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Thursday, 15 August 2024

Casting & Network Topology

Casting: Transmitting the data in the form of packets over the internet is called casting.

Types of Castings
The different types of casting are as follows −
Unicast − Transmitting data from one host to another host (one-one)
Broad cast − Transmitting data from one host to many host (one-all)
Multicast − Transmitting data from one host to a particular group of host (one-many).

Unicast
Transmitting data from one source host to one destination host is called a unicast. It is called as a one to one transmission.
For example − source Host IP Address 192.168.20.1 sending data to destination Host having IP Address 192.122.140.34
Packet = Data + Source Host IP + Destination Host IP
                              192.168.20.1      192.122.140.34
Broadcast
Transmitting data from one source host to all other hosts present in the same or other network is called broadcast. It is called a one to all transmission.

Broadcast is classified into two types, which are as follows −
Limited Broadcast − Transmitting data from one source host to all other hosts present in the same network is called a limited broadcast.

In Limited Broad casting if the destination address is 255.255.255.255 then the packet will be sent to all the hosts in the network.
Limited Broadcast address of any network
=255.255.255.255
= 11111111.11111111.11111111.11111111
For example − If the source IP address is 11.1.2.3 sending data to all other hosts present in the same network, then the destination address is 255.255.255.255.
Packet = Data + Source Host IP + Destination Host IP
                                 11.1.2.3              255.255.255.255

Direct Broadcast − Transmitting data from source host to all other hosts present in different networks then it is called as direct broadcast.

In direct broadcast Host ID bits are all set to 1, Network ID is the IP address where all destination hosts are present.
For example − Source IP address is 11.1.2.3  sending data to all other nodes present at different network having IP address 24.0.0.0
Therefore source address= 11.1.2.3
Destination address= 24.255.255.255
Packet = Data + Source Host IP + Destination Host IP
                                 11.1.2.3            24.255.255.255
Multicast
Transmitting data from one source host to a particular group of hosts that are interested in receiving the data is called Multicast. It is also called one to many transmissions.
For example − Sending messages on whatsapp to particular groups, video conferences, and sending email to groups of people.

Network topology:
Network topology is the layout of a network. It consists of two parts; 
physical and logical. 
The physical part describes the physical layout of a network while the logical part describes how the data flows in that network. 
Both, physical and logical parts are also known as the physical topology and the logical topology.

Physical part (topology) + Logical part (topology) = Network topology

Types Of Network Topologies:


BUS Topology
Bus topology is a network type in which every computer and network device is connected to single cable. When it has exactly two endpoints, then it is called Linear Bus topology.
    
When a computer transmits data in this topology, all computers see that data over the wire, but only that computer accepts the data to which it is addressed. It is just like an announcement that is heard by all but answered only by the person to whom the announcement is made.
For example, if in the above network, PC-A sends data to the PC-C then all computers of the network receive this data but only the PC-C accepts it. The following image shows this process.
  
If PC-C replies, only the PC-A accepts the return data. The following image shows this process.
Features of Bus Topology
1.It transmits data only in one direction.
2.Every device is connected to a single cable

Advantages of Bus Topology
1.It is cost effective.
2.Cable required is least compared to other network topology.
3.Used in small networks.
4.It is easy to understand.
5.Easy to expand joining two cables together.

Disadvantages of Bus Topology
1.Cables fails then whole network fails.
2.If network traffic is heavy or nodes are more the performance of the network decreases.
3.Cable has a limited length.
4.It is slower than the ring topology.

RING Topology
It is called ring topology because it forms a ring as each computer is connected to another computer, with the last one connected to the first. Exactly two neighbours for each device.
                            
Features of Ring Topology
1.A number of repeaters are used for Ring topology with large number of nodes, because if someone wants to send some data to the last node in the ring topology with 100 nodes, then the data will have to pass through 99 nodes to reach the 100th node. Hence to prevent data loss repeaters are used in the network.
2.The transmission is unidirectional, but it can be made bidirectional by having 2 connections between each Network Node, it is called Dual Ring Topology.
3.In Dual Ring Topology, two ring networks are formed, and data flow is in opposite direction in them. Also, if one ring fails, the second ring can act as a backup, to keep the network up.
4.Data is transferred in a sequential manner that is bit by bit. Data transmitted, has to pass through each node of-hg  the network, till the destination node.

