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Class B private IP address range network mask

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Hello, all.

 

I need to know the following:

 

1. What is the default network mask for Class B private IP address range of 172.16.0.0 - 172.31.255.255.

2. Provide source of information please. Don't say "I know, it's obvious, everybody knows". If you know - provide proof: RFC, etc.

 

BTW, I looked up RFC1918 and it wasn't clear to me. If you could explain - would be perfect.

 

Thanks a lot

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"Private Address Space

 

The Internet Assigned Numbers Authority (IANA) has reserved the

following three blocks of the IP address space for private internets:

 

10.0.0.0 - 10.255.255.255 (10/8 prefix)

172.16.0.0 - 172.31.255.255 (172.16/12 prefix)

192.168.0.0 - 192.168.255.255 (192.168/16 prefix)

 

We will refer to the first block as "24-bit block", the second as

"20-bit block", and to the third as "16-bit" block. Note that (in

pre-CIDR notation) the first block is nothing but a single class A

network number, while the second block is a set of 16 contiguous

class B network numbers, and third block is a set of 256 contiguous

class C network numbers."

 

i think you are confused with the prefix. The prefix over here mentions how long is the network. In other words it defines the network range.

 

The default network mask for any Class B IP address is 255.255.0.0 irrespective of it's a Private or Public IP.

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Hello, all.

 

I need to know the following:

 

1. What is the default network mask for Class B private IP address range of 172.16.0.0 - 172.31.255.255.

2. Provide source of information please. Don't say "I know, it's obvious, everybody knows". If you know - provide proof: RFC, etc.

 

BTW, I looked up RFC1918 and it wasn't clear to me. If you could explain - would be perfect.

 

Thanks a lot

 

Class A default mask is 255.0.0.0 or /8

Class B default mask is 255.255.0.0 or /16

Class C default mask is 255.255.255.0 or /24

 

The /x prefix describes how many bits in the subnet mask are set to 1. Thus identifying the Network Portion of the IP address. The remainder of the bits are 0 and are known as the host portion of the IP address.

 

This is taken from RFC 1918, the industry standard for Private IP address ranges

3. Private Address Space

 

The Internet Assigned Numbers Authority (IANA) has reserved the

following three blocks of the IP address space for private internets:

 

10.0.0.0 - 10.255.255.255 (10/8 prefix)

172.16.0.0 - 172.31.255.255 (172.16/12 prefix)

192.168.0.0 - 192.168.255.255 (192.168/16 prefix)

 

Next time say Please.

 

 

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Hello,

 

The default network mask for Class B private IP address is 255.255.0.0

You can not see private IP address of classe B who is not in this range: 172.16.0.0 - 172.31.255.255

This default netmask is used in the network who use 16 bits like bit of network address.

If you have for example your network address uses 14 bits to calculate your subnet you do like this

11111111.11111100.00000000.00000000

And you convert in decimal

255.252.0.0

For your second i don t understand.

Ok thanks

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Hello,

 

The default network mask for Class B private IP address is 255.255.0.0

You can not see private IP address of classe B who is not in this range: 172.16.0.0 - 172.31.255.255

This default netmask is used in the network who use 16 bits like bit of network address.

If you have for example your network address uses 14 bits to calculate your subnet you do like this

11111111.11111100.00000000.00000000

And you convert in decimal

255.252.0.0

For your second i don t understand.

Ok thanks

 

The mask is actually 255.240.0.0 or a /12 which gives the increment/block size of 16 in the second octet

 

So the range is 172.16.0.1 - 172.31.255

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The mask is actually 255.240.0.0 or a /12 which gives the increment/block size of 16 in the second octet

 

So the range is 172.16.0.1 - 172.31.255

 

Quite amusing )

Even though the question has been declared as "closed" couple of posts ago people just seem wanting to help me out on this! ))

Thanks much guys.

 

Mezilla, would you mind telling what is your experience with networks and what is the source of information regarding /12?

Thanks

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Quite amusing )

Even though the question has been declared as "closed" couple of posts ago people just seem wanting to help me out on this! ))

Thanks much guys.

 

Mezilla, would you mind telling what is your experience with networks and what is the source of information regarding /12?

Thanks

 

Hi, well i have spent the past 6 months studying for my CCNA, so i pretty much know subnetting inside out. I have learn't alot about how networks work in this time. I am not sure if that answer is exactly right,

but if you do the math, a /12 means 255.240.0.0....11111111.11110000.00000000.00000000

 

Since the second octet is a 240, that means an increment of 16 (256-240). So this would mean subnets of 172.0.0.0 , 172.16.0.0 , 172.32.0.0 and so on

 

So the private address range for class B networks falls in the 172.16.0.0 subnet...172.16.0.1 - 172.31.255.255

 

Correct me if i am wrong

 

 

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Hello, all.

 

I need to know the following:

 

1. What is the default network mask for Class B private IP address range of 172.16.0.0 - 172.31.255.255.

2. Provide source of information please. Don't say "I know, it's obvious, everybody knows". If you know - provide proof: RFC, etc.

 

BTW, I looked up RFC1918 and it wasn't clear to me. If you could explain - would be perfect.

