transmit data packets, although they each use different means and paths to reach their destinations.

 

 


 

Advantages of OSI Model

•        The OSI model helps users and operators of computer networks:

•        Determine the required hardware and software to build their network.

•        Understand and communicate the process followed by components communicating across a network. 

•        Perform troubleshooting, by identifying which network layer is causing an issue and focusing efforts on that layer.

The OSI model helps network device manufacturers and networking software vendors:

•        Create devices and software that can communicate with products from any other vendor, allowing open interoperability

•        Define which parts of the network their products should work with.

•        Communicate to users at which network layers their product operates – for example, only at the application layer, or across the stack.

 

 


 

IP address

 

•        IP address is a numerical identifier that uniquely identifies the devices in a computer network. Two types of IP address are widely used in IP network.

•        IP version 4

•        IP version 6

•        IP v4 address is a 32-bit logical address.

•        It is written in decimal format. The 32-bit address length is divided into 4 equal parts called an octet. Each octet contains 8 bit and is separated by a dot.

•        For example, 192.168.5.10 is an IP v4 address.

 

Features of IP v4 Address

•        IPv4 is a 32-bit length address.

•        It is divided into 4 equal parts.

•        Each part consists of 8 bits and is called Octet.

•        Each octet is separated by dot notation.

•        It is normally written in a human-readable numbering system ie decimal number.

•        232 = 4.7 billion of addresses are available for IPV4.

•        IPv4 consists of two parts: The network part and the host part.

•        The network part shows that the IP address belongs to which network. The host shows the number of different hosts in the same network.

 


IPv4 classes

•        Class A address ranges from 0.0.0.0 to 127.255.255.255

•        Class B address ranges from 128.0.0.0 to 191.255.255.255

•        Class C address ranges from 192.0.0.0 to 223.255.255.255

•        Class D address ranges from 224.0.0.0 to 239.255.255.255

•        Class E address ranges from 240.0.0.0 to 255.255.255.255

 


 

Subnet Mask

The subnet mask is the 32-bit length of series of binary 0s (zeroes) and 1s (ones) that distinguishes the network part and

the host part of an IP address.  Series of  1s denote the network portion and 0s denote the host portion.

When we assign an IP address to any host in a network, a subnet mask is also given to it. For example,

IP address is 192.168.5.10

The subnet mask is 255.255.255.0

If we convert subnet mask to binary bits, then it looks  like this:

11111111.11111111.11111111.00000000

The series of 1s is called the network bits and the 0s are called host bit.

Network bit will remain unchanged for every IP assigned to any host in the same network and the network address is derived

by ANDing the binary equivalent of IP address and the subnet mask.

These series of 0s can be varied from 0s to 1s for all the hosts within the same network.

Hence, in the above example,

The number of networks is given by = 2n, where n denotes the number of network bits.

and the number of hosts per network is given by=2h-2 where h is the number host bit

 

IPV6

•        IPv6 is a 128-bits address having an address space of 2128, which is way bigger than IPv4. IPv6 use Hexa-Decimal format separated by colon (:).

1.        There are 8 groups, and each group represents 2 Bytes (16-bits). 

2.        Each Hex-Digit is of 4 bits (1 nibble)

3.        Delimiter used – colon (:)

•        FE80:CD00:0000:0CDE:1257:0000:211E:729C

 



ipv4 subnetting

In this blog post, we dive deep into the art of subnetting IPv4 addresses, a crucial skill for network administrators and engineers. We start with the fundamentals of subnetting, explaining how IP addresses are divided into network and host portions. The post includes a variety of practice questions, each accompanied by detailed explanations to help you master the concepts. Whether you're preparing for certification exams or just brushing up on your skills, this guide will provide you with the knowledge and confidence you need to tackle subnetting challenges.


#Subnetting #IPv4 #Networking #IP Addressing #CCNA #Network Administration #IT_Certification #Practice Questions #NetworkEngineering #Subnetting #Explained


(Solutions Provided at End)

Question #1

What is the range of assignable IP addresses for a subnet containing an IP address of 172.16.1.10 /19?

a. 172.16.0.1 – 172.16.31.254

b. 172.16.0.1 – 172.16.63.254

c. 172.16.0.0 – 172.16.31.255

d. 172.16.0.1 – 172.16.31.255

e. 172.16.0.0 – 172.16.63.254

Question #2

You are assigning IP addresses to hosts in the 192.168.4.0 /26 subnet. Which two of the following IP addresses are assignable IP addresses that reside in that subnet?

a. 192.168.4.0

b. 192.168.4.63

c. 192.168.4.62

d. 192.168.4.32

e. 192.168.4.64

Question #3

A host in your network has been assigned an IP address of 192.168.181.182 /25. What is the subnet to which the host belongs?

a. 192.168.181.128 /25

b. 192.168.181.0 /25

c. 192.168.181.176 /25

d. 192.168.181.192 /25

e. 192.168.181.160 /25

Question #4

You are working with a Class B network with the private IP address of 172.16.0.0 /16. You need to maximize the number of broadcast domains, where each broadcast domain can accommodate 1000 hosts. What subnet mask should you use?

a. /22


b. /23

c. /24

d. /25

e. /26

Question #5

What is the directed broadcast address of a subnet containing an IP address of 172.16.1.10 /19?

a. 172.16.15.255

b. 172.16.31.255

c. 172.16.255.255

d. 172.16.95.255

e. 172.16.0.255

Question #6

A customer is using a Class C network of 192.168.10.0 subnetted with a 28-bit subnet mask. How many subnets can be created by using this subnet mask?

a. 32

b. 16

c. 30

d. 8

e. 14

Question #7

Given a subnet of 172.16.56.0 /21, identify which of the following IP addresses belong to this subnet. (Select 2.)

a. 172.16.54.129

b. 172.16.62.255

c. 172.16.61.0

d. 172.16.65.255

e. 172.16.64.1

Question #8

What is the subnet address of the IP address 192.168.5.55 with a subnet mask of 255.255.255.224?

a. 192.168.5.0 /27

b. 192.168.5.16 /27

c. 192.168.5.32 /27

d. 192.168.5.48 /27

e. 192.168.5.64 /27


Question #9

You are working for a company that will be using the 192.168.1.0 /24 private IP address space for IP addressing inside their organization.

