Interactive Problems, Computer Networking: A Top Down Approach

Subnet Addressing

Consider the router and the three attached subnets below (A, B, and C). The number of hosts is also shown below. The subnets share the 24 high-order bits of the address space: 172.22.36.0/24

Assign subnet addresses to each of the subnets (A, B, and C) 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.

Solution

1. The address 172.22.36.0/24 is private.

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 23 hosts, so it will need at least 25 addresses (for the subnet ID and broadcast address). The least number of bits that satisfy this is 5 bits. Knowing that, we take the prior subnet and add 32, the result of which is 172.22.36.192/27

4. The broadcast address of subnet A (172.22.36.192/27) is 172.22.36.223, because it is the last address in the IP range.

5. The first IP address of subnet A (172.22.36.192/27) is 172.22.36.193, found by adding 1 to the subnet address.

6. The last IP address of subnet A (172.22.36.192/27) is 172.22.36.222, found by subtracting 1 from the broadcast address (172.22.36.223).

7. Similar to the prior subnet, subnet B has 121 hosts, so it will need at least 123 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 172.22.36.0/25

8. The broadcast address of subnet B (172.22.36.0/25) is 172.22.36.127, because it is the last address in the IP range.

9. The first IP address of subnet B (172.22.36.0/25) is 172.22.36.1, found by adding 1 to the subnet address.

10. The last IP address of subnet B (172.22.36.0/25) is 172.22.36.126, found by subtracting 1 from the broadcast address (172.22.36.127).

11. Similar to the prior two subnets, subnet C has 43 hosts, so it will need at least 45 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 172.22.36.128/26

12. The broadcast address of subnet C (172.22.36.128/26) is 172.22.36.191, because it is the last address in the IP range.

13. The first IP address of subnet C (172.22.36.128/26) is 172.22.36.129, found by adding 1 to the subnet address.

14. The last IP address of subnet C (172.22.36.128/26) is 172.22.36.190, found by subtracting 1 from the broadcast address (172.22.36.191).

Interactive end-of-chapter exercises

Subnet Addressing

Question List

Solution

Question 1 of 14

Question 2 of 14

Question 3 of 14

Question 4 of 14

Question 5 of 14

Question 6 of 14

Question 7 of 14

Question 8 of 14

Question 9 of 14

Question 10 of 14

Question 11 of 14

Question 12 of 14

Question 13 of 14

Question 14 of 14