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Difficulty: MediumIPv4 Addressing and Subnetting

A network administrator is configuring a interface on a syslog server with the IPv4 address 192.168.45.138/26192.168.45.138/26. What is the network ID (subnet address) of the subnetwork to which this server belongs?

Answer: 192.168.45.128 / 192.168.45.128/26

Answer

192.168.45.128
With a /26/26 prefix, the subnet mask is 255.255.255.192255.255.255.192. The block size in the fourth octet is 256−192=64256 - 192 = 64. Subnets in this range start at .0.0, .64.64, .128.128, and .192.192. Since the host address fourth octet is 138138, it resides within the 192.168.45.128/26192.168.45.128/26 subnet (covering addresses 192.168.45.128192.168.45.128 through 192.168.45.191192.168.45.191). The network ID is 192.168.45.128192.168.45.128.

Step-by-Step Solution

1
Determine the subnet mask block size from the CIDR prefix length.
A /26/26 prefix leaves 32−26=632 - 26 = 6 host bits in the fourth octet. The block size (increment value) is 26=642^6 = 64.
Knowing the block size allows identification of the subnet boundaries in the relevant octet.
2
Calculate the subnet boundaries in the fourth octet.
The fourth octet subnet ranges start at increments of 6464: .0.0, .64.64, .128.128, and .192.192.
Subnet IDs always begin on multiples of the block size starting from 0.
3
Locate the range containing the host IP fourth octet (138138).
The fourth octet value 138138 falls between 128128 and 191191. Therefore, the network ID is 192.168.45.128192.168.45.128.
The network ID is the lowest address (boundary start) of the subnet containing the host.

Key Concept

Determining Network Address from IPv4 Host Address and CIDR Prefix
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Topics

IPv4 Addressing and Subnetting

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