Subnetting Example - Question

What is the last valid host on the subnetwork 192.168.156.0 /26

well if its a /26 then its a 192 subnet right? so how is there a 156.0 subnet when the subnets on a 192 are 0, 64, 128, 192, 256

I am seeing this as a class C address (because its /26 not /1

, but even if its class B I still dont see how there would be a 156.0 subnet - can anybody help explain

The answer is 192.168.156.62 - but why?

thnx

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Comments

trackit

you are right in that /26 means 192 and increment in 64 but you are forgetting that the action is going on in 4th octet, not third. So 192.168.156.0 subnetted with /26 means:

192.168.156.0 - 192.168.156.63
192.168.156.64 - 192.168.156.127
etc

so, as the network in intrest is 192.168.156.0 - 192.168.156.63
192.168.156.63 is broadcast so that leavs 192.168.156.62 as last valid host address

amtt81

OH! Right!

so the subnet IDs would be:

192.168.156.0
192.168.156.64
192.168.156.128
192.168.156.192

/26 is 192 mask
with 4 subnets, each with 62 hosts - so that too would tell me that the last valid host in any of the 4 subnets would be a .62

right?

thnx again

trackit

yap, exactly!

no problem

EDIT:

just wanted to make sure you got it right

"with 4 subnets, each with 62 hosts - so that too would tell me that the last valid host in any of the 4 subnets would be a .62"

not sure what you meant by that, but the subnets are:

192.168.156.0 (valid hosts 192.168.156.1 - 192.168.156.62)
192.168.156.64 (valid hosts 192.168.156.65 - 192.168.156.126)
192.168.156.128 (valid hosts 192.168.156.129 - 192.168.156.190)
192.168.156.192 (valid hosts 192.168.156.193 - 192.168.156.254)

amtt81

Oh, I see where I was wrong with that "the last host would always be" thing. Thanks for the edit - now I get EVERYTHING on this one. I am still new to this, so I appreciate your help immensely.