) Consider a router that interconnects four subnets: Subnet 1, Subnet 2, Subnet 3 and Subnet 4. Suppose all of the interfaces in each of these four subnets are required to have the prefix 223.1.17/24. Also suppose that Subnet 1 is required to support at least 30 interfaces, Subnet 2 is to support at least 100 interfaces, Subnet 3 is to support at least 10 interfaces, Subnet 4 is to support at least 10 interfaces. Provide four network addresses (of the form a. b.c. d/x) that satisfy these constraints. Show all your work.

Check the explanation

Explanation:

223.1.17/24 indicates that out of 32-bits of IP address 24 bits have been assigned as subnet part and 8 bits for host id.

The binary representation of 223.1.17 is 11011111 00000001 00010001 00000000

Given that, subnet 1 has 63 interfaces. To represent 63 interfaces, we need 6 bits (64 = 26)

So its addresses can be from 223.1.17.0/26 to 223.1.17.62/26

Subnet 2 has 95 interfaces. 95 interfaces can be accommodated using 7 bits up to 127 host addresses can represented using 7 bits (127 = 27)

and hence, the addresses may be from 223.1.17.63/25 to 223.1.17.157/25

Subnet 3 has 16 interfaces. 4 bits are needed for 16 interfaces (16 = 24)

So the network addresses may range from 223.1.17.158/28 to 223.1.17.173/28

this seems to be in the wrong category. but i'll take some free points my dude

posting this in wrong category will result in people seeing it who won't know the answer.