Subnetting Question

Blang008

Can somebody explain this?

Question: What is the first valid host on the subnetwork that the node 172.24.196.247 255.255.254.0 belongs to?

Answer: 172.24.196.1

mike3

Blang008 wrote:

Can somebody explain this?

Question: What is the first valid host on the subnetwork that the node 172.24.196.247 255.255.254.0 belongs to?

Answer: 172.24.196.1

It's a class B IP address. So the network is 172.24.195.0 and the second is 172.24.197.0. So the first node on the network 172.24.195.0 is 172.24.196.1. You can have any IP of 172.24.196.1-254 because IP address falls in the range of 196.


I'm not sure how else to explain this. It's not a class C IP address.

Does that make sense?

dynamik

AND the bits to get the (sub)network number
10101100.00011000.11000100.11110111 - 172.24.196.247
11111111.11111111.11111110.00000000 - 255.255.254.0
=========================================
10101100.00011000.11000100.00000000 - 172.24.196.0

172.24.196.0 is the network address, so 172.24.196.1 is the first usable address.

nel

niter3 wrote:

Blang008 wrote:

Can somebody explain this?

Question: What is the first valid host on the subnetwork that the node 172.24.196.247 255.255.254.0 belongs to?

Answer: 172.24.196.1

It's a class B IP address. So the network is 172.24.195.0 and the second is 172.24.197.0. So the first node on the network 172.24.195.0 is 172.24.196.1. You can have any IP of 172.24.196.1-254 because IP address falls in the range of 196.


I'm not sure how else to explain this. It's not a class C IP address.

Does that make sense?

The network address is actually 172.24.196.0, meaning the first host is 196.1. the broadcast is 172.24.197.255. Last host is obviously 197.254.

To get the answer i used odoms decimal method (which dynamik will slate me for )

Think of it this way any octet which has a subnet mask of 255 or 0's leave it intact (thats the first 2 octets - 172.24) the third octet being 254 is the one to concentrate on. i subtracted this from 256 to get 2. you then add multiples of 2 until you get to the nearest mutiple of 2 to the IP address you have - 196. So here you have 172.24.196.0. To work out the broadcast add 2 to 196 and you get 198.0. Now subtract 1 from 198.0 to get the broadcast 197.255. and so on.

Thats probably a rubbish explanation but i hope it makes some sense!

EDIT: Woah dynamik how the hell did you get in there before me

mike3

nel wrote:

niter3 wrote:

Blang008 wrote:

Can somebody explain this?

Question: What is the first valid host on the subnetwork that the node 172.24.196.247 255.255.254.0 belongs to?

Answer: 172.24.196.1

It's a class B IP address. So the network is 172.24.195.0 and the second is 172.24.197.0. So the first node on the network 172.24.195.0 is 172.24.196.1. You can have any IP of 172.24.196.1-254 because IP address falls in the range of 196.


I'm not sure how else to explain this. It's not a class C IP address.

Does that make sense?

The network address is actually 172.24.196.0, meaning the first host is 196.1. the broadcast is 172.24.197.255. Last host is obviously 197.254.

To get the answer i used odoms decimal method (which dynamik will slate me for )

Think of it this way any octet which has a subnet mask of 255 or 0's leave it intact (thats the first 2 octets - 172.24) the third octet being 254 is the one to concentrate on. i subtracted this from 256 to get 2. you then add multiples of 2 until you get to the nearest mutiple of 2 to the IP address you have - 196. So here you have 172.24.196.0. To work out the broadcast add 2 to 196 and you get 198.0. Now subtract 1 from 198.0 to get the broadcast 197.255. and so on.

Thats probably a rubbish explanation but i hope it makes some sense!

EDIT: Woah dynamik how the hell did you get in there before me

That's right. I just caught what I wrote.. opps..