## Subnetting in a Computer Network

Subnetting: Refers break large networks into smaller which provides security, low collision.

It is 32 bit address

Total address space = 2^32

Example: 192.168.0.0;

Here all the separated dot (.) are consist of 8 bits.

167.199.170.82/27 here /27 slash notation or CIDR(classless interdomain routing); 27 is prefix length.

Example: If a IP contain  slash notation that ip provide us three information

Total number of address, n = 2^(32-n)

First address by letting ‘0’ (32-n) rightmost bit.

Last address by letting ‘1’ (32-n) rightmost bit.

If you do not have basics on subnet please read my previous articles https://draftsbook.com/ip-and-its-properties-address-space-subnetting-with-real-time-example/

### Example 1: The above example Given IP address: 167.199.170.82/27 find total number of address What is first address? What is last address?

Solution:

Given address: 167.199.170.82/27; prefix length, CIDR, n=27

total number of address 2^(32-n) =2^5 = 32 address

Given address, Its binary form 10100111.11000111.10101010.01010010

As (32-n) = 32-27 = 5; So, the last octet right last 5 bits will let ‘0’

That is the now the binary form 10100111.11000111.10101010.01000000 which decimal form is

167.199.170.64

So, first address 167.199.170.64/27.

OR,

Address: 167.199.170.82= 10100111.11000111.10101010.010 10010(as range 128-191 so Class B)
Netmask: 255.255.255.224 = 27 11111111.11111111.11111111.111 00000(as default mask Class B)
Wildcard: 0.0.0.31 =00000000.00000000.00000000.000 11111
=>
Network Address: 167.199.170.64/27 =10100111.11000111.10101010.010 00000 (Class B ‘AND ‘ operation on given address and mask)

Given IP binary form 10100111.11000111.10101010.01010010

As (32-n) = 32-27 = 5; So, the last octet right last 5 bits 1

That is the now the binary form 10100111.11000111.10101010.01011111 which decimal form is

167.199.170.95

So, Last address 167.199.170.95/27.

### Example 2: An address several prefix lengths can place on a address block 230.8.24.56/16.

As (32-16)=16 so first bit address right 16 bit will 0

SO binary form 230.8.24.56 = 011100110.00000100.00011000.00111000

First address = 230.8.0.0

To find last address last rightmost 16 bit will 1

SO binary form 230.8.24.56= 011100110.00000100.11111111.11111111

=230.8.255.255

A rule: Number of requested address N need to be power of 2

N= 2^32-n

Or, n= 32- log2^N

### Example 3: A company start adreess 14.24.74.0/24 they need subblock. Each subblock 10 address, 6 subblock 60 and 3rd subblock 120 adress?

Solution:

Subblock address always power of 2. For 1st block address has 2^(32-24)=256, which need to divided.

32-24=8 that is rightmost 8 bit will 0; so 1st address 14.24.74.0/24;

32-24=8 that is rightmost 8 bit will 1; so last address 14.24.74.255/24;

As question 120 is not a power of 2 so immediate power 128(2^7)

So, n=32-log2^128=32-7=25; therefore so 1st address 14.24.74.0/25 and so 1st address 14.24.74.127/25;

#### (b) as 60 blocks not power of 2 immediate large 64 n=32-log2^64 =32-4=26

so 1st address 14.24.74.128/26; last address 14.24.74.191/26;

#### © 10 also not power of  2 immediate large 16; n=32-log2^16=32-4=28

so 1st address 14.24.74.192/28; last address 14.24.74.207/28;

Total 208 address used rest 48 for reserve.

