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Difficulty: HardIPv4 Addressing and Subnetting

A network administrator is provisioning a dedicated IPv4 subnet for a new VoIP deployment from an available 192.168.10.0/24192.168.10.0/24 network block. The subnet must accommodate 6060 IP phones, 22 router interfaces operating VRRP for default gateway redundancy, and 11 switch management interface. Which of the following CIDR prefixes represents the smallest subnet that fulfills all required host assignments while preserving the maximum amount of address space for future allocation?

  1. /25/25Answer
  2. B
    /26/26
  3. C
    /27/27
  4. D
    /24/24

Answer

The smallest subnet prefix that satisfies the host requirements while maximizing remaining address space is /25/25.
The scenario requires 6363 usable host IP addresses (6060 IP phones + 22 VRRP router addresses + 11 switch management address). The number of usable hosts in an IPv4 subnet is determined by 2h−22^h - 2, where hh is the number of host bits. A /26/26 subnet has 66 host bits, yielding 26−2=622^6 - 2 = 62 usable addresses, which is 1 address short of the required total. A /25/25 subnet has 77 host bits, providing 27−2=1262^7 - 2 = 126 usable host IP addresses. Therefore, /25/25 is the smallest subnet mask (longest CIDR prefix) that meets the requirement while conserving IP space.

Step-by-Step Solution

1
Calculate the total number of required usable host IP addresses.
Total usable host IPs required = 60 (phones)+2 (VRRP interfaces)+1 (switch management)=6360 \text{ (phones)} + 2 \text{ (VRRP interfaces)} + 1 \text{ (switch management)} = 63 addresses.
Every active host device and redundant gateway interface requires a distinct usable IP address within the subnet.
2
Apply the usable host capacity formula for IPv4 subnetting.
Usable host capacity = 2h−22^h - 2, where h=32−prefix lengthh = 32 - \text{prefix length}.
The network ID (all host bits 0) and broadcast address (all host bits 1) are reserved and cannot be assigned to hosts.
3
Evaluate candidate CIDR prefix lengths against the required 63 usable addresses.
For /26/26 (h=6h=6): 26−2=64−2=622^6 - 2 = 64 - 2 = 62 usable addresses (insufficient by 1 IP). For /25/25 (h=7h=7): 27−2=128−2=1262^7 - 2 = 128 - 2 = 126 usable addresses (sufficient).
Selecting /26/26 by counting total addresses (6464) instead of usable addresses (6262) is a common subnetting calculation error.
4
Select the smallest subnet (longest prefix mask) that accommodates the minimum host requirement.
/25/25 is the smallest subnet mask that fulfills the 63 usable IP address requirement.
Although /24/24 also fits 63 hosts, it leaves 191 unused IP addresses in the subnet, wasting space compared to /25/25.

Key Concept

Subnet Usable Host Capacity Calculation (2h−22^h - 2 Rule) and CIDR Prefix Selection
Estimated Time:2m 0s
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Topics

IPv4 Addressing and Subnetting

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