Please, explain to me this subnetting!

What is the first valid host on the subnetwork that the node 172.23.195.117/26 belongs to?
Method #1
/26 = 192
256-192=64

192/64 = 3
3* 64 = 192 <-- our subnet

1st ip : 172.23.192.1
last ip: 172.23.255.1
BC:172.23.255.255

Method #2

64.0 , 192.0, 255.0
x , 192.1
x ,254.254
x ,254.255

The right answer: 172.23.195.65

HOW ?!

Thanks,

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Comments

gosh1976

if you have a /26 subnet mask you wouldn't be doing anything to the third octet and since you are borrowing 2 bits form the last octet then you would know that the "chunk" size of the networks will be 64...

so the network id's would be:
x.x.x.0
x.x.x.64
x.x.x.128

172.23.195.117/26 is on the second network so you take the network id 172.23.195.64 and add one to get the first host 172.23.195.65

miller811

marvix wrote: »

What is the first valid host on the subnetwork that the node 172.23.195.117/26 belongs to?
Method #1
/26 = 192
256-192=64

192/64 = 3
3* 64 = 192 <-- our subnet

1st ip : 172.23.192.1
last ip: 172.23.255.1
BC:172.23.255.255

Method #2

64.0 , 192.0, 255.0
x , 192.1
x ,254.254
x ,254.255

The right answer: 172.23.195.65

HOW ?!

Thanks,

you are heading in the right direction but took a wrong turn

172.23.195.117/26 means the interesting octet is the last one

so 172.23.195.0 is your stating point
/26 mean as you stated 256-192 = 64, so you block size is 64.
172.23.195.0 can be broken into 4 different subnets

172.23.195.0 - 172.23.195.63
172.23.195.64 - 172.23.195.127
172.23.195.128 - 172.23.195.191
172.23.195.192 - 172.23.195.255

so you fit in the second subnet
172.23.195.64
172.23.195.65 is the first available host
172.23.195.126 is the last available host
172.23.195.127 is the broadcast address

so the broadcast is one address below the next subnet
the last host is one below the broadcast

marvix

Thanks!