# 216 and Subnetting

Discussion in 'MCSE' started by Mike, Nov 12, 2003.

1. ### MikeGuest

I have 6 tests under my belt and I saved 216 for the end. I have been
studying subnetting and thought I had it down. As I take practice tests it
seems that is not the case.

So anyone that knows or can point me in the right direction/website/anything
your help would be much appreciated. I really want to know this not just
for the test but for my job in general.

I will use my last practice question as an example of how I am confused.

The question wants to know which addresses can be divided into at least 4
subnets with at least 100 hosts per subnet.

A. 10.25.0.0/22
B. 10.25.0.0/23
C. 10.25.0.0/24
D. 10.25.0.0/25

# My Calculations:

/22 = 255.255.252.0 = 14 subnet bits 10 host bits
/23 = 255.255.254.0 = 15 subnet bits 9 host bits
/24 = 255.255.255.0 = 16 subnet bits 8 host bits
/25 = 255.255.255.128 = 17 subnet bits 7 host bits

using 2^n -2

A = 2^14 -2 = 16382 subnets and 2^10-2 =1022 hosts
B = 2^15 -2 = 32766 subnets and 2^9-2= 510 hosts
C = 2^16 -2 = 65534 subnets and 2^8-2= 254 hosts
D = 2^17 -2 = 131070 subnets and 2^7-2= 128 hosts

I thought the answer should be all of them. Apparently I was wrong.
The test says that only A and B are correct. It says that C can be divided
into 2 or 4 subnets and D can be divided into 4 subnets with only 30 hosts.

I know it is a CIDR address but I guess I just don't quite get that concept
or at least the math behind it. Is there a foormula for this type of
calculation? Am I just way off with subnetting and need to read the book
again? I don't quite understand where you start and stop counting the bits
with CIDR.

Mike, Nov 12, 2003

2. ### Andy FosterGuest

On Tue, 11 Nov 2003 22:50:15 -0500, Mike wrote:

<snip>
>
> A. 10.25.0.0/22
> B. 10.25.0.0/23
> C. 10.25.0.0/24
> D. 10.25.0.0/25
>
> # My Calculations:
>
> /22 = 255.255.252.0 = 14 subnet bits 10 host bits /23 = 255.255.254.0 =
> 15 subnet bits 9 host bits /24 = 255.255.255.0 = 16 subnet bits
> 8 host bits /25 = 255.255.255.128 = 17 subnet bits 7 host bits
>

Forget classes for a moment. Just because 10.x.x.x is class A doesn't
necessarily give you /8 - the possible answers give you the ranges/masks.
eg. 10.25.0.0/22 gives you the range 10.25.0.0 - 10.25.3.255 (or 10 host
bits starting from 10.25.0.0) - the rest is not yours.
100 hosts need 7 bits, and 4 subnets need 2 bits, so you need 9 host bits
to play with - so a mask or /23 (32 - 9) or less is required.

HTH

Andy

Andy Foster, Nov 12, 2003

3. ### PaulGuest

Andy

Could you let me know how you get the following range in

10.25.0.0/22 gives you the range 10.25.0.0 - 10.25.3.255

I don't understand how you are calculating 10.25.0.0 -
10.25.3.255 with /22

Thanks

Paul
>-----Original Message-----
>On Tue, 11 Nov 2003 22:50:15 -0500, Mike wrote:
>
><snip>
>>
>> A. 10.25.0.0/22
>> B. 10.25.0.0/23
>> C. 10.25.0.0/24
>> D. 10.25.0.0/25
>>
>> # My Calculations:
>>
>> /22 = 255.255.252.0 = 14 subnet bits 10 host

bits /23 = 255.255.254.0 =
>> 15 subnet bits 9 host bits /24 = 255.255.255.0 =

16 subnet bits
>> 8 host bits /25 = 255.255.255.128 = 17 subnet bits 7

host bits
>>

>
>Forget classes for a moment. Just because 10.x.x.x is

class A doesn't
>necessarily give you /8 - the possible answers give you

>eg. 10.25.0.0/22 gives you the range 10.25.0.0 -

10.25.3.255 (or 10 host
>bits starting from 10.25.0.0) - the rest is not yours.
>100 hosts need 7 bits, and 4 subnets need 2 bits, so you

need 9 host bits
>to play with - so a mask or /23 (32 - 9) or less is

required.
>
>HTH
>
>Andy
>.
>

Paul, Nov 12, 2003
4. ### EricGuest

www.learntosubnet.com

>-----Original Message-----
>Andy
>
>Could you let me know how you get the following range in
>
>10.25.0.0/22 gives you the range 10.25.0.0 - 10.25.3.255
>
>I don't understand how you are calculating 10.25.0.0 -
>10.25.3.255 with /22
>
>Thanks
>
>Paul
>>-----Original Message-----
>>On Tue, 11 Nov 2003 22:50:15 -0500, Mike wrote:
>>
>><snip>
>>>
>>> A. 10.25.0.0/22
>>> B. 10.25.0.0/23
>>> C. 10.25.0.0/24
>>> D. 10.25.0.0/25
>>>
>>> # My Calculations:
>>>
>>> /22 = 255.255.252.0 = 14 subnet bits 10 host

