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Difficulty: HardIPv4 Addressing and Subnetting

An organization is assigned the IPv4 address block 172.28.0.0/19172.28.0.0/19 to provision subnetting for a multi-building campus network. The system architecture mandates dividing this entire block into equal-sized subnets where each subnet must support at least 120120 usable host IP addresses. What is the maximum number of such equal-sized subnets that can be created from this /19/19 prefix?

Answer: 64 subnets

Answer

64 subnets can be created.
To host at least 120 usable IP addresses, 7 host bits are required (27−2=1262^7 - 2 = 126 usable hosts). A block with 7 host bits corresponds to a /25/25 prefix. Subnetting a /19/19 block into /25/25 subnets borrows 25−19=625 - 19 = 6 bits, resulting in 26=642^6 = 64 subnets.

Step-by-Step Solution

1
Determine the required host bits
7 host bits (h=7h = 7)
The formula for usable IPv4 hosts per subnet is 2h−22^h - 2. For 120 hosts, 2h−2≥120  ⟹  2h≥1222^h - 2 \ge 120 \implies 2^h \ge 122. The smallest power of two satisfying this is 27=1282^7 = 128 (128−2=126128 - 2 = 126 usable host addresses).
2
Determine the required subnet prefix length
/25 prefix length
Subtracting 7 host bits from total 32 bits (32−7=2532 - 7 = 25) gives a subnet mask of /25/25 (255.255.255.128255.255.255.128).
3
Calculate the number of subnets derived from the parent prefix
64 subnets
The parent prefix is /19/19 and the subnet prefix is /25/25. The number of borrowed subnet bits is 25−19=625 - 19 = 6 bits. The total number of subnets created is 26=642^6 = 64.

Key Concept

VLSM and Subnet Sizing Calculation
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Topics

IPv4 Addressing and Subnetting

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