Quick IP Subnet question

[twelve]

I recently had my ISP set up a few extra IP addresses for me and they sent me the info like so:

192.168.1.8/29
255.255.255.248

I changed the actual address so no one gets any bad ideas, but how do I determine exactly which IP addresses I have now? I think there should be 8 useable.

[Ron Prestenback]

Put it in binary to understand what to do:

11111111.11111111.11111111.11111000 --- /29 or 255.255.255.248

192.168.1 -- will be the same for all your clients, so let's forget about that part

11111000 --- /29 from above (subnet)
00001000 --- 8 (network address) so

all client addresses must match the network address only where there is a "1" in the subnet, so

00001001 --- 9
00001011 --- 11
00001101 --- 13
00001110 --- 14

These are your valid addresses for your network.
192.168.1.9
192.168.1.11
192.168.1.13
192.168.1.14

Your broadcast addresses for this network are:
192.168.1.8 or 00001000
192.168.1.15 or 00001111

Notice for broadcast address, our "client" part of the octet, i.e. the last three digits, are all 0's or all 1's ---- and that is why they are not valid client addresses..

Ron

[twelve]

Thanks... um... you don't do the binary conversion in your head do you? I would think that would hurt.

[Ron Prestenback]

HAHAHA

Yes I do it in my head, but not the way you think...I do from memory (I know how many ones equal 248, etc.) from doing it so many times...but, if you need help with it, the windows calculator converts easily.

Just make sure you're in scientific view, and type in a number, then put click on the BIN radio button, and that number will instantly be converted to BIN, and vice versa. If the radio button is on BIN you can type 11111000 then click on DEC and it will be converted to 248.

Ron

[widget]

.10 is valid also.
An easy way for me to figure out approx. how many addys you can use is take number of host bits left over (in your case 3, from your mask) and put into equation 2 to the x minus 2 ...viola you got 6 addys to use for nodes.

[Ron Prestenback]

Yep, you're right...I missed .10 (11111010)

And for your formula, let me clarify (I always like to clarify...<shrug>)

2 ^ x - 2 = [number of subnets]
x = number of subnet bits that are 0 so
2 ^ 3 = 8, except that we must subtract the two broadcase addresses, no matter what subnet mask we have, so that leaves us with 6 addresses.

Thanks for catching that for me!

Ron