Section 8

• A problem with classful addresses was that if a company had more than 254 hosts they would need to be assigned a Class B network.
• They would have much less than the 65534 hosts allocated, so this was a huge amount of the global address space being wasted.

CIDR removed the fixed /8, /16 and /24 requirements for the address classes, and allowed them to be split into smaller networks, like 175.10.0.0/20 . This also restricts issues to the local part of the network and reduces CPU load.

Let’s say that you work for a company with 4 departments in two different offices. You could buy four different Class C IP ranges, or buy a single one and subnet it.

To subnet the network into smaller subnets, we need to “borrow” host bits and add them to the network portion of the address, moving the network address line to the right.

Let’s say we’ve been allocated Class C 200.15.10.0/24. To calculate the number of available subnets, the formula is $2^{subnet-bits}$. On the other hand, to get the number of available hosts we need to make $2^{host-bits}-2$. We subtract 2 because the network address and broadcast address cannot be assigned to hosts.

As we subtract 2 bits from the host address, we used to subtract two from the network address. This wasn’t really practical and is not done anymore. In CISCO routers, the command ip subnet-zero overrides this behavior and is enabled by default.

Following the same example, we’ll use a submask of /31, which is the max you can use. This borrows 7 bits out of 8, and gives us 128 subnets with 2 hosts each. This breaks the breaks the standard rules for IP addressing, but it’s supported for point to point link, which don’t need a broadcast address.

We now use a /30 subnet mask, or 255.255.255.252. This let’s us have 2 bits per ahost address, meaning: $2^2-2=2$. It borrows 6 bits for the network address, meaning we got $2^6=64$.

We see that the /31 is useful if you need to maximise use of your address space, but /30 is more standard and used.

With this setup we get $2^3-2=6$ for the hosts, and $2^5=32$ for the network address. This means that valid host addresses are going to be:

• 200.15.10.1 to 200.15.10.6 (0 for network and 7 for broadcast).
• 200.15.10.9 to 200.15.10.14 (8 for network and 15 for broadcast).

1. What are the network address, broadcast address, and valid host addresses for the IP address 192.22.45.173/26?
2. What’s the subnet mask in the dotted decimal notation?

The number of host bits is six, so we got $2^6 -2=62$ host addresses. Going from 0, we got:

• .1 to .62 (.0 network and .63 broadcast).
• .65 to .126 (.64 network and .127 broadcast).
• .129 to .190 (.128 network and .191 broadcast).

The subnet mask borrowed the last two bits in the last octet, so we got: $128+64=192$. Then, the submask is 255.255.255.192.

Early routing protocols only supported fixed length subnet masking, where all subnets had to be the same size. Now, you can have subnets of different sizes according to how many hosts you have.

Subnetting considerations:

• How many locations do we have in the network?
• How many hosts are in each location?
• What are the IP addressing requirements for each location?
• What size is appropriate for each subnet?

For designing the subnets, you first find the largest segment and allocate a suitable subnet size for it and the start of the address space, then rinse and repeat.

Let’s say we got an office in New York with two departments: Engineering, with 28 hosts, and Sales with 14. Then we got Boston with other two departments: Engineering with 28 hosts, and Sales with 7. Design the subnets for this setting.

For the Eng. Dept. we can calculate that the submask is going to be:

$$ x = \frac{log(30)}{log(2)} \approx 5 $$

Where $x$ is the bit size of the host address. This means we can have a submask of /27 or 255.255.255.224. It’ll leave us with:

$$ \text{hosts }= 2^5 - 2 = 30 $$

We’ll get these subnets for both Eng. Dept.:

• .1 to .30 (.0 and .31 reserved) for NY Eng Dept.
• .33 to .62 (.32 and .63 reserved) for Boston Eng Dept.

The next largest subnet is NY sales with 14. It’ll require the following bit size for the host address:

$$ x = \frac{log(16)}{log(2)} = 4 $$

With this, we know the subnet mask is /28 or 255.255.255.240, which means 16 addresses. We’ll have:

• .65 to .78 (.64 and .79 reserved) for the NY Sales Dept.

