Basic addressing scheme question

Lincoln Hawk

I have a router that is directly connected to a switch (from fa0/0), and the switch is connected to a PC. This network is 172.16.0.0/16. On the other side of the router, fa0/1, I have the same set up (router is connected to a switch, switch to a PC) and this network is 172.17.0.0/16.

I have instructions here to assign the first ip address to the router interface. I’m unsure if the first ip address for network 172.16.0.0/16 will be 172.16.1.0 or 172.16.0.1.

Could someone please explain this answer? Thanks in advance.

Neeko

Lincoln Hawk wrote: »

I have a router that is directly connected to a switch (from fa0/0), and the switch is connected to a PC. This network is 172.16.0.0/16. On the other side of the router, fa0/1, I have the same set up (router is connected to a switch, switch to a PC) and this network is 172.17.0.0/16.

I have instructions here to assign the first ip address to the router interface. I’m unsure if the first ip address for network 172.16.0.0/16 will be 172.16.1.0 or 172.16.0.1.

Could someone please explain this answer? Thanks in advance.

It would be 172.16.0.1. The numbers are just representations of binary which work right to left in powers of 2, much like our regular number system does with powers of 10. So the far right of each octet is 1, meaning the first usable IP when you have two octets for hosts would be 172.16.0.1

kryolla

if you look at it in binary

172.16 | 0.0 /16
Network | Host

2 to the 16 will give you 65,536

Instead of looking at 0.0 which is 2 octets look at it combined into 1

x x x x x 1024 512 256 | 128 64 32 16 8 4 2 1

so for 0 it will be all zeros which is your network and all 1's is you broadcast and you already know what 1 thru 255 is, now what is 256, 257th HOST and so on. Hopefully I didn't confuse you even more.