Easy Peasy Subnet/Client Question, Help Though

[Daff_Duck]

Say I had a subnet of 255.255.255.0/30

The way I understand it to be is that I would have 6 open nodes and 2 closed

And the equation I have been given is 6 squared =36 -2 =34 providing 34 subnets

As for the clients well there are 2 closed nodes and 2 squared is 4 -2 =2 clients thus there are 2 clients to every node... Can someone take a look at this and cofirm if I am right or if it isnt then can you elaborate? At present I as mentioned in a previous post I am studying the MTA Network Admin course and the abbreviation of network subnets is causing a small problem for me.

Thanks to anyone who takes the time to look and point me in the right direction.

 

[Mr.Stu]

255.255.255.0 is a /24

255.255.255.252 is /30

192.168.1.0/30 would give you:

192.168.1.0 network address
192.168.1.1 first useable host
192.168.1.2 second useable host
192.168.1.3 broadcast address

Useable hosts = (2^#hostbits) - 2

Subtract additional one if you need a default gateway.

 

[Daff_Duck]

255.255.255.0 is a /24

255.255.255.252 is /30

192.168.1.0/30 would give you:

192.168.1.0 network address
192.168.1.1 first useable host
192.168.1.2 second useable host
192.168.1.3 broadcast address

Useable hosts = (2^#hostbits) - 2

Subtract additional one if you need a default gateway.

Thanks for the info but I was looking to work out the total number of subnets and hosts for the /30 (mainly to see if I was on the right path)

 

[snowbhoy]

http://www.skullbox.net/subnetcalculator.php if you want a cheat sheet.

All down to bits, when we talk of a /xx that's the number of bits of subnet, the remaining bits left from the 32 available are for the host (So a /30 leaves 2 bits of host addresses)

Every subnet has to have a value for the network (First address in range) and a value for broadcast in that network (Last value in range)

/32 has no remaining bits, so it's a host address (Special case)
/31 has 1 bit available, with two possible values - 0 or 1. But we need one for network & one for host means it's unusable (Well used to be, but now we can, but let's not complicate it)
/30 has 2 bits, 00, 01, 10, 11 so had four values available, take two off for network and broadcast means you have two usable host addresses.
/29 has 3 bits, eight possible values -2 (Network and broadcast) leaves 6 usable host addresses

etc etc

That make more sense?

 

[Daff_Duck]

http://www.skullbox.net/subnetcalculator.php if you want a cheat sheet.

All down to bits, when we talk of a /xx that's the number of bits of subnet, the remaining bits left from the 32 available are for the host (So a /30 leaves 2 bits of host addresses)

Every subnet has to have a value for the network (First address in range) and a value for broadcast in that network (Last value in range)

/32 has no remaining bits, so it's a host address (Special case)
/31 has 1 bit available, with two possible values - 0 or 1. But we need one for network & one for host means it's unusable (Well used to be, but now we can, but let's not complicate it)
/30 has 2 bits, 00, 01, 10, 11 so had four values available, take two off for network and broadcast means you have two usable host addresses.
/29 has 3 bits, eight possible values -2 (Network and broadcast) leaves 6 usable host addresses

etc etc

That make more sense?

Thank You! It makes perfect sense, ever thought of writing a book as you make a lot more sense than various web pages I have looked at. Once again cheers