Velocity Reviews > MCSE > 216 and Subnetting

# 216 and Subnetting

Mike
Guest
Posts: n/a

 11-13-2003
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" <(E-Mail Removed)> wrote in message
news:23f001c3a937\$272ef4b0\$(E-Mail Removed)...
> 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
Guest
Posts: n/a

 11-13-2003
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" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
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" <(E-Mail Removed)> wrote in message
news:23f001c3a937\$272ef4b0\$(E-Mail Removed)...
> 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
> >>.
> >>

> >.
> >

Consultant
Guest
Posts: n/a

 11-13-2003
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" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> 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" <(E-Mail Removed)> wrote in message
> news:(E-Mail Removed)...
>
> 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
> ------------------------------------------------------------------------
>
>
>
>

Mike
Guest
Posts: n/a

 11-13-2003
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" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> 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" <(E-Mail Removed)> wrote in message
> news:(E-Mail Removed)...
> 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" <(E-Mail Removed)> wrote in message
> news:23f001c3a937\$272ef4b0\$(E-Mail Removed)...
> > 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
> > >>.
> > >>
> > >.
> > >

>
>
>

Brat
Guest
Posts: n/a

 11-13-2003
You need to pay attention here... MIKE asked for help not Andy

"TheXman" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> 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" <(E-Mail Removed)> wrote in message
> news:(E-Mail Removed)...
>
> 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
Guest
Posts: n/a

 11-13-2003
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" <(E-Mail Removed)> wrote in message
news:bp09n8\$q57\$(E-Mail Removed)...
You need to pay attention here... MIKE asked for help not Andy

"TheXman" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> 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" <(E-Mail Removed)> wrote in message
> news:(E-Mail Removed)...
>
> 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
> ------------------------------------------------------------------------
>
>
>
>