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TOPIC: 255.255.240/20

255.255.240/20 12 years 11 months ago #993

  • fasmy98
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im wondering the /20 is the subnet or the newtwork available?

i found that "172.16.0.0/16 0 has 16 bits reserved for representing the host..If network 172.16.0.0/16 is subnetted by borrowing three bits
In our example above, the first subnet, subnet 172.16.0.0/19, is called subnet zero"

so,is it after subnet or before subnet???
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Re: 255.255.240/20 12 years 11 months ago #994

  • tfs
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You will want to read the sections under the Networking menu. Look under General at Subnetting and Supernetting - CIDR for a complete explantion. Very clear.

What you are dealing with is Classless Interdomain Routing (CIDR). In your first example, you probably mean 255.255.240.0/20. This could be borrowing 12 bits if a Class A address or 4 bits if a class B. Can't tell without seeing the address.

In your 2nd example, 172.16.0.0/16 is not subnetted - you are just using the newer format (the / format). This is a class B address (private, by the way) - the mask would be 255.255.0.0. You then subnet this address by borrowing 3 bits from the Host ID (not the network ID) and you have 172.16.0.0/19 (or 255.255.224.0). You now have 13 bits on the Host side to work with.

Again, look at the Subnetting tutorials here and it will become very clear.

Hope this helps,

Tom.
Thanks,

Tom
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Re: 255.255.240/20 12 years 11 months ago #995

  • Chris
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Fasmy98,

The /20 represents the subnet bits that are locked to "1", or in simple english, it means 20 subnet bits.

I'll analyse for you the 172.16.0.0/19 to help you understand whats happening.

172.16.0.0 with 19 subnet bits

19 subnet bits= 1111 1111 . 1111 1111 . 1110 0000 . 0000 0000
In decimal, that means 255.255.224.0

So, 19 bits= 255.255.224.0

Now, we need to find how many bits are used for the subnets.

Since we have a Class B network (172.X.X.X), the default subnet mask for it will be 255.255.0.0.

Default mask= 255.255.0.0
Our mask is = 255.255.224.0

Lets look at it in binary to give us a better idea:
Default Mask= 1111 1111 . 1111 1111 . 0000 0000 . 0000 0000
Our Mask is_= 1111 1111 . 1111 1111 . 1110 0000 . 0000 0000

Now you can clearly see the 3 bits used --^

These 3 bits give us 2^3-2=6 useable subnets
Let's take a look at them:

172.16.0.0 -Subnet Zero - not used according to Cisco
172.16.32.0 - First Valid Subnet
172.16.64.0
172.16.96.0
172.16.128.0
172.16.160.0
172.16.192.0 - Last Valid Subnet
172.16.224.0 - Subnet broadcast - not used according to Cisco

Now take notice that according to Cisco, the first and last subnets, that is, subnet zero and the broadcast subnet, are NOT valid.

In practice though, you can use them without a problem!

Let's now take a look at the available Hosts per Subnet. We will use the first one as our example:

Subnet : 172.16.32.0
Bits assigned to Network+Subnet=19
Remaining bits for hosts= 32-19= 13
Valid number of hosts= 2^13-2= 8190 hosts PER subnet.

First valid IP for this subnet= 172.16.32.1
Last valid IP for this subnet= 172.16.63.254

Broadcast for this subnet= 172.16.63.255

Now what you have here is a full analysis of the 172.16.0.0/19 network and its first useable subnet.

I hope that answers your questions. If anyone spots a mistake, please correct me!

Cheers,
Chris Partsenidis.
Founder & Editor-in-Chief
www.Firewall.cx
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Re: 255.255.240/20 12 years 11 months ago #1024

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ok....now i see it..
that means, the range for the subnet depends on the bit subnet taken from host right???

172.16.32.0
172.16.64.0

this is bcoz it took 3 bit from the host...

:lol:
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Re: 255.255.240/20 12 years 11 months ago #1031

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You got it.

Again, take a look at the tutorials. They really are very good and straight forward (unlike some I've seen).
Thanks,

Tom
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