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subnetting

Posted on 2011-02-10
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Last Modified: 2012-05-11
I am having trouble with the following question, even though it looks simple:
A co needs to have min of 300 subnets each with max of 50 hosts per subnet.Working with class B address space, which of the follwoing subnets will support this addressing scheme?
 
A. 255.255.255.0
B. 255.255.255.128
C. 255.255.252.0
D. 255.255.255.224
E. 255.255.255.192
F. 255.255.248.0  
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Question by:totaram
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12 Comments
 
LVL 11

Expert Comment

by:Renato Montenegro Rustice
ID: 34866643
That would be option B.
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LVL 11

Expert Comment

by:Renato Montenegro Rustice
ID: 34866645
Sorry... E.
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LVL 8

Expert Comment

by:nwtechdesk
ID: 34866678
Answer F: would give you 65,533 hosts....
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LVL 11

Accepted Solution

by:
Renato Montenegro Rustice earned 50 total points
ID: 34866696
The reason why is quite simple.

First I would suggest you to memorize some of those masks.

128 = 10000000
192 = 11000000
224 = 11100000

Well, let's get the 128. It has 7 zeros. This section means hosts. 1 for nets, 0 for hosts. 2^7 = 128. Thats too many. We need only 50.

224 is 2^5 (5 zeros), so it equals 32. Thats not enough.

192 is 2^6 that equals 64. Thats just fine. If you discard 2 addresses (one for the subnet address and one for the broadcast), you will end up with 62 addresses. Thats quite enough.

Is that clear?
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Expert Comment

by:nwtechdesk
ID: 34866710
Sorry.... 255.255.0.0 would be 65,533
255.255.248.0 would be 2,048 hosts
0
 
LVL 11

Expert Comment

by:Renato Montenegro Rustice
ID: 34866734
About the subnetwoks, it usually wont matter (for the testing purposes). Those type of question will always focus on the idea that you must save ip addresses. So, you must always get closer to the minimum host address needed for your network. If you manage to do that, usually there will last enough addresses to the subnets. In your case, 1024 subnets using option E.
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Author Comment

by:totaram
ID: 34867230
rmrutice;
You are right about "192 is 2^6 that equals 64" calculations, but how are we getting 300 subnets here.
Also, we get 62 hosts for class c network and not class B network...
0
 
LVL 11

Assisted Solution

by:Renato Montenegro Rustice
Renato Montenegro Rustice earned 50 total points
ID: 34869992
If it's a class b network you must consider that the remaining bits "only the ones" beyond the first 16 are for dedicated to subnet calculations.

11111111 11111111 11111111 11000000
-------------------------| First 16 bits from left to right
Class B (something from 128.0.0.0 through 191.255.255.255)

You have the remaining ones (10 bits) to calculate the subnets. So, 2^10 = 1024, more than the minimum needed. The remaining 6 zeros will be used for host addresses. Remember that the goal is to save host addresses to maximize the address space.
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LVL 79

Expert Comment

by:lrmoore
ID: 34871248
There is only one answer that gives you the first part of the question - 50 hosts on each subnet... It wouldn't matter if it was a Class A or B major subnet.

The answer is E

If the question had mentioned using Class C address space, the answer would be "none of the above"

Key to taking these tests is not to get too wrapped up in the minutiae. Find the answer that fits 3/4 of the question right away and then determine if the rest of the question really matters. If "none of the above" is not an optional answer, then is has to be one of the given choices, so pick the "best" solution out of the given choices, which you've already narrowed down.
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Author Comment

by:totaram
ID: 34871463
Really the answer is B & E...
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LVL 4

Expert Comment

by:red_nectar
ID: 34874766
Answers B & E are equally good:
B. 255.255.255.128 - 512 Class B Subnets, each with 126 hosts
E. 255.255.255.192 - 1024 Class B Subnets, each with 62 hosts

Wrong Answers:
A. 255.255.255.0 - 256 Class B Subnets, each with 254 hosts
C. 255.255.252.0 -  64 Class B Subnets, each with 1024 hosts
D. 255.255.255.224 -  2048 Class B Subnets, each with 30 hosts
F. 255.255.248.0  -  32 Class B Subnets, each with 2046 hosts
Read http://www.nnk.com.au/index.php?option=com_content&view=article&id=20:subnetting-without-tears&catid=3:articles&Itemid=5
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LVL 11

Expert Comment

by:Renato Montenegro Rustice
ID: 34878432
The author of the question can, of course, test you in both sides: the subnet calculation and the host calculation. Using the method I provided, you will manage to make both calculations. As I said, usually, they want to know if you are able to maximize the use of the address space in order to plan a addressing schema of your network using the best practices.

Private addresses are not plentiful, so you must use them wisely.
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