1. Determine the class of your network.

This is taken from the first octet in the IP address. IPs are separated into classes. Like Texas Chili.

Octet Range

ClassDefault Subnet Mask

Private Addressing

Notes

1 - 126* Class A 255.0.0.0 10.x.x.xUsed by really high-level companies with a lot of computers

*127.0.0.1 LoopbackUsed to test IP connectivity to self

128 - 191 Class B 255.255.0.0172.16.x.x - 172.31.x.x

Used by larger companies, I see a lot of this in mid-level networks

192 - 223 Class C 255.255.255.0192.168.x.x - 192.168.255.x

Used by small companies, home offices, small home networks

224 - 239 Class D n/a n/a Multicast Broadcast only

240 - 255 Class E n/a n/aExperimental: Reserved for future use (I don't know what the hell that means, either)

What is "private addressing?" Well, each class has a separate set of addresses that everyone has agreed on are "not Internet routable." This is very handy for firewalls and NAT. See, there aren't enough IP addresses to go around any more. So you have "outside IPs" (which are most of them), and "inside IPs" which are the private addresses. The most common you will see for home and small office networks are the 192.168.x.x variety. Because this class C address range has a default mask of 255.255.255.0, you can really only have 254 hosts per subnet

2. Determine how many bits you must borrow from the hosts portion.

Use the following equation:

2n - 2 >= Number of subnets required

So if you need 25 subnets:

1. 2n - 2 >= 25 2. 2n - 2 >= 25 3. 25 - 2 >= 25 4. 32 - 2 >= 25

If n = 4 then it would have been 16, less than 25. People would laugh at you. And point.

Note: At this point, you should about expandability. If you are going to need more than 30 (25-2 )sub-networks in the future, you should go one greater, like 62 (26-2). Remember, the more networks you have, the less collision domains you have. But you will need routers to get them to speak to each other. That's another topic for a later time, but keep in mind if you have TOO many hosts all working on the same subnet (like over 100 servers all talking at once), you are going to slow your network down.

3. Make your custom subnet mask.

First, you need to determine your default mask above. To illustrate this better, let's toss our ten-digit fingers out the window, and think of them as nothing more than nonsense labels for the moment. Computers don't have fingers or toes, only switches.

Suppose you had this address:

192.168.35.xIn binary, it looks like this:

11000000.10101000.00100011.x

Looking at the default subnet mask:

255.255.255.0Which the computer sees as:

11111111.11111111.11111111.00000000

Now since n = 5 in the example above, we know we must steal five bits, which is done from left to right

255.255.255.248

This is easier to see in binary:

11111111.11111111.11111111.11111000

As you can see, this makes less sense in our decimal world. "Huh? 248? WTF?!" Easy, tiger. This is simpler than it looks. See, each "bit" from left to right represents a factor of 2some power. Here's how we break down an octet:

Power of: 27 26 25 24 23 22 21 20

Decimal Number 128 64 32 16 8 4 2 1

Binary Number 1 1 1 1 1 0 0 0

So 5 bits stolen is really 128 + 64 + 32 + 16 + 8 = 248 (in decimal). If you think this is fun, wait until I write about Binary to Hexadecimal! But now we are getting off topic again. Focus. Before going to step 4, go and get yourself a cookie and a drink of water. I'll wait.

4. Find out your hosts per network.

Okay, you have the new subnet mask, now what? "What the heck are my goddamn IP address ranges???" Whoa, too much caffeine, there, dude. But rest assured, even with too much Mountain Dew sloshing in your belly, you can do this. First, we have to determine the number of hosts per subnet. This is also better understood in binary:

Now, remember your default mask? Unless you have ADHD, you know it as:

255.255.255.248

But again, it's easie

11111111.11111111.11111111.11111000

r to see in binary

Since the 11111111.11111111.11111111.11111xxx is the network portion, then the remaining bits (which are zeros) are the hosts portion!

Forget decimal for now

Hosts (oooh):

11111111.11111111.11111111.11111000

Now, I may not be a bright man, but even I can tell there are only 3 bits left. So we use a familiar looking equation again to determine how many hosts (anything that needs an IP address to function) we have for each network.

