Question

Consider a router that interconnects IPv4 three subnets: Subnet 1, Subnet 2, Subnet 3. Suppose all of the interfaces in each of these three subnets are required to have the prefix 172.12 /16. Also suppose that Subnet 1 is required to support at least 30000 interfaces, Subnet 2 is to support at least 2000 interfaces, and Subnet 3 is to support at least 10000 interfaces. Provide three network addresses (of the form a.b.c.d/x) that satisfy these constraints

Answer 1

The address given to us is 172.12.0.0 /16, which means we have 16 bits for the host id.

Further, when we have to divide this into 3 subnets with the given configuration, we will first arrange the subnets in descending order based on the number of interfaces required. So we will have

Subnet 1 with 30000 interfaces,

Subnet 3 with 10000 interfaces, and

Subnet 2 with 2000 interfaces,

So now, we have the order in which we will create the subnets. We will divide the given address into our 3 subnets in the following manner,

1. For subnet 1

The number of interfaces required = 30000

The number of bits required to allocate an address to these interfaces = 2¹⁵.

Therefore, out of 16 host id bits, 15 bits will be allocated to Subnet 1 itself.

We will allocate the first bit of the third byte to (0)₂ for this subnet. Consequently, we will get the IP addresses as

Initial IP address = 172.12.0.0 /17

Final IP address = 172.12.127.255 /17

2. For Subnet 3

The number of interfaces required = 10000

The number of bits required to allocate an address to these interfaces = 2¹⁴.

Therefore, out of the remaining 15 host id bits, 14 bits will be allocated to Subnet 2.

We will allocate the first two bits of the third byte to (10)₂ for this subnet. Consequently, we will get the IP addresses as

Initial IP address = 172.12.128.0 /18

Final IP address = 172.12.191.255 /18

3. For Subnet 2

The number of interfaces required = 2000

The number of bits required to allocate an address to these interfaces = 2¹¹.

Therefore, out of the remaining 14 host id bits, 11 bits will be allocated to Subnet 2.

We will allocate the first five bits of the third byte to (11000)₂ for this subnet. Consequently, we will get the IP addresses as

Initial IP address = 172.12.192.0 /21

Final IP address = 172.12.199.255 /21

Therefore, the three network addresses are:-

172.12.0.0 /17 for Subnet 1

172.12.192.0 /21 for Subnet 2

172.12.128.0 /18 for Subnet 3