microsoft.public.cert.exams.mcse - Is the calculator applet available...

Author

Is the calculator applet available...

Anthony Murfet

2002-07-25, 5:25 pm

....when taking the 70-216 and 70-221 exams? I'd hate to have to do
subnetting questions by doing manual binary >< decimal math.

thanks,
Tony Murfet

Laura A. Robinson

2002-07-25, 5:25 pm

circa Thu, 25 Jul 2002 22:47:27 GMT, in
microsoft.public.cert.exam.mcse, Anthony Murfet (tmurfet@hotmail.com)
said,
> ...when taking the 70-216 and 70-221 exams? I'd hate to have to do
> subnetting questions by doing manual binary >< decimal math.
>
If I recall correctly, yes.

Laura
--
One man's mundane and boring existence is another man's Technicolor.
-Tick, Strange Days

David Brownridge

2002-07-25, 8:25 pm

"Anthony Murfet" <tmurfet@hotmail.com> wrote in message
news:3e%%8.3764$Vj3.285138@news0.telusplanet.net...

> ...when taking the 70-216 and 70-221 exams?

Yes it is.

> I'd hate to have to do subnetting questions
> by doing manual binary >< decimal math.

There's a lot to be said for getting familiar enough with binary that
you can do at least subnet conversions faster in your head than by
opening the calculator.

Eg, suppose we want a /28 subnet mask (which allows 2^4 - 2 = 16-2 =
14 host addresses). The last octet has 0 in the low order 4 bits. So
that's

256 - 2^4 = 256 - 16 = 240

and the required mask is 255.255.255.240.

Conversely in the mask 255.255.192.0,

256 - 192 = 64 = 2^6

so the low order 6 bits of the third octet are 0; the mask represents
/18 (which allows 2^14 - 2 = 16384-2 = 16382 host addresses).

I assume you know the first dozen or so powers of 2 by heart.... (If
not, _learn_ them. This is one place where rote memorisation is
called for.)

At the risk of going on and on, an analogy. A couple of lifetimes ago
I had to sit an exam demonstrating a minimal level of proficiency in
translating (scientific writing in) French into readable English. The
exam was open book, in the sense that one could take in a
French-English dictionary. But if one spent too much time looking in
the dictionary, one would run out of time -- so it was a mistake to
rely on the dictionary rather than learning vocabulary.

Anyone working professionally with computers -- particularly on the
network side of things -- needs to be comfortable with binary (and
hexadecimal). And subnet masks are pretty simple.

--
rgds
DVD

(David Brownridge) <mailto

VD@melbpc.org.au>

_Mike_

2002-07-26, 11:25 am

"Anthony Murfet" <tmurfet@hotmail.com> wrote in message
news:3e%%8.3764$Vj3.285138@news0.telusplanet.net...
> ...when taking the 70-216 and 70-221 exams? I'd hate to have to do
> subnetting questions by doing manual binary >< decimal math.

Tony,

Do yourself a favour, don't bother with binary calculations, make
sure your understand them, but once you have grasped the concepts,
use the following method for subnet calculations. Before the exam
starts, write the following table out in the attached spreadsheet
(there are not macros in the spreadsheet and I have up-to-date virus
software)

If you need to calculate the subnet mask for a class B network
150.10.0.0 and lets say you need a minimum of five subnets
Using the attached table, look at the third column on the
left, the closest match to 5 subnets is the third row which gives
you 6 subnets [Row eight on the spreadsheet], this means you
need to take three bits for your subnet mask. (reading back to
the first column on row 8 of the spreadsheet)

Enter three N's in Cells E7, F7, G7. Follow up the third N,
and you get a running total of 224, this means you need a
subnet of 255.255.224.0

Lastly, it means you have 5 bits in third octet for hosts
plus 8 bits in the fourth octet totalling 13 bits, again
following the row for the number 13 [Row 18 in the
spreadsheet] and it shows you can have 8190 hosts. To the
right of the three N's, pit 5 H's completing the row, this
shows in the third octet which bits are being used for the
networks (subnets) and how many bits are used for hosts.

