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Valid host address

seanrseanr Member Posts: 4 ■□□□□□□□□□
Can anyone explain a method for fast calculations of valid host addresses within a subnet ?


255.255.224.0 gives us 16 nets with 4094 host per net.

Lets use 172.128.208.0 as the subnet id.

What is the Host Address Range?

Well, I know the first valid host on this subnet is
172.128.208.1

The way I currently calculate this would just to keep adding 255 until I hit 4094. ugh :)

any shortcuts ?

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    dave0212dave0212 Member Posts: 287
    Would that not give you 8 subnets with 8190 hosts??
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    seanrseanr Member Posts: 4 ■□□□□□□□□□
    dave0212 wrote:
    Would that not give you 8 subnets with 8190 hosts??

    icon_sad.gif I meant .240 sorry
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    dtlokeedtlokee Member Posts: 2,378 ■■■■□□□□□□
    If you are using a Class B address then:

    255.255.224.0 will give you 3 subnet bits and 13 host bits which is 8 subnets and 8190 hosts per subnet. To find the ranges you need to write out all the combinations of subnet bits, then put all "0" or all "1"'s to find the lower and upper bounds respectively.
    The only easy day was yesterday!
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    dave0212dave0212 Member Posts: 287
    I would say the easiest way to calculate quickly is

    16 subnets
    256/16 is 16 so your increment is 16

    Add 16 to the subnet octet and -2 to get your last host

    172.128.208.1 - 172.128.223.254


    That is how i do it in my head but i am pretty quick at mental arithmetic :D
    This week I have achieved unprecedented levels of unverifiable productivity


    Working on
    Learning Python and OSCP
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    seanrseanr Member Posts: 4 ■□□□□□□□□□
    Thanks dave0212, that is a really short method and I like it.
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    dave0212dave0212 Member Posts: 287
    seanr wrote:
    Thanks dave0212, that is a really short method and I like it.

    Glad it helps

    It got me through my CCNA icon_lol.gif
    This week I have achieved unprecedented levels of unverifiable productivity


    Working on
    Learning Python and OSCP
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    seanrseanr Member Posts: 4 ■□□□□□□□□□
    dtlokee wrote:
    If you are using a Class B address then:

    255.255.224.0 will give you 3 subnet bits and 13 host bits which is 8 subnets and 8190 hosts per subnet. To find the ranges you need to write out all the combinations of subnet bits, then put all "0" or all "1"'s to find the lower and upper bounds respectively.

    can you give an example of this ? and off topic , can you use a calculator on the exams ?
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    laidbackfreaklaidbackfreak Member Posts: 991
    seanr wrote:
    and off topic , can you use a calculator on the exams ?

    nope no calculators allowed...... all you get is a wipe board n marker pen....
    if I say something that can be taken one of two ways and one of them offends, I usually mean the other one :-)
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