IP subnetting

If you keep 172.17.32.0/20 as is, you have just one big network (of course).

If you use a /28 mask, then you have the following subnets (relative to the original address and mask):

172.17.32.0/28 .00100000.0000 0000 (last two octets, space at /28) 172.17.32.16/28 .00100000.0001 0000 172.17.32.32/28 .00100000.0010 0000 172.17.32.48/28 .00100000.0011 0000 ... 172.17.32.240/28 .00100000.1111 0000 (16 subnets so far)

then:

172.17.33.0/28 .00100001.0000 0000 ... 172.17.33.240/28 .00100001.1111 0000 (other 16 subnets )

and so on until

172.17.47.0/28 .00101111.0000 0000 ... 172.17.47.240/28 .00101111.1111 0000

So, you have 16*16 subnets (eg, 256) if you use a /28 mask.

You can also see this very quickly if you think that you gain 8 bits for the subnet part (/20 -> /28), thus you have 2^8 = 256 subnets.

"RIckH" ha scritto nel messaggio news: snipped-for-privacy@n2g2000hse.googlegroups.com...

Hi,

2^(28-20) = 256 subnets (with ip subnet-zero)

172.17.32.0/20 172.17.32.0-172.17.47.255

172.17.32.0 (16 subnets) 172.17.32.0-172.17.32.15 172.17.32.16-172.17.32.31 .... .... 172.17.32.240-172.17.32.255

172.17.33.0 (16 subnets) 172.17.33.0-172.17.33.15 172.17.33.16-172.17.33.31 .... .... 172.17.33.240-172.17.33.255 ..... .....

172.17.47.0 (16 subnets) 172.17.47.0-172.17.47.15 172.17.47.16-172.17.47.31 .... .... 172.17.47.240-172.17.47.255

Regards, Gabriele

If you keep 172.17.32.0/20 as is, you have just one big network (of course).

If you use a /28 mask, then you have the following subnets (relative to the original address and mask):

172.17.32.0/28 .00100000.0000 0000 (last two octets, space at /28) 172.17.32.16/28 .00100000.0001 0000 172.17.32.32/28 .00100000.0010 0000 172.17.32.48/28 .00100000.0011 0000 ... 172.17.32.240/28 .00100000.1111 0000 (16 subnets so far)

then:

172.17.33.0/28 .00100001.0000 0000 ... 172.17.33.240/28 .00100001.1111 0000 (other 16 subnets )

and so on until

172.17.47.0/28 .00101111.0000 0000 ... 172.17.47.240/28 .00101111.1111 0000

So, you have 16*16 subnets (eg, 256) if you use a /28 mask.

You can also see this very quickly if you think that you gain 8 bits for the subnet part (/20 -> /28), thus you have 2^8 = 256 subnets.

Ok, strictly speaking, you "gain" 255 subnets (you had one, and now you have

256), but I hope you got the idea.

Thanks, That makes perfect sense now.

I found a site that does a really good job at explaining this (In My humble oppinion)

Site:

the actuall explanation: Cheers, the_Druid