CCNA IP Subnetting Questions Blog

CCNA IP Subnetting Questions

Subnetting is a crucial skill for network engineers and anyone pursuing Cisco Certified Network Associate (CCNA)
certification. It is important to have a solid understanding of IP subnetting concepts to design and manage
efficient and scalable networks. In this blog post, we will explore CCNA IP subnetting questions that will help
you practice and reinforce your knowledge in this area.

Question 1: Subnet Calculation

Given the IP address 192.168.10.0/24, calculate the number of subnet bits required to create 16
subnets. Also, determine the subnet mask for each resulting subnet.

Answer:

To create 16 subnets, we need to borrow 4 subnet bits from the host bits. This leaves us with 4 host bits.
The subnet mask for each resulting subnet will be 255.255.255.240. The subnets will be:

  • Subnet 1: 192.168.10.0/28
  • Subnet 2: 192.168.10.16/28
  • Subnet 3: 192.168.10.32/28
  • Subnet 16: 192.168.10.240/28

Question 2: VLSM Design

You are given the IP network 192.168.0.0/24. Design a Variable Length Subnet Mask (VLSM) scheme
that allows for the following subnets:

  • Subnet A: 40 hosts
  • Subnet B: 25 hosts
  • Subnet C: 10 hosts

Answer:

To accommodate the largest subnet (Subnet A) needing 40 hosts, we need to assign a /26 subnet mask (255.255.255.192)
to it. This allows for 62 usable IP addresses (2^6 – 2).

Subnet B requires 25 hosts, which fits within a /27 subnet (255.255.255.224). This allows for 30 usable IP addresses.

Finally, Subnet C requires only 10 hosts, which fits within a /28 subnet (255.255.255.240). This allows for 14 usable IP addresses.

The resulting subnets will be:

  • Subnet A: 192.168.0.0/26
  • Subnet B: 192.168.0.64/27
  • Subnet C: 192.168.0.96/28

Question 3: Supernetting

You have the following networks:

  • 192.168.10.0/24
  • 192.168.20.0/24
  • 192.168.30.0/24

Can you create a supernet that encompasses all these networks?

Answer:

Yes, we can create a supernet by combining the networks 192.168.10.0/24, 192.168.20.0/24,
and 192.168.30.0/24. To determine the supernet, we need to find the common network prefix:

The binary representation of the different networks are:

  • 192.168.10.0/24: 11000000.10101000.00001010.00000000
  • 192.168.20.0/24: 11000000.10101000.00010100.00000000
  • 192.168.30.0/24: 11000000.10101000.00011110.00000000

As we can see, the first 22 bits are common among all three network addresses. Therefore, the supernet will be
192.168.8.0/22.

Question 4: Classless Inter-Domain Routing (CIDR)

You are assigned the IP block 172.16.0.0/16 and need to allocate subnets to multiple departments.
The departments require the following number of hosts:

  • Department A: 1000 hosts
  • Department B: 500 hosts
  • Department C: 250 hosts

Design the subnet allocation for each department.

Answer:

For Department A, we need a subnet that can accommodate 1000 hosts. The closest subnet mask that fits this
requirement is /22 (255.255.252.0), which allows for 1022 usable IP addresses.

For Department B, we need a subnet that can accommodate 500 hosts. The closest subnet mask that fits this
requirement is /23 (255.255.254.0), which allows for 510 usable IP addresses.

For Department C, we need a subnet that can accommodate 250 hosts. The closest subnet mask that fits this
requirement is /24 (255.255.255.0), which allows for 254 usable IP addresses.

The resulting subnets will be:

  • Department A: 172.16.0.0/22
  • Department B: 172.16.4.0/23
  • Department C: 172.16.6.0/24

Question 5: IPv6 Subnetting

Given the IPv6 address 2001:0db8:85a3:0000:0000:8a2e:0370:7334/64, calculate the number of subnets
that can be created and the number of hosts per subnet.

Answer:

In IPv6, the first 64 bits represent the network prefix, and the remaining 64 bits represent the interface
identifier. Since the network prefix has a length of 64 bits, there are no subnet bits available. Therefore, we
cannot create subnets from this address.

However, within this single network, we can have a maximum of 2^64 (18,446,744,073,709,551,616) unique host
identifiers.

Question 6: Troubleshooting Subnetting

A network administrator has subnetted a network and wants to verify the subnet configuration. The IP address and
subnet mask of a host on the network are 192.168.2.135 and 255.255.255.224 respectively.
Determine the following:

  • The network address of the subnet.
  • The broadcast address of the subnet.
  • The range of usable host IP addresses within the subnet.

Answer:

The network address of the subnet can be obtained by performing a bitwise AND operation between the host IP
address and the subnet mask:

192.168.2.135 AND 255.255.255.224 = 192.168.2.128

The broadcast address of the subnet can be obtained by performing a bitwise OR operation between the host IP
address and the inverted subnet mask:

192.168.2.135 OR 0.0.0.31 = 192.168.2.159

The range of usable host IP addresses within the subnet will be from the network address + 1 to the broadcast
address – 1:

Usable IP range: 192.168.2.129192.168.2.158

Conclusion

In this blog post, we explored various CCNA IP subnetting questions, covering subnet calculations, Variable
Length Subnet Masking (VLSM), supernetting, Classless Inter-Domain Routing (CIDR), IPv6 subnetting, and
troubleshooting subnet configurations. These questions provide important practice opportunities to enhance your
subnetting skills. Remember to understand the underlying concepts and continue practicing subnetting to master
this critical skill for networking professionals.

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