subnet

p.provinop.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 icon_rolleyes.gif

Comments

  • PavlovPavlov Member Posts: 264
    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 (256-12icon_cool.gif

    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
    Pavlov
    A+, Net+, i-Net+, CIW-A
    MCP NT4, MCSA 2K, MCSE 2K
  • p.provinop.provino Member Posts: 16 ■□□□□□□□□□
    Excuse me for my English Is not my mother tongue... icon_redface.gif
    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
  • PavlovPavlov Member Posts: 264
    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.
    Pavlov
    A+, Net+, i-Net+, CIW-A
    MCP NT4, MCSA 2K, MCSE 2K
  • p.provinop.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 icon_wink.gif
    thank you
  • MindrakerMindraker 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 icon_rolleyes.gif

    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!
  • kalikali 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
  • tcpsyntcpsyn 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?
  • 2lazybutsmart2lazybutsmart Member Posts: 1,119
    hmm... 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.
  • tcpsyntcpsyn Member Posts: 28 ■□□□□□□□□□
    I'm laid off.. got lots of time :)
    And we're using windows why?
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