For departments, we still need a subnet for the Boston Sales Dept. We got:

$$ x = \frac{log(9)}{log(2)} \approx 4 $$

With this, we know the subnet mask is /28 or 255.255.255.240, which means 16 addresses. We’ll have:

• .81 to .94 (.80 and .95 reserved) for the Boston Sales Dept.

Lastly, we need a link between both New York and Boston. As we saw, we can do it with a /30 or 255.255.255.252 subnet mask. We’ll have it as:

• .97 to 99 (.96 and .100 reserved) for the link between offices.

If we subnet an Class B on the 4 Octet into /29, we got 3 bits for host addressing. This yields 6 hosts per network, but as we have a Class B /16 address range, we have 13 bits for the network address, meaning we can have 8192 subnets.

For the IP address 135.15.10.138/29, which would be the network address, broadcast address and range of valid IP addresses?

For this, we know addresses go from 8 to 8, so one will begin in 0, 8, 16, etc. We can see that the IP address for that range is going to be starting from 135.15.10.136, so we can get:

• Network address: 135.15.10.136.
• Broadcast address: 135.15.10.143.
• Valid IP addresses: 135.15.10.137-142.

You’ve been asked to subnet the 134.65.0.0 network into six different networks. Which subnet mask do you use?

We see that we need at least 3 bits, so knowing a Class B has as default /16, I’d use /19.

If we have a Class A 60.0.0.0/8, and apply the subnet mask 255.255.255.240, how many subnets we have an how many hosts per subnet?

We can see that $128+64+32+16=240$, so we know that the network bit is 4 and the host bit 4 for the last octet. It means we have $2^{28-8}=1048576$ subnets with $2^4-2=14$ hosts each.

For the IP address 60.15.10.75/28, what’s the network, broadcast and host addresses?

We can see that we got 4 bits for the hosts. This means we get $2^4=16$ block of addresses. We can see that we’ll have the block starting at 64. Then:

• Network address: 60.15.10.64.
• Broadcast address: 60.15.10.79.
• Valid IP addresses:60.15.10.65-78.

We’re allocated Class A 60.0.0.0/8.

If we subnet this into /19 networks, how many subnets do we have, and how many hosts per subnet?

We know we got 13 bits for the host address, and we borrow 11 bits for the network address. We can get:

• Subnets: $2^11=2048$.
• Hosts per subnet: $2^13-2 = 8190$.

For the IP address 60.15.10.75/19, what are the important addresses?

We are working on the third octet. From this, we can see we’ll subtract the bits from the first octet: $19-8=11$. From this, we know that from the 16 bits, 11 will be from the network bits. The last 8 will be from the second octet, so we can see that we got 3 bits from the network in the third octet. The rest are going to be bits for the host addresses. This means we get $2^5=32$ host addresses per network. Then:

• Network address: 60.15.0.0.
• Broadcast address: 60.15.31.255.
• Host addresses: 60.15.1-31.0-255

• Given a network requirement of “x” amount of subnets and “y” amount of hosts per subnet, what network address and subnet mask should be used for each subnet?
• Given a particular IP address and subnet mask, calculate the subnet’s network address, the broadcast address and range of valid host IP addresses.

These were added in the RFC 1918 (request for comments), which specifies private IP address ranges which are not routable on the public internet.

These addresses were originally designed for hosts which should have no internet connectivity. The range are:

• 10.0.0.0 - 10.255.255.255 or 10.0.0.0/8.
• 172.16.0.0 - 172.31.255.255 or 172.16.0.0/12.
• 192.168.0.0-192.168.255.255 or 192.168.0.0/16.

This was the answer in the 90’s when the Internet authorities started to predict an address exhaustion. It uses a 128 bit address, compared to the IPv4 32 bit address. This allows to provide more than $7.9*10^{28}$ times as many addresses as IPv4.

The main problem is that there is not a seamless migration from IPv4 to IPv6, so the companies implemented NAT (network address translation) was implemented as a workaround, where an organization can use private IP address on their inside network, but still grant their hosts internet access by translating them to their outside public IP address, meaning many hosts on the inside can share a few or even a single IP public address.

You can check for more subnetting problems in:

• Subnetting Questions
• Subnetting