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Question

Difficulty: MediumIPv4 Addressing and Subnetting

A network administrator is designing an IPv4 subnet for a new regional warehouse facility. The facility requires static and dynamic IP address assignments for 115115 endpoint devices, including inventory scanners, workstation PCs, and network printers. What is the maximum number of usable host IP addresses available in the smallest CIDR subnet mask that satisfies this host requirement?

Answer: 126 usable host IPs

Answer

The smallest CIDR block accommodating 115 hosts is a /25 prefix (255.255.255.128), which provides 126 usable host IP addresses.
To accommodate 115 hosts, a subnet requires at least 115+2=117115 + 2 = 117 total IP addresses. The smallest power of 2 covering 117 is 27=1282^7 = 128, corresponding to a /25/25 subnet prefix. Subtracting the Network ID and Broadcast address (128−2128 - 2) results in 126 usable host IP addresses.

Step-by-Step Solution

1
Calculate the total required IP space including reserved network and broadcast addresses.
115 required host IPs+2 reserved addresses=117 total IP addresses115 \text{ required host IPs} + 2 \text{ reserved addresses} = 117 \text{ total IP addresses}.
Every IPv4 subnet reserves the first address for the Network ID and the final address for the Broadcast address.
2
Determine the smallest block size (2n2^n) that accommodates the total address requirement.
27=128 total addresses2^7 = 128 \text{ total addresses}, corresponding to host bits n=7n=7 and prefix length 32−7=/2532 - 7 = /25.
26=642^6 = 64 total addresses (6262 usable) is insufficient for 115115 devices.
3
Subtract the 2 reserved addresses to find total usable host capacity.
128−2=126 usable host IP addresses128 - 2 = 126 \text{ usable host IP addresses}.
Usable host capacity formula is 2(32−prefix)−22^{(32 - \text{prefix})} - 2.

Key Concept

IPv4 Subnet Mask Host Capacity Calculation
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Topics

IPv4 Addressing and Subnetting

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