subnetting ?

jbwallace

Ok, so I've been going through these subnetting sites today practicing my subnetting. I ran across this problem though, and I cannot figure out how they came up with their answer. Can one of you guys show me where I'm going wrong, or confirm that they are wrong? The problem is:

What is the last valid host on the subnetwork 192.168.187.32/28?

There answer is: 192.168.187.46

I got 192.168.187.47

jbwallace

OOPS, I just saw the error of my ways. Never mind!!

Kevins51825

They are correct. broadcast is .47.

Slowhand

The answer they gave you is correct, it's all in the wording of the question. The last usable IP address is 192.168.187.46.

192.168.187.47 is the broadcast address and cannot be assigned to a host. So, your usable range is 192.168.187.33 - 192.168.187.46, with 192.168.187.32 being the network address.

Here's a chart that might be easier to refer to, courtesy of SolarWinds' Advanced Subnet Calculator (a handy tool for checking your answers, just don't rely on it instead of learning to subnet):

IP Address       : 192.168.187.32
Address Class    : Classless /28
Network Address  : 192.168.187.32

Subnet Address   : 192.168.187.32
Subnet Mask      : 255.255.255.240
Subnet bit mask  : nnnnnnnn.nnnnnnnn.nnnnnnnn.nnnnhhhh
Subnet Bits      : 28
Host Bits        : 4
Possible Number of Subnets : 1
Hosts per Subnet : 14