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Question 669 - N10-008 discussion

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To find the best subnet mask that meets the requirement of six usable IP addresses, we need to calculate the number of host bits and the number of host addresses for each option. The number of host bits is the number of 0s in the binary representation of the subnet mask, and the number of host addresses is 2^host bits - 2 (the -2 is to exclude the network address and the broadcast address). The option that has the smallest number of host addresses that is greater than or equal to six is the best choice. Here are the calculations for each option:

A.
255.255.255.128 Binary: 11111111.11111111.11111111.10000000 Host bits: 7 Host addresses: 2^7 - 2 = 126 - 2 = 124 This option has too many host addresses for the requirement.
Answers
A.
255.255.255.128 Binary: 11111111.11111111.11111111.10000000 Host bits: 7 Host addresses: 2^7 - 2 = 126 - 2 = 124 This option has too many host addresses for the requirement.
B.
255.255.255.192 Binary: 11111111.11111111.11111111.11000000 Host bits: 6 Host addresses: 2^6 - 2 = 64 - 2 = 62 This option also has too many host addresses for the requirement.
Answers
B.
255.255.255.192 Binary: 11111111.11111111.11111111.11000000 Host bits: 6 Host addresses: 2^6 - 2 = 64 - 2 = 62 This option also has too many host addresses for the requirement.
C.
255.255.255.224 Binary: 11111111.11111111.11111111.11100000 Host bits: 5 Host addresses: 2^5 - 2 = 32 - 2 = 30 This option has the smallest number of host addresses that is greater than or equal to six, so this is the best choice.
Answers
C.
255.255.255.224 Binary: 11111111.11111111.11111111.11100000 Host bits: 5 Host addresses: 2^5 - 2 = 32 - 2 = 30 This option has the smallest number of host addresses that is greater than or equal to six, so this is the best choice.
D.
255.255.255.240 Binary: 11111111.11111111.11111111.11110000 Host bits: 4 Host addresses: 2^4 - 2 = 16 - 2 = 14 This option has fewer host addresses than the requirement, so this is not a valid choice.
Answers
D.
255.255.255.240 Binary: 11111111.11111111.11111111.11110000 Host bits: 4 Host addresses: 2^4 - 2 = 16 - 2 = 14 This option has fewer host addresses than the requirement, so this is not a valid choice.
Suggested answer: C

Explanation:

This subnet mask will allow you to have six usable IP addresses in each subnet, with a minimum of wasted addresses. You can use the following formula to calculate the number of subnets and the subnet ID for each subnet:

Number of subnets = 2^network bits Subnet ID = (subnet number - 1) x number of host addresses + network address

The network bits are the number of 1s in the binary representation of the subnet mask, and the network address is the first address in the range. For example, if your range is 192.168.1.0/27, then the network bits are 27, the network address is 192.168.1.0, and the number of host addresses is 30. Therefore, the number of subnets is 2^27, and the subnet ID for the first subnet is (1 - 1) x 30 + 192.168.1.0 = 192.168.1.0. The subnet ID for the second subnet is (2 - 1) x 30 + 192.168.1.0 = 192.168.1.32, and so on.

Subnet masks are covered in Objective 1.4 of the CompTIA Network+ N10-008 certification exam1.

Subnet masks can be calculated based on binary and CIDR-block notations2.

Subnet masks can be used to determine the number of host bits and host addresses3.

1: CompTIA Network+ Certification Exam Objectives, page 42: IPv4 Subnet Masks -- N10-008 CompTIA Network+ : 1.413: Calculating IPv4 Subnets and Hosts -- N10-008 CompTIA Network+ : 1.44

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asked 02/10/2024
Alexander Goris
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