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I am revising some questions for an exam and one of the questions was provide the ip numbers and a diagram for a company which has 3 branches. Paris/Berlin/Cairo (random locations I know) Now I belive I have figured out the answer but I would appreciate someone checking as my brain is fried and I think if this answer is wrong I won't be able to see that's its wrong.

You have 3 routers which means 6 networks All should be /24, the IP addresses should resemble 10.12."network number"."host number" ie 10.12.59.1, so that its a /24 that means the mask is going to be 255.255.255.0, so my host ranges are, not including broadcast addresses

  • 1-30
  • 33-62
  • 65-94
  • 97-126
  • 129-158
  • 161-190
  • 193-222
  • 225-254

So I could use 6 of them for my 6 networks, 1 for each of the branches and 1 for the 3 networks joining the branches.

The Berlin branch network is extended with 2 additional VLANS. These networks should be sub-netted from 10.12.59.0 with a /27 structure. Now this is where I start to know I am weak in my knowledge. the mask for a CIDR/27 is 255.255.255.224 but I cant figure out how to implment a /27 numbering within a /24 network. Can anyone explain please?

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You're the teacher and you need help with subnetting? –  SpacemanSpiff Apr 13 '12 at 13:18
4  
@spacemanspiff - if he has a doubt or hesitation, would you prefer a teacher not ask for someone to double check? After working with teachers, I've learned their dirty secret...they're not infallible. :-) –  Bart Silverstrim Apr 13 '12 at 13:39

1 Answer 1

This is done with Variable Length Subnetting.

To do VLSN we just look at all our networks and arrange them in size. First the big ones then the smaller ones. You already have your /24 networks so to correct your subnets:

if you have 10.12.0.0 and you need 6 /24 networks and 2 /27, then we need to first accommodate the /24 networks:

So we get:

 Subnet 1:
10.12.0.0 (networkaddress)
10.12.0.1 (first host)
10.12.0.254 (last host)
10.12.0.255 (broadcast address)
Subnet 2:
10.12.1.0 (networkaddress)
10.12.1.1 (first host)
10.12.1.254 (last host)
10.12.1.255 (broadcast address)
Subnet 3:
10.12.2.0 (networkaddress)
10.12.2.1 (first host)
10.12.2.254 (last host)
10.12.2.255 (broadcast address)
Subnet 4:
10.12.3.0 (networkaddress)
10.12.3.1 (first host)
10.12.3.254 (last host)
10.12.3.255 (broadcast address)
Subnet 5:
10.12.4.0 (networkaddress)
10.12.4.1 (first host)
10.12.4.254 (last host)
10.12.4.255 (broadcast address)
Subnet 6:
10.12.5.0 (networkaddress)
10.12.5.1 (first host)
10.12.5.254 (last host)
10.12.5.255 (broadcast address)

Now we need to accommodate two /27 networks in one more /24 networks. Let's first see if two /27 fit into a /24:

/27 = 32 IP's (1 network, 30 hosts, 1 broadcast). 
/24 = 256 IP's (1 network, 254 hosts, 1 broadcast). 

So we can fit in 8 /27 networks basically, but we only need two, our network address to put the /27 in is 10.12.6.0. So now you start as you would do normally:

Subnet 7:
10.12.6.0 (networkaddress)
10.12.6.1 (first host)
10.12.6.30 (last host)
10.12.6.31 (broadcast address)
Subnet 8:
10.12.6.32 (network address)
10.12.6.33 (first host)
10.12.6.62 (last host)
10.12.6.63 (network address)

Good luck with your exam :)

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Sorry Can you just explain how you can fit 4 x /27 networks into a /24 I get confused combining the 2 - Thanks –  Peter davies Apr 15 '12 at 12:51
2  
I did...it's basic math you have a /24 network (10.12.6.0/24) you just look how many times a /27 can fit in there (8 times 8*32=256). So you just subnet within your /24, starting with the /24 first ip. I mean it's not different from subnetting a /24, because that can basically be a subnet within a /23. –  Lucas Kauffman Apr 15 '12 at 13:01

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