They have multiple geographical locations and want to carve up the 192.168.1.0 /24 address space into subnets. Their largest subnet will need 13 hosts.

What subnet mask should you use to accommodate at least 13 hosts per subnet, while maximizing the number of subnets that can be created?

a. 255.255.255.248

b. 255.255.255.224

c. 255.255.255.252

d. 255.255.255.192

e. 255.255.255.240

Question #10

A customer is using a Class C network of 192.168.10.0 subnetted with a 28-bit subnet mask. How many assignable addresses are available in each of the subnets?

a. 32

b. 16

c. 30

d. 8

e. 14

Question #11

An IP address of 192.168.0.100 /27 belongs to which of the following subnets?

a. 192.168.0.92

b. 192.168.0.128

c. 192.168.0.64

d. 192.168.0.96

e. 192.168.0.32

Question #12

What subnet mask should be used to subnet the 192.168.10.0 network to support the number of subnets and IP addresses per subnet shown in the following topology?


a. 255.255.255.0

b. 255.255.255.128

c. 255.255.255.192

d. 255.255.255.224

e. 255.255.255.240


Solutions

Question #1

What is the range of assignable IP addresses for a subnet containing an IP address of 172.16.1.10 /19?

a. 172.16.0.1 – 172.16.31.254

b. 172.16.0.1 – 172.16.63.254

c. 172.16.0.0 – 172.16.31.255

d. 172.16.0.1 – 172.16.31.255

e. 172.16.0.0 – 172.16.63.254

Answer: a

To determine the subnets, assignable IP address ranges, and directed broadcast addresses created by the 19-bit subnet mask we perform the following steps:

Step #1: Identify the interesting octet (i.e. the octet that contains the first zero in the binary subnet mask).

In this question, we have a 19-bit subnet mask, which is written in binary as:

11111111 11111111 11100000 00000000

The interesting octet is the third octet, because the third octet (i.e. 11100000) is the first octet to contain a 0 in the binary.

Step #2: Identify the decimal value in the interesting octet of the subnet mask.

A 19-bit subnet mask can be written in dotted decimal notation as: 255.255.224.0

Since the third octet is the interesting octet, the decimal value in the interesting octet is 224.

Step #3: Determine the block size by subtracting the decimal value of the interesting octet from 256.

Block Size = 256 – 224 = 32

Step #4: Determine the subnets by counting by the block size in the interesting octet, starting at 0.

Placing a zero in the first interesting octet identifies the first subnet as:

172.16.0.0 /19


We then count by the block size (of 32) in the interesting octet (the third octet in this question) to determine the remaining subnets:

172.16.32.0 /19

172.16.64.0 /19

172.16.96.0 /19

172.16.128.0 /19

172.16.160.0 /19

172.16.192.0 /19

172.16.224.0 /19

Step #5: Identify the subnet address, the directed broadcast address, and the usable range of addresses.

Looking through the subnets created by the 19-bit subnet mask reveals that the IP address of 172.16.1.10 resides in the 172.16.0.0 /19 subnet.

The directed broadcast address, where all host bits are set to a 1, is 1 less than the next subnet address.

The next subnet address is 172.16.32.0. So, the directed broadcast address for the 172.16.0.0 /19 subnet is 1 less than 172.16.32.0, which is:

172.16.31.255

The usable IP addresses are all the IP addresses between the subnet address and the directed broadcast address. Therefore, in this example, the assignable IP address range for the 172.16.0.0 /19 network is:

172.16.0.1 – 172.16.31.254

Question #2

You are assigning IP addresses to hosts in the 192.168.4.0 /26 subnet. Which two of the following IP addresses are assignable IP addresses that reside in that subnet?

a. 192.168.4.0

b. 192.168.4.63

c. 192.168.4.62

d. 192.168.4.32

e. 192.168.4.64

Answer: c and d

To determine subnets and usable address ranges created by the 26-bit subnet mask we perform the following steps:


Step #1: Identify the interesting octet (i.e. the octet that contains the first zero in the binary subnet mask).

In this question, we have a 26-bit subnet mask, which is written in binary as:

11111111 11111111 11111111 11000000

The interesting octet is the forth octet, because the forth octet (i.e. 11000000) is the first octet to contain a 0 in the binary.

Step #2: Identify the decimal value in the interesting octet of the subnet mask.

A 26-bit subnet mask can be written in dotted decimal notation as: 255.255.255.192

Since the forth octet is the interesting octet, the decimal value in the interesting octet is 192.

Step #3: Determine the block size by subtracting the decimal value of the interesting octet from 256.

Block Size = 256 – 192 = 64

Step #4: Determine the subnets by counting by the block size in the interesting octet, starting at 0.

Placing a zero in the first interesting octet identifies the first subnet as:

192.168.4.0 /26

We then count by the block size (of 64) in the interesting octet (the forth octet in this question) to determine the remaining subnets:

192.168.4.64 /26

192.168.4.128 /26

192.168.4.192 /26

Step #5:

This question is asking about the 192.168.4.0 /26 subnet. From the above list of subnets, we can determine that the assignable range of IP addresses for this subnet is 192.168.4.1 – 192.168.4.62. We can also determine that 192.168.4.0 is the network address, and 192.168.4.63 is the directed broadcast address.

From the assignable range of IP addresses we have calculated, we can determine that the two assignable IP addresses given as options in this question are:
192.168.4.62 and 192.168.4.32.


Question #3

A host in your network has been assigned an IP address of 192.168.181.182 /25. What is the subnet to which the host belongs?

a. 192.168.181.128 /25

b. 192.168.181.0 /25

c. 192.168.181.176 /25

d. 192.168.181.192 /25

e. 192.168.181.160 /25

Answer: a

To determine subnets and usable address ranges created by the 25-bit subnet mask we perform the following steps:

Step #1: Identify the interesting octet (i.e. the octet that contains the first zero in the binary subnet mask).