Your range of IP's will be in blocks of 32 (You get 32
by following the third Octet upto the row on top labelled
Bits. Your first range of IP's is 150.10.32.001 to
150.10.63.254 (Network ID 150.10.32.0 , subnet
mask 255.255.224.0, broadcast address 150.10.63.255

The 2nd range is 150.10.64.001 to 150.10.95.254 and
so on ....

I had only subnet questions in 216 and none in 221 , but
this method, one you have mastered is far easier and quicker
to use than calculating via binary, even if you have a
calculator. This was a godsend for me in my Cisco exams.

I have rushed the explanation a bit, I am off to pub. Feel
free to ask questions if I have not explained myself clearly.
Once you have mastered this technique you will breeze
through subnet calculations in seconds. If in my rush to
post this before going out, I have some mistakes, I apologise
in advance, this newsgroup has those that will jump in pointing
out my mistakes and I how it is all because I did a bootcamp,
never mind, I have a thick skin and so long as you understand
this method, you are well on your way.

Cheers and good luck
Mike

PS this is based on the assumption you cannot use the first and
last address which is normal what exams are based on.

begin 666 Subnet Mask.xls
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end

RobinS

2002-07-27, 9:25 am

It is available at http://www.telusplanet.net/public/sparkman/netcalc.htm
and is definitely the sweetest thing!!

Trust me!

Peace

"David Brownridge" <DVD@melbpc.org.au> wrote in message
news:ahq9bm$sh1$1@possum.melbpc.org.au...
> "Anthony Murfet" <tmurfet@hotmail.com> wrote in message
> news:3e%%8.3764$Vj3.285138@news0.telusplanet.net...
>
> > ...when taking the 70-216 and 70-221 exams?
>
> Yes it is.
>
>
> > I'd hate to have to do subnetting questions
> > by doing manual binary >< decimal math.
>
> There's a lot to be said for getting familiar enough with binary that
> you can do at least subnet conversions faster in your head than by
> opening the calculator.
>
> Eg, suppose we want a /28 subnet mask (which allows 2^4 - 2 = 16-2 =
> 14 host addresses). The last octet has 0 in the low order 4 bits. So
> that's
>
> 256 - 2^4 = 256 - 16 = 240
>
> and the required mask is 255.255.255.240.
>
> Conversely in the mask 255.255.192.0,
>
> 256 - 192 = 64 = 2^6
>
> so the low order 6 bits of the third octet are 0; the mask represents
> /18 (which allows 2^14 - 2 = 16384-2 = 16382 host addresses).
>
> I assume you know the first dozen or so powers of 2 by heart.... (If
> not, _learn_ them. This is one place where rote memorisation is
> called for.)
>
>
> At the risk of going on and on, an analogy. A couple of lifetimes ago
> I had to sit an exam demonstrating a minimal level of proficiency in
> translating (scientific writing in) French into readable English. The
> exam was open book, in the sense that one could take in a
> French-English dictionary. But if one spent too much time looking in
> the dictionary, one would run out of time -- so it was a mistake to
> rely on the dictionary rather than learning vocabulary.
>
> Anyone working professionally with computers -- particularly on the
> network side of things -- needs to be comfortable with binary (and
> hexadecimal). And subnet masks are pretty simple.
>
> --
> rgds
> DVD
>
> (David Brownridge) <mailto

VD@melbpc.org.au>
>
>
>
>

Laura A. Robinson

2002-07-27, 12:25 pm

circa Sat, 27 Jul 2002 10:58:46 -0400, in
microsoft.public.cert.exam.mcse, RobinS (robin.nayyar@sympatico.ca)
said,
>
> It is available at http://www.telusplanet.net/public/sparkman/netcalc.htm
> and is definitely the sweetest thing!!
>
> Trust me!
>
> Peace
>
And that site will not be available during the exam. The Windows
calculator and that site are quite different.

Laura
--
One man's mundane and boring existence is another man's Technicolor.
-Tick, Strange Days

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