Consider a router that interconnects three subnets: Subnet1, Subnet2, and Subnet3. Suppose all of the interfaces in each of these three subnets are required to have the prefix 223.1.17/24. Also suppose that Subnet #1 is to support at least 60 interfaces Subnet #2 is to support at least 90 interfaces Subnet #3 is to support at least 12 interfaces. Provide three network addresses (of the form a.b.c.d/x) that satisfy these constraints.
What I thought is Subnet 1: 223.1.17.0/26 Subnet 2: 223.1.17.64/25 Subnet 3 : 223.1.17.192/28
Is this correct?
Answer
No. You have the size calculations done correctly but subnet 1 and 2 will overlap. A /25 network must start at either 0 or 128.
The requirements for each network is: 60 rounded up to the nearest power of two is 64 => we need 6 bits Subnet #2 needs 90 addresses so again, rounding up to 128 => we need 7 bits Subnet #3 only has 12 adresses, rounded gives us 16 => 4 bits.
You can fit this in one of two ways (you can permutate subnet 1 and three in more).
subnet 1 223.1.17.128/26
subnet 2 223.1.17.0/25
subnet 3 223.1.17.192/28
or
subnet 1 223.1.17.0/26
subnet 2 223.1.17.128/25
subnet 3 223.1.17.64/28