>bits /23 = 255.255.254.0 =
>>> 15 subnet bits 9 host bits /24 = 255.255.255.0 =

>16 subnet bits
>>> 8 host bits /25 = 255.255.255.128 = 17 subnet bits 7

>host bits
>>>

>>
>>Forget classes for a moment. Just because 10.x.x.x is

>class A doesn't
>>necessarily give you /8 - the possible answers give you

>>eg. 10.25.0.0/22 gives you the range 10.25.0.0 -

>10.25.3.255 (or 10 host
>>bits starting from 10.25.0.0) - the rest is not yours.
>>100 hosts need 7 bits, and 4 subnets need 2 bits, so you

>need 9 host bits
>>to play with - so a mask or /23 (32 - 9) or less is

>required.
>>
>>HTH
>>
>>Andy
>>.
>>

>.
>

Eric, Nov 12, 2003
5. ### TheXmanGuest

If you break down the subnet mask to its bit form, /22
will give you:

10.25.0.0/22

nnnnnnnn.nnnnnnnn.sssssshh.hhhhhhhh = /22
where n=network bits / s=subnet bits / h=host bits

or
11111111.11111111.11111100.00000000 = /22
or
255.255.252.0 = /22

Therefore, in the 3rd octect, you have 6 bits for the
subnets and 10 bits for the hosts. Remember, the formula
to calculate subnets and hosts is 2^bits - 2.

In this example:
Number of subnets = 2^6 - 2 or 62
Number of hosts = 2^10 - 2 or 1022

The first subnet in the range would be 10.25.4.1 -
10.25.7.254.
The last subnet in the range would be 10.25.248.1 -
10.25.251.254.

Sincerely,
Xavier Todd Clarke

----- Original Message -----
From: Paul
Newsgroups: microsoft.public.cert.exam.mcse
Sent: Wednesday, November 12, 2003 6:00 AM
Subject: Re: 216 and Subnetting

Andy

Could you let me know how you get the following range in

10.25.0.0/22 gives you the range 10.25.0.0 - 10.25.3.255

I don't understand how you are calculating 10.25.0.0 -
10.25.3.255 with /22

Thanks

Paul
>-----Original Message-----
>On Tue, 11 Nov 2003 22:50:15 -0500, Mike wrote:
>
><snip>
>>
>> A. 10.25.0.0/22
>> B. 10.25.0.0/23
>> C. 10.25.0.0/24
>> D. 10.25.0.0/25
>>
>> # My Calculations:
>>
>> /22 = 255.255.252.0 = 14 subnet bits 10 host

bits /23 = 255.255.254.0 =
>> 15 subnet bits 9 host bits /24 = 255.255.255.0 =

16 subnet bits
>> 8 host bits /25 = 255.255.255.128 = 17 subnet bits 7

host bits
>>

>
>Forget classes for a moment. Just because 10.x.x.x is

class A doesn't
>necessarily give you /8 - the possible answers give you

>eg. 10.25.0.0/22 gives you the range 10.25.0.0 -

10.25.3.255 (or 10 host
>bits starting from 10.25.0.0) - the rest is not yours.
>100 hosts need 7 bits, and 4 subnets need 2 bits, so you

need 9 host bits
>to play with - so a mask or /23 (32 - 9) or less is

required.
>
>HTH
>
>Andy
>.
>
>-----Original Message-----
>Andy
>
>Could you let me know how you get the following range in
>
>10.25.0.0/22 gives you the range 10.25.0.0 - 10.25.3.255
>
>I don't understand how you are calculating 10.25.0.0 -
>10.25.3.255 with /22
>
>Thanks
>
>Paul
>>-----Original Message-----
>>On Tue, 11 Nov 2003 22:50:15 -0500, Mike wrote:
>>
>><snip>
>>>
>>> A. 10.25.0.0/22
>>> B. 10.25.0.0/23
>>> C. 10.25.0.0/24
>>> D. 10.25.0.0/25
>>>
>>> # My Calculations:
>>>
>>> /22 = 255.255.252.0 = 14 subnet bits 10 host