2n - 2 >= Number of hosts per subnet

So in this case, 23 - 2 = 8 - 2 = 6 hosts per subnet. Wow, that's not a whole lot. But you do have a Class C address, and you were the bozo that wanted 25 networks.

5. Now we determine the subnet ranges!

Bet you thought we'd never get here. There is an incrementing interval we find find out first. That's EASY. It's:

256 - subnet mask = Interval

So, in our example, 256 - 248 = 8! Well, whattia know! This will also be the value of the lowest network, unless you are including zero networks, but for this case, we won't. So, now we start moving the tables around.

Subnet Subnet AddressFirst useable IP Address

Last useable IP address

Broadcast Address

First 192.168.35.8 192.168.35.9 192.168.35.14 192.168.35.15Second 192.168.35.16 192.168.35.17 192.168.35.22 192.168.35.23Third 192.168.35.24 192.168.35.25 192.168.35.30 192.168.35.31Fourth 192.168.35.32 192.168.35.33 192.168.35.38 192.168.35.39... 30th 192.168.35.240 192.168.35.241 192.168.35.246 192.168.35.247

Note we didn't use the 192.168.35.0 or 192.168.35.248 networks. We usually discard the first and last (which is why there are only 30 instead of 32 available in the equation above). This makes sense because the first would have 192.168.35.0 as a subnet, when it's really considered the name of a whole network. And 192.168.35.248 would have 192.168.35.255 as a broadcast address for a subnet when it should be the default broadcast for the whole 192.168.35.0 network. This will confuse people later on. But I'd be a liar in a Viking hat if I said that it hasn't been done before. It's just not a good IDEA. Many subnet calculators out there also include these networks, usually called zero networks.

As a side note, how many hosts SHOULD you have per subnet? Well, the rule of thumb is that you shouldn't have more than 254 hosts in the same subnet because they will all share the same collision domain. Your network will slow to a crawl with all that traffic. Keep in mind, a lot of traffic isn't just a

machine accessing the Internet, a lot of services will constantly send traffic back and forth. If you have ever sniffed traffic on a closed network, even "idle" computers are sending info back and forth, espically if you haven't firewalled them and you have default services enabled. If you have 1000 hosts on the same subent, even if the machines are idle, you will notice some machines will start to slow down as the NIC or software tries to process all the messages.

A lot of people have /24 networks to make sure this doesn't happen, which is 254 hosts, and /24 is an "even break point" of one octet that's easier to do quick math. Routers/firewalls and managed switches (like VLANS) will take care of any translation from subnet to subnet.

More ExamplesWe're going to go through more examples now, so you can understand more.

Example 1

Problem: Given IP range of 10.x.x.x, and you have to create 10 subnets.

1. You have a Class A address. Default mask is 255.0.0.0 2. 24 - 2 = 14, which is > 10. So we need to steal 4 bits from the host portion. 3. That means our subnet mask is 255.(128+64+32+16).0.0 = 255.240.0.0 4. Our host bits left are 20, so 220 - 2 = 1,048,574 hosts per net (that ought to be enough, in fact,

way too many but hey) 5. Our digital interval is 256 - 240 or 16. So we know the first network is 10.16.0.0! You wouldn't

want to use it, since it's a zero network, but I'll include it anyway to show you where it goes.

SubnetSubnet Address

First useable IP Address

Last useable IP address

Broadcast Address

First 10.16.0.0 10.16.0.1 10.31.255.254 10.31.255.255Second 10.32.0.0 10.32.0.1 10.47.255.254 10.47.255.255Third 10.48.0.0 10.48.0.1 10.63.255.254 10.63.255.255...14th 10.224.0.0 10.224.0.1 10.239.255.254 10.239.255.255

Example 2

Problem: Given IP range of 129.50.x.x, and you have to create subnets for each state in the US.

1. You have a Class B address. Default mask is 255.255.0.0 2. 26 - 2 = 62, which is > 50 states. So we need to steal 6 bits from the host portion. 3. That means our subnet mask is 255.255.(128+64+32+16+8+4).0 = 255.255.252.0 4. Our host bits left are 10, so 210 - 2 = 1,024 hosts per net. 5. Our interval is 256 - 252 or 4. So we know the first network is 129.50.4.0!

