koitsu
Premium
join:2002-07-16
Mountain View, CA
1 edit | Re: How do I determine subnet manually? I think this was posted to the wrong forum. Should be over in one of the Networking or Advanced Networking forums. 
Regardless, an attempt answer your question -- and it's hard to answer because I'm not completely sure I understand the question fully:
You'll want to split 141.67.0.0/16 into separate subnets that offer enough IP space to guarantee at least 4000 (usable) IPs per block. That's fairly easy to figure out. Let's start with a /24 (what you call a "Class C") and break it down from there:
/24 = 256 IPs
/23 = 512 IPs
/22 = 1024 IPs
/21 = 2048 IPs
/20 = 4096 IPs
So what you'd have is multiple /20 subnets, guaranteeing 4093 usable IPs out of each of those netblocks (lost 3 IPs go to gateway, broadcast, and network).
How many of those subnets would you have? Well, that's simple: A /16 gives you 65536 IPs, so do some simple math:
65536 / 4096 = 16
You'd have a total of 16 /20 netblocks to use. Thus, your segregated netblocks would be:
141.67.[[16*0 through 16*15]].0/20, e.g.:
141.67.0.0/20
141.67.16.0/20
141.67.32.0/20
141.67.48.0/20
141.67.64.0/20
141.67.80.0/20
141.67.96.0/20
141.67.112.0/20
141.67.128.0/20
141.67.144.0/20
141.67.160.0/20
141.67.176.0/20
141.67.192.0/20
141.67.208.0/20
141.67.224.0/20
141.67.248.0/20
Now, you lose 3 IPs per /20 (one for network, one for broadcast, and one for gateway [optional]). So ultimately you're "losing" 48 IPs due to the subnetting of the /16.
Hope this makes sense.
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