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Difficulty: MediumIPv4 Addressing and Subnetting

A network administrator is designing an IPv4 subnetting scheme for a branch office using the 192.168.10.0/24192.168.10.0/24 network address block. Each departmental VLAN requires support for at least 2828 usable host IP addresses. Which subnet mask will satisfy this host requirement while maximizing the total number of subnets that can be created?

  1. 255.255.255.224255.255.255.224Answer
  2. B
    255.255.255.240255.255.255.240
  3. C
    255.255.255.192255.255.255.192
  4. D
    255.255.255.248255.255.255.248

Answer

The subnet mask 255.255.255.224255.255.255.224 satisfies the requirement.
To support at least 2828 hosts, 55 bits are required for the host portion because 25−2=302^5 - 2 = 30 usable IP addresses. A /27/27 prefix (255.255.255.224255.255.255.224) leaves 55 host bits, yielding 88 subnets from a /24/24 block, which maximizes the subnet count while satisfying the host constraint.

Step-by-Step Solution

1
Calculate the minimum number of host bits (hh) required using the formula 2h−2≥hosts2^h - 2 \ge \text{hosts}.
25−2=30≥282^5 - 2 = 30 \ge 28, so h=5h = 5 host bits are needed.
Two IP addresses per subnet are reserved for the network ID and the broadcast address.
2
Determine the network prefix length from the total number of IPv4 bits (3232).
Prefix length =32−5=/27= 32 - 5 = /27.
Subtracting host bits from 3232 leaves the number of network bits.
3
Convert the CIDR prefix /27/27 into dotted-decimal subnet mask format.
/27=255.255.255.224/27 = 255.255.255.224.
The fourth octet contains three binary 11 s (111000002=22411100000_2 = 224).

Key Concept

Calculating required host bits and dotted-decimal subnet mask to maximize subnets while satisfying host capacity constraints
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Topics

IPv4 Addressing and Subnetting

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