Reputation: 2348
First of all, I apologize for now knowing how to subnet better. It's not something I've had to do up until now.
I have been given a network/cidr of 10.10.32.0/19.
I have 6 subnets that need 255 ips each.
The rest of the IPs I would like to spread across 3 remaining subnets.
This is what i came up with using cidrsubnet
cidrsubnet("10.10.32.0/19", 2, 0) # "10.10.64.0/21" 2048 hosts
cidrsubnet("10.10.32.0/19", 2, 1) # "10.10.72.0/21" 2048 hosts
cidrsubnet("10.10.32.0/19", 2, 2) # "10.10.80.0/21" 2048 hosts
cidrsubnet("10.10.32.0/19", 5, 24) # "10.10.88.0/24" 255 hosts
cidrsubnet("10.10.32.0/19", 5, 25) # "10.10.89.0/24" 255 hosts
cidrsubnet("10.10.32.0/19", 5, 26) # "10.10.90.0/24" 255 hosts
cidrsubnet("10.10.32.0/19", 5, 27) # "10.10.91.0/24" 255 hosts
cidrsubnet("10.10.32.0/19", 5, 28) # "10.10.92.0/24" 255 hosts
cidrsubnet("10.10.32.0/19", 5, 29) # "10.10.93.0/24" 255 hosts
This leaves me with about 512 ips that can't be used. Am I doing this right?
Upvotes: 1
Views: 545
Reputation: 40326
You have 2^11
(32-19) hosts == 8192
Your range begins at 10.10.32.0
and runs through 10.10.63.255
6*256+3*2048 == 7680
leaving the 512
You'll need to rebalance so you don't waste IPs.
Your first cidrsubnet(10.10.32/19,2,0)
is 10.10.32.0--10.10.39.255
Upvotes: 2