Choosing IP Addresses

Hey all ...

I have a strange question ...

When practicing subnetting or if you have to set up a network from scratch for a company (private network like departments within a company) ... How do you choose a network ...

Do you stick to a certain class C network or do you pick your favourite class C and and have fun from there ...

What I mean is that class C is 192 and above in the first octet and you need to have the first 3 octets reserved (255.255.255.*)

With this in mind ... how do you choose between saying 192 or 216 ... etc ...

(really hope this makes sense)

All information would be greatly appreciated.
Regards.

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Comments

krjay

Well I choose 192.168 because it's a reserved private address.

More specifically though: Usually I've found the questions I'm practicing for CCENT to give you the network and ask you to subnet it such as this:

"You are designing a subnet mask for the 172.17.0.0 network. You want 110 subnets with up to 300 hosts on each subnet. Which mask do you choose?"

Emissary_of_Pain

What about the 2nd and 3rd octets ? ...

Isn't anything between 192 and 223 considered reserved ?

networker050184

If you want to use anything besides RFC-1918 space you need to have it allocated from your RIR, which is AfriNIC in Africa.

Most likely you will be using RFC-1918 IPv4 space within your organization though so you can pretty much pick what ever you want in that range. How you pick it really depends on how your network is going to be set up. How many locations and their size, VLANs used etc.

The RFC-1918 space is:

10/8
172.16/12
192.168/16

Emissary_of_Pain

Ok so lets say you need to set up a network from scratch for a company ...

They are going to have 6 Departments each with multiple hosts in each department ... how do you choose which IP to use (lets stick to class C)

Sorry if this all sounds silly but I am used to having the network already in existence and just working with the already determined IP range

networker050184

How big are these six departments?

Personally I'd start like this.

10.1/16 to site 1
10.2/16 to site 2
...
...
10.255/16 to site 255

Now you have a /16 for each site to work with. Then you can start breaking your subnets out of that /16 depending on needs of host per subnet. Say you have three VLANs at site 1 all needing a /24.

VLAN1 - 10.1.0.0/24
VLAN2 - 10.1.1.0/24
VLAN3 - 10.1.2.0/24

krjay

Emissary_of_Pain wrote: »

Ok so lets say you need to set up a network from scratch for a company ...

They are going to have 6 Departments each with multiple hosts in each department ... how do you choose which IP to use (lets stick to class C)

Sorry if this all sounds silly but I am used to having the network already in existence and just working with the already determined IP range

You should probably know a rough estimate of how many hosts per subnet before you are dead set on a class C

sratakhin

It's better to start designing a new network in 10.x.x.x range, unless you are absolutely sure that it will never need more hosts per subnet than a class C network provides.

Emissary_of_Pain

Ah, that makes sense ... thank you ! ...

The way I am practicing (in packet tracer) is that my GF or who ever gives me a random company name and the departments with hosts in each ...

Example:

Metal Car Spares

6 Departments :
Sales -2 networks in sales of 20 hosts and 7 hosts
HR - 15 hosts
Data Capture - 8 Hosts
Accounts - 4 Hosts
Reception - 2 hosts
Cashiers - 3 hosts

1 centralized Mail/File server with all departments having Internet Access

Each department must have expansion room to double in size

d6bmg

Emissary_of_Pain wrote: »

Ah, that makes sense ... thank you ! ...

The way I am practicing (in packet tracer) is that my GF or who ever gives me a random company name and the departments with hosts in each ...

Example:

Metal Car Spares

6 Departments :
Sales -2 networks in sales of 20 hosts and 7 hosts
HR - 15 hosts
Data Capture - 8 Hosts
Accounts - 4 Hosts
Reception - 2 hosts
Cashiers - 3 hosts

1 centralized Mail/File server with all departments having Internet Access

Each department must have expansion room to double in size

Max 28 hosts per dept with the scope to double it?
Using the following convention would work:
192.168.1.0/26
192.168.1.63/26
192.168.1.127/26
192.168.1.191/26
192.168.2.0/26
192.168.2.63/26

will work with basic VLSM. Or you can optimize it for every network out there.
Or you can use any address space from RFC-1918

sratakhin

That's not VLSM

What's the point of giving the cashiers 'department' 62 hosts if they only need 3?

