|
Home > Archive > alt.certification.cisco > August 2003 > Subnetting Question
You are viewing an archived Text-only version of the thread.
To view this thread in it's original format and/or if you want to reply to
this thread please [click here]
| Author |
Subnetting Question
|
|
| Kevin Goad 2003-07-31, 9:25 pm |
| I'm having difficulty with my subnet masking with masks under 24
bits(This is what really screwed me the first time I took my test, I
thought I had subnetting down, apparantly not.). Could someone please
let me know if this is right?
Im trying this: 192.168.0.0/19, and I get this:
Mask: 255.255.224.0
# of Nets: 6
# of Hosts Per Network: 8190
1st Net: 192.168.0.1
2nd Net: 192.168.32.32
3rd Net: 192.168.64.64
4th Net: 192.168.96.96
5th Net: 192.168.128.128
6th Net: 192.168.192.192
1st Host: 192.168.0.2
1st B'cast: 192.168.32.31
For some reason that doesn't look right. Since a network number cannot
be all "0"'s or all "1''s, the third and forth octect of the first host
on the first network would have to be 00100000.00000001, which is
192.168.32.1.
Im not sure what I am doing wrong.
TIA
| |
|
| On Thu, 31 Jul 2003 20:38:26 -0400, Kevin Goad wrote:
> I'm having difficulty with my subnet masking with masks under 24
> bits(This is what really screwed me the first time I took my test, I
> thought I had subnetting down, apparantly not.). Could someone please
> let me know if this is right?
>
> Im trying this: 192.168.0.0/19, and I get this:
>
>
> Mask: 255.255.224.0
> # of Nets: 6
> # of Hosts Per Network: 8190
>
>
> 1st Net: 192.168.0.1
> 2nd Net: 192.168.32.32
> 3rd Net: 192.168.64.64
> 4th Net: 192.168.96.96
> 5th Net: 192.168.128.128
> 6th Net: 192.168.192.192
>
> 1st Host: 192.168.0.2
> 1st B'cast: 192.168.32.31
Almost right.
You have 8 subnets, not 6.
And they are as follows :
1 192.168.0.1 to 192.168.31.254
2 192.168.32.1 to 192.168.63.254
3 192.168.64.1 to 192.168.95.254
4 192.168.96.1 to 192.168.127.254
5 192.168.128.1 to 192.168.159.254
6 192.168.160.1 to 192.168.191.254
7 192.168.192.1 to 192.168.223.254
8 192.168.224.1 to 192.168.255.254
Host address example : 192.168.0.1
Broadcast example : 192.168.31.255
Remember that since the network/host is a 32 bit number, all four octets,
not just one.
The network (or host) portion of the 3rd octet can most certainly be all 1's or 0's,
as long as the 1st and 2nd (4th for host) octets are not also.
Sorry, my explanation is probably confusing.
Hope this helps.
| |
| Kevin Goad 2003-07-31, 10:24 pm |
| Sean wrote:
> On Thu, 31 Jul 2003 20:38:26 -0400, Kevin Goad wrote:
>
>
>>I'm having difficulty with my subnet masking with masks under 24
>>bits(This is what really screwed me the first time I took my test, I
>>thought I had subnetting down, apparantly not.). Could someone please
>>let me know if this is right?
>>
>>Im trying this: 192.168.0.0/19, and I get this:
>>
>>
>>Mask: 255.255.224.0
>># of Nets: 6
>># of Hosts Per Network: 8190
>>
>>
>>1st Net: 192.168.0.1
>>2nd Net: 192.168.32.32
>>3rd Net: 192.168.64.64
>>4th Net: 192.168.96.96
>>5th Net: 192.168.128.128
>>6th Net: 192.168.192.192
>>
>>1st Host: 192.168.0.2
>>1st B'cast: 192.168.32.31
>
>
> Almost right.
> You have 8 subnets, not 6.
> And they are as follows :
> 1 192.168.0.1 to 192.168.31.254
> 2 192.168.32.1 to 192.168.63.254
> 3 192.168.64.1 to 192.168.95.254
> 4 192.168.96.1 to 192.168.127.254
> 5 192.168.128.1 to 192.168.159.254
> 6 192.168.160.1 to 192.168.191.254
> 7 192.168.192.1 to 192.168.223.254
> 8 192.168.224.1 to 192.168.255.254
>
> Host address example : 192.168.0.1
> Broadcast example : 192.168.31.255
>
> Remember that since the network/host is a 32 bit number, all four octets,
> not just one.
> The network (or host) portion of the 3rd octet can most certainly be all 1's or 0's,
> as long as the 1st and 2nd (4th for host) octets are not also.
>
> Sorry, my explanation is probably confusing.
> Hope this helps.
>
>
>
>
Ok, thanks for that. But could you explain how you got those numbers?
I was taught the formula (2^n)-2 to find the Number of networks and
hosts in a subnet, where n equals the number of "1" or "0" bits in the
mask. Maybe you could tell me how to apply this accross the
octects(less than 24 bit bitmasks).
Example(And how I Figure out masks >24 bits):
Mask = 255.255.255.192
192.168.1.1
Since that is 2 bits over the Class C natural mask, then the formula
would be (2^2)-2, which is 2.
So far:
2 Total Networks
And since there are 6 "0" bits left, you would do (2^6)-2, which
is 62.
So Far:
2 Total Networks
62 Hosts on Each Network
Since the first network always starts on 256 minus the last octect
of the mask, we can do 256-192, which is 64.
So Far:
2 Total Networks
62 Hosts Per Network
First Network: 192.168.1.64
Since there are 62 hosts(64 addresses) per network, we can add 64
to get the second and final network.