### Example 4: Find the subnet mask and number of host on each subnet mask at a class B. IP= 172.16.2.1/23(pally sancay bank 2018)

Solution:

Here 32-23=9  that is rightmost 9 bit will 1; Given IP binary form 11111111. 11111111. 11111110.00000000.subnet mask= 255.255.254.0

Valid network = 2^n =23-16=7; 2^7

Valid Host = 2^n – 2

N= 32-23=9=2^9 -2=510

#### Find first address

Binary form on Given IP and mask

10101100.00010000.00000010.00000001

11111111.11111111.11111110.00000000

11111111.11111111.11111110.00000001(Doing ‘OR’ operation )

So, 172.16.2.0/23 first IP address

### OR,

Address:   172.16.2.1            10101100.00010000.0000001 0.00000001

Netmask:   255.255.254.0 = 23(32-23)=9    11111111.11111111.1111111 0.00000000

Wildcard:  0.0.1.255             00000000.00000000.0000000 1.11111111

=>

Network Address:   172.16.2.0/23         10101100.00010000.0000001 0.00000000 (IP and mask and operation)(Class B)

HostMin:   172.16.2.1            10101100.00010000.0000001 0.00000001

HostMax:   172.16.3.254          10101100.00010000.0000001 1.11111110

Hosts/Net: 510                  (Private Internet)

### Solution:

Given ip = 192.16.13.0 [32-30=2] binary form 11000000.00010000.00001101.00000000

As it is 192 so class C and ask 255.255.255.0 binary form 11111111.11111111.11111111.11111100(rightmost 2 ‘0’ bits);

Now apply ‘And’ operation to find first address:192.16.13.0/30

1st address or Network address = 192.16.13.0/30

Or, See the full operation sequentially:

Given IP Address:   192.16.13.0           11000000.00010000.00001101.000000 00(range 192-223 so Class C )

Netmask:   255.255.255.252 = 30  11111111.11111111.11111111.111111 00(default mask Class C )

Wildcard:  0.0.0.3               00000000.00000000.00000000.000000 11(apply ‘AND’ between IP and mask)

=>

Network:   192.16.13.0/30        11000000.00010000.00001101.000000 00 (Class C)

### Last address/ Broadcast address need mask last 2 bits are 1 and apply OR operation:

Given IP Address:   192.16.13.0           11000000.00010000.00001101.000000 00

Netmask:   255.255.255.252 = 30  11111111.11111111.11111111.111111 00

Broadcast: 192.16.13.3           11000000.00010000.00001101.000000 11 (apply OR operation)

HostMin:   192.16.13.1           11000000.00010000.00001101.000000 01

HostMax:   192.16.13.2           11000000.00010000.00001101.000000 10

Hosts/Net: 2

So, last address 192.16.13.3           /30

### Example 6: A block address is granted to a small organization. If one of the address is 205.16.37.39/28. What is the first and last address of the block? [ministry (AP)-2017]

Given IP Address:   205.16.37.39          11001101.00010000.00100101.0010 0111

Netmask:   255.255.255.240 = 28  11111111.11111111.11111111.1111 0000

Wildcard:  0.0.0.15              00000000.00000000.00000000.0000 1111

=>

Network:   205.16.37.32/28       11001101.00010000.00100101.0010 0000 (Class C)

Broadcast: 205.16.37.47          11001101.00010000.00100101.0010 1111

HostMin:   205.16.37.33          11001101.00010000.00100101.0010 0001

HostMax:   205.16.37.46          11001101.00010000.00100101.0010 1110

Hosts/Net: 14

## What is IP and its properties

“IP” stands for Internet Protocol where data is sent by which the method or protocol from one computer to another.  Actually, each computer on the Internet has at least one IP address that uniquely identifies it from all other computers on the Internet. An IP address is a unique global address for a network interface.

IP is an real life example of the postal system. It allows you to address a package and drop it in the system, but there’s no direct link between you and the recipient. TCP/IP, in contrast, creates a connection between two hosts, so that they can send messages back and forth for a period of time.

TCP is one of the main protocols in TCP/IP networks. Whereas the IP protocol deals only with packets, TCP enables two hosts to establish a connection and exchange streams of data. TCP guarantees delivery of data and also guarantees that packets will be delivered in the same order in which they were sent.

Two (2) IP addressing standards are in use today. The IPv4 standard is most familiar to people and supported everywhere on the Internet, but the newer IPv6 standard is gradually replacing it.

• IPv4 addresses consist of four (4) bytes (32 bits)
• IPv6 addresses are sixteen (16) bytes (128 bits) long.

An IPv4 address is a 32-bit or  4 byte address that uniquely and universally defines the connection of a device (for example, a computer or a router) to the Internet.

An IPv4 address consists of four numbers, with a single dot (.) separating each number or set of digits.