In this question, we have a 25-bit subnet mask, which is written in binary as:

11111111 11111111 11111111 10000000

The interesting octet is the forth octet, because the forth octet (i.e. 10000000) is the first octet to contain a 0 in the binary.

Step #2: Identify the decimal value in the interesting octet of the subnet mask.

A 25-bit subnet mask can be written in dotted decimal notation as: 255.255.255.128

Since the forth octet is the interesting octet, the decimal value in the interesting octet is 128.

Step #3: Determine the block size by subtracting the decimal value of the interesting octet from 256.

Block Size = 256 – 128 = 128

Step #4: Determine the subnets by counting by the block size in the interesting octet, starting at 0.

Placing a zero in the first interesting octet identifies the first subnet as:

192.168.181.0 /25

We then count by the block size (of 128) in the interesting octet (the forth octet in this question) to determine the remaining subnets, or in this case just a single additional subnet.


192.168.181.128 /25

Now that we have our two subnets identified, we can determine the subnet in which the IP address of 192.168.181.182 resides.

Since the usable range of IP addresses for the 192.168.181.128 /25 network is 192.168.181.129 – 192.168.181.254 (because 192.168.181.128 is the network address, and 192.168.181.255 is the directed broadcast address), and since 192.168.181.182 is in that range, the subnet to which 192.168.181.182 /25 belongs is:

192.168.181.128 /25

Question #4

You are working with a Class B network with the private IP address of 172.16.0.0 /16. You need to maximize the number of broadcast domains, where each broadcast domain can accommodate 1000 hosts. What subnet mask should you use?

a. /22

b. /23

c. /24

d. /25

e. /26

Answer: a

In addition to testing your knowledge of subnetting, this question is also making sure you understand that a subnet is a broadcast domain. This should not be confused with a collision domain (i.e. each port on a switch is in its own collision domain).

To determine how many host bits are required to support 1000 hosts, we can create a table from the following formula:

Number of Hosts = 2h – 2, where h is the number of host bits

From this formula, we can create the following table:

1 Host Bit => 0 Hosts

2 Host Bits => 2 Hosts

3 Host Bits => 6 Hosts

4 Host Bits => 14 Hosts

5 Host Bits => 30 Hosts

6 Host Bits => 62 Hosts


7 Host Bits => 126 Hosts

8 Host Bits => 254 Hosts

9 Host Bits => 510 Hosts

10 Host Bits => 1022 Hosts

This table tells us that a subnet with 10 host bits will accommodate the requirement of 1000 hosts. If we have 10 host bits, then we have a
22-bit subnet mask (i.e. 32 – 10 = 22). Also, by not using more host bits than we need, we are maximizing the number of subnets that can be created.

Question #5

What is the directed broadcast address of a subnet containing an IP address of 172.16.1.10 /19?

a. 172.16.15.255

b. 172.16.31.255

c. 172.16.255.255

d. 172.16.95.255

e. 172.16.0.255

Answer: b

To determine the subnets, assignable IP address ranges, and directed broadcast addresses created by the 19-bit subnet mask we perform the following steps:

Step #1: Identify the interesting octet (i.e. the octet that contains the first zero in the binary subnet mask).

In this question, we have a 19-bit subnet mask, which is written in binary as:

11111111 11111111 11100000 00000000

The interesting octet is the third octet, because the third octet (i.e. 11100000) is the first octet to contain a 0 in the binary.

Step #2: Identify the decimal value in the interesting octet of the subnet mask.

A 19-bit subnet mask can be written in dotted decimal notation as: 255.255.224.0

Since the third octet is the interesting octet, the decimal value in the interesting octet is 224.

Step #3: Determine the block size by subtracting the decimal value of the interesting octet from 256.


Block Size = 256 – 224 = 32

Step #4: Determine the subnets by counting by the block size in the interesting octet, starting at 0.

Placing a zero in the first interesting octet identifies the first subnet as:

172.16.0.0 /19

We then count by the block size (of 32) in the interesting octet (the third octet in this question) to determine the remaining subnets:

172.16.32.0 /19

172.16.64.0 /19

172.16.96.0 /19

172.16.128.0 /19

172.16.160.0 /19

172.16.192.0 /19

172.16.224.0 /19

Step #5: Identify the subnet address, the directed broadcast address, and the usable range of addresses.

Looking through the subnets created by the 19-bit subnet mask reveals that the IP address of 172.16.1.10 resides in the 172.16.0.0 /19 subnet.

The directed broadcast address, where all host bits are set to a 1, is 1 less than the next subnet address.

The next subnet address is 172.16.32.0. So, the directed broadcast address for the 172.16.0.0 /19 subnet is 1 less than 172.16.32.0, which is:

172.16.31.255

The usable IP addresses are all the IP addresses between the subnet address and the directed broadcast address. Therefore, in this example, the assignable IP address range for the 172.16.0.0 /19 network is:

172.16.0.1 – 172.16.31.254

Question #6

A customer is using a Class C network of 192.168.10.0 subnetted with a 28-bit subnet mask. How many subnets can be created by using this subnet mask?

a. 32

b. 16

c. 30

d. 8


e. 14

Answer: b

The subnet in this question is a Class C network, because there is a 192 in the first octet. A class C network has a
natural mask of 24 bits. However, this network has a 28-bit subnet mask. Therefore, we have 4 borrowed bits, which are network bits added to a network’s natural mask (i.e. 28 – 24 = 4). The number of subnets can be calculated as follows:

Number of Subnets = 2s, where s is the number of borrowed bits.

Therefore, in this question, the number of created subnets is 16:

Number of Subnets = 24 = 16

Question #7

Given a subnet of 172.16.56.0 /21, identify which of the following IP addresses belong to this subnet. (Select 2.)

a. 172.16.54.129

b. 172.16.62.255

c. 172.16.61.0

d. 172.16.65.255

e. 172.16.64.1

Answer: b, c

To determine subnets and usable address ranges created by the 21-bit subnet mask we perform the following steps:

Step #1: Identify the interesting octet (i.e. the octet that contains the first zero in the binary subnet mask).

