subnet

Can you explain for which reason
172.16.0.0/25
has 510 subnet and 126 host fo each subnet
The first is 172.16.0.128 and the others
11111111.11111111.11111111.10000000
How can i find the next network id
EXAMPle 172.16.0.128/25 172.16.0.254/25
and the other 172.16.0.?????
IS right 255.255.255.128
Thank you

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Pavlov

I'm a bit confused by your question. Are you having difficulty understanding the idea of "stealing bits" from octets? Is this part of a bigger question?

If I see 172.16.0.0/25
My mask should be 255.255.255.128

My first host address would be 172.16.0.1
My last host address would be 172.16.0.127

My counter is 128 (256-12

Does this make sense? It's all about stealing the bits to get to /25. You can find more about subnetting at www.learntosubnet.com

p.provino

Excuse me for my English Is not my mother tongue...

Is it right that the subnet should be 510 and
126 hosts for subnet?
If i have a 172.16.0.0/25 network id
Can you tell me the first valid network id range
such us
172.16.0.128/25
Which is the first five network id 172.16.0.???/25
It will be 172.16.0.254/25??
Thank you and forgive me for my language

Pavlov

based on my previous post - I would say your first 5 valid IP addresses are:
172.16.0.1
172.16.0.2
172.16.0.3
172.16.0.4
172.16.0.5

Last 5 would be:
172.16.0.123
172.16.0.124
172.16.0.125
172.16.0.126
172.16.0.127

/25 just indicates that the mask won't be the default of 255.255.255.0 - the /25 forces you to steal one bit from the next octet making your subnet mask 255.255.255.128.

Am I making sense to you - I think I may be adding to confusion.

Let me know if this helps.

p.provino

From your answer I have understood some new
think.
I was confused because I thought that the 172.16.0.0/25 was in this way

11111111.11111111.11111111.10000000
This rapresentation confused me because it is no possible to have 510 subnet if the third octet
was all 1
From your answer I see that the range of network id is however to find in a 172.16.0.0
However the question that I i finally find was

What are the first 10 possible subnets for this subnetted network:172.16.0.0/25
The first network id is 172.16.0.128/25 and there are 2^9-2 possible subnets
What are the ranges of valid host address for the first few subnets?
I don't remember where i see it.
Your explanation was very cleared don't worry

thank you

Mindraker

p.provino wrote:

Can you explain for which reason
172.16.0.0/25
has 510 subnet and 126 host fo each subnet
The first is 172.16.0.128 and the others
11111111.11111111.11111111.10000000
How can i find the next network id
EXAMPle 172.16.0.128/25 172.16.0.254/25
and the other 172.16.0.?????
IS right 255.255.255.128
Thank you

Want the easy way?
11111111.11111111.11111111.10000000

0000000 = 7 zeros.
2^7 = 128.
128 - 2 = 126 hosts.

11111111.1 = 9 ones.
2^9 = 514
514 - 2 = 512 subnets.

Voilà!

kali

Mindraker, several questions about this method:

0000000 = 7 zeros.
2^7 = 128.
128 - 2 = 126 hosts. -- why are you subtracting 2 ?

also:
11111111.1 = 9 ones.
2^9 = 514 --2^9 = 512, NOT 514
514 - 2 = 512 subnets.

FINALLY:
if its 172.16.0.0/16 does that mean that there are 65536 - 2 = 65534 subnets and 65534 hosts?

Thank you

tcpsyn

I believe he's subtracting 192.168.0.0 which is the network id and 192.168.0.255 which is the broadcast id. Neither of which can be used for host IPs.

2lazybutsmart

hmm... makes me wonder how you fished for this 'old' post.

tcpsyn

I'm laid off.. got lots of time