Practical Approch to ClasslessInterDomainRouting(CIDR)







For example: I have an IP address like 10.94.0.0/16 ......

Normally CIDR notation can be written after the last 4th octet after the slash(/)...i.e. /16


An IP address is divided into 4 parts with each part separated by a dot. Each part is called an "Octet".

IP address above has the following octets:
Octet 1: 10
Octet 2: 94
Octet 3: 0
Octet 4: 0
====> 10.94.0.0
This is called dotted decimal format.

Each octet can be translated into binary which is what computer uses when transmitting data on the network.
When written in binary each octet has 8 bits:

Octet 1: 00001010  --> 10 

Octet 2: 01011110  --> 94

Octet 3: 00000000  --> 0

Octet 4: 00000000 --> 0

==> 00001010.01011110.00000000.00000000 ==> 10.94.0.0

The CIDR notation of /16 is used to describe what subnet we use. 
CIDR notation always describes a subnet written in binary.
The number 16 refers to the number bits that are "turned on(1-bits)" meaning how many bits are "1" ---> 16 means sixteen 1's written as below.
Starting from the left of the IP address (in binary format)  turn on the first 16 bits as below to write out the subnet mask.

11111111.11111111.00000000
.00000000

When we translate this back to dotted decimal format we get

Subnet Mask: 255.255.0.0
IP Address : 10.94.0.0

When a computer, router or any layer 3 networking device receives a packet it performs an "AND" with the IP address and the subnet mask to figure out what subnet the IP address is in.


IP Ad:  00001010.01011110.00000000.00000000
Mask:   11111111.11111111.00000000.00000000          
AND         00001010.01011110.00000000.00000000
      

SO Finally Subnet is  10.94.0.0

ip address in CIDR notation is 10.94.xxx.xxx/16.

 The xxx can be any number from 0 to 255.

The subnet (i.e. Network Address) is always the first IP address available and

 the Broadcast Address is always the last address available.

Subnet: 10.94.0.0
Broadcast: 10.94.255.255


So we always have 2 IP addresses we cannot use.

So we can have hosts starting at 10.94.0.1 and ending at 10.94.255.254. This range is  available address range

To calculate the number of hosts

first translate  subnet mask to binary:

255.255.0.0 = 11111111.11111111.00000000.00000000

Count the number of zeros = 16

Now we raise the number 2 to the power of 16  i.e. 2^16 = 65536 (we use 2 because we are dealing with binary, each bit has 2 states, 0 or 1)

Lastly, remember we always have 2 IP addresses we cannot use, for network address and broadcast address (or 3 if we count the Default Gateway too).
So in total we have 65536 -2 = 65534 hosts in a CIDR /16 network

Below describes all the CIDR notations and its available hosts in detail:




IPv4 CIDR blocks

IPv4 CIDR IP/CIDR 0 to last IP addr Mask Hosts Size Notes
a.b.c.d/32 +0.0.0.0 255.255.255.255 1 1/256 C

a.b.c.d/31 +0.0.0.1 255.255.255.254 2 1/128 C d = 0 ... (2n) ... 254
a.b.c.d/30 +0.0.0.3 255.255.255.252 4 1/64 C d = 0 ... (4n) ... 252
a.b.c.d/29 +0.0.0.7 255.255.255.248 8 1/32 C d = 0 ... (8n) ... 248
a.b.c.d/28 +0.0.0.15 255.255.255.240 16 1/16 C d = 0 ... (16n) ... 240
a.b.c.d/27 +0.0.0.31 255.255.255.224 32 ⅛ C d = 0 ... (32n) ... 224
a.b.c.d/26 +0.0.0.63 255.255.255.192 64 ¼ C d = 0, 64, 128, 192
a.b.c.d/25 +0.0.0.127 255.255.255.128 128 ½ C d = 0, 128
a.b.c.0/24 +0.0.0.255 255.255.255.000 256 1 C

a.b.c.0/23 +0.0.1.255 255.255.254.000 512 2 C c = 0 ... (2n) ... 254
a.b.c.0/22 +0.0.3.255 255.255.252.000 1,024 4 C c = 0 ... (4n) ... 252
a.b.c.0/21 +0.0.7.255 255.255.248.000 2,048 8 C c = 0 ... (8n) ... 248
a.b.c.0/20 +0.0.15.255 255.255.240.000 4,096 16 C c = 0 ... (16n) ... 240
a.b.c.0/19 +0.0.31.255 255.255.224.000 8,192 32 C c = 0 ... (32n) ... 224
a.b.c.0/18 +0.0.63.255 255.255.192.000 16,384 64 C c = 0, 64, 128, 192
a.b.c.0/17 +0.0.127.255 255.255.128.000 32,768 128 C c = 0, 128
a.b.0.0/16 +0.0.255.255 255.255.000.000 65,536 256 C = 1 B

a.b.0.0/15 +0.1.255.255 255.254.000.000 131,072 2 B b = 0 ... (2n) ... 254
a.b.0.0/14 +0.3.255.255 255.252.000.000 262,144 4 B b = 0 ... (4n) ... 252
a.b.0.0/13 +0.7.255.255 255.248.000.000 524,288 8 B b = 0 ... (8n) ... 248
a.b.0.0/12 +0.15.255.255 255.240.000.000 1,048,576 16 B b = 0 ... (16n) ... 240
a.b.0.0/11 +0.31.255.255 255.224.000.000 2,097,152 32 B b = 0 ... (32n) ... 224
a.b.0.0/10 +0.63.255.255 255.192.000.000 4,194,304 64 B b = 0, 64, 128, 192
a.b.0.0/9 +0.127.255.255 255.128.000.000 8,388,608 128 B b = 0, 128
a.0.0.0/8 +0.255.255.255 255.000.000.000 16,777,216 256 B = 1 A

a.0.0.0/7 +1.255.255.255 254.000.000.000 33,554,432 2 A a = 0 ... (2n) ... 254
a.0.0.0/6 +3.255.255.255 252.000.000.000 67,108,864 4 A a = 0 ... (4n) ... 252
a.0.0.0/5 +7.255.255.255 248.000.000.000 134,217,728 8 A a = 0 ... (8n) ... 248
a.0.0.0/4 +15.255.255.255 240.000.000.000 268,435,456 16 A a = 0 ... (16n) ... 240
a.0.0.0/3 +31.255.255.255 224.000.000.000 536,870,912 32 A a = 0 ... (32n) ... 224
a.0.0.0/2 +63.255.255.255 192.000.000.000 1,073,741,824 64 A a = 0, 64, 128, 192
a.0.0.0/1 +127.255.255.255 128.000.000.000 2,147,483,648 128 A a = 0, 128
0.0.0.0/0 +255.255.255.255 000.000.000.000 4,294,967,296 256 A





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