What happens at /32?

A normal network requires a network address (the start of the subnet), some host addresses for the devices, and a broadcast address (the last address in the subnet) which can be used by any device to blanket request all devices on a network to respond to a who has query for which device has a specific IP address allocated, that the requesting device wants to contact.

The amount of available device addresses is therefore always 2 less than the total, because of the network and broadcast addresses allocated and required in the traditional network model. But as an IP address is only a virtual address assigned to a real interface, unlike the real burnt in, uniquely defining MAC address, surely we can do virtually anything in programming to an interface like sub-interfacing etc...and we do that...so what happens at /32...?

----------- NET ------------------|HOST- 8 host bits = 2^8 256 (254 devices) 0 subnet bits =2^0=1 subnet of 256

/25 1111 1111 | 1111 1111 | 1111 1111 | 1000 0000 255.255.255.128 128 addresses

----------- NET ------------------- -|<-HOST 7 host bits = 2^7 128 (126 devices) 1 subnet bits =2^1=2 subnets of 128

/26 1111 1111 | 1111 1111 | 1111 1111 | 1100 0000 255.255.255.192 64 addresses

----------- NET ------------------- -- | 6 host bits = 2^6 64 (62 devices) 2 subnet bits =2^2=4 subnets of 64

/27 1111 1111 | 1111 1111 | 1111 1111 | 1110 0000 255.255.255.224 32 addresses

----------- NET ------------------ - ---| 5 host bits = 2^5=32 32 (30 devices) 3 subnet bits =2^3=8 subnets of 32

/28 1111 1111 | 1111 1111 | 1111 1111 | 1111 0000 255.255.255.240 16 addresses

----------- NE-- - -----------------------| 4 host bits = 2^4=16 16 (14 devices) 4 subnet bits =2^4=16 subnets of 16

/29 1111 1111 | 1111 1111 | 1111 1111 | 1111 1000 255.255.255.248 8 addresses

----------- N ---- - -----------------------| 3 host bits = 2^3=8 8 (6 devices) 5 subnet bits =2^5= 32 subnets of 8

NOW this is where the logic starts to get interesting: At this point we still have 1 network address, 2 host addresses and 1 broadcast address. However, as it is a PtP link, why is a broadcast address required for functionality...?

It isnt as far as routers connected across a PtP link are concerned, as there will be a static route or a routing protocol defining the ONLY IP address at the other end of the link at each router, so broadcasts wont be necessary between them will they?? (as they are the only responders available to each other anyway for that broadcast domain)...So from that premise, we can get:

/30 1111 1111 | 1111 1111 | 1111 1111 | 1111 1100 255.255.255.252 4 addresses (2 devices) but Bcast address not required??

----------- NET -----------------------------| 2 host bits = 2^2=4 4 (2 devices) 6 subnet bits =2^6= 64 subnets of 4

/31 1111 1111 | 1111 1111 | 1111 1111 | 1111 1110 255.255.255.254 2 addresses (1 PtP device = 1 net, 1 host address but NO Bcast address!!)

----------- N------- - -----------------------| 1 host bits = 2^1=2 2 (1 device) 7 subnet bits =2^7= 128 subnets of 2

/32 1111 1111 | 1111 1111 | 1111 1111 | 1111 1111 255.255.255.255 1 address (1 PtP device if 1 is a subnet of a larger network!!)

----------- NET -------------------------------| 0 host bits = 2^0=1 1 (1 device) 8 subnet bits=2^8=256 subnets of 1

If you look at your home ADSL modem IP address and subnet mask, you may find it is a /32, yet is obviously working without a traditional Bcast or Network address i.e. is a subnet of 1 address only, your IP address. The ISP address at the other end (your modems Default Gateway) is your Supernet address - the last example in the above list.

This makes sense if the ISP has its usual public address that yours is a tiny subnet of, along with all its other customers. As it is a PtP link between you and the ISP, a Bcast address is not necessary, as the ISP router knows you are the only address on that serial link, so your address is ALSO the subnet address?

What y'all think....?