COMPUTER NETWORK: November 2017

Monday, November 13, 2017

Classful Addressing

In Classful Addressing System, IP Addresses are organized into following 5 classes-

Class A
Class B
Class C
Class D
Class E

1. Class A-

If the 32 bit binary address starts with a bit 0, then IP Address belongs to class A.

In class A IP Address,

The first 8 bits are used for the Network ID.
The remaining 24 bits are used for the Host ID.

Total Number Of IP Addresses-

Total number of IP Addresses available in class A

= Numbers possible due to remaining available 31 bits

= 2³¹

Total Number Of Networks-

Total number of networks available in class A

= Numbers possible due to remaining available 7 bits in the Net ID – 2

= 2⁷ – 2

= 126

(The reason of subtracting 2 is explained later.)

Total Number Of Hosts-

Total number of hosts that can be configured in class A

= Numbers possible due to available 24 bits in the Host ID – 2

= 2²⁴ – 2

(The reason of subtracting 2 is explained later.)

Range Of 1st Octet-

We have-

Minimum value of 1st octet = 00000000 = 0
Maximum value of 1st octet = 01111111 = 127

From here,

Range of 1st octet = [0, 127]
But 2 networks are reserved and unused.
So, Range of 1st octet = [1, 126]

Use-

Class A is used by organizations requiring very large size networks like NASA, Pentagon etc.

2. Class B-

If the 32 bit binary address starts with bits 10, then IP Address belongs to class B.

In class B IP Address,

The first 16 bits are used for the Network ID.
The remaining 16 bits are used for the Host ID.

Total Number Of IP Addresses-

Total number of IP Addresses available in class B

= Numbers possible due to remaining available 30 bits

= 2³⁰

Total Number Of Networks-

Total number of networks available in class B

= Numbers possible due to remaining available 14 bits in the Net ID

= 2¹⁴

Total Number Of Hosts-

Total number of hosts that can be configured in class B

= Numbers possible due to available 16 bits in the Host ID – 2

= 2¹⁶ – 2

Range Of 1st Octet-

We have-

Minimum value of 1st octet = 10000000 = 128
Maximum value of 1st octet = 10111111 = 191

So, Range of 1st octet = [128, 191]

Use-

Class B is used by organizations requiring medium size networks like IRCTC, banks etc.

3. Class C-

If the 32 bit binary address starts with bits 110, then IP Address belongs to class C.

In class C IP Address,

The first 24 bits are used for the Network ID.
The remaining 8 bits are used for the Host ID.

Total Number Of IP Addresses-

Total number of IP Addresses available in class C

= Numbers possible due to remaining available 29 bits

= 2²⁹

Total Number Of Networks-

Total number of networks available in class C

= Numbers possible due to remaining available 21 bits in the Net ID

= 2²¹

Total Number Of Hosts-

Total number of hosts that can be configured in class C

= Numbers possible due to available 8 bits in the Host ID – 2

= 2⁸ – 2

Range Of 1st Octet-

We have-

Minimum value of 1st octet = 11000000 = 192
Maximum value of 1st octet = 110111111 = 223

So, Range of 1st octet = [192, 223]

Use-

Class C is used by organizations requiring small to medium size networks.
For example- engineering colleges, small universities, small offices etc.

4. Class D-

If the 32 bit binary address starts with bits 1110, then IP Address belongs to class D.

Class D is not divided into Network ID and Host ID.

Total Number Of IP Addresses-

Total number of IP Addresses available in class D

= Numbers possible due to remaining available 28 bits

= 2²⁸

Range Of 1st Octet-

We have-

Minimum value of 1st octet = 11100000 = 224
Maximum value of 1st octet = 11101111 = 239

So, Range of 1st octet = [224, 239]

Use-

Class D is reserved for multicasting.
In multicasting, there is no need to extract host address from the IP Address.
This is because data is not destined for a particular host.

5. Class E-

If the 32 bit binary address starts with bits 1111, then IP Address belongs to class E.

Class E is not divided into Network ID and Host ID.

Total Number Of IP Addresses-

Total number of IP Addresses available in class E

= Numbers possible due to remaining available 28 bits

= 2²⁸

Range Of 1st Octet-

We have-

Minimum value of 1st octet = 11110000 = 240
Maximum value of 1st octet = 11111111 = 255

So, Range of 1st octet = [240, 255]

Use-

Class E is reserved for future or experimental purposes.

Classes of IP Address-

All the classes of IP Address are summarized in the following table-

Class of IP Address	Total Number of IP Addresses	1st Octet Decimal Range	Number of Networks available	Hosts per network	Default Subnet Mask
Class A	2³¹	1 – 126	2⁷ – 2	2²⁴ – 2	255.0.0.0
Class B	2³⁰	128 – 191	2¹⁴	2¹⁶ – 2	255.255.0.0
Class C	2²⁹	192 – 223	2²¹	2⁸ – 2	255.255.255.0
Class D	2²⁸	224 – 239	Not defined	Not defined	Not defined
Class E	2²⁸	240 – 254	Not defined	Not defined	Not defined

Sunday, November 12, 2017

Subnetting in Networking

In networking,

The process of dividing a single network into multiple sub networks is called as subnetting.
The sub networks so created are called as subnets.

