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I have to split up a network into 11 subnetworks, and then those networks must be split up further. 

With the address 136.192.0.0\16:

I found that I have 2^(32-16) = 65,536 available addresses. 
I can make 11 groups of 4,096 addresses.
I then get 20 as the subnet mask prefix.

I end up with the following 11 IP Addesses, one for each network. 

Network ID              Subnet Mask

138.192.16.0            255.255.240.0 or \20
138.192.32.0            255.255.240.0 or \20
138.192.48.0            255.255.240.0 or \20
138.192.64.0            255.255.240.0 or \20
138.192.80.0            255.255.240.0 or \20
138.192.96.0            255.255.240.0 or \20
138.192.112.0           255.255.240.0 or \20

138.192.128.0           255.255.240.0 or \20
138.192.144.0           255.255.240.0 or \20
138.192.160.0           255.255.240.0 or \20
138.192.176.0           255.255.240.0 or \20

The first network address is 138.192.16.0\20 I need to split this address into 5 further subnets. 

I have 2^(32-20) = 4096 addresses available with the \20 subnet mask
If I split it up into 5 groups thats 4096/5 = 819.2 but we have to round 
that to a multiple of 2, so we get 5 groups of 512

Because 9 bits are for the host, that means 23 are for the network 
That gives me 255.255.254.0 for the subnet mask.

This is as far as I've gotten beause I am unsure how to proced with producing these 5 Network ID's. Im also uncertain if I'm doing it right.

I did find this resource How do you calculate the prefix, network, subnet, and host numbers? But am unsure of what its doing when subnetting.

Drew Ackerman
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  • You'll have to do your own homework – Ron Trunk Jun 15 '17 at 22:26
  • Seeking help for a concept is not having someone do my homework. I will reformat the question to better convey that I want help learning to submask – Drew Ackerman Jun 15 '17 at 22:32
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    The answer you linked to is one of the better resources. What part don't you understand? – Ron Trunk Jun 15 '17 at 22:38
  • Im uncertain If I have split the networks into sub networks correctly. i have edited the question to hopefully better explain my problem, and show the work i've already done. I dont think it is correct. – Drew Ackerman Jun 15 '17 at 22:49
  • Remove the first paragraph from your question – Ron Trunk Jun 15 '17 at 22:52
  • Question has been edited again – Drew Ackerman Jun 15 '17 at 23:07
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    please use forward slashes for subnet notation. – Teun Vink Jun 16 '17 at 05:25
  • It's easier if you work with binaries. Subnetting in 2 enlarges the mask by one bit with two possibilities for the new bit, in 4 by 2 bits with four possibilties (00, 01, 10, 11), and so on. Subnetting in 11 requires 4 bits (for 16 actual subnets), subnetting those into 5 requires another 3 bits (for 8 actual subnets). Subnets don't need to be the same size, so you could subnet 1.2.0.0/16 into 1.2.128.0/17, 1.2.0.0/18 and 1.2.64.0/18. – Zac67 Jun 26 '17 at 20:54

2 Answers2

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Your overall /16 (2^16 hosts) can be broken up into 16 /20's (2^12 hosts).

Each of your /20's could be turned into 8 /23 (2^9 host) or 16 /24 (2^8 host) subnets - or some combination therein.

Keep in mind that you might need five subnets out of the /20 but that doesn't mean you have to immediately fully consume the entire supernet. Allocation schemes tend to be a lot more extensible and it might actually be that further subnets within a site also make some sense.

With the hope that I'm not doing someone's homework, consider something more like the following - I have 5 user sites that range in size between 100 and 300 users. Each site may grow. Since you're using modern equipment there's not necessarily a requirement to have everyone on the same L2 domain within a site, so a few slightly disjoint ranges aren't the end of the world. That said, in the macro sense you want to maintain clean boundaries for summarization to facilitate routing protocol scalability and ease of application recognition/security/etc.

There are a few ways to do this. One obvious way would be to allocate the /23's contiguously from 0. The remaining 3 /23's would be allocated on a /24 basis (6 total) on an as-needed basis if/when the sites grow.

Alternately, perhaps you start from the premise that the 2 sites that are already at 300 users would be allocated a /22, out of which each gets the lower /23. If/when they grow there's already another /23 ready. The smaller 3 sites each get /23's. This leaves you with an unused /23, which could be broken up into two /24's to use if either of the small sites grows.

Scale this concept out and perhaps you don't allocate the entire /16 into /20's. Maybe you take the 55 sites you're describing (11 supernets of 5 subnets each, right?) and do partial allocations out of individual /21's (say the first /23 to start with for each) while leaving the remaining 9 /21's for other sites.

The other thing to consider is that there's some amount of use in most address schemes for things other than end users. Examples would include management loopbacks, small subnets for out-of-band ports, subnets for servers and, especially, point-to-point links. The kicker with these is that as soon as you've allocated a single /31 or /32 you've already broken up a subnet. This tends to call for setting aside a chunk of the supernet for these kinds of infrastructure functions. Maybe the /21 overall allocation actually starts as the first /23, then a reserved /24 for growth and then a /24 for infrastructure (...which perhaps includes a /25 for local servers, a /26 for OOB management, a /27 for loopbacks, etc).

The point is that you can slice this up an any number of ways depending on your needs. It's all just powers of 2, after all. Put it in a spreadsheet and see what both makes sense now and leaves some room to grow without blowing up the routing table.

rnxrx
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/16 to /20 4bits, 16 subnets

/20 to /24 4bits, 16 subnets with 254(265) hosts each.

Total = 16 x 16 x 256 = 65536 ip's

or

/16 to /20 4bits, 16 subnets

/20 to /23 3bits, 8 subnets with 510(512) hosts each.

Total = 16 x 8 x 512 = 65536 ip's

Pieter
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