SubnetSubnet Address

First useable IP Address

Last useable IP address

Broadcast Address

First 129.50.4.0 129.50.4.1 129.50.7.254 129.50.7.255Second 129.50.8.0 129.50.8.1 129.50.11.254 129.50.11.255Third 129.50.12.0 129.50.12.1 129.50.15.254 129.50.15.255...50th 129.50.200.0 129.50.200.1 129.50.203.254 129.50.203.255

...62nd 129.50.248.0 129.50.248.1 129.50.251.254 129.50.251.255

6. You have 12 subnets left over for expansions into Guam, Puerto Rico, and the US Virgin Islands.

Example 3

Problem: You are a poor contractor, signed up for a great contract with the British Government to set up 2 monitoring stations in each country in the whole world. You got the IP range of 222.100.42.x. Can you do it with separate networks for each country?

1. You have a Class C address. Default mask is 255.255.255.0 2. 28 - 2 = 254, which is > 192 countries. So we need to steal 8 bits from the host portion. 3. That means our subnet mask is 255.255.255.(128+64+32+16+8+4+2+1) = 255.255.255.255 4. Our host bits left are 0, so no can do!

Problem: How about just the countries in Europe?

1. You have a Class C address. Default mask is 255.255.255.0 2. 26 - 2 = 62, which is > 44 European countries (as of this writing). So we need to steal 6 bits

from the host portion. 3. That means our subnet mask is 255.255.255.(128+64+32+16+8+4) = 255.255.255.252 4. Our host bits left are 2, so 22 - 2 = 2 hosts per net... just enough! 5. Our interval is 256 - 252 or 4. So we know the first network is 222.100.42.4!

SubnetSubnet Address

First useable IP Address

Last useable IP address

Broadcast Address

Albania 222.100.42.4 222.100.42.5 222.100.42.6 222.100.42.7Andorra 222.100.42.8 222.100.42.9 222.100.42.10 222.100.42.11Austria 222.100.42.12 222.100.42.13 222.100.42.14 222.100.42.15...Serbia/Montenegro 222.100.42.176 222.100.42.177 222.100.42.178 222.100.42.179...62nd 222.100.42.248 222.100.42.249 222.100.42.250 222.100.42.251

6. You have 12 subnets left over for expansions into some of the Middle Eastern regions.

See, it's not that hard, once you think in binary. Some classes you take might ask you to think in HEX, which is like having 10 fingers and using all the toes on one foot. They go 0,1,2,3...9 and then A,B,C,D,E, and F for 10-15. "Hexadecimal" means "six and ten" or 16, which is 0-15 (remember, computers always start with 0 when counting). In some of the old configs I have seen on Sun/Solaris boxes, I have seen the netmask as ff.ff.ff.C0. That's really 255.255.255.192, because ff = 16 x 16 numbers, or 256, which is 0-255 in range.

And you now know what that is in binary, right? ;-)

Who is the author? Grig "Punkie" Larson is listed elsewhere on this site, but as of 2009, he's a Systems Administrator at a major ISP in Maryland, with a LPIC Level 1, RHCT, and is a Network+ Certified fool. He failed his CCNA at 64%. Dooooh! Who wrote that goddamn test? Retarded Elbonians? But that was in 2001. Time to stop living in the past. Maybe I'll get one in 2009.

IP Address / Route Summarization Example #2.From the previous page, you must have know that IP route summarization can also be referred to as

route aggregation. It helps reduce the number of routing entries in a router IP address routing table for faster lookup of destination.

Let’s look at the example below:

Let’s try and summarize Network 10.1.0.0 through 10.5.0.0. 

First, list everything into binary:

Firstly, to get the network address, follow and match the binary bits, starting on the left and stop where the bits do not match from the example above.

Notice that the first octet are matched, the second octet has no matching bits on, so is the third and last.

so, the summary IP will be 10.0.0.0 = Network address.

Finally, to work out the summary subnets mask; we match the 8 bits of the first octet (see above) which is the network, and five matching zeros in the second octet which is the subnet. 

255.248.0.0

How did we get the 248…?

Remember the bits value = 128  64  32  16  8  4  2  1

                                            0    0    0    0   0  0  0   0

You add the five bits values in the second octet from the left ; 128+64+32+16+8 = 248

How did we get /13…?

Count all the matching bits (see above) from the left up to the last matching bit…kazam!...you get your CIDR .