Subnetting

Jarrel

Well-known member
  • Feb 17, 2020
    350
    1
    522
    Australia
    www.jarrelrivera.com
    For subnetting scenarios that I give to my students, I specify the size of the subnetworks as well (either varied with specifics or equal size).
    In that way, I would have a specific answer that I can use to check if my students have done them correctly or not.

    If you just say make 11 networks, then they can have varied responses which could strain you with marking/checking
     

    Tess Sluijter

    Well-known member
    Apr 1, 2020
    376
    1
    535
    the Netherlands
    www.kilala.nl
    How does one prove if his subnetting is correct?

    for example on 172.16.0.0 and I'm making 11 networks

    How do I prove that what I'm doing is correct? I hope I'm making sense :)
    To be blunt: your question mentions zero information about subnet masks, desired network size, etc. So your question is incomplete. There is nothing to prove.
     

    Rick Butler

    Well-known member
  • Aug 8, 2019
    1,855
    7
    3,369
    Colorado Springs, CO
    www.intellitec.edu
    To be blunt: your question mentions zero information about subnet masks, desired network size, etc. So your question is incomplete. There is nothing to prove.
    Quite right. Without a subnet mask, the general assumption is that we're talking about a Class B address, which suggests the mask is going to be 255.255.0.0. Of course, that's assumptive in the absence of the mask or CIDR notation.

    However, to complete the square, let me suggest that a mask of 255.255.240.0, or a /20 applied to 172.16.0.0 will create 11 networks. (actually it creates 16).
    172.16.0.0 - 172.16.0.15
    172.16.0.16 - 172.16.0.31
    172.16.0.32 - 172.16.0.47
    172.16.0.48 - 172.16.0.63
    172.16.0.64 - 172.16.0.79
    172.16.0.80 - 172.16.0.95
    172.16.0.96 - 172.16.0.111
    172.16.0.112 - 172.16.0.127
    172.16.0.128 - 172.16.0.143
    172.16.0.144 - 172.16.0.159
    172.16.0.160 - 172.16.0.175
    172.16.0.176 - 172.16.0.191
    172.16.0.192 - 172.16.0.207
    172.16.0.208 - 172.16.0.223
    172.16.0.224 - 172.16.0.239
    172.16.0.240 - 172.16.0.255

    Additionally, remember that when CompTIA asks about what the mask needs to be to get "this many networks/hosts", they are looking for the bare minimum. Naturally, /21 /22, etc will also give you 11 networks, but in questions like this, they are looking for the line. So a /20 will give you the 11 networks you need, but a /19 will not (that only produces 8).

    /r
     
    Agree with the above. Sharing on related note, this is a good practice site or can be used to generate scenarios.


    Lee
     
    A nice way to illustrate subnetting is defining a particular need, like a router to router network, where all you want to end up with is two valid host IP addresses for the subnetwork (255.255.255.252). Example:

    172.16.0.0 = Network Address
    172.16.0.1 = Host 1
    172.16.0.2 = Host 2
    172.16.0.3 = Broadcast Address
     

    Nate G

    Well-known member
  • Aug 12, 2020
    51
    47
    Quite right. Without a subnet mask, the general assumption is that we're talking about a Class B address, which suggests the mask is going to be 255.255.0.0. Of course, that's assumptive in the absence of the mask or CIDR notation.

    However, to complete the square, let me suggest that a mask of 255.255.240.0, or a /20 applied to 172.16.0.0 will create 11 networks. (actually it creates 16).
    172.16.0.0 - 172.16.0.15
    172.16.0.16 - 172.16.0.31
    172.16.0.32 - 172.16.0.47
    172.16.0.48 - 172.16.0.63
    172.16.0.64 - 172.16.0.79
    172.16.0.80 - 172.16.0.95
    172.16.0.96 - 172.16.0.111
    172.16.0.112 - 172.16.0.127
    172.16.0.128 - 172.16.0.143
    172.16.0.144 - 172.16.0.159
    172.16.0.160 - 172.16.0.175
    172.16.0.176 - 172.16.0.191
    172.16.0.192 - 172.16.0.207
    172.16.0.208 - 172.16.0.223
    172.16.0.224 - 172.16.0.239
    172.16.0.240 - 172.16.0.255

    Additionally, remember that when CompTIA asks about what the mask needs to be to get "this many networks/hosts", they are looking for the bare minimum. Naturally, /21 /22, etc will also give you 11 networks, but in questions like this, they are looking for the line. So a /20 will give you the 11 networks you need, but a /19 will not (that only produces 8).

