Velocity Reviews > MCSA > 70-291 - Range of Addresses

# 70-291 - Range of Addresses

Armand Collin
Guest
Posts: n/a

 04-05-2007
Hi everyone,

Please see this description from MS-Press 70-291 Self Paced:

For example, Figure 2-12 shows the address space for 10.0.0.0/12. Because
this address is a Class A address, the default number of 1-bits in the
subnet mask is 8; the mask has been extended 4 bits. Thus, 4 bits remain for
the subnet ID and 20 bits remain for the host ID. On such a network, the
range of addresses on the first subnet (ID 000) is 10.0.0.1-10.15.255.254.

I understood how to calculate the number of subnets and the number of hosts
per subnet, but how do you get this range - 10.0.0.1-10.15.255.254 ??

Armand

Nickoftime@Rocketmail.com
Guest
Posts: n/a

 04-05-2007
On Apr 5, 9:22 am, "Armand Collin" <(E-Mail Removed)>
wrote:
> Hi everyone,
>
> Please see this description from MS-Press 70-291 Self Paced:
>
> For example, Figure 2-12 shows the address space for 10.0.0.0/12. Because
> this address is a Class A address, the default number of 1-bits in the
> subnet mask is 8; the mask has been extended 4 bits. Thus, 4 bits remain for
> the subnet ID and 20 bits remain for the host ID. On such a network, the
> range of addresses on the first subnet (ID 000) is 10.0.0.1-10.15.255.254.
>
> I understood how to calculate the number of subnets and the number of hosts
> per subnet, but how do you get this range - 10.0.0.1-10.15.255.254 ??
>
>
> Armand

I am by no means an expert on this as I am studying myself for this
same exam but I think you need to look at the range like this for it
to make sense. It looks like the all 0's range.

..{0000}0000.00000000.00000001 - .{0000}1111.11111111.11111110

I know in the past there were limits to the using a subnet 0 and all
ones subnet and some devices would not allow you to use these subnets
but now, at least on Cisco devices you can use them when the ip subnet
zero command is configured.

But again I am by no means an expert.

Nick

TheITGirl
Guest
Posts: n/a

 04-05-2007

"Armand Collin" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> Hi everyone,
>
> Please see this description from MS-Press 70-291 Self Paced:
>
> For example, Figure 2-12 shows the address space for 10.0.0.0/12. Because
> this address is a Class A address, the default number of 1-bits in the
> subnet mask is 8; the mask has been extended 4 bits. Thus, 4 bits remain
> for the subnet ID and 20 bits remain for the host ID. On such a network,
> the range of addresses on the first subnet (ID 000) is
> 10.0.0.1-10.15.255.254.
>
> I understood how to calculate the number of subnets and the number of
> hosts per subnet, but how do you get this range - 10.0.0.1-10.15.255.254
> ??
>
>
> Armand
>
>
>

Hello Armand

It may help to break the addresses down into binary.

The network portion of the address, which will be the first 12 bits, is
always going to be the same, so each address in the network will begin
000001010 0000xxxx xxxxxxxx xxxxxxxx.

The minimum value that the remaining 20 host bits can take is 0000 00000000
00000001 (I believe Microsoft work on the assumption that you don't use
subnet zero). The maximum value that the host bits can take is 1111
11111111 11111110 (remember that you cannot have all ones in the host

If you now add the the network bits to the minimum and maximum value for the
host bits, the range of addresses in binary will be 00001010 00000000
00000000 00000001 (10.0.0.1) to 00001010 00001111 11111111 11111110
(10.15.255.254).

Does that make it any easier?

IT Girl MCDST

CBIC
Guest
Posts: n/a

 04-05-2007

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

> Hello Armand
>
> It may help to break the addresses down into binary.
>
> The network portion of the address, which will be the first 12 bits, is
> always going to be the same, so each address in the network will begin
> 000001010 0000xxxx xxxxxxxx xxxxxxxx.
>
> The minimum value that the remaining 20 host bits can take is 0000
> 00000000 00000001 (I believe Microsoft work on the assumption that you
> don't use subnet zero). The maximum value that the host bits can take is
> 1111 11111111 11111110 (remember that you cannot have all ones in the host
>
> If you now add the the network bits to the minimum and maximum value for
> the host bits, the range of addresses in binary will be 00001010 00000000
> 00000000 00000001 (10.0.0.1) to 00001010 00001111 11111111 11111110
> (10.15.255.254).
>
> Does that make it any easier?
>

That's a good explanation. Way better than I could do. It's one thing to
know how to subnet and quite another to explain it. Nice job.

Michael D. Alligood
Guest
Posts: n/a

 04-05-2007
Great Job! I have not performed subnetting in a while, and that returned
me back to the good 'ol days. Excellent job in your explanation...

