IP Addressing

[isdollsm1]

Just a questions on this. This seems simple but doesn't make sense to me. I have seen engineers who have an IP of 192.168.100.1/26 for an SVI address. My question is if you are going to use a /26 why would you use .1 as your SVI IP? Shouldn't you use 192.168.100.65/26? Doesn't this defeat the purpose of using a /26? Isn't 192.168.100.1 on another network? Thanks for any response

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[Joseph W. Doherty]

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Posting

Unclear you confusion.  192.168.100.1/26 is indeed in a different network from 192.168.100.65/26.  192.168.100.1/26 is the first IP in its /26 network, as 192.168.100.65/26 is the first in the next /26 network.

[johnd2310]

Hi,

Your question needs clarification. 192.168.100.1/26 is the first address on the 192.168.100.0/26 network and so is a valid address to use. 192.168.100.65/26 is the first address on the 192.168.100.64/26 network. so you can have the following configuration:

interface vlan 10

ip address 192.168.100.1 255.255.255.192

interface vlan 20

ip address 192.168.100.65 255.255.255.192

[isdollsm1]

Would that be the first address because the address is a class C and should be a /24 and since we are moving over the bits to a /26 the 192.168.100.0 would be covered under this?

VLAN 10

192.168.100.1 255.255.255.192 is the SVI for VLAN10

Should I start my dhcp at 192.168.100.10 or start at 192.168.100.65. It seems I should start with the .65 since I am using a /26.

Thanks

[johnd2310]

Hi,

192.168.100.65 is in a different network to 192.168.100.1 Since SVI for vlan 10 is 192.168.100.1 your dhcp range would be from 192.168.100.2 - 192.168.100.62

Have a look at the following doc that explains a little about subnets:

http://www.cisco.com/c/en/us/support/docs/ip/routing-information-protocol-rip/13788-3.html

Thanks

John

[devils_advocate]

The default subnet mask for a Class C address is /24 which will give you 192.168.100.1 through 192.168.100.254 as usable addresses.

If you choose not to use the default mask of /24, you can split that /24 subnet up into four /26 subnets. Exactly four /26 subnets make up a /24 subnet, although that may not be the best way of thinking about it if you are learning :)

If you want to use 192.168.100.0 as your example and you wanted to use a /26 mask, you would end up with the following subnets:

192.168.100.0/26
192.168.100.1 - 192.168.100.62 for hosts
192.168.100.63 is broadcast address

192.168.100.64/25
192.168.100.65 - 192.168.100.126
192.168.100.127 is broadcast address

192.168.100.128/26
192.168.100.129 - 192.168.100.190
192.168.100.191 is broadcast address

192.168.100.192/26
192.168.100.193 - 192.168.100.254
192.168.100.255 is broadcast address

If you choose to split the Class C 192.168.100.0/24 network up into four smaller /26 networks, you would get the subnets above. Obviously you would not longer be able to use 192.168.100.0 as a /24. 

/24 is just the 'default' subnet mask for Class C addresses but you can make a subnet smaller or bigger by changing the subnet mask. Using a /26 is making it smaller and allowing you to have four smaller subnets instead of one larger /24 subnet.

Hopefully that makes some sense...