In this question, we have a 21-bit subnet mask, which is written in binary as:

11111111 11111111 11111000 00000000

The interesting octet is the third octet, because the third octet (i.e. 11111000) is the first octet to contain a 0 in the binary subnet mask.

Step #2: Identify the decimal value in the interesting octet of the subnet mask. A 21-bit subnet mask can be written in dotted decimal notation as: 255.255.248.0


Since the third octet is the interesting octet, the decimal value in the interesting octet is 248.

Step #3: Determine the block size by subtracting the decimal value of the interesting octet from 256.

Block Size = 256 – 248 = 8

Step #4: Determine the subnets by counting by the block size in the interesting octet, starting at 0.

Placing a zero in the first interesting octet identifies the first subnet as:

172.16.0.0 /21

We then count by the block size (of 8) in the interesting octet (the third octet in this question) to determine the remaining subnets:

172.16.8.0 /21 172.16.16.0 /21 172.16.24.0 /21 172.16.32.0 /21 172.16.40.0 /21 172.16.48.0 /21 172.16.56.0 /21 172.16.64.0 /21 ... SUBNETS OMITTED ...

We can stop counting after we pass the subnet we are being asked about. Specifically, in this question, we’re being asked about 172.16.56.0 /21.

Step #5: Identify the subnet address, the directed broadcast address, and the usable range of addresses.

The subnet address, where all host bits are set to a 0, is given:

172.16.56.0 /24

The directed broadcast address, where all host bits are set to a 1, is 1 less than the next subnet address.

The next subnet address is 172.16.64.0. So, the directed broadcast address for the 172.16.54.0 /21 subnet is 1 less than 172.16.64.0, which is: 172.16.63.255

The usable IP addresses are all the IP addresses between the subnet address and the directed broadcast address. Therefore, in this example, the usable IP address range for the 172.16.56.0 /21 network is:


172.16.56.1 – 172.16.63.254

The only IP addresses in this question that reside in this range are:

172.16.62.255 172.16.61.0

WARNING:
Many CCNA R&S candidates look at IP addresses like these and immediately assume they are not usable IP addresses, because they have a 0 or a 255 in the forth octet. They argue that 172.16.61.0 is a subnet address and that 172.16.62.255 is a directed broadcast address.

While that would only be true of the subnet mask were 24-bits, remember that, by definition, a subnet address has all of its host bits set to a 0, and a directed broadcast address has all of its host bits set to a 1. In this question, we have 11 host bits (i.e. 32 – 21 = 11), not 8 host bits. So, 172.16.62.255 and 172.16.61.0 are actually usable IP addresses.

Question #8

What is the subnet address of the IP address 192.168.5.55 with a subnet mask of 255.255.255.224?

a. 192.168.5.0 /27

b. 192.168.5.16 /27

c. 192.168.5.32 /27

d. 192.168.5.48 /27

e. 192.168.5.64 /27

Answer: c

To determine subnets and usable address ranges created by the 27-bit subnet mask we perform the following steps:

Step #1: Identify the interesting octet (i.e. the octet that contains the first zero in the binary subnet mask).

In this question, we have a 27-bit subnet mask, which is written in binary as:

11111111 11111111 11111111 11100000

The interesting octet is the forth octet, because the forth octet (i.e. 11100000) is the first octet to contain a 0 in the binary.

Step #2: Identify the decimal value in the interesting octet of the subnet mask.

A 27-bit subnet mask can be written in dotted decimal notation as: 255.255.255.224


Since the forth octet is the interesting octet, the decimal value in the interesting octet is 224.

Step #3: Determine the block size by subtracting the decimal value of the interesting octet from 256.

Block Size = 256 – 224 = 32

Step #4: Determine the subnets by counting by the block size in the interesting octet, starting at 0.

Placing a zero in the first interesting octet identifies the first subnet as:

192.168.5.0 /27

We then count by the block size (of 32) in the interesting octet (the forth octet in this question) to determine the remaining subnets:

192.168.5.32 /27

192.168.5.64 /27

192.168.5.96 /27

192.168.5.128 /27

192.168.5.160 /27

192.168.5.192 /27

192.168.5.224 /27

Now that we have all of our subnets identified, we can determine the subnet in which the IP address of 192.168.5.55 resides.

Since the usable range of IP addresses for the 192.168.5.32 /27 network is 192.168.5.33 – 192.168.5.62 (because 192.168.5.32 is the network address, and 192.168.5.63 is the directed broadcast address), and since 192.168.5.55 is in that range, the subnet to which 192.168.5.55 /27 belongs is:

192.168.5.32 /27

Question #9

You are working for a company that will be using the 192.168.1.0 /24 private IP address space for IP addressing inside their organization.

They have multiple geographical locations and want to carve up the 192.168.1.0 /24 address space into subnets. Their largest subnet will need 13 hosts.

What subnet mask should you use to accommodate at least 13 hosts per subnet, while maximizing the number of subnets that can be created?

a. 255.255.255.248


b. 255.255.255.224

c. 255.255.255.252

d. 255.255.255.192

e. 255.255.255.240

Answer: e

We can determine the maximum number of hosts allowed in a subnet by raising the number 2 to the power of the number of host bits and then subtracting 2. So, the formula looks like this:

Maximum Number of Hosts per Subnet = 2h – 2, where h is the number of host bits.

Why are we subtracting two? Well, there are two IP addresses in the subnet that cannot be assigned. These addresses are: (1) the network address, where all of the host bits are set to a 0 and (2) the directed broadcast address, where all of the host bits are set to a 1.

