Member Posts: 7 ■□□□□□□□□□
So Class B private range is 172.16.0.0 -172.31.0.0 which is mask of 12
So it gives you 2^12 networks which is 4096
and
It also gives you 2^20 which is 1048576. -2 hosts

Does it mean that each of those 4096 networks contain 1048574 hosts

• Member Posts: 999 ■■■■□□□□□□
If you are give a /12 that is 1 network. 172.16.0.0 - 172.31.0.0
You can subnet the /12 to be multiple subnets to give you more networks

The first 12 bits in a /12 are NOT yours to play with, if I give you 172.16.0.0/12 you can't use 8.10.20.0
| - Not yours - ||
Yours
|
|xxxxxxxx.xxxx||0000.00000000.00000000|

If i'm given 172.16.0.0/12, I can use a subnet mask of /24 and have 4096 networks with 254 hosts
| - Not yours - || - Networks - || - Hosts - |
|xxxxxxxx.xxxx||1111.11111111||.00000000|
A.A.S. in Networking Technologies
A+, Network+, CCNA
• Member Posts: 7 ■□□□□□□□□□
Sorry I get it now

172.16.0.0's host bits are 0.0 which is 65536 - 2 = 65534 host addresses
and for 172.31.0.0 - 172.16.0.0, 16 contagious networks we have 16 x 65534 = 1048544 hosts between those networks
• Member Posts: 7 ■□□□□□□□□□
I was talking about RFC 1918, which defines non-routable class A, B and C
This is a bit confusing as we have mask incrementing in 4
10.0.0.0 - 10.255.255.255 /8
172.16.0.0 - 172.31.0.0 /12 (which is 8 + 4)
192.168.0.0 - 192.168.255.255 /16 (which is 12 + 4)
• Member Posts: 86 ■■□□□□□□□□
I think you're confused about the masks. Assuming classful addressing is being used, your 10. network is a /8, but your 172.16-32 is a /16, and 192.168 is a /24. Anything between 8, 16, 24, and 32 is subnetting, which is classless, and your mask no longer conforms to A,B,C because you are borrowing bits from the host portion to create subnets (2 subnets for each bit borrowed) and reducing the number of hosts addresses available.
• Member Posts: 999 ■■■■□□□□□□
Private network - Wikipedia, the free encyclopedia

[TH="bgcolor: #F2F2F2, align: center"]RFC1918 name[/TH]
[TH="bgcolor: #F2F2F2, align: center"]IP address range[/TH]
[TH="bgcolor: #F2F2F2, align: center"]number of addresses[/TH]
[TH="bgcolor: #F2F2F2, align: center"]largest CIDR block (subnet mask)[/TH]
[TH="bgcolor: #F2F2F2, align: center"]host id size[/TH]
[TH="bgcolor: #F2F2F2, align: center"]classful description[/TH]

24-bit block
10.0.0.0 - 10.255.255.255
16,777,216
10.0.0.0/8 (255.0.0.0)
24 bits
8 bits
single class A network

20-bit block
172.16.0.0 - 172.31.255.255
1,048,576
172.16.0.0/12 (255.240.0.0)
20 bits
12 bits
16 contiguous class B networks

16-bit block
192.168.0.0 - 192.168.255.255
65,536
192.168.0.0/16 (255.255.0.0)
16 bits
16 bits
256 contiguous class C networks

A.A.S. in Networking Technologies
A+, Network+, CCNA
• Member Posts: 1,340 ■■■■□□□□□□
You could subnet 172.16.0.0/12 into anything from /13 to /30. That said, I have never seen it broken into anything larger than /16. I personally use a /23 at my Condo and a different /23 at my Parent's House - I renumbered their network [so the two networks wouldn't overlap] when I setup a VPN.
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