--

Michael D. Alligood
MCSA, MCDST, MCP,
A+, Network+, i-Net+,

The Classroom Blog

CertGuard Member
The Stronghold for Excellence in I.T. Certification
www.CertGuard.com

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

> "Armand Collin" <(E-Mail Removed)> wrote in message
> news:(E-Mail Removed)...
> > Hi everyone,
> >
> > Please see this description from MS-Press 70-291 Self Paced:
> >
> > For example, Figure 2-12 shows the address space for 10.0.0.0/12. Because
> > this address is a Class A address, the default number of 1-bits in the
> > subnet mask is 8; the mask has been extended 4 bits. Thus, 4 bits remain
> > for the subnet ID and 20 bits remain for the host ID. On such a network,
> > the range of addresses on the first subnet (ID 000) is
> > 10.0.0.1-10.15.255.254.
> >
> > I understood how to calculate the number of subnets and the number of
> > hosts per subnet, but how do you get this range - 10.0.0.1-10.15.255.254
> > ??
> >
> >
> > Armand
> >
> >
> >

> Hello Armand
>
> It may help to break the addresses down into binary.
>
> The network portion of the address, which will be the first 12 bits, is
> always going to be the same, so each address in the network will begin
> 000001010 0000xxxx xxxxxxxx xxxxxxxx.
>
> The minimum value that the remaining 20 host bits can take is 0000 00000000
> 00000001 (I believe Microsoft work on the assumption that you don't use
> subnet zero). The maximum value that the host bits can take is 1111
> 11111111 11111110 (remember that you cannot have all ones in the host
>
> If you now add the the network bits to the minimum and maximum value for the
> host bits, the range of addresses in binary will be 00001010 00000000
> 00000000 00000001 (10.0.0.1) to 00001010 00001111 11111111 11111110
> (10.15.255.254).
>
> Does that make it any easier?
>
> IT Girl MCDST

Armand Collin
Guest
Posts: n/a

 04-05-2007

It's clear now.

Regards,
Armand

"CBIC" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
>
> "TheITGirl" <(E-Mail Removed)> wrote in message
> news:(E-Mail Removed)...
>
>> Hello Armand
>>
>> It may help to break the addresses down into binary.
>>
>> The network portion of the address, which will be the first 12 bits, is
>> always going to be the same, so each address in the network will begin
>> 000001010 0000xxxx xxxxxxxx xxxxxxxx.
>>
>> The minimum value that the remaining 20 host bits can take is 0000
>> 00000000 00000001 (I believe Microsoft work on the assumption that you
>> don't use subnet zero). The maximum value that the host bits can take is
>> 1111 11111111 11111110 (remember that you cannot have all ones in the
>>
>> If you now add the the network bits to the minimum and maximum value for
>> the host bits, the range of addresses in binary will be 00001010 00000000
>> 00000000 00000001 (10.0.0.1) to 00001010 00001111 11111111 11111110
>> (10.15.255.254).
>>
>> Does that make it any easier?
>>

>
> That's a good explanation. Way better than I could do. It's one thing to
> know how to subnet and quite another to explain it. Nice job.
>

TheITGirl
Guest
Posts: n/a

 04-05-2007

"Michael D. Alligood" <(E-Mail Removed)> wrote in message
news:ua%(E-Mail Removed)...
> Great Job! I have not performed subnetting in a while, and that returned
> me back to the good 'ol days. Excellent job in your explanation...
>
> --
>
> Michael D. Alligood
> MCSA, MCDST, MCP,
> A+, Network+, i-Net+,
>
> The Classroom Blog
>
> CertGuard Member
> The Stronghold for Excellence in I.T. Certification
> www.CertGuard.com
>

Thanks Michael <head swells with pride>

I don't do any subnetting at all in my job, but took a networking course as
part of my degree and the binary side of things just "clicked".

Regards.

IT Girl MCDST

TheITGirl
Guest
Posts: n/a

 04-05-2007

"CBIC" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
>
> "TheITGirl" <(E-Mail Removed)> wrote in message
> news:(E-Mail Removed)...
>
>> Hello Armand
>>
>> It may help to break the addresses down into binary.
>>
>> The network portion of the address, which will be the first 12 bits, is
>> always going to be the same, so each address in the network will begin
>> 000001010 0000xxxx xxxxxxxx xxxxxxxx.
>>
>> The minimum value that the remaining 20 host bits can take is 0000
>> 00000000 00000001 (I believe Microsoft work on the assumption that you
>> don't use subnet zero). The maximum value that the host bits can take is
>> 1111 11111111 11111110 (remember that you cannot have all ones in the
>>
>> If you now add the the network bits to the minimum and maximum value for
>> the host bits, the range of addresses in binary will be 00001010 00000000
>> 00000000 00000001 (10.0.0.1) to 00001010 00001111 11111111 11111110
>> (10.15.255.254).
>>
>> Does that make it any easier?
>>

>
> That's a good explanation. Way better than I could do. It's one thing to
> know how to subnet and quite another to explain it. Nice job.
>

Thanks CBIC

IT Girl MCDST

catwalker63
Guest
Posts: n/a

 04-05-2007
TheITGirl piffled away vaguely:

<question and exceedingly excellent explanation snipped>

--

Catwalker
MCNGP #43
www.mcngp.com
"I have a gun. It's loaded. Shut up."