Consider a router that interconnects three subnets: Subnet 1, Subnet 2, and Subnet 3. Suppose all of the interfaces in each of the three subnets are required to have the prefix 223.1.17/24. Also suppose that Subnet 1 is required to support upto 125 interfaces, and Subnect 2 and 3 are each required to support up to 60 interfaces. Provide three network addresses (of the form a.b.c.d/x) that satisfy these constraints.
Try an answer before revealing the guidance below.
Key Concepts
- CIDR prefix
- Host bits
- Address block alignment
- /25 and /26
Answer Approach
- Find the host bits needed for 125 interfaces and for 60 interfaces.
- Convert each to a prefix length.
- Allocate the largest block first, then fit the others into the remaining space without overlap.
Full Answer
Answer status: Draft answer (unofficial). Revision notes, not an official marking scheme.
125 interfaces need 7 host bits (128 addresses) → /25; 60 interfaces need 6 host bits (64 addresses) → /26. One valid allocation: Subnet 1: 223.1.17.0/25; Subnet 2: 223.1.17.128/26; Subnet 3: 223.1.17.192/26.
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