In the actual exam, if you are given scratch paper or access to a note taking application, you might want to write out a table such as the following for your reference:

1 Host Bit: 2
1 – 2 = 0

2 Host Bits: 22 – 2 = 2

3 Host Bits: 23 – 2 = 6

4 Host Bits: 24 – 2 = 14

5 Host Bits: 25 – 2 = 30

6 Host Bits: 26 – 2 = 62

7 Host Bits: 27 – 2 = 126

8 Host Bits: 28 – 2 = 254

In this question, we’re asked to determine a subnet mask that accommodates at least 13 hosts per subnet. By looking at the reference table we created, we can see that 4 host bits (which support 14 hosts) would work, while 3 host bits (which supports only 6 hosts) would not be enough.

So, we need a subnet with 4 host bits, which are enough host bits to meet the design goal, but not more than we need. Using more host bits than we need would violate the requirement to maximize the number of subnets.

A subnet mask with 4 host bits has 28 network bits (i.e. 32 – 4 = 28), and therefore a 28-bit subnet mask. A 28-bit subnet mask can be written as:

255.255.255.240


Question #10

A customer is using a Class C network of 192.168.10.0 subnetted with a 28-bit subnet mask. How many assignable addresses are available in each of the subnets?

a. 32

b. 16

c. 30

d. 8

e. 14

Answer: e

An IPv4 address contains a total of 32 bits. Since, in this question, we have 28 subnet bits, the number of host bits is 4 (i.e. 32 – 28 = 4). The number of assignable IP addresses in a subnet can be calculated as follows:

Number of Assignable IP Addresses = 2h – 2, where h is the number of host bits.

Therefore, in this question, each subnet has 14 assignable IP addresses:

Number of Assignable IP Addresses = 24 – 2 = 16 – 2 = 14

Question #11

An IP address of 192.168.0.100 /27 belongs to which of the following subnets?

a. 192.168.0.92

b. 192.168.0.128

c. 192.168.0.64

d. 192.168.0.96

e. 192.168.0.32

Answer: d

To determine the subnets created by the 27-bit subnet mask we perform the following steps:

Step #1: Identify the interesting octet (i.e. the octet that contains the first zero in the binary subnet mask).

In this question, we have a 19-bit subnet mask, which is written in binary as:

11111111 11111111 11111111 11100000


The interesting octet is the forth octet, because the forth octet (i.e. 11100000) is the first octet to contain a 0 in the binary.

Step #2: Identify the decimal value in the interesting octet of the subnet mask.

A 27-bit subnet mask can be written in dotted decimal notation as: 255.255.255.224

Since the forth octet is the interesting octet, the decimal value in the interesting octet is 224.

Step #3: Determine the block size by subtracting the decimal value of the interesting octet from 256.

Block Size = 256 – 224 = 32

Step #4: Determine the subnets by counting by the block size in the interesting octet, starting at 0.

Placing a zero in the first interesting octet identifies the first subnet as:

192.168.0.0 /27

We then count by the block size (of 32) in the interesting octet (the forth octet in this question) to determine the remaining subnets:

192.168.0.32 /27

192.168.0.64 /27

192.168.0.96 /27

192.168.0.128 /27

192.168.0.160 /27

192.168.0.192 /27

192.168.0.224 /27

Step #5: Identify the subnet address of the IP address 192.168.0.100 /27.

Looking through the subnets created by the 27-bit subnet mask reveals that the IP address of 192.168.0.100 resides in the
192.168.0.96 subnet.

Question #12

What subnet mask should be used to subnet the 192.168.10.0 network to support the number of subnets and IP addresses per subnet shown in the following topology?


a. 255.255.255.0

b. 255.255.255.128

c. 255.255.255.192

d. 255.255.255.224

e. 255.255.255.240

Answer: c

To meet the design requirements, four subnets must be created, and each subnet must accommodate a maximum of 50 IP addresses.

We can begin by creating a listing of how many subnets are created from different numbers of borrowed bits, using the formula:

Number of Subnets Created = 2n, where n is the number of borrowed bits

1 borrowed bits => 2 subnets

2 borrowed bits => 4 subnets

3 borrowed bits => 8 subnets

4 borrowed bits => 16 subnets

5 borrowed bits => 32 subnets

6 borrowed bits => 64 subnets

7 borrowed bits => 128 subnets

From this, we can see we need at least 2 borrowed bits to accommodate 4 subnets. However, we need to make sure the subnet will accommodate 50 IP addresses. To determine this, we can use the formula:

Number of IP Addresses = 2h – 2, where h is the number of host bits


If we have 2 borrowed bits (i.e. the minimum number of borrowed bits required for 4 subnets), we have 6 host bits (i.e. 8 – 2 = 6). From the above formula, we can determine the number of IP addresses supported by 6 host bits.

Number of IP Addresses = 26 – 2 = 62

Since 6 host bits meet our requirement of at least 50 IP addresses per subnet, we can use a 26-bit subnet mask (i.e. 2 bits added to the Class C default mask (also known as the natural mask) of 24 bits). A 26-bit subnet mask can be written as:

255.255.255.192

Project: Smart Home Automation System

Project Title: Smart Home Automation System

Project Description:

The Smart Home Automation System is an advanced Java project designed to simulate and manage a smart home environment. This project integrates various components of object-oriented programming (OOP) concepts, multi-threading, networking, database management, GUI development, design patterns, and more. The system enables users to control and monitor home appliances, manage security systems, and optimize energy consumption through a centralized Java-based application.

Key Features:

  • User Authentication: Secure login and registration system with role-based access control.
  • Device Management: Add, remove, and control various smart devices like lights, fans, thermostats, and security cameras.
  • Real-Time Monitoring: Live data streaming from sensors (temperature, motion, etc.) with instant alerts for anomalies.
  • Energy Management: Track and optimize energy consumption with intelligent suggestions and automation rules.
  • Scheduling and Automation: Create schedules for device operations and automate repetitive tasks using a rules engine.
  • Remote Access: Control and monitor the smart home from anywhere via a networked application or web interface.
  • Database Integration: Store user data, device states, logs, and energy usage metrics using an SQL or NoSQL database.
  • Multithreading: Efficient handling of multiple device operations simultaneously using Java's multithreading capabilities.
  • Design Patterns: Implement design patterns such as Singleton, Factory, Observer, and MVC to structure the project effectively.
  • GUI Development: Build an intuitive user interface using JavaFX or Swing, featuring real-time data visualization and control panels.
  • Networking: Establish communication between devices using sockets, and manage remote connections via REST APIs or WebSockets.
  • Security: Encrypt sensitive data and use secure protocols for communication to protect user privacy and system integrity.
  • Logging and Debugging: Implement logging mechanisms to monitor system activities and troubleshoot issues effectively.