 

Thanks a lot

 

Sybex - CCNA Study Guide, 6th Edition, pg 115, Subnet Mask:

Not all networks need subnets, meaning they use the default subnet mask. This is basically

the same as saying that a network doesn’t have a subnet address. Table 3.1 shows the default

subnet masks for Classes A, B, and C. These default masks cannot change. In other words, you

can’t make a Class B subnet mask read 255.0.0.0. If you try, the host will read that address

as invalid and usually won’t even let you type it in. For a Class A network, you can’t change

the first byte in a subnet mask; it must read 255.0.0.0 at a minimum. Similarly, you cannot

assign 255.255.255.255, as this is all 1s—a broadcast address. A Class B address must start

with 255.255.0.0, and a Class C has to start with 255.255.255.0.

 

Class Format Default Subnet Mask

A network.node.node.node 255.0.0.0

B network.network.node.node 255.255.0.0

C network.network.network.node 255.255.255.0

 

 

 

 

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Hi, well i have spent the past 6 months studying for my CCNA, so i pretty much know subnetting inside out. I have learn't alot about how networks work in this time. I am not sure if that answer is exactly right,

but if you do the math, a /12 means 255.240.0.0....11111111.11110000.00000000.00000000

 

Since the second octet is a 240, that means an increment of 16 (256-240). So this would mean subnets of 172.0.0.0 , 172.16.0.0 , 172.32.0.0 and so on

 

So the private address range for class B networks falls in the 172.16.0.0 subnet...172.16.0.1 - 172.31.255.255

 

Correct me if i am wrong

 

RFC1918 says:

...

172.16.0.0 - 172.31.255.255 (172.16/12 prefix)...

 

...

the second block is a set of 16 contiguous

class B network numbers

...

 

I bolded out what actually confuses me the most.

If we use 255.240.0.0 mask then block size is 16, hence network numbers are 0, 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208, 224, 240 (totaling 16 networks).

But interval 172.16.0.0 - 172.31.255.255 covers only one of listed networks!

In this interval according to mask 255.240.0.0 host bits are used in second octet - from bit 3 to 0, that makes up the "31" (16 - network number, and 1111b = 15d, 16 + 15 = 31).

 

So, for "16 contiguous class B network numbers" to be true the network mask should be 255.255.0.0!

But what does "/12 prefix" mean??!

 

Guys, I'm totally lost. ((

 

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RFC1918 says:

 

 

I bolded out what actually confuses me the most.

If we use 255.240.0.0 mask then block size is 16, hence network numbers are 0, 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208, 224, 240 (totaling 16 networks).

But interval 172.16.0.0 - 172.31.255.255 covers only one of listed networks!

In this interval according to mask 255.240.0.0 host bits are used in second octet - from bit 3 to 0, that makes up the "31" (16 - network number, and 1111b = 15d, 16 + 15 = 31).

 

So, for "16 contiguous class B network numbers" to be true the network mask should be 255.255.0.0!

But what does "/12 prefix" mean??!

 

Guys, I'm totally lost. ((

 

basically when you are looking at mask such as /24 or /27 , then that just means the first 24 or 27 bits of the mask are 1s

 

/24 also means 255.255.255.0 or 11111111.11111111.11111111.00000000...since each octet is 8 bits

 

/27 also means 255.255.255.224 or 11111111.11111111.11111111.11100000..to get the 224 it is just (128+64+32)

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Thankfully I understand that part.

 

What I can't understand is why RFC1918 folks (BTW, they're unavailable for comments - I got an "address not found" reply from mail server for all of them :) ) say there's the "/12 prefix" range which by default should have "/16" mask.

/12 = 255.240.0.0 = 11111111.11110000.00000000.00000000

/16 = 255.255.0.0 = 11111111.11111111.00000000.00000000

These can't be both used at the same time for 172.16.0.0 - 172.31.255.255 range.

Do you follow? Does anyone follow? :)

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Thankfully I understand that part.

 

What I can't understand is why RFC1918 folks (BTW, they're unavailable for comments - I got an "address not found" reply from mail server for all of them :) ) say there's the "/12 prefix" range which by default should have "/16" mask.

/12 = 255.240.0.0 = 11111111.11110000.00000000.00000000

/16 = 255.255.0.0 = 11111111.11111111.00000000.00000000

These can't be both used at the same time for 172.16.0.0 - 172.31.255.255 range.

Do you follow? Does anyone follow? :)

 

Hi,

 

First, keep in mind that the default mask for a class B address, either private or public, is /16 (255.255.0.0)

 

Then understand that the /12 mask in the RFC definition of private addresses is not a default mask for a given network. It's in fact used to represent the range of the class B private addresses, namely it summarizes a block of 16 contiguous class B addresses with a /16 mask into one address, so:

 

172.16.0.0 /16

172.17.0.0 /16

172.18.0.0 /16

...

...

172.31.0.0 /16

 

All the preceding network addresses can be summarized into one address, which is 172.16.0.0/12 because they all have the same first 12 bits.

It's just a matter of shortening the representaion. You'll better understand what I said when you'll study summarization

 

HTH

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I asked the same question on another forum: Todd Lammle forum.

 

Last answer also (like amine00) states that "/12 prefix" is not the subnet mask.

It's a block size identifier, so to speak.

Just shows that in this private range only bits 3-0 in second octet should be used, not bits 7-0.

But the subnet mask is still /16 as for any class B network.

 

So, same rules apply as for regular class B network, but the network count is only 2^4 = 16.

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