Advantages of Ring Topology
1.Transmitting network is not affected by high traffic or by adding more nodes, as only the nodes having tokens can transmit data.
2.Cheap to install and expand

Disadvantages of Ring Topology
1.Troubleshooting is difficult in ring topology.
2.Adding or deleting the computers disturbs the network activity.
3.Failure of one computer disturbs the whole network.


STAR Topology
In this topology, all computers connect to a centralized networking device. Usually, a networking switch or a Hub (in earlier days) is used as the centralized device. Each computer in the network uses its own separate twisted pair cable to connect to the switch. Twisted pair cable uses RJ-45 connectors on both ends.
The following image shows an example of the star topology.
                                        
To transmit data, the star topology uses the same concept which the bus topology uses. It means, if you build a network using the star topology, then that network will use the bus topology to transmit the data.

Features of Star Topology
1.Every node has its own dedicated connection to the hub.
2.Hub acts as a repeater for data flow.
3.Can be used with twisted pair, Optical Fibre or coaxial cable.

Advantages of Star Topology
1.Fast performance with few nodes and low network traffic.
2.Hub can be upgraded easily.
3.Easy to troubleshoot.                                                                              
4.Easy to setup and modify.                                                                       
5.Only that node is affected which has failed, rest of the nodes can work smoothly.

Disadvantages of Star Topology
1.Cost of installation is high.
2.Expensive to use.
3.If the hub fails then the whole network is stopped because all the nodes depend on the hub.
4.Performance is based on the hub that is it depends on its capacity

MESH Topology
It is a point-to-point connection to other nodes or devices. All the network nodes are connected to each other. Mesh has n(n-1)/2 physical channels to link n devices.
                           
Required connections = n * (n-1)/2
Here, n is the number of end devices or locations.
For example, to make a fully meshed network of 4 end devices, we need 4*(4-1)/2 = 6 connections.
There are two techniques to transmit data over the Mesh topology, they are :
1.Routing
2.Flooding

MESH Topology: Routing
In routing, the nodes have a routing logic, as per the network requirements. Like routing logic to direct the data to reach the destination using the shortest distance. Or, routing logic which has information about the broken links, and it avoids those node etc. We can even have routing logic, to re-configure the failed nodes.

MESH Topology: Flooding
In flooding, the same data is transmitted to all the network nodes, hence no routing logic is required. The network is robust, and the its very unlikely to lose the data. But it leads to unwanted load over the network.
 
Types of Mesh Topology
1.Partial Mesh Topology : In this topology some of the systems are connected in the same fashion as mesh topology but some devices are only connected to two or three devices.
2.Full Mesh Topology : Each and every nodes or devices are connected to each other.

Features of Mesh Topology
1.Fully connected.
2.Robust.
3.Not flexible.

Advantages of Mesh Topology
1.Each connection can carry its own data load.
2.It is robust.
3.Fault is diagnosed easily.
4.Provides security and privacy.

Disadvantages of Mesh Topology
1.Installation and configuration is difficult.
2.Cabling cost is more.
3.Bulk wiring is required.

TREE Topology
It has a root node and all other nodes are connected to it forming a hierarchy. It is also called hierarchical topology. It should at least have three levels to the hierarchy.
                         
Features of Tree Topology
1.Ideal if workstations are located in groups.
2.Used in Wide Area Network.

Advantages of Tree Topology
1.Extension of bus and star topologies.
2.Expansion of nodes is possible and easy.
3.Easily managed and maintained.
4.Error detection is easily done.

Disadvantages of Tree Topology
1.Heavily cabled.
2.Costly.
3.If more nodes are added maintenance is difficult.
4.Central hub fails, network fails.

