# Appropriate Subnet Mask for Eight LANs with 5 to 26 Hosts

### Question

Which subnet mask would be appropriate for a network address range to be subnetted for up to eight LANs, with each LAN containing 5 to 26 hosts?

### Explanations

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A. B. C. D. E.

D

To determine the appropriate subnet mask for a network address range to be subnetted for up to eight LANs, with each LAN containing 5 to 26 hosts, we need to calculate the maximum number of hosts that can be accommodated in each LAN and then choose a subnet mask that can accommodate this range of hosts.

To calculate the maximum number of hosts that can be accommodated in each LAN, we need to determine the number of host bits required. The formula to calculate the number of host bits required is:

2^n - 2 >= h

Where n is the number of host bits and h is the number of hosts required.

For example, to accommodate 5 to 26 hosts, we need at least 5 host bits, as 2^5 - 2 = 30, which is greater than 26. Therefore, we need a subnet mask that provides at least 5 host bits.

To determine the appropriate subnet mask, we can use the following formula:

subnet mask = 2^(32 - n) - 1

Where n is the number of host bits required.

Using the example of 5 to 26 hosts, we need at least 5 host bits, so the subnet mask would be:

subnet mask = 2^(32 - 5) - 1 = 31 = 11111111.11111111.11111111.11100000

Option A. 0.0.0.240: This subnet mask provides only 4 host bits, which is not sufficient to accommodate 5 to 26 hosts.

Option B. 255.255.255.252: This subnet mask provides only 2 host bits, which is not sufficient to accommodate 5 to 26 hosts.

Option C. 255.255.255.0: This subnet mask provides 8 host bits, which is more than sufficient to accommodate 5 to 26 hosts in each LAN. However, it would create only one LAN, which is not in line with the requirement of up to 8 LANs.

Option D. 255.255.255.224: This subnet mask provides 5 host bits, which is sufficient to accommodate 5 to 26 hosts in each LAN. Therefore, this is the correct option.

Option E. 255.255.255.240: This subnet mask provides only 4 host bits, which is not sufficient to accommodate 5 to 26 hosts.

Therefore, the correct answer is option D, 255.255.255.224.