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Difficulty: Very hardIPv4 Addressing and Subnetting

A network administrator is assigned the IPv4 address block 172.28.64.0/21172.28.64.0/21 to provision subnets for a multi-department enterprise deployment. Using Variable Length Subnet Masking (VLSM), subnets must be allocated sequentially starting from the lowest available IP address in order of capacity requirements (from largest to smallest):

- Subnet A: Requires capacity for up to 400400 usable host IP addresses.
- Subnet B: Requires capacity for up to 180180 usable host IP addresses.
- Subnet C: Requires capacity for up to 6060 usable host IP addresses.
- Subnet D: Requires capacity for exactly 22 usable host IP addresses.

Each subnet must be allocated using the smallest valid CIDR prefix length that accommodates its requirement. What is the total number of unallocated IP addresses remaining in the 172.28.64.0/21172.28.64.0/21 block after all four subnets are allocated?

Answer: 1212 IP addresses

Answer

The total number of unallocated IP addresses remaining in the 172.28.64.0/21172.28.64.0/21 parent block after provisioning all four subnets is 1212.
The parent block 172.28.64.0/21172.28.64.0/21 contains 20482048 total IP addresses. Allocating the subnets sequentially requires: Subnet A (400400 hosts) →/23\rightarrow /23 (512512 IPs), Subnet B (180180 hosts) →/24\rightarrow /24 (256256 IPs), Subnet C (6060 hosts) →/26\rightarrow /26 (6464 IPs), and Subnet D (22 hosts) →/30\rightarrow /30 (44 IPs). Total allocated IP space is 512+256+64+4=836512 + 256 + 64 + 4 = 836 IP addresses. Subtracting 836836 from 20482048 leaves 12121212 total unallocated IP addresses.

Step-by-Step Solution

1
Calculate total IP address capacity of the parent block
172.28.64.0/21172.28.64.0/21 contains 232−21=211=20482^{32-21} = 2^{11} = 2048 total IP addresses
A /21/21 prefix uses 21 network bits, leaving 11 host bits for addressing.
2
Determine prefix length and block size for Subnet A
Prefix: /23/23, Block size: 512512 total IPs (172.28.64.0/23172.28.64.0/23)
To support 400400 hosts, 9 host bits are needed (29−2=510≥4002^9 - 2 = 510 \ge 400), giving a /23/23 prefix (512512 IPs).
3
Determine prefix length and block size for Subnet B
Prefix: /24/24, Block size: 256256 total IPs (172.28.66.0/24172.28.66.0/24)
To support 180180 hosts, 8 host bits are needed (28−2=254≥1802^8 - 2 = 254 \ge 180), giving a /24/24 prefix (256256 IPs).
4
Determine prefix length and block size for Subnet C
Prefix: /26/26, Block size: 6464 total IPs (172.28.67.0/26172.28.67.0/26)
To support 6060 hosts, 6 host bits are needed (26−2=62≥602^6 - 2 = 62 \ge 60), giving a /26/26 prefix (6464 IPs).
5
Determine prefix length and block size for Subnet D
Prefix: /30/30, Block size: 44 total IPs (172.28.67.64/30172.28.67.64/30)
To support 22 hosts (point-to-point link), 2 host bits are needed (22−2=2≥22^2 - 2 = 2 \ge 2), giving a /30/30 prefix (44 IPs).
6
Calculate remaining unallocated IP space
2048−(512+256+64+4)=2048−836=12122048 - (512 + 256 + 64 + 4) = 2048 - 836 = 1212 IP addresses
Subtracting the sum of all allocated CIDR block sizes from the total parent address space yields the remaining unallocated IP addresses.

Key Concept

Variable Length Subnet Masking (VLSM) allocation and host requirement calculations
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Topics

IPv4 Addressing and Subnetting

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