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subnet calculation

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vivendi

Programmer
May 12, 2005
34
First of all, i had no idea where to post this, so sorry if this is in the wrong place. But i hope someone can help me anyway.

I'm making a test for cisco (CCNA01), but im at a question where
i have to calculate subnets.
But im not sure how to do this.

The question is as follows:
A company with a Class B license needs to have a minimum of 1000
subnets with each subnet capable of accommodating 50 hosts. Which mask
below is the appropriate one?


A - 255.255.0.0
B - 255.255.240.0
C - 255.255.255.0
D - 255.255.255.192
E - 255.255.255.224


I know the answer is D (255.255.255.192), but i would really like to
know how i could find out why D is the correct answer.
Could someone please help me out with the calculations.


Thanks for any help.
 
Honestly, this is not my strong suit - understanding the "easy" ways of doing this... but, I can tell you this:

A subnet mask creates a "fence" of 1's to define the network. 255 = 11111111 in binary. The 0's are not fenced, they are grouped together. So a subnet mask that's:
255.255.255.0 looks like this in binary 11111111.11111111.11111111.00000000
That's 8 zeros at the end - 2 to the power of 8. 256 - 1 for broadcast, -1 for network, 254 hosts.

Look at it this way:
0-0-0-0-0-0-0-0
128-64-32-16-8-4-2-1

So a mask of 255.255.255.192 looks like:
11111111.11111111.11111111.11000000
using the above values, add the last 6 zeros together - 32+16+8+4+2+1 = 63 (really 62 available hosts).

At least, that's my understanding of how it works.
Then
 
I agree with lwcomputing:
I took this test and Cisco is looking for the answer that calculates the smallest subnet that will encompass the hosts, IE, the most efficient subnet so the portion of the question you are concerned with is primarily (capable of accommodating 50 hosts).
They suggest that you write out your formulas on the white board provided just to ease calculations, as lwcomputing did above for the binary, also the 2-to-the-power of 1, 2, 3, etc for calculation your subnets, etc.
 
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