So Far:
2 Total Networks
62 Hosts Per Network
First Network: 192.168.1.64
2nd Network: 192.168.1.128
Could somebody take a look at this, and tell me how I can apply it to
masks <24 bits, to get the answers that Sean came up with? I'm lost. 
| |
|
|
On Thu, 31 Jul 2003, Kevin Goad wrote:
> I'm having difficulty with my subnet masking with masks under 24
> bits(This is what really screwed me the first time I took my test, I
> thought I had subnetting down, apparantly not.). Could someone please
> let me know if this is right?
>
> Im trying this: 192.168.0.0/19, and I get this:
>
192.168.0.0 is a private class C with a natural mask of 255.255.255.0
(/24). But this is a minor point so let's put that aside for now.
>
> Mask: 255.255.224.0 # of Nets: 6 # of Hosts Per Network: 8190
>
You are assuming it's a class B with a natural /16 mask. You are
subnetting using /19 thus borrowing 3 bits, leaving 13 bits for hosts.
2**n -2 = 2**3 -2 = 6 netwoks.
2**2 -2 = 2**13 -2 = 8190 hosts per net.
>
> 1st Net: 192.168.0.1
> 2nd Net: 192.168.32.32
> 3rd Net: 192.168.64.64
> 4th Net: 192.168.96.96
> 5th Net: 192.168.128.128
> 6th Net: 192.168.192.192
>
Wrong! /19 subnet mask is 255.255.224.0
Subtract this from 224 from 256, you have 32 (the increment of your
networks on the third octet). Thus,
1st Net: 192.168.0.0 (can't use all 0's)
first host - 192.168.0.1
last host - 192.168.31.254
broadcast - 192.168.31.255
2nd Net: 192.168.32.0
first host - 192.168.32.1
last host - 192.168.63.254
broadcast - 192.168.63.255
3rd Net: 192.168.64.0
first host - 192.168.64.1
last host - 192.168.95.254
broadcast - 192.168.95.255
4th Net: 192.168.96.0
5th Net: 192.168.128.0
6th Net: 192.168.160.0
7th Net: 192.168.192.0
8th Net: 192.168.224.0
first host - 192.168.224.1
last host - 192.168.255.255 (since this is all 1's, can't use)
Doan
> 1st Host: 192.168.0.2
> 1st B'cast: 192.168.32.31
>
>
>
> For some reason that doesn't look right. Since a network number cannot
> be all "0"'s or all "1''s, the third and forth octect of the first host
> on the first network would have to be 00100000.00000001, which is
> 192.168.32.1.
>
> Im not sure what I am doing wrong.
>
> TIA
>
>
| |
| Kevin Goad 2003-08-01, 12:25 am |
| Doan wrote:
> On Thu, 31 Jul 2003, Kevin Goad wrote:
>
>
>>I'm having difficulty with my subnet masking with masks under 24
>>bits(This is what really screwed me the first time I took my test, I
>>thought I had subnetting down, apparantly not.). Could someone please
>>let me know if this is right?
>>
>>Im trying this: 192.168.0.0/19, and I get this:
>>
>
> 192.168.0.0 is a private class C with a natural mask of 255.255.255.0
> (/24). But this is a minor point so let's put that aside for now.
>
>
>>Mask: 255.255.224.0 # of Nets: 6 # of Hosts Per Network: 8190
>>
>
> You are assuming it's a class B with a natural /16 mask. You are
> subnetting using /19 thus borrowing 3 bits, leaving 13 bits for hosts.
> 2**n -2 = 2**3 -2 = 6 netwoks.
> 2**2 -2 = 2**13 -2 = 8190 hosts per net.
>
>
>>1st Net: 192.168.0.1
>>2nd Net: 192.168.32.32
>>3rd Net: 192.168.64.64
>>4th Net: 192.168.96.96
>>5th Net: 192.168.128.128
>>6th Net: 192.168.192.192
>>
>
> Wrong! /19 subnet mask is 255.255.224.0
> Subtract this from 224 from 256, you have 32 (the increment of your
> networks on the third octet). Thus,
>
> 1st Net: 192.168.0.0 (can't use all 0's)
> first host - 192.168.0.1
> last host - 192.168.31.254
> broadcast - 192.168.31.255
> 2nd Net: 192.168.32.0
> first host - 192.168.32.1
> last host - 192.168.63.254
> broadcast - 192.168.63.255
> 3rd Net: 192.168.64.0
> first host - 192.168.64.1
> last host - 192.168.95.254
> broadcast - 192.168.95.255
> 4th Net: 192.168.96.0
> 5th Net: 192.168.128.0
> 6th Net: 192.168.160.0
> 7th Net: 192.168.192.0
> 8th Net: 192.168.224.0
> first host - 192.168.224.1
> last host - 192.168.255.255 (since this is all 1's, can't use)
>
> Doan
>
>
>>1st Host: 192.168.0.2
>>1st B'cast: 192.168.32.31
>>
>>
>>
>>For some reason that doesn't look right. Since a network number cannot
>>be all "0"'s or all "1''s, the third and forth octect of the first host
>>on the first network would have to be 00100000.00000001, which is
>>192.168.32.1.
>>
>>Im not sure what I am doing wrong.
>>
>>TIA
>>
>>
>
>
Wow, somebody give this man a raise. If I have understood you
correctly, then in a 9 through 24 bit mask, you only increment the
networks where the octect splits the host and network.
IE. second octect's mask is 242, I only have to worry about
incrementing the second octect, and all the networks will start x.x.0.1
If its really that easy, I'm gonna hate myself.
| |
| Kevin Goad 2003-08-01, 12:25 am |
| Kevin Goad wrote:
> Doan wrote:
>
>> On Thu, 31 Jul 2003, Kevin Goad wrote:
>>
>>
>>> I'm having difficulty with my subnet masking with masks under 24
>>> bits(This is what really screwed me the first time I took my test, I
>>> thought I had subnetting down, apparantly not.). Could someone please
>>> let me know if this is right?
