Subnetting help

[Jammin1984]

Right, im going over subnets again after doing them last year, i thought i had remembered it until the Cisco testerererer cd hit me with this: (I have put this in code tags as it allows spacing)

        Given the IP address 172.156.100.100 & a subnet mask of 255.255.255.224 what is the broadcast address:

        A:172.156.100.127

        B: 172.156.100.255

        C: 172.156.32

        D: 172.156.100.128
 
So I have done my working like this:

IP = 172.156.100.100 /27

.                            | N | N | N | S |  H  |
Broadcast:                               011 11111    = 172.156.100.127

But the correct answer is D: 172.156.100.128??

I have tried to work through the example given again using ANDing:

Subnet mask = 225.225.225.224
IP                 = 172.156.100.100

Only showing 'The intresting Octet':

Subent =    224   =   11100000
IP     =    100   =   01100100
                       ------------
Network address =     01100000 = 96

Therefore;

Broadcast       =     01111111 = 127

Could the Cisco CD have a little error, or a i doing this the wrong way?

I have worked through another example from the same CD:

What is the broadcast for IP subnet 192.168.100.32, given a mask of 255.255.255.244.

So, 192.168.100.32 /27

        | S  |    H    |
224      111  00000
         001  00000    - Network
         001  11111    - Broadcast = 192.168.100.63

Which, according to the CD is correct. Is the CD wrong? This is confusing me a lil.

Thanks much

[shoum]

May be you are right.There will be some mistake in cisco cd.

Put your IP address and Subnet Mask.The result will be the same.

http://www.subnetonline.com/subcalc/subnet8.html

[sharkyro]

Yes, the CD is wrong. The broadcast address for 172.156.100.100/27 is 172.156.100.127... 172.156.100.128 is the network address for the following subnet.