Example-

Following diagram shows the subnetting of a big single network into 4 smaller subnets-

Advantages-

The two main advantages of subnetting a network are-

It improves the security.
The maintenance and administration of subnets is easy.

Subnet ID-

Each subnet has its unique network address known as its Subnet ID.
The subnet ID is created by borrowing some bits from the Host ID part of the IP Address.
The number of bits borrowed depends on the number of subnets created.

Types of Subnetting-

Subnetting of a network may be carried out in the following two ways-

Fixed Length Subnetting
Variable Length Subnetting

1. Fixed Length Subnetting-

Fixed length subnetting also called as classful subnetting divides the network into subnets where-

All the subnets are of same size.
All the subnets have equal number of hosts.
All the subnets have same subnet mask.

2. Variable Length Subnetting-

Variable length subnetting also called as classless subnetting divides the network into subnets where-

All the subnets are not of same size.
All the subnets do not have equal number of hosts.
All the subnets do not have same subnet mask.

Subnetting Examples-

Now, we shall discuss some examples of subnetting a network-

Example-01:

Consider-

We have a big single network having IP Address 200.1.2.0.
We want to do subnetting and divide this network into 2 subnets.

Clearly, the given network belongs to class C.

Also Read- Classes of IP Address

For creating two subnets and to represent their subnet IDs, we require 1 bit.

So,

We borrow one bit from the Host ID part.
After borrowing one bit, Host ID part remains with only 7 bits.

If borrowed bit = 0, then it represents the first subnet.
If borrowed bit = 1, then it represents the second subnet.

IP Address of the two subnets are-

200.1.2.00000000 = 200.1.2.0
200.1.2.10000000 = 200.1.2.128

For 1st Subnet-

IP Address of the subnet = 200.1.2.0
Total number of IP Addresses = 2⁷ = 128
Total number of hosts that can be configured = 128 – 2 = 126
Range of IP Addresses = [200.1.2.00000000, 200.1.2.01111111] = [200.1.2.0, 200.1.2.127]
Direct Broadcast Address = 200.1.2.01111111 = 200.1.2.127
Limited Broadcast Address = 255.255.255.255

For 2nd Subnet-

IP Address of the subnet = 200.1.2.128
Total number of IP Addresses = 2⁷ = 128
Total number of hosts that can be configured = 128 – 2 = 126
Range of IP Addresses = [200.1.2.10000000, 200.1.2.11111111] = [200.1.2.128, 200.1.2.255]
Direct Broadcast Address = 200.1.2.11111111 = 200.1.2.255
Limited Broadcast Address = 255.255.255.255

Example-02:

Consider-

We have a big single network having IP Address 200.1.2.0.
We want to do subnetting and divide this network into 4 subnets.

Clearly, the given network belongs to class C.

For creating four subnets and to represent their subnet IDs, we require 2 bits.

So,

We borrow two bits from the Host ID part.
After borrowing two bits, Host ID part remains with only 6 bits.

If borrowed bits = 00, then it represents the 1st subnet.
If borrowed bits = 01, then it represents the 2nd subnet.
If borrowed bits = 10, then it represents the 3rd subnet.
If borrowed bits = 11, then it represents the 4th subnet.

IP Address of the four subnets are-

200.1.2.00000000 = 200.1.2.0
200.1.2.01000000 = 200.1.2.64
200.1.2.10000000 = 200.1.2.128
200.1.2.11000000 = 200.1.2.192

For 1st Subnet-

IP Address of the subnet = 200.1.2.0
Total number of IP Addresses = 2⁶ = 64
Total number of hosts that can be configured = 64 – 2 = 62
Range of IP Addresses = [200.1.2.00000000, 200.1.2.00111111] = [200.1.2.0, 200.1.2.63]
Direct Broadcast Address = 200.1.2.00111111 = 200.1.2.63
Limited Broadcast Address = 255.255.255.255

For 2nd Subnet-

IP Address of the subnet = 200.1.2.64
Total number of IP Addresses = 2⁶ = 64
Total number of hosts that can be configured = 64 – 2 = 62
Range of IP Addresses = [200.1.2.01000000, 200.1.2.01111111] = [200.1.2.64, 200.1.2.127]
Direct Broadcast Address = 200.1.2.01111111 = 200.1.2.127
Limited Broadcast Address = 255.255.255.255

For 3rd Subnet-

IP Address of the subnet = 200.1.2.128
Total number of IP Addresses = 2⁶ = 64
Total number of hosts that can be configured = 64 – 2 = 62
Range of IP Addresses = [200.1.2.10000000, 200.1.2.10111111] = [200.1.2.128, 200.1.2.191]
Direct Broadcast Address = 200.1.2.10111111 = 200.1.2.191
Limited Broadcast Address = 255.255.255.255

For 4th Subnet-

IP Address of the subnet = 200.1.2.192
Total number of IP Addresses = 2⁶ = 64
Total number of hosts that can be configured = 64 – 2 = 62
Range of IP Addresses = [200.1.2.11000000, 200.1.2.11111111] = [200.1.2.192, 200.1.2.255]
Direct Broadcast Address = 200.1.2.11111111 = 200.1.2.255
Limited Broadcast Address = 255.255.255.255

Example-03:

Consider-

We have a big single network having IP Address 200.1.2.0.
We want to do subnetting and divide this network into 3 subnets.

