### Subnet Addressing

Consider the router and the two attached subnets below (A and B). The number of hosts is also shown below. The subnets share the 24 high-order bits of the address space: 134.174.58.0/24

Assign subnet addresses to each of the subnets (A and B) so that the amount of address space assigned is minimal, and at the same time leaving the largest possible contiguous address space available for assignment if a new subnet were to be added. Then answer the questions below.

### Question List

1. Is the address space public or private?

2. How many hosts can there be in this address space?

3. What is the subnet address of subnet A? (CIDR notation)

4. What is the broadcast address of subnet A?

5. What is the starting address of subnet A?

6. What is the ending address of subnet A?

7. What is the subnet address of subnet B? (CIDR notation)

8. What is the broadcast address of subnet B?

9. What is the starting address of subnet B?

10. What is the ending address of subnet B?

### Solution

1. The address 134.174.58.0/24 is public.

2. Maximum number of hosts = 2^x - 2 = 2^8 - 2 = 254. The reason we have to subtract 2 from the final number is because there are always 2 addresses allocated for each address block: the subnet ID (the first address) and the broadcast address (the last address); for example, if you have 5 bits for hosts, you can have 30 hosts, because 2 of the addresses are for the subnet ID and the broadcast address which when added equals 32, which is 2^5.

3. Subnet A has 125 hosts, so it will need at least 127 addresses (for the subnet ID and broadcast address). The least number of bits that satisfy this is 7 bits. Knowing that, we take the prior subnet and add 128, the result of which is 134.174.58.0/25

4. The broadcast address of subnet A (134.174.58.0/25) is 134.174.58.127, because it is the last address in the IP range.

5. The first IP address of subnet A (134.174.58.0/25) is 134.174.58.1, found by adding 1 to the subnet address.

6. The last IP address of subnet A (134.174.58.0/25) is 134.174.58.126, found by subtracting 1 from the broadcast address (134.174.58.127).

7. Similar to the prior subnet, subnet B has 50 hosts, so it will need at least 52 addresses (for the subnet ID and broadcast address). The least number of bits that satisfy this is 6 bits. Knowing that, we take the prior subnet and add 64, the result of which is 134.174.58.128/26

8. The broadcast address of subnet B (134.174.58.128/26) is 134.174.58.191, because it is the last address in the IP range.

9. The first IP address of subnet B (134.174.58.128/26) is 134.174.58.129, found by adding 1 to the subnet address.

10. The last IP address of subnet B (134.174.58.128/26) is 134.174.58.190, found by subtracting 1 from the broadcast address (134.174.58.191).

That's incorrect

That's correct

The answer was: public

The answer was: 254

The answer was: 134.174.58.0/25

The answer was: 134.174.58.127

The answer was: 134.174.58.1

The answer was: 134.174.58.126

The answer was: 134.174.58.128/26

The answer was: 134.174.58.191

The answer was: 134.174.58.129

The answer was: 134.174.58.190