# This stuff is getting challenging!

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The pace is picking up.   This week is Chapter 6 of the Cisco
courseware.  It starts with binary math (which I already knew), then
quickly moves on to subnetting and the /24, /27, etc. stuff.

The chapter 5 test was pretty difficult, with next-hop questions that
were tricky.

Wow....the cert exams are gonna suck!

Re: This stuff is getting challenging!

Subnetting is a difficult subject for many people to grasp.  There are many
sources of this information, but I encourage you continue searching until
you find a description that helps you understand the best.

Think "computer".  Look at IP addresses and imaging how they look in binary
like the computer and networking devices see them, not in our decimal
numbering system.  As you compare a host IP address through its subnet mask,
you can see what binary bits sepecify the network portion of an address and
the host specific portion.

you make up:
(1) What class is the IP address
(2) What is the size of the subnet which it is in (number of IPs & number
of usable hosts)
(3) What is the lowest IP address in that subnet - the subnet IP (or subnet
ID, depending on the source)
(4) What is the highest IP address in that subnet - the broadcast IP
Once you can answer those four questions rapidly for any IP address and
254, 252, 248, 240, 224, 192, and 128, then you have conquered IP address
subnetting.
Also, you need to know how to covert numbers from decimal to binary and
binary to decimal rather quickly for 8 bit binary numbers and decimal
numbers between 0 and 255.  Once on the job, you will still be doing this
but not notice because you are so quick calculating it in your head that you
do not covert to binary any longer.

**********
BINARY METHOD:
**********
192.168.50.3
255.255.240.0
You convert to binary:
11000000.10101000.00110010.00000011
11111111.11111111.11110000.00000000
If you break the IP address into its network portion host portion based on
where the 1's and 0's in the subnet mask are seperating the address:
11000000.10101000.00110000.00000000 = 192.168.48.0
00000000.00000000.00000010.00000011 = 0.0.2.3
CHECK:  Add these two decimal format IP address portions together, and you
end up with the original address:
192.168.48.0
0.0.2.3
+_____________
192.168.50.3
Now that you have verified your binary math, convert all of the host bits in
192.168.48.0 from 0's to 1's:
11000000.10101000.00111111.11111111 = 192.168.63.255
Figure up the number of binary bits in the host portion and figure up 2 to
the nth power for that value:
2^12 = 4096
Now you have determined that the IP address is in a subnet with the
following information:
Number of IP addresses in the subnet: 4096 with only 4094 host

**********
MY METHOD:
**********
192.168.50.3
255.255.240.0
Perform some odd subnet math subtracting each octet from 256 and then
multiplying the differences together:
256   256   256   256
255   255   240    0
-_________________
1 x   1 x  16 x 256 = 4096
Now take that "funny number" in the subnet mask subtracted from 256 and
start counting ranges from 0 by that number and then look for that the value
in the corresponding octet of the IP address:
256 - 240 (3rd octet of subnet mask) = 16
Couting by 16s, looking for 50 (3rd octet of IP address):  0...15,
16...31, 32...47, 48...63 FOUND 50!
Insert those low and high ranges with a 0 and 255 after them, respectively:
192.168.*48*.*0*
192.168.*63*.*255*
That is it.  You just determined the low and high IP address in a range and
the size of the range - no binary math!

Slash format subnet masks keep us from having to say two-fifty-five over and
over.  Just take the slash value and subtract it from 32 then figure up 2 to
the nth power.  For example, 192.168.50.3//20 would require finding 32-20
which is 12 then figure out 2^12 which is 4096.  I do not do exponent math,
I just memorize the powers of 2 and count on my fingers.  It is just like
the sizes of computer RAM we look for deals on all of the time.  Two (1),
four (2) , eight (3), sixteen (4), thirty-two (5), sixty-four (6),
one-twenty-eight (7), two-fifty-six (8), five-twelve (9),
one-thousand-twenty-four (10), two-thousand-fourty-eight (11),
four-thousand-ninety-six (12).  Done!

-----
Scott Perry
Indianapolis, IN
-----

Re: This stuff is getting challenging!

On Thu, 25 Sep 2008 11:28:38 -0400, "Scott Perry"

This is the method presented to us last night in a video by Odom. Very
quick and easy.  However, it was presented to us BEFORE the binary
method, so it was sort of "Follow this formula and you'll get the

I'm not like that.  I need to know why something works before I
method...very easy.

But this math is all well and good...we still haven't been given
context.  Why do this?  What is the situation in which you have this
information and you need to do these calculations?

I'd love to have a part-time internship and see this stuff in action.

Re: This stuff is getting challenging!

And this is just the first of 4 courses to cover off the CCNA :-)

But subnetting is an essential skill.

At least with the current courses you weren't first hit with all that
"Class" stuff that others before you have to suffer. Now's it's just a

/22 means you can't touch the first (leftmost) 22 bits, so you have 10 to
work with. From these 10 you work out how many you need to create your
required number of subnets and how many to provide the number of hosts you
need in each. This way is easier to teach and it's great to see students
actually getting the concept sooner.

But like all things maths related you must practice and practice and
practice - you won't become proficient at it by theorising about it.. When
those minutes are ticking by in the cert exam this is one process you need
to be just able to do with the minimum of effort.

Aubrey

Re: This stuff is getting challenging!

wrote:

My instructor said that the first class covers 50% of CCNA material.

Subnetting is fun.

I know how to do those with a few reading and exmaples the instructor
showed but I kow that I have to practise to practise  a lot more to be
efficient time-wise.

This is probably the only type of uestions where one needs to be fast
(aside from knowing the concepts), right? Most other questions would
be either you know it or you don't, I think.

Re: This stuff is getting challenging!

On Sep 25, 6:34=A0am, Mitch@_._ wrote:

When you have time, can you give an example of that tricky next-hop
questions?  It is probbaly the same one that I saw. I think that once
we understandd the exact function of each application layer, it
wouldn't look as tricky.  When I took Chapter 6, I didn't get to read
the material as much as I would have wanted because I was working and
was just tight with time. But later, I understood what was what.