>bits /23 = 255.255.254.0 =
>>> 15 subnet bits 9 host bits /24 = 255.255.255.0 =

>16 subnet bits
>>> 8 host bits /25 = 255.255.255.128 = 17 subnet bits 7

>host bits
>>>

>>
>>Forget classes for a moment. Just because 10.x.x.x is

>class A doesn't
>>necessarily give you /8 - the possible answers give you

>>eg. 10.25.0.0/22 gives you the range 10.25.0.0 -

>10.25.3.255 (or 10 host
>>bits starting from 10.25.0.0) - the rest is not yours.
>>100 hosts need 7 bits, and 4 subnets need 2 bits, so you

>need 9 host bits
>>to play with - so a mask or /23 (32 - 9) or less is

>required.
>>
>>HTH
>>
>>Andy
>>.
>>

>.
>

TheXman, Nov 12, 2003
6. ### Andy FosterGuest

On Wed, 12 Nov 2003 08:08:04 -0800, TheXman wrote:

> If you break down the subnet mask to its bit form, /22 will give you:
>
> 10.25.0.0/22
>
> nnnnnnnn.nnnnnnnn.sssssshh.hhhhhhhh = /22 where n=network bits / s=subnet
> bits / h=host bits
>
> or
> 11111111.11111111.11111100.00000000 = /22 or
> 255.255.252.0 = /22
>
>
> Therefore, in the 3rd octect, you have 6 bits for the subnets and 10 bits
> for the hosts. Remember, the formula to calculate subnets and hosts is
> 2^bits - 2.
>

16 bits in the 3rd octet ?
For you to have 16 bits to play with, you would have to be given
10.25.0.0/16.

> In this example:
> Number of subnets = 2^6 - 2 or 62
> Number of hosts = 2^10 - 2 or 1022
>

Wrong!

> The first subnet in the range would be 10.25.4.1 - 10.25.7.254.
> The last subnet in the range would be 10.25.248.1 - 10.25.251.254.
>

Wrong!
10.25.4.1 - 10.25.7.254 is the usable host range for 10.25.4.0/22.
That has nothing to do with the original question.

HTH

Andy

Andy Foster, Nov 12, 2003
7. ### Andy FosterGuest

On Wed, 12 Nov 2003 04:00:06 -0800, Paul wrote:

> Andy
>
> Could you let me know how you get the following range in your calculation:
>
> 10.25.0.0/22 gives you the range 10.25.0.0 - 10.25.3.255
>
> I don't understand how you are calculating 10.25.0.0 - 10.25.3.255 with
> /22
>

As suggested elsewhere www.learntosubnet.com

This isn't how I calculated it, but the longhand is as follows....

Converting 10.25.0.0 into binary gives
00001010.00011001.00000000.00000000
/22 (aka 255.255.252.0) into binary gives
11111111.11111111.11111100.00000000

The range of addresses (not hosts) described by 10.25.0.0/22 in binary is
00001010.00011001.00000000.00000000 (10.25.0.0) to
00001010.00011001.00000011.11111111 (10.25.3.255)

This would give a host range of 10.25.0.1 to 10.25.3.254 if it were used
as a single subnet, (which it isn't)

HTH

Andy

Andy Foster, Nov 12, 2003
8. ### YOUNGMANGuest

To Andy Foster -

It must be wonderful to be so sure of your own abilities . . . howeve

TheXman quite clearly stated that for the address 10.25.0.0/22 - "i
the 3rd octect, you have 6 bits for the subnets and 10 bits
for the hosts"

This was not edited afterwards as you quoted it in your own post.

However you then misquoted and mocked it

> 16 bits in the 3rd octet ?
> Wrong!

Wrong!

TheXman is quite right in stating that in the example given the Numbe
of subnets = 2^6 - 2 or 62 and the Number of hosts = 2^10 - 2 or 1022

That's why, surprisingly enough, the address range given was correc
for the example given i.e 10.25.0.0/22!

HTH

;

YOUNGMA
-----------------------------------------------------------------------
Posted via http://www.examnotes.ne
-----------------------------------------------------------------------

YOUNGMAN, Nov 13, 2003
9. ### TheXmanGuest

Thank you YOUNGMAN. Can you believe the nerve of this Andy Foster
character? First, he ask for help understanding IP subnetting. Then, he
mocks correct answers as if he knows IP subnetting. Instead of defending my
answer, I decided it would be more pleasurable to let Andy make a FOOL of
himself when he shows others his failed understanding of IP subnetting. You
can only teach people who want to learn. Obviously, Andy Foster already
knows everything. (lol).
--
Sincerely,
TheXman

"YOUNGMAN" <> wrote in message
news:...

To Andy Foster -

It must be wonderful to be so sure of your own abilities . . . however

TheXman quite clearly stated that for the address 10.25.0.0/22 - "in
the 3rd octect, you have 6 bits for the subnets and 10 bits
for the hosts"

This was not edited afterwards as you quoted it in your own post.