Technologies Used:

  • Java SE 11+
  • JavaFX or Swing
  • MySQL / MongoDB
  • Multithreading and Concurrency
  • Socket Programming
  • REST API / WebSocket
  • Design Patterns (Singleton, Factory, Observer, MVC)
  • Encryption and Security Protocols

#JavaProject #SmartHomeAutomation #ObjectOrientedProgramming #Multithreading #JavaFX #Swing #Networking  #DatabaseManagement #DesignPatterns #HomeSecuritySystem #EnergyManagement #IoT #RealTimeMonitoring #JavaCoding #AdvancedJava #JavaSE #JavaDevelopment #JavaGUI #RESTAPI #WebSocket

Here is a comprehensive implementation of a Smart Home Automation System in Java, covering various advanced concepts like multithreading, OOP, GUI, database management, and networking.

Requirements:

  • Java SE 11 or later
  • JavaFX SDK (for GUI)
  • MySQL or MongoDB (for database)
  • Maven (for dependency management)
  • IntelliJ IDEA / Eclipse (for development)
  • JDBC Driver (for MySQL, if using MySQL)

Project Structure:

  1. Model Layer: Contains classes for devices, users, and other entities.
  2. Controller Layer: Manages the system's logic.
  3. View Layer: Handles GUI interactions.
  4. Database Layer: Manages database operations.
  5. Networking Layer: Manages communication between devices and clients.

1. Model Layer: Device, User, and Sensor Classes


package com.smarthome.model; public abstract class Device { private String id; private String name; private boolean status; public Device(String id, String name) { this.id = id; this.name = name; this.status = false; } public String getId() { return id; } public String getName() { return name; } public boolean isStatus() { return status; } public void setStatus(boolean status) { this.status = status; } public abstract void operate(); } class Light extends Device { public Light(String id, String name) { super(id, name); } @Override public void operate() { setStatus(!isStatus()); System.out.println(getName() + " turned " + (isStatus() ? "ON" : "OFF")); } } class Thermostat extends Device { private double temperature; public Thermostat(String id, String name) { super(id, name); this.temperature = 22.0; } public double getTemperature() { return temperature; } public void setTemperature(double temperature) { this.temperature = temperature; System.out.println(getName() + " set to " + temperature + "°C"); } @Override public void operate() { // Custom operation for Thermostat } } class User { private String username; private String password; private String role; public User(String username, String password, String role) { this.username = username; this.password = password; this.role = role; } public String getUsername() { return username; } public String getPassword() { return password; } public String getRole() { return role; } } class Sensor { private String id; private String type; private double value; public Sensor(String id, String type) { this.id = id; this.type = type; } public String getId() { return id; } public String getType() { return type; } public double getValue() { return value; } public void setValue(double value) { this.value = value; } }

2. Controller Layer: SmartHomeController


package com.smarthome.controller; import com.smarthome.model.Device; import com.smarthome.model.User; import java.util.HashMap; import java.util.Map; public class SmartHomeController { private Map<String, Device> devices; private Map<String, User> users; public SmartHomeController() { devices = new HashMap<>(); users = new HashMap<>(); } public void addDevice(Device device) { devices.put(device.getId(), device); } public void addUser(User user) { users.put(user.getUsername(), user); } public void operateDevice(String deviceId) { Device device = devices.get(deviceId); if (device != null) { device.operate(); } else { System.out.println("Device not found"); } } public User authenticate(String username, String password) { User user = users.get(username); if (user != null && user.getPassword().equals(password)) { return user; } return null; } public void listDevices() { devices.values().forEach(device -> { System.out.println(device.getName() + " is " + (device.isStatus() ? "ON" : "OFF")); }); } }

3. View Layer: JavaFX GUI


package com.smarthome.view; import com.smarthome.controller.SmartHomeController; import com.smarthome.model.Light; import com.smarthome.model.Thermostat; import com.smarthome.model.User; import javafx.application.Application; import javafx.scene.Scene; import javafx.scene.control.*; import javafx.scene.layout.GridPane; import javafx.stage.Stage; public class SmartHomeApp extends Application { private SmartHomeController controller; public static void main(String[] args) { launch(args); } @Override public void start(Stage primaryStage) { controller = new SmartHomeController(); controller.addUser(new User("admin", "admin123", "admin")); controller.addDevice(new Light("1", "Living Room Light")); controller.addDevice(new Thermostat("2", "Living Room Thermostat")); primaryStage.setTitle("Smart Home Automation System"); GridPane grid = new GridPane(); grid.setHgap(10); grid.setVgap(10); Label userLabel = new Label("Username:"); TextField userTextField = new TextField(); grid.add(userLabel, 0, 0); grid.add(userTextField, 1, 0); Label passLabel = new Label("Password:"); PasswordField passField = new PasswordField(); grid.add(passLabel, 0, 1); grid.add(passField, 1, 1); Button loginButton = new Button("Login"); grid.add(loginButton, 1, 2); loginButton.setOnAction(e -> { User user = controller.authenticate(userTextField.getText(), passField.getText()); if (user != null) { showDashboard(primaryStage); } else { showAlert("Invalid credentials"); } }); Scene scene = new Scene(grid, 300, 200); primaryStage.setScene(scene); primaryStage.show(); } private void showDashboard(Stage stage) { GridPane grid = new GridPane(); grid.setHgap(10); grid.setVgap(10); Label deviceLabel = new Label("Devices:"); grid.add(deviceLabel, 0, 0); ListView<String> deviceList = new ListView<>(); controller.listDevices().forEach(device -> deviceList.getItems().add(device.getName())); grid.add(deviceList, 1, 0); Button operateButton = new Button("Operate"); grid.add(operateButton, 1, 1); operateButton.setOnAction(e -> { String selectedDevice = deviceList.getSelectionModel().getSelectedItem(); controller.operateDevice(selectedDevice); }); Scene dashboardScene = new Scene(grid, 400, 300); stage.setScene(dashboardScene); } private void showAlert(String message) { Alert alert = new Alert(Alert.AlertType.ERROR); alert.setContentText(message); alert.showAndWait(); } }