Each of the four numbers can range from 0 to 255.

An address space is the total number of addresses used by the protocol. If a protocol uses N bits to define an address, the address space is  2N  because each bit can have two different values (0 or 1) and N bits can have 2N  values.

IPv4 uses 32-bit addresses, which means that the address space is 232  or 4,294,967,296 (more than 4 billion).

• There are two prevalent notations to show an IPv4 address:

(1) Binary notation and

(2) Dotted decimal notation.

Binary Notation:

01110101   10010101   00011101   00000010

Dotted-Decimal Notation:   117.  149.  29.  2

IPv4 addressing, at its inception, used the concept of classes. This architecture is called classful addressing. In classful addressing, the address space is divided into five classes: A, B, C, D, and E.

 Class Binary Decimal A 0 0-127 B 1o 128-191 C 110 192-223 D 1110 224-239 E 1111 240-255

 Class Start address End Address A 0.0.0.0 127.255.255.255 B 128.0.0.0 191.255.255.255 C 192.0.0.0 223.255.255.255 D 224.0.0.0 239.255.255.255 E 240.0.0.0 255.255.255.255

Exercise

1. 193.14.56.22

The first byte is 193 (between 192 and 223); the class is C.

1. 14.23.120.8

The first byte is 14 (between 0 and 127); the class is A.

In classful addressing, an IP address of class A, B and C is divided into two parts : Netid and Hostid.

The netid and hostid are of varying lengths that is varies on depending on the class of the address.

Netid(n): The part of an IP address that identifies the network.

Hostid(32-n): The part of an IP address that identifies a host in a network.

 Class Net Id Host Id Start address End Address A 8 24 0.0.0.0 127.255.255.255 B 16 16 128.0.0.0 191. 255.255.255 C 24 8 192.0.0.0 223. 255.255.255 D Not define – 224.0.0.0 239. 255.255.255 E Not define – 240.0.0.0 255. 255.255.255

• If the first decimal number in IP address is 0 to 127, then it is a class A address.
• Class A IP addresses use the 1st 8 bits (1st Octet) to designate the Network address.
• The 1st bit which is always a 0, is used to indicate the address as a Class A address & the remaining 7 bits are used to designate the Network.
• The other 3 octets contain the Host address.
• There are 128 (27) Class A Network Addresses, but because addresses with all zeros aren’t used & address 127 is a special purpose address, 126 Class A Networks are available.
• formula to compute the number of hosts available in any of the class addresses, where “n” represents the number of bits in the host portion:
• (2n – 2) = Number of available hosts
• For a Class A network, there are:
• 224 – 2 or 16,777,214 hosts.
• Half of all IP addresses are Class A addresses.
• You can use the same formula to determine the number of Networks in an address class.
• , a Class A address uses 7 bits to designate the network, so (27 – 2) = 126 or there can be 126 Class A Networks.

Class B IP Addresses

• If the first decimal number in IP address is 128 to 191, then it is a class B address.
• Class B addresses use the 1st 16 bits (two octets) for the Network address.
• The last 2 octets are used for the Host address.
• The 1st 2 bit, which are always 10, designate the address as a Class B address & 14 bits are used to designate the Network. This leaves 16 bits (two octets) to designate the Hosts.
• So how many Class B Networks can there be?

Using our formula, (214 – 2), there can be 16,382 Class B Networks & each Network can have (216 – 2) Hosts, or 65,534 Hosts

Class C IP address

• If the first three bits of the address are 1 1 0, it is a class C network address.
• The first three (3) bits are class identifiers.
• The next 21 bits are for the network address.
• The last eight (8) bits  identify the host.
• There are millions of class C network numbers.
• However, each class C network can have 254 hosts.
 Class Number of Blocks/ Networks Block size/ Address per block Start address End Address Application A 128(27) 16,777,216(224) 0.0.0.0 127.255.255.255 Unicast B 16384(214) 65536(216) 128.0.0.0 191. 255.255.255 Unicast C 2097152(221) 256(28) 192.0.0.0 223. 255.255.255 Unicast D 224.0.0.0 239. 255.255.255 Multicast E 240.0.0.0 255. 255.255.255 Multicast