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Home > Archive > CCNA > May 2003 > Subnetting 101-Lab 05/07/2003
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Subnetting 101-Lab 05/07/2003
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| anchor40 2003-05-07, 8:54 am |
| code:
Here's today's question.
Given the network 172.16.31.0, you must create 6 subnets to
support 24 users each. How many are available for future growth?
Please remember, don't reply to this thread with the answer,
so that everyone can have the opportunity to try it for
themselves, and the answer will be posted tomorrow.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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Now for the answer to the 5/6/03 question.
Which of the following statements are true?
A. 172.16.168.0/27 is a valid host address
B. 172.16.168.255/23 is a valid host address
C. 172.16.168.0/23 is not a valid host address
D. 172.16.168.255/26 is a valid host address
Answer: Both B and C are correct. 172.16.168.1-172.16.169.254 are
valid hosts on the /23 network.
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| soccer4net 2003-05-07, 10:06 am |
| Should we assume /24 because of the .31 in the third octet? | |
| anchor40 2003-05-07, 1:14 pm |
| Who proof read these? They should be flogged!!!
Oh, wait, er, uh, never mind... 
Yes, it should have been 172.16.31.0/24 for the provided network.
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| PooPooPlatter 2003-05-07, 2:27 pm |
| So I got C right, but I don't understand B...
B. 172.16.168.255/23 is a valid host address
I thought it means if 172.16.168.255 is a valid host address.... I thought 255 is for boardcast...then its not a host address
Please explain. | |
| anchor40 2003-05-07, 2:48 pm |
| The thing to remember is that "255" is the broadcast address ONLY if it makes the host address all 1s, as in a /24 network.
Look at the bits:
A /23 means use the left 23 bits for network, and the right 9 bits for hosts.
code:
172.16.168.0/23 <network ID (all 0's)
172. 16.1010 1000.0000 0000
255.255.1111 1110.0000 0000
--| SN |---------- <--subnet
| Hosts |
172.16.168.255/23
172. 16.1010 1000.1111 1111
255.255.1111 1110.0000 0000
--| SN |---------- <--subnet
| Hosts |
The host bits for this network are 011111111, not all 1's.
The /23 mask provides 510 host addresses, and the valid IP hosts for this network are 172.16.168.1 through 172.16.169.254.
The secret - when in doubt, convert the decimal values to binary and map out the range of valid hosts, as well as the range of valid networks.
HTH... | |
| PooPooPlatter 2003-05-07, 10:06 pm |
| WOW, thank god you are posting these great questions. I don't think the Sybex book teaches it.... or maybe I missed it.
Very tricky, tricky  | |
| anchor40 2003-05-08, 9:11 am |
| Thanks for the kind words! 
I know how confusing the topic can be, and the cert exams take full advantage by wording the questions very creatively. If I can help you practice with some tricky questions, then the easy ones will be a snap!
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| kmartinkf 2003-05-09, 7:27 am |
| Giving subnetting questions is a great idea!!
On a CCNA exam, one is most likely to see a classful address being used, ie. 172.16.0.0/16. Thisquestion would most likely be asked using a class C address -- 192.168.1.0/24. | |
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| quote:
Here's today's question.
Given the network 172.16.31.0, you must create 6 subnets to
support 24 users each. How many are available for future growth?
I hope he'll post the answer soon. | |
| anchor40 2003-05-14, 8:56 am |
| Hi, DXU76.
To find the answer to a question, look for the thread with the next date, as I include the answer with the next question.
Here's the link for this question:
http://www.examnotes.net/article1009458.html
If you can't find the thread, you can always PM me, and I'll help you out.
Again, I call it [bold]"daily"[/bold] because while I will try to get it out every day, I can't guarantee I will be able to. Thanks!
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