Finding network addresses [closed]

Question

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?

Burdette

Answer 1

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

Trever

Thanks! But why is it that a/25 network should start at either 0 or 128? – Burdette

a/25 means you have remaining 7 bits. So we can divide that 256 addresses to 2 parts 0 to 127 and 128 to 255 – Adler