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Difficulty: HardIPv4 Addressing and Subnetting

A network administrator is designing an IP addressing scheme for a new corporate branch facility using the assigned block 10.45.128.0/1910.45.128.0/19. The engineering division requires a dedicated subnet capable of supporting at least 1,1001,100 usable host IP addresses. Which CIDR prefix represents the smallest possible subnet allocation that satisfies this host requirement while conserving IP space, and what is the last usable host IP address in that subnet?

  1. /21/21 prefix with a last usable host address of 10.45.135.25410.45.135.254Answer
  2. B
    /22/22 prefix with a last usable host address of 10.45.131.25410.45.131.254
  3. C
    /21/21 prefix with a last usable host address of 10.45.135.25510.45.135.255
  4. D
    /21/21 prefix with a last usable host address of 10.45.143.25410.45.143.254

Answer

The /21/21 prefix with a last usable host address of 10.45.135.25410.45.135.254
The correct selection uses a /21/21 prefix, which grants 11 host bits (211−2=2,0462^{11} - 2 = 2,046 usable host addresses), satisfying the requirement for at least 1,100 hosts while conserving address space. With a 3rd octet block size of 8, the subnet range is 10.45.128.010.45.128.0 to 10.45.135.25510.45.135.255, making 10.45.135.25410.45.135.254 the final usable host IP address.

Step-by-Step Solution

1
Determine the minimum number of host bits required for 1,100 hosts.
Using the formula 2h−2≥11002^h - 2 \ge 1100, h=10h=10 yields 1,0221,022 usable hosts (insufficient), while h=11h=11 yields 2,0462,046 usable hosts. Therefore, 1111 host bits are needed.
Two addresses (network ID and broadcast address) must be subtracted from the total address count.
2
Calculate the required CIDR prefix length.
32−11=2132 - 11 = 21, so a /21/21 prefix is required.
An IPv4 address consists of 32 total bits.
3
Determine the block size and subnet range starting from 10.45.128.010.45.128.0.
With 1111 host bits (33 bits in the 3rd octet and 88 bits in the 4th octet), the 3rd octet block increment is 23=82^3 = 8. The subnet spans 10.45.128.010.45.128.0 to 10.45.135.25510.45.135.255.
The 3rd octet starts at 128 and increments by 8, giving a range of 128 through 135.
4
Identify the broadcast address and last usable host address.
Broadcast address is 10.45.135.25510.45.135.255, and the last usable host IP is 10.45.135.25410.45.135.254.
The last usable host IP is immediately prior to the subnet broadcast address.

Key Concept

Calculating required subnet mask size based on host capacity constraints and determining subnet host address bounds.
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Topics

IPv4 Addressing and Subnetting

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