4. Database Layer: Database Connection


package com.smarthome.database;

import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.SQLException;

public class DatabaseConnection {
    private static final String URL = "jdbc:mysql://localhost:3306/smarthome";
    private static final String USER = "root";
    private static final String PASSWORD = "password";

    public static Connection getConnection() throws SQLException {
        return DriverManager.getConnection(URL, USER, PASSWORD);
    }
}

5. Networking Layer: Socket Communication

package com.smarthome.networking;

import java.io.*;
import java.net.ServerSocket;
import java.net.Socket;

public class SmartHomeServer {
    private ServerSocket serverSocket;

    public SmartHomeServer(int port) throws IOException {
        serverSocket = new ServerSocket(port);
    }

    public void start() {
        while (true) {
            try {
                Socket clientSocket = serverSocket.accept();
                new ClientHandler(clientSocket).start();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
    }

    private static class ClientHandler extends Thread {
        private Socket clientSocket;
        private BufferedReader in;
        private PrintWriter out;

        public ClientHandler(Socket socket) {
            this.clientSocket = socket;
        }

        public void run() {
            try {
                in = new BufferedReader(new InputStreamReader(clientSocket.getInputStream()));
                out = new PrintWriter(clientSocket.getOutputStream(), true);

                String inputLine;
                while ((inputLine = in.readLine()) != null) {
                    // Process input and send response
                    out.println("Received: " + inputLine);
                }
            } catch (IOException e) {
                e.printStackTrace();
            } finally {
                try {
                    in.close();
                    out.close();
                    clientSocket.close();
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
        }
    }

    public static void main(String[] args) {
        try {
            SmartHomeServer server = new SmartHomeServer(8080);
            server.start();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}


Setting Up the Project:

  1. Install Java SE 11+: Ensure that you have Java Development Kit (JDK) 11 or later installed.

  2. Set Up JavaFX: Download and configure JavaFX SDK for your IDE (IntelliJ IDEA or Eclipse).

  3. Install MySQL: Install MySQL server and create a database named smarthome. Use the provided DatabaseConnection class for managing connections.

  4. Configure JDBC Driver: Ensure that the MySQL JDBC driver is added to your project’s classpath.

  5. Run the Application:

    • Start the SmartHomeServer to listen for incoming client connections.
    • Run the SmartHomeApp to launch the GUI and start controlling the smart devices.
  6. Maven Dependencies: If using Maven, include the following dependencies in your pom.xml:


<dependencies>
    <dependency>
        <groupId>mysql</groupId>
        <artifactId>mysql-connector-java</artifactId>
        <version>8.0.26</version>
    </dependency>
    <dependency>
        <groupId>org.openjfx</groupId>
        <artifactId>javafx-controls</artifactId>
        <version>16</version>
    </dependency>
</dependencies>


Expanding the Project:

  • Add More Devices: Implement additional device types (e.g., security cameras, smart locks).
  • Enhance Networking: Develop a mobile app or web client to remotely access the smart home system.
  • Improve Security: Implement more sophisticated encryption and user management features.
  • Integrate AI: Incorporate machine learning models to optimize energy usage or predict user behavior.

Python Project: Audio Transcription and Text-to-Speech Conversion Using Wav2Vec2 and Pyttsx3

 Explore an advanced Python project that combines audio transcription and text-to-speech synthesis using state-of-the-art tools like Librosa, PyTorch, and Hugging Face's Transformers library. This script demonstrates how to load and resample audio files, transcribe speech to text using Facebook's Wav2Vec2 model, and convert text back to speech with customizable voice options using pyttsx3. Perfect for anyone interested in speech processing, AI-driven voice technology, or natural language processing projects. Ideal for enhancing your Python skills and diving into real-world applications of AI in audio analysis.



import librosa

from scipy.signal import resample

import torch

from transformers import Wav2Vec2ForCTC, Wav2Vec2Tokenizer

import pyttsx3

from scipy.signal import resample


# Load audio file

audio_file = "directory of audio file"

audio, sr = librosa.load(audio_file, sr=None)



def resample_audio(audio, orig_sr, target_sr):

    duration = audio.shape[0] / orig_sr

    target_length = int(duration * target_sr)

    resampled_audio = resample(audio, target_length)

    return resampled_audio


# Example usage:

# resampled_audio = resample_audio(audio, 48000, 16000)


# Resample if necessary

if sr != 16000:

    audio = resample_audio(audio, sr, 16000)

    sr = 16000


print("Audio loaded and resampled successfully.")


# Load Wav2Vec2 model and tokenizer

tokenizer = Wav2Vec2Tokenizer.from_pretrained("facebook/wav2vec2-base-960h")

model = Wav2Vec2ForCTC.from_pretrained("facebook/wav2vec2-base-960h")


print("Model loaded successfully.")


# Tokenize input

input_values = tokenizer(audio, return_tensors="pt").input_values


# Perform inference

with torch.no_grad():

    logits = model(input_values).logits


# Get predicted ids

predicted_ids = torch.argmax(logits, dim=-1)


# Decode the ids to text

transcription = tokenizer.batch_decode(predicted_ids)[0]

print("Transcription: ", transcription)


# Text-to-Speech

def text_to_speech(text, voice_gender='female', rate=150):

    engine = pyttsx3.init()

    voices = engine.getProperty('voices')

    

    if voice_gender == 'male':

        engine.setProperty('voice', voices[0].id)

    else:

        engine.setProperty('voice', voices[1].id)

    

    engine.setProperty('rate', rate)

    engine.say(text)

    engine.runAndWait()


# Example usage

long_text = "hi what happened"

text_to_speech(long_text, voice_gender='male', rate=150)  # For male voice

text_to_speech(long_text, voice_gender='female', rate=180)  # For female voice


print("Text-to-Speech conversion completed.")