>>>
>>> Im trying this: 192.168.0.0/19, and I get this:
>>>
>>
>> 192.168.0.0 is a private class C with a natural mask of 255.255.255.0
>> (/24). But this is a minor point so let's put that aside for now.
>>
>>
>>> Mask: 255.255.224.0 # of Nets: 6 # of Hosts Per Network: 8190
>>>
>>
>> You are assuming it's a class B with a natural /16 mask. You are
>> subnetting using /19 thus borrowing 3 bits, leaving 13 bits for hosts.
>> 2**n -2 = 2**3 -2 = 6 netwoks.
>> 2**2 -2 = 2**13 -2 = 8190 hosts per net.
>>
>>
>>> 1st Net: 192.168.0.1
>>> 2nd Net: 192.168.32.32
>>> 3rd Net: 192.168.64.64
>>> 4th Net: 192.168.96.96
>>> 5th Net: 192.168.128.128
>>> 6th Net: 192.168.192.192
>>>
>>
>> Wrong! /19 subnet mask is 255.255.224.0
>> Subtract this from 224 from 256, you have 32 (the increment of your
>> networks on the third octet). Thus,
>>
>> 1st Net: 192.168.0.0 (can't use all 0's)
>> first host - 192.168.0.1
>> last host - 192.168.31.254
>> broadcast - 192.168.31.255
>> 2nd Net: 192.168.32.0
>> first host - 192.168.32.1
>> last host - 192.168.63.254
>> broadcast - 192.168.63.255
>> 3rd Net: 192.168.64.0
>> first host - 192.168.64.1
>> last host - 192.168.95.254
>> broadcast - 192.168.95.255
>> 4th Net: 192.168.96.0
>> 5th Net: 192.168.128.0
>> 6th Net: 192.168.160.0
>> 7th Net: 192.168.192.0
>> 8th Net: 192.168.224.0
>> first host - 192.168.224.1
>> last host - 192.168.255.255 (since this is all 1's, can't use)
>>
>> Doan
>>
>>
10.0.0.0/10
255.192.0.0
2 Networks
1st Network: 10.64.0.0
Host1: 10.64.0.1
BCast: 10.127.255.254
2nd Network: 10.128.0.0
Host1: 10.128.0.1
BCast: 10.191.255.254
So would this be correct?
Thanks
| |
|
|
"Kevin Goad" <None@none.com> wrote in message
news:VrScnfAvi_aFerSiXTWJkQ@co
mcast.com...
> Kevin Goad wrote:
>
> > Doan wrote:
> >
> >> On Thu, 31 Jul 2003, Kevin Goad wrote:
> >>
> >>
> >>> I'm having difficulty with my subnet masking with masks under 24
> >>> bits(This is what really screwed me the first time I took my test, I
> >>> thought I had subnetting down, apparantly not.). Could someone please
> >>> let me know if this is right?
> >>>
> >>> Im trying this: 192.168.0.0/19, and I get this:
> >>>
> >>
> >> 192.168.0.0 is a private class C with a natural mask of 255.255.255.0
> >> (/24). But this is a minor point so let's put that aside for now.
> >>
> >>
> >>> Mask: 255.255.224.0 # of Nets: 6 # of Hosts Per Network: 8190
> >>>
> >>
> >> You are assuming it's a class B with a natural /16 mask. You are
> >> subnetting using /19 thus borrowing 3 bits, leaving 13 bits for hosts.
> >> 2**n -2 = 2**3 -2 = 6 netwoks.
> >> 2**2 -2 = 2**13 -2 = 8190 hosts per net.
> >>
> >>
> >>> 1st Net: 192.168.0.1
> >>> 2nd Net: 192.168.32.32
> >>> 3rd Net: 192.168.64.64
> >>> 4th Net: 192.168.96.96
> >>> 5th Net: 192.168.128.128
> >>> 6th Net: 192.168.192.192
> >>>
> >>
> >> Wrong! /19 subnet mask is 255.255.224.0
> >> Subtract this from 224 from 256, you have 32 (the increment of your
> >> networks on the third octet). Thus,
> >>
> >> 1st Net: 192.168.0.0 (can't use all 0's)
> >> first host - 192.168.0.1
> >> last host - 192.168.31.254
> >> broadcast - 192.168.31.255
> >> 2nd Net: 192.168.32.0
> >> first host - 192.168.32.1
> >> last host - 192.168.63.254
> >> broadcast - 192.168.63.255
> >> 3rd Net: 192.168.64.0
> >> first host - 192.168.64.1
> >> last host - 192.168.95.254
> >> broadcast - 192.168.95.255
> >> 4th Net: 192.168.96.0
> >> 5th Net: 192.168.128.0
> >> 6th Net: 192.168.160.0
> >> 7th Net: 192.168.192.0
> >> 8th Net: 192.168.224.0
> >> first host - 192.168.224.1
> >> last host - 192.168.255.255 (since this is all 1's, can't
use)[
color=darkred]
> >>
> >> Doan
> >>
> >>
>
> 10.0.0.0/10
> 255.192.0.0
>
> 2 Networks
>
> 1st Network: 10.64.0.0
> Host1: 10.64.0.1
> BCast: 10.127.255.254
> 2nd Network: 10.128.0.0
> Host1: 10.128.0.1
> BCast: 10.191.255.254
>
> So would this be correct?
>
>
> Thanks
>[/color]
The broadcast addresses should be
10.127.255.255
10.191.255.255
One less than the next network.
| |
| Kevin Goad 2003-08-01, 1:25 am |
|
> The broadcast addresses should be
> 10.127.255.255
> 10.191.255.255
> One less than the next network.
