# subnetting ccent exam

so let's say you're taking the CCENT exam and it asks a question like:  Which ip addresses falls withing the range of the given ip?

164.36.250.3
255.255.248.0

a) 164.36.247.1 /21
b) 164.36.250.0 /21
c) 164.36.248.1 /21
d) 164.36.255.255 /21

On the CCENT/CCNA ICND1 Official Exam Cert cd it tells you to use the 'magic' number to get multiples of 8.  The cd states that you should be able to do this in 15 seconds.  How is this possible, is there another formula I can use?  Please do the math before responding.  Thank you.
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Commented:
with this subnet you would be ranging from 164.36.248.1 to 164.36.255.254 and you'd get 2046 possible hosts. So the answer should be C. The subnet mask of the given IP tells you this. The first 2 values are locking you into communicating with the 164.36.x.x networks, while the third is allowing you to communicate with anything above 164.36.248.0. The only 2 valid IP's I can see are A and C, and A is outside the range of IP's in the designated subnet. You just have to identify the scope. 164.36.248.0/21. From there it is easy to see which addresses fall within your scope.
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Commented:
I teach (among others) the ICND1 course regularly, and the subnets usually cause the most frustration and grief... :P

I don't think you'll do it in 15 secs right away, takes some experience to see the patterns, but if you follow the "eight easy steps" from the ICND1 course you should be fine.

First, write down the octet being subnet'ed in binary (1st octet: /1-8 mask, 2nd: /9 -16, 3rd: /17-24 and 4th: /25-32).

In this case it's /21, so it's the third octet:

250: 11111010

Then write the subnet mask for the third octet in binary:

Two first octets are up to 16, so we get: 17, 18, 19, 20, 21: 11111000.

Next to each other:

250: 11111010
248: 11111000

Then repeat the "most significant bits four times" (bits coverd by the mask):

11111000
11111000
11111000
11111000

Then find network and broadcast address by setting all host-bits to 0 and 1:

network:
11111000.00000000 = 248.0 (last two octets)
11111111.11111111 = 255.255

Then find first and last host by having all 0's but the last and all 1's but the last:

11111000.00000001:  = 248.1
11111111.11111110: = 255.254.

So, first host is 163.36.248.1 and last host is 163.36.255.254.

a) 164.36.247.1 /21 = wrong, it's before 248.1
b) 164.36.250.0 /21 = right, it's between 248.1 and 255.254 (don't get fooled by the 0, look at the binary below)
c) 164.36.248.1 /21 = right again

250.0 in binary:
11111010.00000000 <- not all hostbits are 0 = valid host.
11111000.00000000 <- network mask to identify hostbits.

They'll try to trick you on the test with host addresses ending with 0 or 255, which in a /24 isn't valid, but in most other prefixes may be perfectly legal.

With practice those "eight easy steps" as described in the ICND1 course will give you the answer to any question like this, but most likely it'll take slightly longer than 15 seconds. In time you'll see the patterns from the masks and do it in your head without writing anything down, but before you're there: take your time and write it down like in the course, and you got free points on the test (provided you don't spend too much time and run out).

And yes, there are loads of other ways of doing it, every time I do the course someone has an "easier way" of doing it, and I respect that, but I find the "eight easy steps" from the ICND1 course to be foolproof (more or less) for beginners.
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Author Commented:
awesome explanation and method hodepine.  this was exactly what I was looking for.

so if I have 192.168.33.133
255.255.255.224

I understand mostly everything except this step below:
----------------------------------------------------------------------------------------
Next to each other:

133: 10000101
224: 11100000

What do I do after this?  I'm confused on how to repeat the "most significant bits four times" part.
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Commented:
The most significant bits are those coverd by the mask.

133: 10000101
224: 11100000

The most significant bits are the three first bits of the address:

100

If the question were, say 199 for the last octet with a mask of /28:

10000111
11110000

Significant bits:

1000

So, if I repeat these four times and add all 0's for net, all 1's for broadcast, all 0's but the last for the first host and all 1's but the last for last host I get this for your example (dash included to indicate significant bits):

100-00000 (all zeros, network) -> 128
100-00001 (all zeros but the last, first host) -> 129
100-11110 (all ones but the last, last host) -> 158
100-11111 (all ones, broadcast) -> 159

The next step (the final 8th step) which I didn't mention last time:

Find the next network by adding to the significant bits:

101-00000 - 160.

You'll find that:

The first host is always the one after the network address, and the last host is always one less than the broadcast (it's really obvious when you think about it), and networks increment with the value of the rightmost bit of the significant bits.

In your example, the third bit from the left (128, 64, 32, etc) has a value of 32, your first network was 128, next will be 128 + 32 = 160, the one after will be 160+32 = 192, etc.

So, once you get the hang of this, you'll find your hosts and your networks pretty quickly. And when you do get a question like this on the exam, take your time and write it down, you'll have all the right answers in front of you with this method.

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