However you then misquoted and mocked it

> 16 bits in the 3rd octet ?
> Wrong!

Wrong!

TheXman is quite right in stating that in the example given the Number
of subnets = 2^6 - 2 or 62 and the Number of hosts = 2^10 - 2 or 1022

That's why, surprisingly enough, the address range given was correct
for the example given i.e 10.25.0.0/22!

HTH

YOUNGMAN
------------------------------------------------------------------------
Posted via http://www.examnotes.net
------------------------------------------------------------------------

TheXman, Nov 13, 2003
10. ### =?Windows-1252?Q?Frisbee=AE_MCNGP?=Guest

TheXman wrote:
> Thank you YOUNGMAN. Can you believe the nerve of this Andy Foster
> character? First, he ask for help understanding IP subnetting.
> Then, he mocks correct answers as if he knows IP subnetting. Instead
> of defending my answer, I decided it would be more pleasurable to let
> Andy make a FOOL of himself when he shows others his failed
> understanding of IP subnetting. You can only teach people who want
> to learn. Obviously, Andy Foster already knows everything. (lol).

help, he replied to the original poster (Mike).

FWIW, both you and YOUNGMAN are complete dicks.

--
Fris "or maybe incomplete ones" beeĀ® MCNGP #13

http://www.mcngp.tk
The MCNGP Team - We're here to help

http://groups.yahoo.com/group/certaholics
Certaholics - We're here if you're beyond help

=?Windows-1252?Q?Frisbee=AE_MCNGP?=, Nov 13, 2003
11. ### MikeGuest

OK. That makes sense. However, if 10.25.0.0 /22 = 6subnet bits and 10 host
bits why doesn't 10.25.0.0 /25 = 9 subnet bits and 7 host bits? Do you
start counting your bits over again when you move to the next octet because
the mask would now be 255.255.255.128? Would 10.25.0.0 /25 really = 1
subnet bit and 7 host bits?

"TheXman" <> wrote in message
news:23f001c3a937\$272ef4b0\$...
> If you break down the subnet mask to its bit form, /22
> will give you:
>
> 10.25.0.0/22
>
> nnnnnnnn.nnnnnnnn.sssssshh.hhhhhhhh = /22
> where n=network bits / s=subnet bits / h=host bits
>
> or
> 11111111.11111111.11111100.00000000 = /22
> or
> 255.255.252.0 = /22
>
>
> Therefore, in the 3rd octect, you have 6 bits for the
> subnets and 10 bits for the hosts. Remember, the formula
> to calculate subnets and hosts is 2^bits - 2.
>
> In this example:
> Number of subnets = 2^6 - 2 or 62
> Number of hosts = 2^10 - 2 or 1022
>
> The first subnet in the range would be 10.25.4.1 -
> 10.25.7.254.
> The last subnet in the range would be 10.25.248.1 -
> 10.25.251.254.
>
> Sincerely,
> Xavier Todd Clarke
>
> ----- Original Message -----
> From: Paul
> Newsgroups: microsoft.public.cert.exam.mcse
> Sent: Wednesday, November 12, 2003 6:00 AM
> Subject: Re: 216 and Subnetting
>
>
> Andy
>
> Could you let me know how you get the following range in
>
> 10.25.0.0/22 gives you the range 10.25.0.0 - 10.25.3.255
>
> I don't understand how you are calculating 10.25.0.0 -
> 10.25.3.255 with /22
>
> Thanks
>
> Paul
> >-----Original Message-----
> >On Tue, 11 Nov 2003 22:50:15 -0500, Mike wrote:
> >
> ><snip>
> >>
> >> A. 10.25.0.0/22
> >> B. 10.25.0.0/23
> >> C. 10.25.0.0/24
> >> D. 10.25.0.0/25
> >>
> >> # My Calculations:
> >>
> >> /22 = 255.255.252.0 = 14 subnet bits 10 host

> bits /23 = 255.255.254.0 =
> >> 15 subnet bits 9 host bits /24 = 255.255.255.0 =

> 16 subnet bits
> >> 8 host bits /25 = 255.255.255.128 = 17 subnet bits 7

> host bits
> >>

> >
> >Forget classes for a moment. Just because 10.x.x.x is

> class A doesn't
> >necessarily give you /8 - the possible answers give you

> >eg. 10.25.0.0/22 gives you the range 10.25.0.0 -

> 10.25.3.255 (or 10 host
> >bits starting from 10.25.0.0) - the rest is not yours.
> >100 hosts need 7 bits, and 4 subnets need 2 bits, so you

