Question

Consider the router and the two attached subnets below (A and B). The number of hosts is also shown below. The subnets share the 23 high-order bits of the address space: 192.168.60.0/23 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 ?

Answer 1

The address space 192.168.60.0/23 is a private address space.

The address space allows for a maximum of 510 hosts.

Subnet A's CIDR notation: 192.168.60.0/24.

Broadcast address of subnet A: 192.168.60.255.

Starting address of subnet A: 192.168.60.1.

Ending address of subnet A: 192.168.60.254.

Subnet B's CIDR notation: 192.168.61.0/24.

Broadcast address of subnet B: 192.168.61.255.

Starting address of subnet B: 192.168.61.1.

Ending address of subnet B: 192.168.61.254.

How to explain

The given address space 192.168.60.0/23 is part of the private IP address range. The /23 subnet mask provides 510 usable host addresses.

To optimize address space usage and allow for additional subnets in the future, subnet A and subnet B are divided into /24 subnets.

Subnet A uses addresses ranging from 192.168.60.0 to 192.168.60.255, and subnet B uses addresses from 192.168.61.0 to 192.168.61.255, each allowing for 254 usable hosts within the subnet.

The Complete Question

Given a router and two attached subnets (A and B) sharing the 23 high-order bits of the address space (192.168.60.0/23), optimize the assignment of subnet addresses for minimal space usage and maximal available contiguous address space if a new subnet is added. Provide the CIDR notation for each subnet and answer the following questions:

Determine whether the address space is public or private.

Calculate the maximum number of hosts allowed in this address space.

Assign the subnet address of subnet A in CIDR notation.

Specify the broadcast address of subnet A.

Identify the starting address of subnet A.

Identify the ending address of subnet A.

Assign the subnet address of subnet B in CIDR notation.

Specify the broadcast address of subnet B.

Identify the starting address of subnet B.

Identify the ending address of subnet B.

Answer 2

The given address space is 192.168.60.0/23. It is a private address space and can accommodate 512 hosts. The subnet addresses, broadcast addresses, and starting and ending addresses for both subnet A and subnet B are provided.

Step-by-step explanation:

The given address space is 192.168.60.0/23.

1. The address space is private.

2. The number of hosts that can be accommodated in this address space is 512. This can be calculated by subtracting 2 from the total number of addresses in the subnet, where the total number of addresses is 2^(32-prefix length).

3. The subnet address of subnet A in CIDR notation is 192.168.60.0/24. This means that the first 24 bits are fixed and the remaining 8 bits can vary to accommodate different hosts.

4. The broadcast address of subnet A can be calculated by setting all the host bits to '1' in the subnet address. So, the broadcast address of subnet A is 192.168.60.255.

5. The starting address of subnet A is the subnet address itself. So, the starting address of subnet A is 192.168.60.0.

6. The ending address of subnet A can be calculated by setting all the host bits to '1' and subtracting 1 from the resulting address. So, the ending address of subnet A is 192.168.60.254.

7. The subnet address of subnet B in CIDR notation is 192.168.61.0/24. Since the first 23 bits are the same as subnet A, the subnet address of subnet B differs in the 24th bit.

8. The broadcast address of subnet B can be calculated similarly as in subnet A. So, the broadcast address of subnet B is 192.168.61.255.

9. The starting address of subnet B is the subnet address itself. So, the starting address of subnet B is 192.168.61.0.

10. The ending address of subnet B can be calculated similar to subnet A. So, the ending address of subnet B is 192.168.61.254.