Question

Consider a router that interconnects three subnets: Subnet 1, Subnet 2, and Subnet 3. Suppose all of the interfaces in each of these three subnets should use addresses in a large block defined by prefix 128.16.0.0/12. You need to further divide this large address block into three smaller non-overlapping address blocks for these three subnets. They need to further meet the following conditions:

a. Subnet 1 is required to support at least 300 interfaces, and the last address for this subnet is 128.31.255.255.
b. Subnet 2 is to support at least 120 interfaces, and the last address for this subnet is 128.16.1.127.
c. Subnet 3 is to support at least 400 interfaces, and the last address for this subnet is 128.17.7.255.
Provide three network prefixes (of the form a.b.c.d/x) for the three subnets that satisfy the above constraints and also minimize the numbers of addresses for these subnets.

Answer 1

To satisfy the address allocation requirements for the three subnets, the network prefixes would be 128.16.0.0/23 for Subnet 1, 128.16.1.0/25 for Subnet 2, and 128.16.2.0/23 for Subnet 3.

Step-by-step explanation:

A student needs to divide a large address block 128.16.0.0/12 into three smaller non-overlapping address blocks for three subnets, each with a specific number of required interfaces and a stipulated last address. The goal is to assign network prefixes to meet the requirements while minimizing the number of addresses.

For Subnet 1, which needs to support at least 300 interfaces, the smallest subnet that can accommodate this is a /23, which allows for 512 addresses (2^(32-23) = 512). Thus, the network prefix for Subnet 1 would be 128.16.0.0/23, as it ends at 128.31.255.255.

For Subnet 2, with a requirement of at least 120 interfaces, the smallest subnet is a /25, which supports 128 addresses (2^(32-25) = 128). The network prefix for Subnet 2 is 128.16.1.0/25, which covers addresses up to 128.16.1.127.

Subnet 3 requires support for at least 400 interfaces. A /23 subnet also supports up to 512 addresses. Therefore, the network prefix for Subnet 3 must start at the next available address following Subnet 2's range and end at the specified last address, 128.17.7.255. The correct network prefix for Subnet 3 would be 128.16.2.0/23.