> need 9 host bits
> >to play with - so a mask or /23 (32 - 9) or less is

> required.
> >
> >HTH
> >
> >Andy
> >.
> >
> >-----Original Message-----
> >Andy
> >
> >Could you let me know how you get the following range in
> >
> >10.25.0.0/22 gives you the range 10.25.0.0 - 10.25.3.255
> >
> >I don't understand how you are calculating 10.25.0.0 -
> >10.25.3.255 with /22
> >
> >Thanks
> >
> >Paul
> >>-----Original Message-----
> >>On Tue, 11 Nov 2003 22:50:15 -0500, Mike wrote:
> >>
> >><snip>
> >>>
> >>> A. 10.25.0.0/22
> >>> B. 10.25.0.0/23
> >>> C. 10.25.0.0/24
> >>> D. 10.25.0.0/25
> >>>
> >>> # My Calculations:
> >>>
> >>> /22 = 255.255.252.0 = 14 subnet bits 10 host

> >bits /23 = 255.255.254.0 =
> >>> 15 subnet bits 9 host bits /24 = 255.255.255.0 =

> >16 subnet bits
> >>> 8 host bits /25 = 255.255.255.128 = 17 subnet bits 7

> >host bits
> >>>
> >>
> >>Forget classes for a moment. Just because 10.x.x.x is

> >class A doesn't
> >>necessarily give you /8 - the possible answers give you

> >>eg. 10.25.0.0/22 gives you the range 10.25.0.0 -

> >10.25.3.255 (or 10 host
> >>bits starting from 10.25.0.0) - the rest is not yours.
> >>100 hosts need 7 bits, and 4 subnets need 2 bits, so you

> >need 9 host bits
> >>to play with - so a mask or /23 (32 - 9) or less is

> >required.
> >>
> >>HTH
> >>
> >>Andy
> >>.
> >>

> >.
> >

Mike, Nov 13, 2003
12. ### TheXmanGuest

Mike:

You're right, 10.25.0.0/25 would be 9 subnet bits and 7 host bits. Where
you may be getting a little mixed up is that you are ignoring your network
bits. I hope this helps, if you still need more help understanding, please

10.25.0.0/25

The /25 bit mask would look like this:

nnnnnnnn nnnnnnnn ssssssss shhhhhhh
where: n=network bits / s=subnet bits / h=host bits
or
255.255.255.128

Therefore:
number of subnets = 2^9 - 2 or 510
number of hosts = 2^7 - 2 or 126

--
TheXman

"Mike" <> wrote in message
news:...
OK. That makes sense. However, if 10.25.0.0 /22 = 6subnet bits and 10 host
bits why doesn't 10.25.0.0 /25 = 9 subnet bits and 7 host bits? Do you
start counting your bits over again when you move to the next octet because
the mask would now be 255.255.255.128? Would 10.25.0.0 /25 really = 1
subnet bit and 7 host bits?

"TheXman" <> wrote in message
news:23f001c3a937\$272ef4b0\$...
> If you break down the subnet mask to its bit form, /22
> will give you:
>
> 10.25.0.0/22
>
> nnnnnnnn.nnnnnnnn.sssssshh.hhhhhhhh = /22
> where n=network bits / s=subnet bits / h=host bits
>
> or
> 11111111.11111111.11111100.00000000 = /22
> or
> 255.255.252.0 = /22
>
>
> Therefore, in the 3rd octect, you have 6 bits for the
> subnets and 10 bits for the hosts. Remember, the formula
> to calculate subnets and hosts is 2^bits - 2.
>
> In this example:
> Number of subnets = 2^6 - 2 or 62
> Number of hosts = 2^10 - 2 or 1022
>
> The first subnet in the range would be 10.25.4.1 -
> 10.25.7.254.
> The last subnet in the range would be 10.25.248.1 -
> 10.25.251.254.
>
> Sincerely,
> Xavier Todd Clarke
>
> ----- Original Message -----
> From: Paul
> Newsgroups: microsoft.public.cert.exam.mcse
> Sent: Wednesday, November 12, 2003 6:00 AM
> Subject: Re: 216 and Subnetting
>
>
> Andy
>
> Could you let me know how you get the following range in
>
> 10.25.0.0/22 gives you the range 10.25.0.0 - 10.25.3.255
>
> I don't understand how you are calculating 10.25.0.0 -
> 10.25.3.255 with /22
>
> Thanks
>
> Paul
> >-----Original Message-----
> >On Tue, 11 Nov 2003 22:50:15 -0500, Mike wrote:
> >
> ><snip>
> >>
> >> A. 10.25.0.0/22
> >> B. 10.25.0.0/23
> >> C. 10.25.0.0/24
> >> D. 10.25.0.0/25
> >>
> >> # My Calculations:
> >>
> >> /22 = 255.255.252.0 = 14 subnet bits 10 host