HYBRID Topology
This topology is a mix of two or more topologies. For example, there are two networks; one is built from the star topology and another is built from the bus topology. If we connect both networks to build a single large network, the topology of the new network will be known as the hybrid topology.
You are not restricted to the bus and star topologies. You can combine any topology with another topology. In modern network implementations, the hybrid topology is mostly used to mix the wired network with the wireless network.
The following image shows an example of the hybrid network topology.
  

Features of Hybrid Topology
•It is a combination of two or topologies
•Inherits the advantages and disadvantages of the topologies included

Advantages of Hybrid Topology
•Reliable as Error detecting and trouble shooting is easy.
•Effective.
•Scalable as size can be increased easily.
•Flexible.

Disadvantages of Hybrid Topology
•Complex in design.
•Costly

Line Configuration in Computer Networks

Line Configuration in Computer Networks

A Network is nothing but a connection made through connection links between two or more devices. Devices can be a computer, printer or any other device that is capable to send and receive data. There are two ways to connect the devices :
1.Point-to-Point connection
2.Multipoint connection

Point-To-Point Connection

It is a protocol which is used as a communication link between two devices. It is simple to establish. The most common example for Point-to-Point connection (PPP) is a computer connected by telephone line. We can connect the two devices by means of a pair of wires or using a microwave or satellite link.

Example: Point-to-Point connection between remote control and Television for changing the channels.

  


MultiPoint Connection

It is also called Multidrop configuration. In this connection two or more devices share a single link.

There are two kinds of Multipoint Connections :

If the links are used simultaneously between many devices, then it is spatially shared line configuration.

If user takes turns while using the link, then it is time shared (temporal) line configuration.

 


Subnetting


 
Question: How many networks and valid hosts for the given IP 192.168.10.0/25
Given IP :11000000.10101000.00001010.00000000
Subnet   :11111111.11111111.11111111.10000000
-----------------------------------------------------------
Network ID
NID : 192.168.10.0

Broadcast ID
BID : 192.168.10.127

Valid Host ID
VHID : 192.168.10.1 to 192.168.10.126

To Find the Number of Bits in 
Network ID :24
Subnet ID :1
Host ID :7

Number of Networks = 2^n where n no of network bits
                               =2^1
                               =2

Number of Valid Hosts = 2^h - 2 where h no of host bits
                                 =2^7 - 2
                                 =128 - 2
                                 =126

N/w NO

NETWORK ID

HOST ADDRESS

BROADCAST ID

1

192.168.10.0

192.168.10.1 - 192.168.10.126

192.168.10.127

2

192.168.10.128

192.168.10.129 - 192.168.10.254

192.168.10.255

Experimental set up of subnetting using Cisco Packet Racer:

Question: How many networks and valid hosts for the given IP 172.16.0.0/19
Given IP :10101100.00010000.00000000.00000000
Subnet   :11111111.11111111.11100000.00000000
-----------------------------------------------------------
Network ID
NID : 172.16.0.0

Broadcast ID
BID : 172.16.31.255

Valid Host ID
VHID : 172.16.0.1 to 172.16.31.254

To Find the Number of Bits in 
Network ID :16
Subnet ID :3
Host ID :13

Number of Networks = 2^n
                               =2^3
                               =8

Number of Valid Hosts = 2^h - 2
                                 =2^13 - 2
                                 =8192 - 2
                                 =8190

N/w NO

NETWORK ID

HOST ADDRESS

BROADCAST ID

1

172.16.0.0

172.16.0.0 - 172.16.31.254

172.16.31.255

2

172.16.32.0

172.16.32.1 - 172.16.63.254

172.16.63.255

3

172.16.64.0

172.16.64.1 - 172.16.95.254

172.16.95.255

4

172.16.96.0

172.16.96.1 – 172.16.127.254

172.16.127.255

5

172.16.128.0

172.16.128.1 – 172.16.159.254

172.16.159.255

6

172.16.160.0

172.16.160.1 – 172.16.191.254

172.16.191.255

7

172.16.192.0

172.16.192.1 – 172.16.223.254

172.16.223.255

8

172.16.224.0

172.16.224.1 – 172.16.255.254

172.16.255.255


Practice Questions
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 = 2^4 = 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: 21 – 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 = 2^4 – 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

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