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Difficulty: MediumIPv4 Addressing and Subnetting

A network administrator is allocated the IPv4 address block 10.75.12.0/2310.75.12.0/23 to provision subnets for remote branch kiosks. Each kiosk requires a subnetwork capable of supporting at least 2828 usable host IP addresses. What is the maximum number of equal-sized subnets meeting this host requirement that can be created from the allocated address block?

Answer: 16 subnets

Answer

The maximum number of equal-sized subnets that can be created is 16.
To support at least 2828 usable host addresses, each subnet must have 5 host bits (25−2=302^5 - 2 = 30 usable addresses). An allocation of 5 host bits yields a /27/27 mask (32−5=2732 - 5 = 27). Dividing the parent /23/23 block into /27/27 subnets borrows 27−23=427 - 23 = 4 subnet bits, yielding 24=162^4 = 16 maximum usable subnets.

Step-by-Step Solution

1
Calculate the minimum number of host bits required per subnet
5 host bits are needed because 25−2=302^5 - 2 = 30 usable hosts, which satisfies the requirement of at least 2828 usable host IP addresses (24−2=142^4 - 2 = 14 is insufficient).
Subnet sizing requires accounting for 2 reserved addresses (network ID and broadcast address).
2
Determine the required CIDR subnet prefix length
The subnet prefix length is /27/27 (32−5=2732 - 5 = 27).
Subtracting the 5 host bits from the 32-bit IPv4 address space leaves 27 bits for the network identifier.
3
Calculate the total number of /27/27 subnets that fit into the assigned /23/23 address block
16 subnets (227−23=24=162^{27 - 23} = 2^4 = 16).
Dividing a /23/23 address block into /27/27 subnets requires borrowing 4 bits from the host portion for subnetting.

Key Concept

IPv4 Subnet Sizing and Subnet Count Calculation
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Topics

IPv4 Addressing and Subnetting

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