> bits /23 = 255.255.254.0 =
> >> 15 subnet bits 9 host bits /24 = 255.255.255.0 =

> 16 subnet bits
> >> 8 host bits /25 = 255.255.255.128 = 17 subnet bits 7

> host bits
> >>

> >
> >Forget classes for a moment. Just because 10.x.x.x is

> class A doesn't
> >necessarily give you /8 - the possible answers give you

> >eg. 10.25.0.0/22 gives you the range 10.25.0.0 -

> 10.25.3.255 (or 10 host
> >bits starting from 10.25.0.0) - the rest is not yours.
> >100 hosts need 7 bits, and 4 subnets need 2 bits, so you

> need 9 host bits
> >to play with - so a mask or /23 (32 - 9) or less is

> required.
> >
> >HTH
> >
> >Andy
> >.
> >
> >-----Original Message-----
> >Andy
> >
> >Could you let me know how you get the following range in
> >
> >10.25.0.0/22 gives you the range 10.25.0.0 - 10.25.3.255
> >
> >I don't understand how you are calculating 10.25.0.0 -
> >10.25.3.255 with /22
> >
> >Thanks
> >
> >Paul
> >>-----Original Message-----
> >>On Tue, 11 Nov 2003 22:50:15 -0500, Mike wrote:
> >>
> >><snip>
> >>>
> >>> A. 10.25.0.0/22
> >>> B. 10.25.0.0/23
> >>> C. 10.25.0.0/24
> >>> D. 10.25.0.0/25
> >>>
> >>> # My Calculations:
> >>>
> >>> /22 = 255.255.252.0 = 14 subnet bits 10 host

> >bits /23 = 255.255.254.0 =
> >>> 15 subnet bits 9 host bits /24 = 255.255.255.0 =

> >16 subnet bits
> >>> 8 host bits /25 = 255.255.255.128 = 17 subnet bits 7

> >host bits
> >>>
> >>
> >>Forget classes for a moment. Just because 10.x.x.x is

> >class A doesn't
> >>necessarily give you /8 - the possible answers give you

> >>eg. 10.25.0.0/22 gives you the range 10.25.0.0 -

> >10.25.3.255 (or 10 host
> >>bits starting from 10.25.0.0) - the rest is not yours.
> >>100 hosts need 7 bits, and 4 subnets need 2 bits, so you

> >need 9 host bits
> >>to play with - so a mask or /23 (32 - 9) or less is

> >required.
> >>
> >>HTH
> >>
> >>Andy
> >>.
> >>

> >.
> >

TheXman, Nov 13, 2003
13. ### ConsultantGuest

you are the fool in this thread. shouldn't you be reading a cominc book or
something? wait a sec, xman & youngman, i see some sick connection here.
take it away

"TheXman" <> wrote in message
news:...
> Thank you YOUNGMAN. Can you believe the nerve of this Andy Foster
> character? First, he ask for help understanding IP subnetting. Then, he
> mocks correct answers as if he knows IP subnetting. Instead of defending

my
> answer, I decided it would be more pleasurable to let Andy make a FOOL of
> himself when he shows others his failed understanding of IP subnetting.

You
> can only teach people who want to learn. Obviously, Andy Foster already
> knows everything. (lol).
> --
> Sincerely,
> TheXman
>
>
>
> "YOUNGMAN" <> wrote in message
> news:...
>
> To Andy Foster -
>
> It must be wonderful to be so sure of your own abilities . . . however
>
> TheXman quite clearly stated that for the address 10.25.0.0/22 - "in
> the 3rd octect, you have 6 bits for the subnets and 10 bits
> for the hosts"
>
> This was not edited afterwards as you quoted it in your own post.
>
> However you then misquoted and mocked it
>
> > 16 bits in the 3rd octet ?
> > Wrong!

>
>
> Wrong!
>
>
> TheXman is quite right in stating that in the example given the Number
> of subnets = 2^6 - 2 or 62 and the Number of hosts = 2^10 - 2 or 1022
>
> That's why, surprisingly enough, the address range given was correct
> for the example given i.e 10.25.0.0/22!
>
> HTH
>
>
>
>
>
>
>
>
>
> YOUNGMAN
> ------------------------------------------------------------------------
> Posted via http://www.examnotes.net
> ------------------------------------------------------------------------
>
>
>
>

Consultant, Nov 13, 2003
14. ### MikeGuest

That does help. I think I am starting to get it. This practice test is
throwing me off. It says that 10.25.0.0/25 can be up to 126 hosts OR it can
divided into 4 subnets of 30 hosts and I 'm sitting here thinking it can be
510 subnets with 126 hosts per subnet. I don't understand where they come
up with the 4 subnets of 30 hosts from.