    /r
    Thank you for this.
     
    • Like
    Reactions: Rick Butler

    Nate G

    Well-known member
  • Aug 12, 2020
    51
    47
    Agree with the above. Sharing on related note, this is a good practice site or can be used to generate scenarios.


    Lee
    Thank you.
     

    Ramasankar

    Well-known member
    Nov 19, 2019
    26
    13
    Quite right. Without a subnet mask, the general assumption is that we're talking about a Class B address, which suggests the mask is going to be 255.255.0.0. Of course, that's assumptive in the absence of the mask or CIDR notation.

    However, to complete the square, let me suggest that a mask of 255.255.240.0, or a /20 applied to 172.16.0.0 will create 11 networks. (actually it creates 16).
    172.16.0.0 - 172.16.0.15
    172.16.0.16 - 172.16.0.31
    172.16.0.32 - 172.16.0.47
    172.16.0.48 - 172.16.0.63
    172.16.0.64 - 172.16.0.79
    172.16.0.80 - 172.16.0.95
    172.16.0.96 - 172.16.0.111
    172.16.0.112 - 172.16.0.127
    172.16.0.128 - 172.16.0.143
    172.16.0.144 - 172.16.0.159
    172.16.0.160 - 172.16.0.175
    172.16.0.176 - 172.16.0.191
    172.16.0.192 - 172.16.0.207
    172.16.0.208 - 172.16.0.223
    172.16.0.224 - 172.16.0.239
    172.16.0.240 - 172.16.0.255

    Additionally, remember that when CompTIA asks about what the mask needs to be to get "this many networks/hosts", they are looking for the bare minimum. Naturally, /21 /22, etc will also give you 11 networks, but in questions like this, they are looking for the line. So a /20 will give you the 11 networks you need, but a /19 will not (that only produces 8).

    /r
    With a /20 I assume the networks have to be
    172.16.0.0 - 172.16.15.255
    172.16.16.0 - 172.16.31.255
    172.16.32.0 - 172.16.47.255
    172.16.48.0 - 172.16.63.255
    172.16.64.0 - 172.16.79.255
    172.16.80.0 - 172.16.95.255
    172.16.96.0 - 172.16.111.255
    172.16.112.0 - 172.16.127.255
    172.16.128.0 - 172.16.143.255
    172.16.144.0 - 172.16.159.255
    172.16.160.0 - 172.16.175.255
    172.16.176.0 - 172.16.191.255
    172.16.192.0 - 172.16.207.255
    172.16.208.0 - 172.16.223.255
    172.16.224.0 - 172.16.239.255
    172.16.240.0 - 172.16.2545.255
     
    With a /20 I assume the networks have to be
    172.16.0.0 - 172.16.15.255
    172.16.16.0 - 172.16.31.255
    172.16.32.0 - 172.16.47.255
    172.16.48.0 - 172.16.63.255
    172.16.64.0 - 172.16.79.255
    172.16.80.0 - 172.16.95.255
    172.16.96.0 - 172.16.111.255
    172.16.112.0 - 172.16.127.255
    172.16.128.0 - 172.16.143.255
    172.16.144.0 - 172.16.159.255
    172.16.160.0 - 172.16.175.255
    172.16.176.0 - 172.16.191.255
    172.16.192.0 - 172.16.207.255
    172.16.208.0 - 172.16.223.255
    172.16.224.0 - 172.16.239.255
    172.16.240.0 - 172.16.255.255
    You would be right. I actually ran a /28 not a /20. It was an early morning without coffee. Nice work. /r