Valid host address

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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Comments

dave0212

Would that not give you 8 subnets with 8190 hosts??

seanr

dave0212 wrote:

Would that not give you 8 subnets with 8190 hosts??

I meant .240 sorry

dtlokee

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.

dave0212

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

seanr

Thanks dave0212, that is a really short method and I like it.

dave0212

seanr wrote:

Thanks dave0212, that is a really short method and I like it.

Glad it helps

It got me through my CCNA

seanr

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 ?

laidbackfreak

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....