>
>
Yea, I knew that. Stupid mistake on my part. Thanks for everyone's help.
| |
| JasonW 2003-08-01, 1:25 am |
| Forgive me if I'm missing the point of your question, but here is how I
approach it.
If you use the (2^n)-2 formula, then that precludes the use of the zero
subnet and the broadcast (hence the minus two).
I find that the easiest way for me to figure subnets is to draw out the
octets in binary starting with the first divided octet, like:
192.168.---|-----.--------
There are three slots to the left of the interval, so (2^3)-2 is the number
of useable subnets (6). There are 13 slots to the right of the interval, so
(2^13)-2 is the number of useable hosts per subnet (8190).
The interval is just to the left of the division, in this case 32. So the
first useable subnet, if zero subnet is not enabled, is 192.168.32.0 (the
first host would be 192.168.32.1). The range of the first subnet will run
from 192.168.32.1 to 192.168.63.254 or graphically:
192.168.001|00000.00000001 to 192.168.001|11111.11111110
The next subnet (64) changes the bits on the left of the interval:
192.168.010|00000.00000000
(or 192.168.64.0, the first host would be 192.168.64.1)
Does that help?
-JasonW
"Kevin Goad" <None@none.com> wrote in message
news:igydnQEGv4Z0WrSiU-KYvw@comcast.com...
> Sean wrote:
> > On Thu, 31 Jul 2003 20:38:26 -0400, Kevin Goad wrote:
> >
> >
> >>I'm having difficulty with my subnet masking with masks under 24
> >>bits(This is what really screwed me the first time I took my test, I
> >>thought I had subnetting down, apparantly not.). Could someone please
> >>let me know if this is right?
> >>
> >>Im trying this: 192.168.0.0/19, and I get this:
> >>
> >>
> >>Mask: 255.255.224.0
> >># of Nets: 6
> >># of Hosts Per Network: 8190
> >>
> >>
> >>1st Net: 192.168.0.1
> >>2nd Net: 192.168.32.32
> >>3rd Net: 192.168.64.64
> >>4th Net: 192.168.96.96
> >>5th Net: 192.168.128.128
> >>6th Net: 192.168.192.192
> >>
> >>1st Host: 192.168.0.2
> >>1st B'cast: 192.168.32.31
> >
> >
> > Almost right.
> > You have 8 subnets, not 6.
> > And they are as follows :
> > 1 192.168.0.1 to 192.168.31.254
> > 2 192.168.32.1 to 192.168.63.254
> > 3 192.168.64.1 to 192.168.95.254
> > 4 192.168.96.1 to 192.168.127.254
> > 5 192.168.128.1 to 192.168.159.254
> > 6 192.168.160.1 to 192.168.191.254
> > 7 192.168.192.1 to 192.168.223.254
> > 8 192.168.224.1 to 192.168.255.254
> >
> > Host address example : 192.168.0.1
> > Broadcast example : 192.168.31.255
> >
> > Remember that since the network/host is a 32 bit number, all four
octets,
> > not just one.
> > The network (or host) portion of the 3rd octet can most certainly be all
1's or 0's,
> > as long as the 1st and 2nd (4th for host) octets are not also.
> >
> > Sorry, my explanation is probably confusing.
> > Hope this helps.
> >
> >
> >
> >
>
> Ok, thanks for that. But could you explain how you got those numbers?
> I was taught the formula (2^n)-2 to find the Number of networks and
> hosts in a subnet, where n equals the number of "1" or "0" bits in the
> mask. Maybe you could tell me how to apply this accross the
> octects(less than 24 bit bitmasks).
>
> Example(And how I Figure out masks >24 bits):
>
>
> Mask = 255.255.255.192
> 192.168.1.1
> Since that is 2 bits over the Class C natural mask, then the formula
> would be (2^2)-2, which is 2.
> So far:
> 2 Total Networks
>
> And since there are 6 "0" bits left, you would do (2^6)-2, which
> is 62.
> So Far:
> 2 Total Networks
> 62 Hosts on Each Network
>
> Since the first network always starts on 256 minus the last octect
> of the mask, we can do 256-192, which is 64.
> So Far:
> 2 Total Networks
> 62 Hosts Per Network
> First Network: 192.168.1.64
>
> Since there are 62 hosts(64 addresses) per network, we can add 64
> to get the second and final network.
> So Far:
> 2 Total Networks
> 62 Hosts Per Network
> First Network: 192.168.1.64
> 2nd Network: 192.168.1.128
>
>
> Could somebody take a look at this, and tell me how I can apply it to
> masks <24 bits, to get the answers that Sean came up with? I'm lost.
>
>
>
>
>
>
| |
| Kevin Goad 2003-08-01, 2:24 am |
| JasonW wrote:
> Forgive me if I'm missing the point of your question, but here is how I
> approach it.
> If you use the (2^n)-2 formula, then that precludes the use of the zero
> subnet and the broadcast (hence the minus two).
> I find that the easiest way for me to figure subnets is to draw out the
> octets in binary starting with the first divided octet, like:
> 192.168.---|-----.--------
>
> There are three slots to the left of the interval, so (2^3)-2 is the number
> of useable subnets (6). There are 13 slots to the right of the interval, so
> (2^13)-2 is the number of useable hosts per subnet (8190).
>
> The interval is just to the left of the division, in this case 32. So the
> first useable subnet, if zero subnet is not enabled, is 192.168.32.0 (the
> first host would be 192.168.32.1). The range of the first subnet will run
> from 192.168.32.1 to 192.168.63.254 or graphically:
> 192.168.001|00000.00000001 to 192.168.001|11111.11111110
>
> The next subnet (64) changes the bits on the left of the interval:
> 192.168.010|00000.00000000
> (or 192.168.64.0, the first host would be 192.168.64.1)
>
> Does that help?
