subnet
p.provino
Member Posts: 16 ■□□□□□□□□□
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
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
Comments
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Pavlov Member Posts: 264I'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.comPavlov
A+, Net+, i-Net+, CIW-A
MCP NT4, MCSA 2K, MCSE 2K -
p.provino Member Posts: 16 ■□□□□□□□□□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 Member Posts: 264based 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.Pavlov
A+, Net+, i-Net+, CIW-A
MCP NT4, MCSA 2K, MCSE 2K -
p.provino Member Posts: 16 ■□□□□□□□□□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 Member Posts: 25 ■□□□□□□□□□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 Member Posts: 4 ■□□□□□□□□□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 Member Posts: 28 ■□□□□□□□□□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.And we're using windows why?
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2lazybutsmart Member Posts: 1,119hmm... makes me wonder how you fished for this 'old' post.Exquisite as a lily, illustrious as a full moon,
Magnanimous as the ocean, persistent as time.