IP Subnetting for Network Administrators - Best Practices | Exam N10-007

Dividing a Public IP Network - Efficiently Allocate Addresses | CompTIA Network+

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Question

A network administrator needs to implement a new IP subnet containing 29 hosts.

It is possible that the number of hosts in that subnet could eventually double.

The company only has a single, unused public IP network left to work with: 164.10.12.0/24

Which of the following would be the BEST way to divide this network without wasting addresses?

A.

164.10.12.0/24 B.

164.10.12.64/29 C.

164.10.12.128/26 D.

164.10.12.191/28

C.

Explanations

A network administrator needs to implement a new IP subnet containing 29 hosts.

It is possible that the number of hosts in that subnet could eventually double.

The company only has a single, unused public IP network left to work with: 164.10.12.0/24

Which of the following would be the BEST way to divide this network without wasting addresses?

A.

164.10.12.0/24

B.

164.10.12.64/29

C.

164.10.12.128/26

D.

164.10.12.191/28

C.

To divide the network 164.10.12.0/24 into subnets with 29 hosts and the potential to double, we need to calculate the subnet mask required for the new subnets.

To support 29 hosts, we need to borrow 5 bits for host addresses, leaving 3 bits for network addresses. This means the subnet mask will be 255.255.255.224 or /27.

The next step is to find the network address and broadcast address for each subnet. To do this, we can use the formula:

Network address = (IP address) AND (Subnet mask) Broadcast address = (Network address) OR (Inverse subnet mask)

Using this formula, we can calculate the network and broadcast addresses for each subnet.

Option A: 164.10.12.0/24 - This is the original network address and cannot be used for the new subnet.

Option B: 164.10.12.64/29 - This subnet mask is correct for 29 hosts. The network address is 164.10.12.64 and the broadcast address is 164.10.12.71.

Option C: 164.10.12.128/26 - This subnet mask does not support 29 hosts, as it only allows for 64 hosts. The next subnet would be 164.10.12.192/26.

Option D: 164.10.12.191/28 - This subnet mask only supports 16 hosts, which is not enough for the requirement of 29 hosts.

Therefore, the best option for dividing the network without wasting addresses is option B, 164.10.12.64/29.