>
> -JasonW
>
>
Yes, that helps my understanding of it. Now im just working on speed,
and the quickest possible way to do it.
BTW, does anyone have a formula/quick way to find out which network an
address is on, given a mask and Address?
Thanks.
| |
|
| On Thu, 31 Jul 2003, Kevin Goad wrote:
> Doan wrote:
> > On Thu, 31 Jul 2003, Kevin Goad wrote:
> >
> >
> >>I'm having difficulty with my subnet masking with masks under 24
> >>bits(This is what really screwed me the first time I took my test, I
> >>thought I had subnetting down, apparantly not.). Could someone please
> >>let me know if this is right?
> >>
> >>Im trying this: 192.168.0.0/19, and I get this:
> >>
> >
> > 192.168.0.0 is a private class C with a natural mask of 255.255.255.0
> > (/24). But this is a minor point so let's put that aside for now.
> >
> >
> >>Mask: 255.255.224.0 # of Nets: 6 # of Hosts Per Network: 8190
> >>
> >
> > You are assuming it's a class B with a natural /16 mask. You are
> > subnetting using /19 thus borrowing 3 bits, leaving 13 bits for hosts.
> > 2**n -2 = 2**3 -2 = 6 netwoks.
> > 2**2 -2 = 2**13 -2 = 8190 hosts per net.
> >
> >
> >>1st Net: 192.168.0.1
> >>2nd Net: 192.168.32.32
> >>3rd Net: 192.168.64.64
> >>4th Net: 192.168.96.96
> >>5th Net: 192.168.128.128
> >>6th Net: 192.168.192.192
> >>
> >
> > Wrong! /19 subnet mask is 255.255.224.0
> > Subtract this from 224 from 256, you have 32 (the increment of your
> > networks on the third octet). Thus,
> >
> > 1st Net: 192.168.0.0 (can't use all 0's)
> > first host - 192.168.0.1
> > last host - 192.168.31.254
> > broadcast - 192.168.31.255
> > 2nd Net: 192.168.32.0
> > first host - 192.168.32.1
> > last host - 192.168.63.254
> > broadcast - 192.168.63.255
> > 3rd Net: 192.168.64.0
> > first host - 192.168.64.1
> > last host - 192.168.95.254
> > broadcast - 192.168.95.255
> > 4th Net: 192.168.96.0
> > 5th Net: 192.168.128.0
> > 6th Net: 192.168.160.0
> > 7th Net: 192.168.192.0
> > 8th Net: 192.168.224.0
> > first host - 192.168.224.1
> > last host - 192.168.255.255 (since this is all 1's, can't use)
> >
> > Doan
> >
> >
> >>1st Host: 192.168.0.2
> >>1st B'cast: 192.168.32.31
> >>
> >>
> >>
> >>For some reason that doesn't look right. Since a network number cannot
> >>be all "0"'s or all "1''s, the third and forth octect of the first host
> >>on the first network would have to be 00100000.00000001, which is
> >>192.168.32.1.
> >>
> >>Im not sure what I am doing wrong.
> >>
> >>TIA
> >>
> >>
> >
> >
>
>
> Wow, somebody give this man a raise. If I have understood you
> correctly, then in a 9 through 24 bit mask, you only increment the
> networks where the octect splits the host and network.
> IE. second octect's mask is 242, I only have to worry about
> incrementing the second octect, and all the networks will start x.x.0.1
>
> If its really that easy, I'm gonna hate myself.
>
You got it!
Doan
| |
|
| On Fri, 1 Aug 2003, JasonW wrote:
> Forgive me if I'm missing the point of your question, but here is how I
> approach it.
> If you use the (2^n)-2 formula, then that precludes the use of the zero
> subnet and the broadcast (hence the minus two).
> I find that the easiest way for me to figure subnets is to draw out the
> octets in binary starting with the first divided octet, like:
> 192.168.---|-----.--------
>
> There are three slots to the left of the interval, so (2^3)-2 is the number
> of useable subnets (6). There are 13 slots to the right of the interval, so
> (2^13)-2 is the number of useable hosts per subnet (8190).
>
> The interval is just to the left of the division, in this case 32. So the
> first useable subnet, if zero subnet is not enabled, is 192.168.32.0 (the
> first host would be 192.168.32.1). The range of the first subnet will run
> from 192.168.32.1 to 192.168.63.254 or graphically:
> 192.168.001|00000.00000001 to 192.168.001|11111.11111110
>
> The next subnet (64) changes the bits on the left of the interval:
> 192.168.010|00000.00000000
> (or 192.168.64.0, the first host would be 192.168.64.1)
>
> Does that help?
>
> -JasonW
>
I think this is a good explanation. When first learning subnet, please
do it in binary. It is the easiest way to understand it. Once you
understand it in binary, converting it to decimal and learning the short
cut is a piece of cake! Subtracting your "interesting octet" subnet
mask from 256 give the the increment (256-224 = 32). Your first non-zero
subnet will be that (32) and your last useable one will be your subnet
mask (224).
Doan
| |
|
| Will somebody please tell me why everyone is subtracting 2 from the toatl
number of networks available????
192.168.x.y/24
x = 0 to 255
y = 1 to 254
Everybody agree with that? Or not?
The NETWORK ADDRESS or HOST ADDRESS must not be all 0's or 1's
This does not mean within one octet.
Example:
24 bit subnet (class c)
The network address is 24 bits long = 2^24-2 networks
Everyone still ok here?
Now, try a 19 bit subnet on a class B (this means 3 subnet bits)
The network address is 19 bits long = 2^19 - 2 networks
Your subnets are just a subset of those (2^19 -2) networks.
If it were otherwise, subnetting would be a massive waste of ip space.
Everybody got it?
