Well, since I'm not a qualified psychiatrist I don't want to comment on your mental
health, but there is a correct answer here. If we go to binary it all becomes clear.
The /19 means that the first 19 bits are network/subnetwork bits and the remaining 13
bits are host bits. Therefore our mask would look like this:
11111111.11111111.11100000.00000000
Since the first two octects are identical, we can dispense with them and concentrate on
the first 3 bits of the third octet and the remaining host bits.
Our subnet addresses would be as follows:
X.X.00000000.00000000 = X.X.0.0 Hosts = X.X.0.1 - X.X.31.254 Bdcst = X.X.31.255*
*Assumes Subnet Zero Allowed
X.X.00100000.00000000 = X.X.32.0 Hosts = X.X.32.1 - X.X.63.254 Bdcst = X.X.63.255
X.X.01000000.00000000 = X.X.64.0 Hosts = X.X.64.1 - X.X.95.254 Bdcst = X.X.95.255
X.X.01100000.00000000 = X.X.96.0 Etcetera, Etcetera....
.
.
Etcetera (You get the picture)
>From this you can see that:
answer A is the "wire" address of the 32 subnet
answer B is the "wire" address of the 64 subnet
answer C is the Broadcast address of the 32 subnet
answer D falls within the valid host range for the 64 subnet and is correct.
BTW, a good source for learning IP Addressing & subnetting is: www.learntosubnet.com
HTH,
Prof. Tom Lisa, CCAI
Community College of Southern Nevada
Cisco Regional Networking Academy
Bruce wrote:
> Q. Which one of the following is a valid host using the address of
> 172.16.0.0 /19?
>
> a. 172.16.32.0
>
> b. 172.16.64.0
>
> c. 172.16.63.255
>
> d. 172.16.80.255
>
> Which one and why?
>
> (I say none of them. Am I going mad?)
>
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