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Difficulty: HardIPv4 Addressing and Subnetting

A network engineer is provisioning subnets from the allocated summary block 172.25.192.0/20172.25.192.0/20 to expand an enterprise corporate network. The design specification requires Subnet A to accommodate at least 1,000 usable host endpoints and to be allocated starting from the very first available boundary of the 172.25.192.0/20172.25.192.0/20 block. Additionally, all assigned addresses must strictly comply with RFC 1918 private IPv4 space. What is the last usable IPv4 host address on Subnet A?

  1. 172.25.195.254Answer
  2. B
    172.25.195.255
  3. C
    172.25.193.254
  4. D
    172.32.195.254

Answer

172.25.195.254
To support at least 1,000 usable hosts, a subnet requires 10 host bits because 210−2=1,0222^{10} - 2 = 1,022 usable addresses (29−2=5102^9 - 2 = 510 is insufficient). Subtracting 10 host bits from 32 total bits yields a /22/22 subnet mask (255.255.252.0255.255.252.0). Allocating a /22/22 subnet from the start of the block 172.25.192.0/20172.25.192.0/20 results in a subnet spanning 172.25.192.0172.25.192.0 to 172.25.195.255172.25.195.255. The last usable IPv4 address is one less than the broadcast address, which is 172.25.195.254172.25.195.254. Furthermore, 172.25.192.0/20172.25.192.0/20 is fully within the RFC 1918 private IP range (172.16.0.0/12172.16.0.0/12).

Step-by-Step Solution

1
Determine required host bits (hh) and prefix length
2h−2≥1,000  ⟹  210−2=1,022≥1,0002^h - 2 \ge 1,000 \implies 2^{10} - 2 = 1,022 \ge 1,000, so h=10h = 10 host bits. The prefix length is 32−10=/2232 - 10 = /22.
Subnet sizing requires finding the smallest power of 2 minus 2 for network and broadcast overhead that satisfies the host capacity requirement.
2
Verify RFC 1918 private addressing compliance
The block 172.25.192.0/20172.25.192.0/20 resides within the Class B private address space of 172.16.0.0/12172.16.0.0/12 (172.16.0.0172.16.0.0 to 172.31.255.255172.31.255.255).
RFC 1918 mandates that enterprise private networks use designated non-routable public IP blocks.
3
Calculate the subnet boundary and broadcast address
A /22/22 prefix has a block size of 44 in the third octet (210−8=42^{10-8} = 4). Starting at 172.25.192.0172.25.192.0, the network spans 172.25.192.0172.25.192.0 to 172.25.195.255172.25.195.255. The broadcast address is 172.25.195.255172.25.195.255.
Adding the block size minus 1 to the starting third octet gives the highest octet boundary for the subnet.
4
Identify the last usable host address
The last usable IP address is one position below the broadcast address: 172.25.195.254172.25.195.254.
The usable host range is bounded between the network ID (+1+1) and the broadcast ID (−1-1).

Key Concept

VLSM host calculation and RFC 1918 private IPv4 address boundaries
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Topics

IPv4 Addressing and Subnetting

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