You don't have 2^3 - 2 135.10.x.x networks
You have 2^3 subnets in the 135.10.x.x network
I really hope I'm making sense. Because I see quite a few people in this
thread doing their math wrong.
-Sean
| |
| John Agosta 2003-08-01, 4:24 pm |
|
"Sean" <codeboy@frozenreality.com> wrote in message
news an.2003.08.01.17.10.51.553042@frozenreality.com...
> Will somebody please tell me why everyone is subtracting 2 from the toatl
> number of networks available????
>
> 192.168.x.y/24
> x = 0 to 255
> y = 1 to 254
>
> Everybody agree with that? Or not?
>
> The NETWORK ADDRESS or HOST ADDRESS must not be all 0's or 1's
> This does not mean within one octet.
>
> Example:
> 24 bit subnet (class c)
> The network address is 24 bits long = 2^24-2 networks
>
> Everyone still ok here?
> Now, try a 19 bit subnet on a class B (this means 3 subnet bits)
>
> The network address is 19 bits long = 2^19 - 2 networks
> Your subnets are just a subset of those (2^19 -2) networks.
> If it were otherwise, subnetting would be a massive waste of ip space.
>
> Everybody got it?
>
> You don't have 2^3 - 2 135.10.x.x networks
> You have 2^3 subnets in the 135.10.x.x network
>
> I really hope I'm making sense. Because I see quite a few people in this
> thread doing their math wrong.
>
> -Sean
>
There are two different worlds that the people who visit this group come
from.
The 'real' world.
And the world of Cisco.
In the real world the amount of subnets you can have is "2 to the power of
[number of subnet bits]"
If we have 13 subnet bits, then we are allowed 2^13 subnets = 8192
In the Cisco world, specifically, the CCNA world, one is taught that the all
ones subnet and the all zeros subnet are not allowed.
Therefore, for the purpose of answering a CCNA exam question correctly, the
amount of subnets available is 8190.
Why does Cisco teach this within the CCNA curriculum? Who knows?
In the CCNP curriculum, the student is told:
"Oh yeah, remember always having to subtract 2 from your subnet space? You
don't
really have to do that after all!"
So now everyone should be happy......
| |
|
|
> Why does Cisco teach this within the CCNA curriculum? Who knows?
> In the CCNP curriculum, the student is told:
>
> "Oh yeah, remember always having to subtract 2 from your subnet space? You
> don't
> really have to do that after all!"
>
> So now everyone should be happy......
It's been a long time since I took the CCNA exam I guess...
Thanks for shining some light on that though, I was a little worried.
| |
|
|
Kevin Goad wrote:
> Yes, that helps my understanding of it. Now im just working on speed,
> and the quickest possible way to do it.
>
> BTW, does anyone have a formula/quick way to find out which network an
> address is on, given a mask and Address?
>
> Thanks.
here's how I do it.
if it's a /19, that's nineteen bits of subnetting, right?
so
(8) (16) (19)
11111111.11111111.11100000.00000000
count up the value.
we know that 8 bits is 255...
the 'interesting octet' --
128 64 32 16 8 4 2 1
1 1 1 0 0 0 0 0
128 + 64 + 32 = 224
here's the magic...
256
-224
------
32
the range of the subnets is 32.
so, we'll use 192.168.0.0
the first range is
192.168.0.0 to 192.168.31.255 first address is the network, last is the
broadcast.
192.168.32.0 to 192.168.63.255 first address is the network, last is the
broadcast.
192.168.64.0 to 192.168.95.255 first address is the network, last is the
broadcast.
--skip
| |
| Kevin Goad 2003-08-01, 9:24 pm |
| Doan wrote:
> On Fri, 1 Aug 2003, JasonW wrote:
>
>
>>Forgive me if I'm missing the point of your question, but here is how I
>>approach it.
>>If you use the (2^n)-2 formula, then that precludes the use of the zero
>>subnet and the broadcast (hence the minus two).
>>I find that the easiest way for me to figure subnets is to draw out the
>>octets in binary starting with the first divided octet, like:
>>192.168.---|-----.--------
>>
>>There are three slots to the left of the interval, so (2^3)-2 is the number
>>of useable subnets (6). There are 13 slots to the right of the interval, so
>>(2^13)-2 is the number of useable hosts per subnet (8190).
>>
>>The interval is just to the left of the division, in this case 32. So the
>>first useable subnet, if zero subnet is not enabled, is 192.168.32.0 (the
>>first host would be 192.168.32.1). The range of the first subnet will run
>>from 192.168.32.1 to 192.168.63.254 or graphically:
>>192.168.001|00000.00000001 to 192.168.001|11111.11111110
>>
>>The next subnet (64) changes the bits on the left of the interval:
>>192.168.010|00000.00000000
>>(or 192.168.64.0, the first host would be 192.168.64.1)
>>
>>Does that help?
>>
>>-JasonW
>>
>
> I think this is a good explanation. When first learning subnet, please
> do it in binary. It is the easiest way to understand it. Once you
> understand it in binary, converting it to decimal and learning the short
> cut is a piece of cake! Subtracting your "interesting octet" subnet
> mask from 256 give the the increment (256-224 = 32). Your first non-zero
> subnet will be that (32) and your last useable one will be your subnet
> mask (224).
>
> Doan
>
I think I have ran into another problem. What if it is a Class A
Address, with a bitmask such as 17 or greater? Then wouldnt that make
not only the "interesting" octect the one where the mask changes"(3rd
octect), but the 2nd octect as well?
I came up with this, but I have no idea if im on the right track or not.
68.49.90.0/23
32766 Networks
510 Hosts on Each Network
Thanks
| |
| JasonW 2003-08-01, 10:25 pm |
| Yes, those numbers are right if you have 68._._._/23
The question doesn't make sense to me if you specify numbers for the octets
(68.49.90.0), since that implies those are static.