Emissary_of_Pain

I agree ... that is a massive waste of host space ... I know that you should plan for growth but that is a bit excessive considering I was told to allow for double growth ...

These are the subnets I landed up going with ...

Network 1 : 192.168.5.0 - .31/27
: 192.168.5.32 - .63/27
Network 2 : 192.168.5.64 - .95/27
Network 3 : 192.168.5.96 - .127/27
Network 4 : 192.168.5.128 - .143/28
Network 5 : 192.168.5.144 - .147/30
Network 6 : 192.168.5.148 - .155/29

Net 7 : 192.168.5.156 - .159/30
Net 8 : 192.168.5.160 - .163/30
Net 9 : 192.168.5.164 - .167/30
Net 10 : 192.168.5.168 - .171/30
Net 11 : 192.168.5.172 - .175/30
Net 12 : 192.168.5.176 - .179/30

Emissary_of_Pain

Random Question ... busy keying in the IP addresses into Packet Tracer and I ran into a very strange problem ...

Network 6 - 192.168.5.148 - 192.168.5.155/29

Now to my understanding I have : .149 for the fa0/0 on the router ... 1 host is .150 and it is fine

BUT

When I try key in the next host as either .151 or .152 it says that the IP is invalid for the subnet range ...

WHY ?!?!?!?! ... ... .153 works perfectly so why wont 151 or 152 ?

Diggs

A /29 (255.255.255.24

subnet mask give you a block of 8 IP addresses, one of which is used for broadcasts and the other the network ID.

192.168.5.0
192.168.5.8
192.168.5.16
...

ptilsen

Emissary, the part you are missing about VLSM is that it's only "variable" to the extent that you can define lengths. You cannot arbitrarily set start points of subnets based on the end point of the previous subnet without regard to length. No matter how your subnets are configured, 192.168.5.153/29 is interpreted as being a host in subnet 198.168.5.152/29, with .152 being the node address and .151 being the broadcast address of the previous subnet (if it's configured properly).

You cannot have network subnets like this:
1. /27
2. /28
3. /30
4. /29

The interpretation of the subnet is set in stone by the mask. You don't get to define start points arbitrarily. If you use 192.168.5.13/29, for example, it will be interpreted as 192.168.5.8/29.

So, in your network, you have a problem between network 5 and 6. Network 4, a /28, can be followed by two /29s, four /30s, etc. or by one /29, then two /30s. It cannot be followed by a /30 and then a /29 at the IP after the end of the /30. Either you leave a /30 unused between Network 5 and 6, or you put your larger subnet first.

Emissary_of_Pain

So would I just "abandon" Network 5's IP that I assigned and create a Network "13" with a /30 range ? ... would that work then ? ...

Is it conflicting because network 5 is already on the network ? ...

(see attached image for network layout using packet tracer)

Untitled.jpg

d6bmg

sratakhin wrote: »

That's not VLSM What's the point of giving the cashiers 'department' 62 hosts if they only need 3?

I didn't go for VLSM, as I think OP is preparing for CCNET and VLSM isn't in CCNET syll.
Or I have made a wrong assumption..

Emissary_of_Pain

I am preparing for my CCNA actually ...

pdosev

Hi
The first thing you need to do is to start with the largest to the smallest network. In your case the subnets are:

hosts network address
40 192.168.0.0 -192.168.0.63 / 26
30 192.168.0.64 – 192.168.0.95 /27
16 192.168.0.96 – 192.168.0.127 / 27
14 192.168.0.128 – 192.168.0.143 / 28
8 192.168.0.144 - 192.168.0.159 / 28
6 192.168.0.160 - 192.168.0.167 / 29
4 192.168.0.168 - 192.168.0.175 / 29

But you can't use 192.168.0.40 - 192.168.0.56 /28 because you have 4 bits for host that means each subnet is having 16 spare addresses(14 host space+1 broadcast and network adr) so in this case they are members of either 192.168.0.32 or 192.168.0.48 /28 networks.
Hope I've helped