With 1124 subnets required of 16915127 what is the range of

With 1124 subnets required of 16.9.15.127 what is the range of the 23rd usable network:

Can I see the calculation pls?

Solution

Given that 1124 subnets required:

1. Convert CIDR Subnet Mask Format To Dotted 4-Tuple Format

Note that when deals with /24 network or longer prefix, you only focus on the last octet. With Class B network, it is similar concept with focusing on the 3rd octet.

Class B network is between /16 and /23 CIDR. As mentioned, you only focus on the 3rd octet where the 1st two and last octets are constant. In other word, only the 3rd octet is changing as follows.
/24: 255.255.255.0
/23: 255.255.254.0
/22: 255.255.252.0
/21: 255.255.248.0
/20: 255.255.240.0
/19: 255.255.224.0
/18: 255.255.192.0
/17: 255.255.128.0
/16: 255.255.0.0

With Class B, there is a similar formula to convert CIDR format into dotted 4-tuple format by holding on these
* The \"longest\" Class B network (the /24) always has 255 as the 3rd octet
* The /24 always has a single Class C network
* The next larger Class B network is always double size of the current Class B network. In other word, the next larger Class B network has double quantities of Class C network than the current Class B network
* The 1st two octet are always 255 and the last octet is always 0 where the 3rd octet is changing

As illustration, let\'s say you like to convert /18 CIDR into dotted 4-tuple format.

The 3rd octet on the Class B network

1

2

3

4

5

6

7

8

9

CIDR     current 3rd        previous number        new current 3rd

         octet number     of Class C network        octet number<p></p><p>/24:         255       -           0           =        255

/23:         255       -           1           =        254

/22:         254       -       2 (1 x 2)      =        252

/21:         252       -       4 (2 x 2)      =        248

/20:         248       -       8 (4 x 2)      =        240

/19:         240       -      16 (8 x 2)      =        224

/18:         224       -      32 (16 x 2)      =        192

/17 192 - 64(32*2) = 128

/16 192 - 128(64*2) = 0

Number of classes required are 3.
Since (256 - 254) x 256 =   2 x 256 =   512 IP addresses within the subnet
Class A 0-512 subnets
Class B 513-1024 subnets
Class C 1025-1124 subnets

  
Maximum subnets: 512
Maximum IP addresses: 510   

Network address:   16.9.14.0 / 23
Usable IP addresses:   16.9.14.1 - 16.9.15.254
Broadcast:   16.9.15.255