(Regarding the math on network portion, I'm finding conflicting information
regarding the (2^n)-2 formula as it applies to the number of networks. Some
insist that you can't use all ones in the network portion, but that doesn't
make sense to me. It is only all ones in the host portion that would define
it as a broadcast, as far as I can see.)
-JasonW
"Kevin Goad" <None@none.com> wrote in message
news:jTSdnZJUUZntl7aiXTWJiQ@co
mcast.com...
> Doan wrote:
(snip)
> I think I have ran into another problem. What if it is a Class A
> Address, with a bitmask such as 17 or greater? Then wouldnt that make
> not only the "interesting" octect the one where the mask changes"(3rd
> octect), but the 2nd octect as well?
>
> I came up with this, but I have no idea if im on the right track or not.
>
>
> 68.49.90.0/23
>
> 32766 Networks
> 510 Hosts on Each Network
>
>
> Thanks
>
| |
| Kevin Goad 2003-08-01, 11:25 pm |
| JasonW wrote:
> Yes, those numbers are right if you have 68._._._/23
>
> The question doesn't make sense to me if you specify numbers for the octets
> (68.49.90.0), since that implies those are static.
>
> (Regarding the math on network portion, I'm finding conflicting information
> regarding the (2^n)-2 formula as it applies to the number of networks. Some
> insist that you can't use all ones in the network portion, but that doesn't
> make sense to me. It is only all ones in the host portion that would define
> it as a broadcast, as far as I can see.)
>
> -JasonW
>
Sorry, I was figuring out some subnet problems using my own IP, and
forgot to take the second and third octect out.
But with the mask being /17-/23 in a class A address, how do I figure
out what to increment the second octect by?
| |
| Ceyko 2003-08-02, 11:24 am |
| Old school (RFCs) rule, do the MS exams still follow old or new RFCs in
regards to subtracting the first and last?
Ceyko
"John Agosta" <j_agosta@wideopenwest.kom> wrote in message
news:reOcnYWX3v9oWLeiU-KYvw@wideopenwest.com...
>
> "Sean" <codeboy@frozenreality.com> wrote in message
> newsan.2003.08.01.17.10.51.553042@frozenreality.com...
> > Will somebody please tell me why everyone is subtracting 2 from the
toatl
> > number of networks available????
> >
> > 192.168.x.y/24
> > x = 0 to 255
> > y = 1 to 254
> >
> > Everybody agree with that? Or not?
> >
> > The NETWORK ADDRESS or HOST ADDRESS must not be all 0's or 1's
> > This does not mean within one octet.
> >
> > Example:
> > 24 bit subnet (class c)
> > The network address is 24 bits long = 2^24-2 networks
> >
> > Everyone still ok here?
> > Now, try a 19 bit subnet on a class B (this means 3 subnet bits)
> >
> > The network address is 19 bits long = 2^19 - 2 networks
> > Your subnets are just a subset of those (2^19 -2) networks.
> > If it were otherwise, subnetting would be a massive waste of ip space.
> >
> > Everybody got it?
> >
> > You don't have 2^3 - 2 135.10.x.x networks
> > You have 2^3 subnets in the 135.10.x.x network
> >
> > I really hope I'm making sense. Because I see quite a few people in
this
> > thread doing their math wrong.
> >
> > -Sean
> >
>
> There are two different worlds that the people who visit this group come
> from.
>
> The 'real' world.
> And the world of Cisco.
>
> In the real world the amount of subnets you can have is "2 to the power of
> [number of subnet bits]"
>
> If we have 13 subnet bits, then we are allowed 2^13 subnets = 8192
>
> In the Cisco world, specifically, the CCNA world, one is taught that the
all
> ones subnet and the all zeros subnet are not allowed.
> Therefore, for the purpose of answering a CCNA exam question correctly,
the
> amount of subnets available is 8190.
>
> Why does Cisco teach this within the CCNA curriculum? Who knows?
> In the CCNP curriculum, the student is told:
>
> "Oh yeah, remember always having to subtract 2 from your subnet space?
You
> don't
> really have to do that after all!"
>
> So now everyone should be happy......
>
>
>
>
x-- 100 Proof News - http://www.100ProofNews.com
x-- 3,500+ Binary NewsGroups, and over 90,000 other groups
x-- Access to over 800 Gigs/Day - $8.95/Month
x-- UNLIMITED DOWNLOAD
| |
|
|
| JasonW 2003-08-02, 3:24 pm |
| The second octet doesn't really increment until the third octet is full,
using regular binary math.
Using the same number, 68._._._/23 draw it out in binary again:
68.00000000.0000000|0.00000000 (the | marks the interval at the 23rd bit)
This makes the variable subnet portion equal to 15 bits (the bits on the
left side of the | mark), not including the static first octet.
In binary, you increment these bits as if they
were all in one stream instead of being divided into octets. So after
the third octet is all 1's, the next increment puts a 1 in the last bit of
the 2nd octet and makes the third octet all 0's:
68.00000000.0000001|0.00000000 (Subnet #1, 68.0.2.0)
68.00000000.0000001|1.11111111 (broadcast address of subnet #1)
68.00000000.0000010|0.00000000 (Subnet #2, 68.0.4.0)
....
68.00000000.1111111|1.11111111 (broadcast address of subnet #127)
68.00000001.0000000|0.00000000 (subnet #128 counting from the interval)
68.00000001.0000000|1.11111111 (broadcast address of subnet #128)
68.00000001.0000001|0.00000000 (subnet #129, 68.1.2.0)
....