Here, the subnetting will be performed in two steps-

Dividing the given network into 2 subnets
Dividing one of the subnets further into 2 subnets

Step-01: Dividing Given Network into 2 Subnets-

The subnetting will be performed exactly in the same way as performed in Example-01.

After subnetting, we have-

Step-02: Dividing One Subnet into 2 Subnets-

We perform the subnetting of one of the subnets further into 2 subnets.
Consider we want to do subnetting of the 2nd subnet having IP Address 200.1.2.128.

For creating two subnets and to represent their subnet IDs, we require 1 bit.

So,

We borrow one more bit from the Host ID part.
After borrowing one bit, Host ID part remains with only 6 bits.

If 2nd borrowed bit = 0, then it represents one subnet.
If 2nd borrowed bit = 1, then it represents the other subnet.

IP Address of the two subnets are-

200.1.2.10000000 = 200.1.2.128
200.1.2.11000000 = 200.1.2.192

Finally, the given single network is divided into 3 subnets having IP Address-

200.1.2.0
200.1.2.128
200.1.2.192

For 1st Subnet-

IP Address of the subnet = 200.1.2.0
Total number of IP Addresses = 2⁷ = 128
Total number of hosts that can be configured = 128 – 2 = 126
Range of IP Addresses = [200.1.2.00000000, 200.1.2.01111111] = [200.1.2.0, 200.1.2.127]
Direct Broadcast Address = 200.1.2.01111111 = 200.1.2.127
Limited Broadcast Address = 255.255.255.255

For 2nd Subnet-

IP Address of the subnet = 200.1.2.128
Total number of IP Addresses = 2⁶ = 64
Total number of hosts that can be configured = 64 – 2 = 62
Range of IP Addresses = [200.1.2.10000000, 200.1.2.10111111] = [200.1.2.128, 200.1.2.191]
Direct Broadcast Address = 200.1.2.10111111 = 200.1.2.191
Limited Broadcast Address = 255.255.255.255

For 3rd Subnet-

IP Address of the subnet = 200.1.2.192
Total number of IP Addresses = 2⁶ = 64
Total number of hosts that can be configured = 64 – 2 = 62
Range of IP Addresses = [200.1.2.11000000, 200.1.2.11111111] = [200.1.2.192, 200.1.2.255]
Direct Broadcast Address = 200.1.2.11111111 = 200.1.2.255
Limited Broadcast Address = 255.255.255.255

Disadvantages of Subnetting-

Point-01:

Subnetting leads to loss of IP Addresses.

During subnetting,

We have to face a loss of IP Addresses.
This is because two IP Addresses are wasted for each subnet.
One IP address is wasted for its network address.
Other IP Address is wasted for its direct broadcasting address.

Point-02:

Subnetting leads to complicated communication process.

After subnetting, the communication process becomes complex involving the following 4 steps-

Identifying the network
Identifying the sub network
Identifying the host
Identifying the process

PRACTICE PROBLEMS BASED ON SUBNETTING IN NETWORKING-

Problem-01:

Suppose a network with IP Address 192.16.0.0. is divided into 2 subnets, find number of hosts per subnet.

Also for the first subnet, find-

Subnet Address
First Host ID
Last Host ID
Broadcast Address

Solution-

Given IP Address belongs to class C.
So, 24 bits are reserved for the Net ID.
The given network is divided into 2 subnets.
So, 1 bit is borrowed from the host ID part for the subnet IDs.
Then, Number of bits remaining for the Host ID = 7.
Thus, Number of hosts per subnet = 2⁷ = 128.

For 1st Subnet-

Subnet Address = First IP Address = 192.16.0.00000000 = 172.16.0.0
First Host ID = 192.16.0.00000001 = 192.16.0.1
Last Host ID = 192.16.0.01111110 = 192.16.0.126
Broadcast Address = Last IP Address = 192.16.0.01111111 = 172.16.0.127

Problem-02:

What is not true about subnetting?

It is applied for a single network
It is used to improve security
Bits are borrowed from network portion
Bits are borrowed from Host portion

Solution-

Clearly, Option (C) is correct.

Problem-03:

In a class B, network on the internet has a subnet mask of 255.255.240.0. What is the maximum number of hosts per subnet?

4096
4094
4092
4090

Solution-

Number of bits reserved for network ID in the given subnet mask = 20.
So, Number of bits reserved for Host ID = 32 – 20 = 12 bits.
Thus, Number of hosts per subnet = 2¹² – 2 = 4094.
In class B, 16 bits are reserved for the network.
So, Number of bits reserved for subnet ID = 20 – 16 = 4 bits.
Number of subnets possible = 2⁴ = 16.
Thus, Option (B) is correct.