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IP VLAN Overlapping
Hi All,
I've been asked to setup a network with 9 VLANS and have been allocated an IP address of 10.0.0.0/21
I can create the first VLAN '10' as 10.0.0.1/21
When I go to create VLAN '20' as 10.0.1.1/21
I receive a 10.0.0.0 overlaps with Vlan '10'
I can only then add VLAN '20' as 10.0.8.1, although if I'm only allocated '10.0.0.0/21' would this still work? What would be best practice?
I'm using a CISCO 3960 Switch
I've been asked to setup a network with 9 VLANS and have been allocated an IP address of 10.0.0.0/21
I can create the first VLAN '10' as 10.0.0.1/21
When I go to create VLAN '20' as 10.0.1.1/21
I receive a 10.0.0.0 overlaps with Vlan '10'
I can only then add VLAN '20' as 10.0.8.1, although if I'm only allocated '10.0.0.0/21' would this still work? What would be best practice?
I'm using a CISCO 3960 Switch
Asked:
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There are a heap of ways this could be done. In the name of simplicity, I would probably use a /16 mask, and use the 2nd octet to select the VLAN.
ie, 10.1.0.1/16 = VLAN100
10.2.0.1/16 = VLAN200
10.3.0.1/16 = VLAN300.
That gives you 256 VLANS, each with 64K IPs.
ie, 10.1.0.1/16 = VLAN100
10.2.0.1/16 = VLAN200
10.3.0.1/16 = VLAN300.
That gives you 256 VLANS, each with 64K IPs.
Hi Malmensa,
Thanks for the reply.
I've only been allocated 10.0.0.1/21. What cab we do to work around this solution?
Thanks for the reply.
I've only been allocated 10.0.0.1/21. What cab we do to work around this solution?
You won;t get an even split, but if you did them as /25, you get as below, you could throw a couple of /24's in to to give you a few bigger subnets,..It really depends how many devices you need in each, or if there's a physical layout that might help, for example if splitting it by locations, one might be big so give it a /24
1) 10.0.0.1-126
2) 10.0.0.127-256
3) 10.0.1.1-126
4) 10.0.1.127-256
5) 10.0.2.1-126
6) 10.0.2.127-256
7) 10.0.3.1-126
8) 10.0.3.127-256
9) 10.0.4.1-126
10) 10.0.4.127-256
...
1) 10.0.0.1-126
2) 10.0.0.127-256
3) 10.0.1.1-126
4) 10.0.1.127-256
5) 10.0.2.1-126
6) 10.0.2.127-256
7) 10.0.3.1-126
8) 10.0.3.127-256
9) 10.0.4.1-126
10) 10.0.4.127-256
...
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Thanks for your input Andy - Mind if I explain something to you here?
Say there was 6 Buildings, 1 switch and I've been allocated 10.0.0.0/21
Would you then setup each building as the following ?
Building 1 10.0.0.0/21
Building 2 10.0.8.0/21
Building 3 10.0.24.0/21
Building 4 = 10.0.32.0/21
Building 5 = 10.0.40.0/21
Building 6 = 10.0.48.0/21
Say there was 6 Buildings, 1 switch and I've been allocated 10.0.0.0/21
Would you then setup each building as the following ?
Building 1 10.0.0.0/21
Building 2 10.0.8.0/21
Building 3 10.0.24.0/21
Building 4 = 10.0.32.0/21
Building 5 = 10.0.40.0/21
Building 6 = 10.0.48.0/21
Nope.
10.0.0.1/21 is a block of 2046 useable addresses, from 10.0.0.1 to 10.0.7.254
If you want to keep subnets within that range, you need to implement some smaller subnets.
As Andy said, you COULD use /25, and have 16 subnets with 126 useable addresses each. /24 would give you 8 subnets of 254.
If you have different numbers of hosts for each subnet, then it is possible to use different masks.
How many hosts are required in each subnet?
10.0.0.1/21 is a block of 2046 useable addresses, from 10.0.0.1 to 10.0.7.254
If you want to keep subnets within that range, you need to implement some smaller subnets.
As Andy said, you COULD use /25, and have 16 subnets with 126 useable addresses each. /24 would give you 8 subnets of 254.
If you have different numbers of hosts for each subnet, then it is possible to use different masks.
How many hosts are required in each subnet?
Malmensa,
6 buildings, lets just say there are 65 hosts per building and have been allocated 10.0.0.0/21
6 buildings, lets just say there are 65 hosts per building and have been allocated 10.0.0.0/21
65 is a difficult number. A /26 would give you 62 hosts per subnet.
Using /25 you can get 16 subnets of 126 useable IPs each.
Andy nearly had it.
Your actual subnets will be
10.0.0.1/25
10.0.0.129/25
10.0.1.1/25
10.0.1.129/25
10.0.2.1/25
10.0.2.129/25
10.0.3.1/25
10.0.3.129/25
10.0.4.1/25
Using /25 you can get 16 subnets of 126 useable IPs each.
Andy nearly had it.
Your actual subnets will be
10.0.0.1/25
10.0.0.129/25
10.0.1.1/25
10.0.1.129/25
10.0.2.1/25
10.0.2.129/25
10.0.3.1/25
10.0.3.129/25
10.0.4.1/25
Actually, your subnet addresses would be:
10.0.0.0/25
10.0.0.128/25
10.0.1.0/25
10.0.1.128/25
10.0.2.0/25
10.0.2.128/25
10.0.3.0/25
10.0.3.128/25
10.0.4.0/25
Unless you know that you'll never have more than 62 hosts per subnet, I would go with a /24.
10.0.0.0/24
10.0.1.0/24
10.0.2.0/24
10.0.3.0/24
10.0.4.0/24
10.0.5.0/24
Each subnet can have 254 hosts and you've got two unused subnets.
10.0.0.0/25
10.0.0.128/25
10.0.1.0/25
10.0.1.128/25
10.0.2.0/25
10.0.2.128/25
10.0.3.0/25
10.0.3.128/25
10.0.4.0/25
Unless you know that you'll never have more than 62 hosts per subnet, I would go with a /24.
10.0.0.0/24
10.0.1.0/24
10.0.2.0/24
10.0.3.0/24
10.0.4.0/24
10.0.5.0/24
Each subnet can have 254 hosts and you've got two unused subnets.
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You'll need to break down the /21 you have into smaller subnets, The number of hosts you require in each network might help you determine the exact split. some might be small subnets, others might be larger..