I've been given a Class C address, let's say 188.8.131.52. I have a topology of 4 routers, each with a serial connection to two of the other routers (like a square). To each router is connected two switches (through ETH), each switch carrying workgroups of 14 hosts. I've been asked to implement a subnetting scheme, in Cisco Packet Tracer.
Now, according to my lectures, I need to select a mask to accommodate a greater number of hosts than this value of 14, plus a further 2 addresses for the Network and Broadcast addresses (all 0s and all 1s). So I need to use the last 4 bits (2^4=16), or a /28 mask.
When using a /28 mask, the first subnet in my network will be from 184.108.40.206 (Network Address of subnet 1) to 220.127.116.11 (Broadcast Address of subnet 1) and on the next address (18.104.22.168), a new subnet will begin. Right?
So for the 14 hosts on my first subnet, I assigned the IPs from 22.214.171.124 to 126.96.36.199 (all 14 hosts have an individual IP). But for each host I also need to assign an address for gateway. And this gateway address is supposed to be the address I assigned to the ETH interface of the Router that is connected to that particular subnet's switch.
My question is: what address can I assign to the ETH interface of the Router? For the first subnet, the addresses ranging from 188.8.131.52 to 184.108.40.206 are already assigned to hosts, 220.127.116.11 is the Network Address (can't be assigned) and 18.104.22.168 is the Broadcast Address (also can't be assigned).