Subnetting in a Computer Network

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

IPV4 address tips:

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

Find First 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)

Find Last Address

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

Broadcast Address: 167.199.170.95 =10100111.11000111.10101010.010 11111( ‘OR ‘ operation on given address and mask)

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 3^rd subblock 120 adress?

Solution:

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

32-24=8 that is rightmost 8 bit will 0; so 1^st 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 1^st address 14.24.74.0/25 and so 1^st 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 1^st 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 1^st 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)

Broadcast Address: 172.16.3.255          10101100.00010000.0000001 1.11111111

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)

Example 5: Find Network Address, Broadcast Address, Network, valid host ip= 192.16.13.0/30. [NESCO -2018]

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

1^st 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

Find valid host = 2^n -2 =2^2-2 = 4-2 =2

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