#PythonProject
#AudioTranscription
#TextToSpeech
#Wav2Vec2
#PyTorch
#Librosa
#NLP
#SpeechRecognition
#VoiceSynthesis
#AIinPython
#NaturalLanguageProcessing
#SpeechToText
#Pyttsx3
#MachineLearning
#DeepLearning
#AudioProcessing
#PythonAI
#TransformersLibrary
#PythonCoding
#PythonTutorial

A Comprehensive Guide to SD-WAN Deployment: Migrating from Traditional WAN to Software-Defined WAN

 In today's rapidly evolving technological landscape, organizations are increasingly turning to Software-Defined Wide Area Network (SD-WAN) solutions to enhance their network performance, reduce costs, and streamline operations. This comprehensive guide will walk you through the key steps and considerations involved in migrating from a traditional WAN architecture to SD-WAN, ensuring a smooth and efficient transition.

1. The Importance of Controller Deployment

The first crucial step in any SD-WAN deployment is setting up the controllers. Controllers act as the central management and control plane of the SD-WAN architecture, ensuring seamless communication and coordination across the network.

  • Deployment Sequence: Typically, organizations start by deploying the controllers, followed by the migration of main data centers and hub sites. Finally, remote sites such as campuses and branches are transitioned. This sequence allows hub sites to route traffic between SD-WAN and non-SD-WAN sites during the migration period.

2. Controllers Deployment Options

One of the primary advantages of SD-WAN is the flexibility in controller deployment. Organizations can choose from several options based on their specific needs and compliance requirements:

  • Cisco-Hosted Cloud: The most popular option, with over 90% of customers opting for this model. Cisco handles provisioning, backup, and disaster recovery, offering SD-WAN control plane as a Software-as-a-Service (SaaS).

  • Public Cloud: Organizations can host controllers in public clouds like Azure and AWS, managed either by a service provider or in-house.

  • On-Premises: Suitable for organizations with strict compliance requirements, such as financial and government institutions. In this model, the organization is responsible for backups and disaster recovery.

3. Secure Controller Connections

Once deployed, controllers must establish secure connections. Organizations can choose between Transport Layer Security (TLS) using TCP transport or Datagram Transport Layer Security (DTLS) using UDP transport, with DTLS being the default.

4. WAN Edge Routers Onboarding

The secure onboarding of WAN edge devices is a critical aspect of SD-WAN deployment. Cisco SD-WAN uses a whitelisting model for authenticating and trusting vEdge devices. Each device is uniquely identified by its Chassis ID and certificate serial number.

  • Controllers Reachability: Ensuring WAN edge routers have reachability to all controllers via available transports is vital. This involves establishing control connections over each provisioned transport, starting with the vBond orchestrator.

  • Common Implementations for Controller Reachability:

    • MPLS routed through a data center or regional hub.
    • Public IP addresses of controllers redistributed into the MPLS cloud.
    • Control plane connection through the Internet, although this is not recommended due to lack of redundancy.

5. Joining the Overlay Fabric

The process of joining a WAN edge device to the SD-WAN overlay fabric involves several steps:

  • IP Reachability: The vEdge device obtains an IP address, default gateway, and DNS information via DHCP.
  • Zero-Touch Provisioning: The device reaches the ZTP server to get information about the vBond orchestrator and organization name.
  • Authentication: The device authenticates with its root-certificate and serial number.
  • Connection to Management Plane: The Edge establishes a secure connection to vManage and downloads the configuration.
  • Connection to Control Plane: The device connects to the vSmart controllers and joins the SD-WAN overlay fabric.

6. SD-WAN Operation, Administration, and Management (OAM)

SD-WAN offers significant advantages in terms of operation, administration, and management:

  • Centralized Management: Simplifies operations and reduces change and deployment times.
  • Transport-Independent Overlay: Allows the use of any combination of transports in an active/active fashion, reducing bandwidth costs.
  • Sophisticated Security: Provides comprehensive control plane encryption and a zero-trust security model.
  • Application Visibility: Enables real-time analysis, enforcement of service-level agreements (SLA), and tracking of performance metrics.

Conclusion

Migrating to SD-WAN can transform your network infrastructure, offering enhanced performance, reduced costs, and simplified management. By understanding the key steps and deployment options, you can ensure a successful transition to a modern, software-defined network architecture. Whether you opt for a Cisco-hosted cloud, public cloud, or on-premises deployment, the flexibility and benefits of SD-WAN make it a compelling choice for organizations of all sizes.

 #SDWAN #SDWANDeployment #SoftwareDefinedWAN #Networking #CiscoSDWAN #WANEdgeRouters #NetworkSecurity #CloudNetworking #ITInfrastructure #TechGuide #NetworkingSolutions #Cisco #SDWANMigration #NetworkManagement #DigitalTransformation #TechBlog

Featured Post

Day 41 — BGP Confederations: Sub-AS Design, External View and Migration

1. Opening Confederations are another way to scale BGP inside a large administrative domain. They divide the domain into member autonomous systems while presenting a single confederation identifier to external peers. They are powerful, but their operational model is more complex than simply 'using private ASNs inside.' The engineering goal is not to memorize another BGP command. It is to understand what information each speaker is allowed to propagate, what path information can be hidden, and what failure domain is created by the chosen control-plane architecture . 2. Concept and standards behavior RFC 5065 defines AS_CONFED_SEQUENCE and AS_CONFED_SET and how member-AS relationships are represented. Confederation external sessions have eBGP-like properties inside the confederation, while the confederation is presented externally as one AS. Modern guidance must also account for the fact that RFC 9774 prohibits new origination of AS_SET/AS_CONFED_SET in ordinary aggregation c...