"TheXman" <> wrote in message
news:...
> Mike:
>
> You're right, 10.25.0.0/25 would be 9 subnet bits and 7 host bits. Where
> you may be getting a little mixed up is that you are ignoring your network
> bits. I hope this helps, if you still need more help understanding, please
>
> 10.25.0.0/25
>
> The /25 bit mask would look like this:
>
> nnnnnnnn nnnnnnnn ssssssss shhhhhhh
> where: n=network bits / s=subnet bits / h=host bits
> or
> 255.255.255.128
>
> Therefore:
> number of subnets = 2^9 - 2 or 510
> number of hosts = 2^7 - 2 or 126
>
> --
> TheXman
>
>
> "Mike" <> wrote in message
> news:...
> OK. That makes sense. However, if 10.25.0.0 /22 = 6subnet bits and 10

host
> bits why doesn't 10.25.0.0 /25 = 9 subnet bits and 7 host bits? Do you
> start counting your bits over again when you move to the next octet

because
> the mask would now be 255.255.255.128? Would 10.25.0.0 /25 really = 1
> subnet bit and 7 host bits?
>
>
> "TheXman" <> wrote in message
> news:23f001c3a937\$272ef4b0\$...
> > If you break down the subnet mask to its bit form, /22
> > will give you:
> >
> > 10.25.0.0/22
> >
> > nnnnnnnn.nnnnnnnn.sssssshh.hhhhhhhh = /22
> > where n=network bits / s=subnet bits / h=host bits
> >
> > or
> > 11111111.11111111.11111100.00000000 = /22
> > or
> > 255.255.252.0 = /22
> >
> >
> > Therefore, in the 3rd octect, you have 6 bits for the
> > subnets and 10 bits for the hosts. Remember, the formula
> > to calculate subnets and hosts is 2^bits - 2.
> >
> > In this example:
> > Number of subnets = 2^6 - 2 or 62
> > Number of hosts = 2^10 - 2 or 1022
> >
> > The first subnet in the range would be 10.25.4.1 -
> > 10.25.7.254.
> > The last subnet in the range would be 10.25.248.1 -
> > 10.25.251.254.
> >
> > Sincerely,
> > Xavier Todd Clarke
> >
> > ----- Original Message -----
> > From: Paul
> > Newsgroups: microsoft.public.cert.exam.mcse
> > Sent: Wednesday, November 12, 2003 6:00 AM
> > Subject: Re: 216 and Subnetting
> >
> >
> > Andy
> >
> > Could you let me know how you get the following range in
> >
> > 10.25.0.0/22 gives you the range 10.25.0.0 - 10.25.3.255
> >
> > I don't understand how you are calculating 10.25.0.0 -
> > 10.25.3.255 with /22
> >
> > Thanks
> >
> > Paul
> > >-----Original Message-----
> > >On Tue, 11 Nov 2003 22:50:15 -0500, Mike wrote:
> > >
> > ><snip>
> > >>
> > >> A. 10.25.0.0/22
> > >> B. 10.25.0.0/23
> > >> C. 10.25.0.0/24
> > >> D. 10.25.0.0/25
> > >>
> > >> # My Calculations:
> > >>
> > >> /22 = 255.255.252.0 = 14 subnet bits 10 host

> > bits /23 = 255.255.254.0 =
> > >> 15 subnet bits 9 host bits /24 = 255.255.255.0 =

> > 16 subnet bits
> > >> 8 host bits /25 = 255.255.255.128 = 17 subnet bits 7

> > host bits
> > >>
> > >
> > >Forget classes for a moment. Just because 10.x.x.x is

> > class A doesn't
> > >necessarily give you /8 - the possible answers give you

> > >eg. 10.25.0.0/22 gives you the range 10.25.0.0 -

> > 10.25.3.255 (or 10 host
> > >bits starting from 10.25.0.0) - the rest is not yours.
> > >100 hosts need 7 bits, and 4 subnets need 2 bits, so you

