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Difficulty: Very hardIPv4 Addressing and Subnetting

A network administrator is performing Variable Length Subnet Masking (VLSM) on an enterprise network segment. A router interface is currently assigned the IP address 10.140.77.218/2110.140.77.218/21. Upon inspecting the existing IP allocations, the administrator determines that subnets 10.140.72.0/2310.140.72.0/23 and 10.140.74.0/2310.140.74.0/23 are already assigned to production VLANs. What is the maximum number of additional /26/26 subnets that can be created using only the remaining unallocated address space within this parent /21/21 block?

  1. 16Answer
  2. B
    32
  3. C
    14
  4. D
    8

Answer

16 additional /26 subnets can be created.
The host IP address 10.140.77.218/2110.140.77.218/21 resides in the network block 10.140.72.0/2110.140.72.0/21, spanning 10.140.72.010.140.72.0 through 10.140.79.25510.140.79.255. The two allocated /23/23 subnets consume IPs 10.140.72.010.140.72.0 through 10.140.75.25510.140.75.255. This leaves 10.140.76.010.140.76.0 through 10.140.79.25510.140.79.255 unallocated, which equals a contiguous /22/22 block (10.140.76.0/2210.140.76.0/22). Dividing a /22/22 space into /26/26 subnets yields 2(26−22)=24=162^{(26-22)} = 2^4 = 16 subnets.

Step-by-Step Solution

1
Determine the network boundary of the parent /21 subnet.
The host IP 10.140.77.218/2110.140.77.218/21 belongs to the network 10.140.72.0/2110.140.72.0/21, covering IP range 10.140.72.010.140.72.0 to 10.140.79.25510.140.79.255.
A /21 mask uses a third-octet block size of 8 (256−248=8256 - 248 = 8). Since 77÷8=977 \div 8 = 9 remainder 55, the network start is 9×8=729 \times 8 = 72.
2
Calculate the address space consumed by existing allocations.
Subnet 10.140.72.0/2310.140.72.0/23 consumes 10.140.72.010.140.72.0 - 10.140.73.25510.140.73.255, and subnet 10.140.74.0/2310.140.74.0/23 consumes 10.140.74.010.140.74.0 - 10.140.75.25510.140.75.255.
Each /23 subnet covers 2 third-octet values (224−23=22^{24-23} = 2).
3
Identify the unallocated address space remaining in the /21 parent block.
The remaining contiguous range is 10.140.76.010.140.76.0 through 10.140.79.25510.140.79.255, which forms a single 10.140.76.0/2210.140.76.0/22 subnet.
The unallocated space spans 4 third-octet blocks (76,77,78,7976, 77, 78, 79), equivalent to a /22 prefix (256−252=4256 - 252 = 4).
4
Calculate how many /26 subnets fit into the remaining /22 block.
Number of /26 subnets = 2(26−22)=24=162^{(26 - 22)} = 2^4 = 16.
Subnet division is determined by exponentiation of the bit prefix difference between the parent block prefix and the target subnet prefix.

Key Concept

VLSM Subnet Allocation and CIDR Block Subdivisions
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Topics

IPv4 Addressing and Subnetting

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