How to calculate IP ranges

siu07 picture siu07 · May 3, 2011 · Viewed 9.3k times · Source

Hi would someone be able to assist with the following question? The question is from a past paper in preparation for an exam.

Consider a router that interconnects three subnets: Subnet 1, Subnet 2, Subnet 3. Suppose all of the interfaces in each of these subnets are required to have the prefix 223.1.17/24. Also suppose that subnet 1 is required to support up to 125 different hosts, and subnets 2 and 3 are each required to support up to 60 different hosts.

Provide three network addresses (of the form a.b.c.d/x) that define the beginning of the IP address range for each subnet, and explain your reasoning.

I think the answer is the following, but I'm not sure.

Subnet 1: 223.1.17.1/25

Subnet 2: 223.1.17.128/26

Subnet 3: 223.1.17.193/26

Regards.

Answer

Dennis Röttger picture Dennis Röttger · May 3, 2011

Not quite, Network addresses are always the first Addresses in a Subnet, so the answers would be:

223.1.17.0/25, beginning of IP Address Range: 223.1.17.1 (until .126)

223.1.17.128/26, beginning of IP Address Range: 223.1.17.129 (until .190)

222.1.17.192/26, beginning of IP Address Range: 223.1.17.193 (until .254)

Other than that, your CIDR-Subnet Length is correct, 1 needs at least 126 Hosts (- BC and NA), which justifies /25 -> 24 bits for Class C, 1 bit for Subnet and 7 bit(=2^7 = 128 - Broadcast - Network Address = 126) for hosts, No. 2 and 3 need at least 62 Hosts (-BC and NA) each.