> > need 9 host bits
> > >to play with - so a mask or /23 (32 - 9) or less is

> > required.
> > >
> > >HTH
> > >
> > >Andy
> > >.
> > >
> > >-----Original Message-----
> > >Andy
> > >
> > >Could you let me know how you get the following range in
> > >
> > >10.25.0.0/22 gives you the range 10.25.0.0 - 10.25.3.255
> > >
> > >I don't understand how you are calculating 10.25.0.0 -
> > >10.25.3.255 with /22
> > >
> > >Thanks
> > >
> > >Paul
> > >>-----Original Message-----
> > >>On Tue, 11 Nov 2003 22:50:15 -0500, Mike wrote:
> > >>
> > >><snip>
> > >>>
> > >>> A. 10.25.0.0/22
> > >>> B. 10.25.0.0/23
> > >>> C. 10.25.0.0/24
> > >>> D. 10.25.0.0/25
> > >>>
> > >>> # My Calculations:
> > >>>
> > >>> /22 = 255.255.252.0 = 14 subnet bits 10 host
> > >bits /23 = 255.255.254.0 =
> > >>> 15 subnet bits 9 host bits /24 = 255.255.255.0 =
> > >16 subnet bits
> > >>> 8 host bits /25 = 255.255.255.128 = 17 subnet bits 7
> > >host bits
> > >>>
> > >>
> > >>Forget classes for a moment. Just because 10.x.x.x is
> > >class A doesn't
> > >>necessarily give you /8 - the possible answers give you
> > >>eg. 10.25.0.0/22 gives you the range 10.25.0.0 -
> > >10.25.3.255 (or 10 host
> > >>bits starting from 10.25.0.0) - the rest is not yours.
> > >>100 hosts need 7 bits, and 4 subnets need 2 bits, so you
> > >need 9 host bits
> > >>to play with - so a mask or /23 (32 - 9) or less is
> > >required.
> > >>
> > >>HTH
> > >>
> > >>Andy
> > >>.
> > >>
> > >.
> > >

>
>
>

Mike, Nov 13, 2003
15. ### BratGuest

You need to pay attention here... MIKE asked for help not Andy

"TheXman" <> wrote in message
news:...
> Thank you YOUNGMAN. Can you believe the nerve of this Andy Foster
> character? First, he ask for help understanding IP subnetting. Then, he
> mocks correct answers as if he knows IP subnetting. Instead of defending

my
> answer, I decided it would be more pleasurable to let Andy make a FOOL of
> himself when he shows others his failed understanding of IP subnetting.

You
> can only teach people who want to learn. Obviously, Andy Foster already
> knows everything. (lol).
> --
> Sincerely,
> TheXman
>
>
>
> "YOUNGMAN" <> wrote in message
> news:...
>
> To Andy Foster -
>
> It must be wonderful to be so sure of your own abilities . . . however
>
> TheXman quite clearly stated that for the address 10.25.0.0/22 - "in
> the 3rd octect, you have 6 bits for the subnets and 10 bits
> for the hosts"
>
> This was not edited afterwards as you quoted it in your own post.
>
> However you then misquoted and mocked it
>
> > 16 bits in the 3rd octet ?
> > Wrong!

>
>
> Wrong!
>
>
> TheXman is quite right in stating that in the example given the Number
> of subnets = 2^6 - 2 or 62 and the Number of hosts = 2^10 - 2 or 1022
>
> That's why, surprisingly enough, the address range given was correct
> for the example given i.e 10.25.0.0/22!
>
> HTH
>
>
>
>
>
>
>
>
>
> YOUNGMAN
> ------------------------------------------------------------------------
> Posted via http://www.examnotes.net
> ------------------------------------------------------------------------
>
>
>
>

Brat, Nov 13, 2003
16. ### TheXmanGuest

Thank you for the correction. I apologize for saying that Andy was
requesting help. Nevertheless, Andy was incorrect and a bit of an A-hole.

--
TheXman

"Brat" <> wrote in message
news:bp09n8\$q57\$...
You need to pay attention here... MIKE asked for help not Andy

"TheXman" <> wrote in message
news:...
> Thank you YOUNGMAN. Can you believe the nerve of this Andy Foster
> character? First, he ask for help understanding IP subnetting. Then, he
> mocks correct answers as if he knows IP subnetting. Instead of defending

my
> answer, I decided it would be more pleasurable to let Andy make a FOOL of
> himself when he shows others his failed understanding of IP subnetting.

You
> can only teach people who want to learn. Obviously, Andy Foster already
> knows everything. (lol).
> --
> Sincerely,
> TheXman
>
>
>
> "YOUNGMAN" <> wrote in message
> news:...
>
> To Andy Foster -
>
> It must be wonderful to be so sure of your own abilities . . . however
>
> TheXman quite clearly stated that for the address 10.25.0.0/22 - "in
> the 3rd octect, you have 6 bits for the subnets and 10 bits
> for the hosts"
>
> This was not edited afterwards as you quoted it in your own post.
>
> However you then misquoted and mocked it
>
> > 16 bits in the 3rd octet ?
> > Wrong!

>
>
> Wrong!
>
>
> TheXman is quite right in stating that in the example given the Number
> of subnets = 2^6 - 2 or 62 and the Number of hosts = 2^10 - 2 or 1022
>
> That's why, surprisingly enough, the address range given was correct
> for the example given i.e 10.25.0.0/22!
>
> HTH
>
>
>
>
>
>
>
>
>
> YOUNGMAN
> ------------------------------------------------------------------------
> Posted via http://www.examnotes.net
> ------------------------------------------------------------------------