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Difficulty: Very hardIPv4 Addressing and Subnetting

A network architect is designing a VLSM address allocation plan starting at the beginning of the 192.168.100.0/23192.168.100.0/23 address block. Subnets must be provisioned sequentially without wasting address space in the following order:
1. Subnet A: Requires support for 120120 usable host IP addresses.
2. Subnet B: Requires support for 5050 usable host IP addresses.
3. Subnet C: Requires a point-to-point link supporting 22 router interfaces.
4. Subnet D: Requires support for 2525 usable host IP addresses.

Assuming all subnets are properly aligned to their required CIDR boundary offsets, what is the broadcast IP address of Subnet D?

  1. A
    192.168.100.227192.168.100.227
  2. B
    192.168.100.224192.168.100.224
  3. 192.168.100.255192.168.100.255Answer
  4. D
    192.168.100.223192.168.100.223

Answer

The broadcast IP address of Subnet D is 192.168.100.255192.168.100.255.
Subnet A requires 120 hosts, fitting into a /25 block (192.168.100.0−192.168.100.127192.168.100.0 - 192.168.100.127). Subnet B requires 50 hosts, fitting into a /26 block (192.168.100.128−192.168.100.191192.168.100.128 - 192.168.100.191). Subnet C requires 2 hosts, fitting into a /30 block (192.168.100.192−192.168.100.195192.168.100.192 - 192.168.100.195). The next available IP address is 192.168.100.196192.168.100.196. Subnet D requires 25 hosts, requiring a /27 subnet (block size 32). Because /27/27 subnets must start on a boundary divisible by 32 (0,32,64,96,128,160,192,2240, 32, 64, 96, 128, 160, 192, 224), Subnet D must start at 192.168.100.224/27192.168.100.224/27. The range for Subnet D is 192.168.100.224192.168.100.224 through 192.168.100.255192.168.100.255, making 192.168.100.255192.168.100.255 the correct broadcast address.

Step-by-Step Solution

1
Calculate prefix size and address range for Subnet A
Subnet A needs 120120 usable hosts (27−2=126≥1202^7 - 2 = 126 \ge 120), requiring a /25/25 mask (block size 128128). Range: 192.168.100.0/25192.168.100.0/25 (192.168.100.0192.168.100.0 to 192.168.100.127192.168.100.127).
Determines the ending address of Subnet A.
2
Calculate prefix size and address range for Subnet B
Subnet B starts at 192.168.100.128192.168.100.128. It needs 5050 usable hosts (26−2=62≥502^6 - 2 = 62 \ge 50), requiring a /26/26 mask (block size 6464). Range: 192.168.100.128/26192.168.100.128/26 (192.168.100.128192.168.100.128 to 192.168.100.191192.168.100.191).
Determines the ending address of Subnet B.
3
Calculate prefix size and address range for Subnet C
Subnet C starts at 192.168.100.192192.168.100.192. It needs 22 usable hosts (22−2=22^2 - 2 = 2), requiring a /30/30 mask (block size 44). Range: 192.168.100.192/30192.168.100.192/30 (192.168.100.192192.168.100.192 to 192.168.100.195192.168.100.195).
Determines the next available unaligned address (192.168.100.196192.168.100.196).
4
Determine prefix size and boundary alignment for Subnet D
Subnet D needs 2525 usable hosts (25−2=30≥252^5 - 2 = 30 \ge 25), requiring a /27/27 mask (block size 3232). The next unaligned IP is 192.168.100.196192.168.100.196. Since a /27/27 network address must be divisible by 3232, the next valid network boundary is 192.168.100.224192.168.100.224.
Enforces natural CIDR boundary alignment rules for variable-length subnets.
5
Calculate the broadcast address of Subnet D
Network ID: 192.168.100.224/27192.168.100.224/27. Range: 192.168.100.224192.168.100.224 to 192.168.100.255192.168.100.255. Broadcast address = 192.168.100.255192.168.100.255.
Adding the block size minus one (32−1=3132 - 1 = 31) to the network address gives the broadcast address.

Key Concept

VLSM Subnet Sizing and Bit-Boundary Alignment
Estimated Time:3m 0s
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Topics

IPv4 Addressing and Subnetting

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