|
Home > Archive > CCDA/CCDP > April 2003 > IP addressing question
You are viewing an archived Text-only version of the thread.
To view this thread in it's original format and/or if you want to reply to
this thread please [click here]
| Author |
IP addressing question
|
|
| meijin 2003-04-25, 3:03 pm |
| The following is a practice question from one of the new CCDA packages I have. This question does not have an explanation, so I am in hopes that someone here can explain it to me. It is as follows:
In which network is the IP address 144.10.63.0 a valid node address that can be used on a workstation or other network device?
A. 144.10.62.0123
B. 144.10.62.0124
C. 144.10.63.0124
D. 144.10.63.0 t25
Answer: A
Anyone want to 'splain it to me?
Thanks! | |
| anchor40 2003-04-25, 4:15 pm |
| For 144.10.63.0 to be a valid host:
A. 144.10.62.0 /23
512 hosts from .62.1 through .63.254 see below for more details
B. 144.10.62.0 /24
not even on this network
C. 144.10.63.0 /24
It's the network ID for this network - .1 would be the first valid host
D. 144.10.63.0 /25
126 hosts, but also network ID for this network; first valid is .1
For A, this CIDR block creates a pool of addresses. Look at the bits:
a.b.0011 111x.xxxx xxxx a.b.62.0 /25
OR
a.b.0011 1110.0000 0000 through
a.b.0011 1111.1111 1111
That means a.b.0011 1111.0000 0000 (63.0) is right in the middle of that pool of bits.
Clear as mud?? | |
| meijin 2003-04-25, 4:26 pm |
| Thanks for the response!
Mud, yes!
I guess it is the format of the IP addresses that is throwing me the most. | |
| anchor40 2003-04-25, 4:41 pm |
| I guess the easist way for this question to be determined is to look at each answer and jot down the valid hosts.
When you normally think of subnetting, you typically look at a /24 (255.255.255.0) and break it into smaller chunks:
/24 = .1 to .254
/25 = .1 to .126 AND .129 to .254
/26 = .1 to .62, .65 to .126, .129 to .190 AND .193 to .254
etc..
For answers B,C,and D, any .0 address is not valid (all network IDs)
The easist way I know to handle this is to get comfortable with binary, look at the masks provided, and determine the bits that matter. Odds are you will have a question along this line (more than 254 hosts on a network).
For this example, anything from 62.1 through 63.254 is a valid host address. Write out the binary for the last two octets, and you'll be able to visualize the pool of bits available for hosts.
In reality, only mask ranges less than the number 24 could possible have a .0 host address.
Drop me a note if you want me to take this from a different angle.
 |
|
|
|
|