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Difficulty: MediumIPv4 Addressing and Subnetting

A network engineer is dividing the enterprise IPv4 address block 10.45.0.0/2110.45.0.0/21 into equal-sized subnets using a prefix mask of /26/26. What is the total number of usable host IP addresses available across all of the newly created /26/26 subnets?

Answer: 1984 hosts

Answer

1984 total usable host IP addresses
Subnetting a /21/21 block into /26/26 subnets borrows 5 bits, creating 25=322^5 = 32 subnets. Each /26/26 subnet has 232−26−2=622^{32-26} - 2 = 62 usable host IP addresses. Multiplying 32 subnets by 62 usable host IPs per subnet equals 1984 total usable host IP addresses across the allocated block.

Step-by-Step Solution

1
Calculate the total number of subnets created from the /21/21 parent block.
32 subnets (226−21=25=322^{26-21} = 2^5 = 32)
Borrowing 5 additional subnet bits (26−2126 - 21) yields 25=322^5 = 32 equal-sized subnets.
2
Calculate the usable host IP address capacity per /26/26 subnet.
62 usable host IP addresses (232−26−2=64−2=622^{32-26} - 2 = 64 - 2 = 62)
Each /26/26 subnet has 6 host bits (26=642^6 = 64 total IPs), minus 2 addresses reserved for the network ID and broadcast address.
3
Compute total usable host capacity across all subnets.
1984 total usable hosts (32×62=198432 \times 62 = 1984)
Multiplying the total number of subnets by the usable capacity of a single subnet yields the aggregate usable host addresses.

Key Concept

Subnetting and Aggregate Host Capacity Calculation
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Topics

IPv4 Addressing and Subnetting

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