Question

Consider a router that interconnects two IPv4 subnets: Subnet 1 and Subnet 2. Suppose all of the interfaces in each of these three subnets are required to have the prefix 10.159.9/24. Also suppose that Subnet 1 is required to support at least 100 interfaces and Subnet 2 is to support at least 30 interfaces. Provide two network addresses (of the form а.b.c.d/X) that satisfy these constraints.

Answer 1

Since Subnet 1 has to support at least 100 interfaces, hence it needed a total of 100+2 = 102 IP addresses( we need 2 extra addresses i.e. first and last addresses which is reserved), and subnet 2 need 30+2 = 32 IP addresses.

Hence the number of unmasked bits needed in Subnet 1 = log₂102=7 bits, number of unmasked bits needed in Subnet 2 =  log₂32=5 bits.

Now in prefix 10.159.9/24, since we have the first 24 bits masked, hence we have 32-24 = 8 bits left.

So out of 8 bits, we have kept 1 bit to identify the subset. Hence 8^th bit from the right will be 0 for Subnet 1 and 1 for Subnet 2. Then 7 bits will be addressed to the interface within the subnet.

Hence the network address of Subnet 1 will be its first IP address = 10.159.9.0/25 (now we have 25 bits masked in each subnet) and the network address of Subnet 1 will be its first IP address = 10.159.9.128/25 (note that 8^th bit is 1).

Hence the network address 10.159.9.0/25 and  10.159.9.128/25 satisfied this constraint.

Answer 2

Since Subnet 1 has to support at least 100 interfaces, hence it needed total 100+2 = 102 IP addresses( we need 2 extra address i.e. first and last address which are reserved), and subnet 2 need 30+2 = 32 IP addresses. Hence number of unmasked bits needed in Subnet 1 = log, 102-7 bits , number of unmasked bits needed in Subnet 2 = log2 32 = 5 bits . Now in prefix 10.159.9/24, since we have first 24 bits masked, hence we have 32-24 = 8 bits left. So out of 8 bits, we have keep 1 bit to identify the subset. Hence 8th bit from right will be 0 for Subnet 1 and 1 for Subnet 2. Then 7 bits will be address of interface within the subnet. Hence the network address of Subnet 1 will be its first IP address = 10.159.9.0/25 (now we have 25 bits masked in each subnet) and network address of Subnet 1 will be its first IP address = 10.159.9.128/25 (note that 8th bit is 1). Hence the network address 10.159.9.0/25 and 10.159.9.128/25 satisfied this constraints. Please comment for any clarification.