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I've been given a Class C address, let's say 192.10.20.0. 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 192.10.20.0 (Network Address of subnet 1) to 192.10.20.15 (Broadcast Address of subnet 1) and on the next address (192.10.20.16), a new subnet will begin. Right?

So for the 14 hosts on my first subnet, I assigned the IPs from 192.10.20.1 to 192.10.20.14 (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 192.10.20.1 to 192.10.20.14 are already assigned to hosts, 192.10.20.0 is the Network Address (can't be assigned) and 192.10.20.15 is the Broadcast Address (also can't be assigned).

closed as off-topic by Ron Maupin Jun 24 '18 at 1:16

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Network classes are dead and buried since 1998, check RTF 2317.

A router can use any free IP address of a subnet. The host parts with all zeroes or all ones are used for the network and the broadcast address respectively, of course. 14 hosts plus one router don't fit in a /28 subnet.

In real life, you'd usually use a subnet mask of /24 with private addresses, so you'd have plenty of space for your small numbers. Public addresses are scarce nowadays, so you'd probably use IPv6 to start with.

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The router interface, (or sub-interface), will need an IP within the range of your new host subnet (I assume this is a new vlan). So, when building a new subnet, you don’t actually exclude 2 addresses for the net ID and broadcast, you actually exclude 3 addresses; the third being the router’s interface (or sub-interface) IP. So you’ll actually need a /27.

Then you’ll have to add ip helpers, or dhcp relays to the router interface (or sub-interface). Then your dhcp server will see the request coming from the router interface IP (let’s say you used 192.10.20.1/27), and it will hand out the address in the range of 192.10.20.2 - 192.10.20.30. This is assuming that you’re using dhcp and will build the scope to exclude the three ips for net ID, broadcast, and router interface.

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