68.00000001.1111111|1.11111111 (broadcast address of subnet #255)
68.00000010.0000000|0.00000000 (subnet #256, 68.2.0.0)
-JasonW
"Kevin Goad" <None@none.com> wrote in message
news:I92cnTsqxoWRubaiXTWJiQ@co
mcast.com...
>
> Sorry, I was figuring out some subnet problems using my own IP, and
> forgot to take the second and third octect out.
>
> But with the mask being /17-/23 in a class A address, how do I figure
> out what to increment the second octect by?
>
| |
| Kevin Goad 2003-08-02, 4:24 pm |
| JasonW wrote:
> The second octet doesn't really increment until the third octet is full,
> using regular binary math.
>
> Using the same number, 68._._._/23 draw it out in binary again:
> 68.00000000.0000000|0.00000000 (the | marks the interval at the 23rd bit)
>
> This makes the variable subnet portion equal to 15 bits (the bits on the
> left side of the | mark), not including the static first octet.
>
> In binary, you increment these bits as if they
> were all in one stream instead of being divided into octets. So after
> the third octet is all 1's, the next increment puts a 1 in the last bit of
> the 2nd octet and makes the third octet all 0's:
>
> 68.00000000.0000001|0.00000000 (Subnet #1, 68.0.2.0)
> 68.00000000.0000001|1.11111111 (broadcast address of subnet #1)
> 68.00000000.0000010|0.00000000 (Subnet #2, 68.0.4.0)
> ...
> 68.00000000.1111111|1.11111111 (broadcast address of subnet #127)
> 68.00000001.0000000|0.00000000 (subnet #128 counting from the interval)
> 68.00000001.0000000|1.11111111 (broadcast address of subnet #128)
> 68.00000001.0000001|0.00000000 (subnet #129, 68.1.2.0)
> ...
> 68.00000001.1111111|1.11111111 (broadcast address of subnet #255)
> 68.00000010.0000000|0.00000000 (subnet #256, 68.2.0.0)
>
> -JasonW
>
Wow, I understand now. Do you have an easy way to find out which
network one of these types of Address & masks are on?
Another Question:
Is it possible to subnet an address with a bitmask of less than its
default mask? Or was my instructor just having a brain fart that day,
and having me subnet a class c address with a 20 bit bitmask?
| |
|
| Ok maybe I missed something but ur not subnetting a class C 192 address with
a /19 mask. You are supernetting.
--
Take the NS out of the reply-address to respond
"Kevin Goad" <None@none.com> wrote in message
news:RyudnUsdDpbmK7SiXTWJjA@co
mcast.com...
> I'm having difficulty with my subnet masking with masks under 24
> bits(This is what really screwed me the first time I took my test, I
> thought I had subnetting down, apparantly not.). Could someone please
> let me know if this is right?
>
> Im trying this: 192.168.0.0/19, and I get this:
>
>
> Mask: 255.255.224.0
> # of Nets: 6
> # of Hosts Per Network: 8190
>
>
> 1st Net: 192.168.0.1
> 2nd Net: 192.168.32.32
> 3rd Net: 192.168.64.64
> 4th Net: 192.168.96.96
> 5th Net: 192.168.128.128
> 6th Net: 192.168.192.192
>
> 1st Host: 192.168.0.2
> 1st B'cast: 192.168.32.31
>
>
>
> For some reason that doesn't look right. Since a network number cannot
> be all "0"'s or all "1''s, the third and forth octect of the first host
> on the first network would have to be 00100000.00000001, which is
> 192.168.32.1.
>
> Im not sure what I am doing wrong.
>
> TIA
>
| |
| Kevin Goad 2003-08-05, 12:24 am |
| RC wrote:
> Ok maybe I missed something but ur not subnetting a class C 192 address with
> a /19 mask. You are supernetting.
>
Yes, you did miss something. The entire thread. It's ok.
| |
|
| Yes but you didn't answer my question, isn't this supernetting and not
subnetting?
--
Take the NS out of the reply-address to respond
"Kevin Goad" <None@none.com> wrote in message
news:Ur-cndvgLKKJubKiXTWJkg@comcast.com...
> RC wrote:
> > Ok maybe I missed something but ur not subnetting a class C 192 address
with
> > a /19 mask. You are supernetting.
> >
> Yes, you did miss something. The entire thread. It's ok.
>
| |
| Kevin Goad 2003-08-06, 9:24 pm |
| RC wrote:
> Yes but you didn't answer my question, isn't this supernetting and not
> subnetting?
>
Well, I think it is.
By the way, I passed my CCNA today with a score of 924. Thanks everyone.
| |
| JasonW 2003-08-07, 2:25 pm |
| Congrats!
-JasonW
"Kevin Goad" <None@none.com> wrote in message
news:6secnRFynvBCAKyiXTWJjg@co
mcast.com...
> RC wrote:
> > Yes but you didn't answer my question, isn't this supernetting and not
> > subnetting?
> >
>
> Well, I think it is.
>
> By the way, I passed my CCNA today with a score of 924. Thanks everyone.
>
| |
| FireSword 2003-08-27, 11:24 pm |
| It is easy if you can convert to binary and know how to do an AND operation.
Example:
192.168.10.135/27
The mask is 255.255.255.224
now you compare the IP to the Mask, every octet that has 255 on its mask
goes unchanged.
So you know the the network starts with: 192.168.10
now convert the last octet to binary:
128 64 32 16 8 4 2 1
1 0 0 0 0 1 1 1
now convert the mask to binary:
128 64 32 16 8 4 2 1
1 1 1 0 0 0 0 0
Now do the AND operation between ip and mask for every bit:
1 AND 1 = 1
1 AND 0 = 0
0 AND 1 = 0
0 AND 0 = 0
10000111
11100000
-----------
10000000
Now convert your result back to decimal.
10000000 = 128
you already had 192.168.10 and now 128, the the Network is 192.168.10.128
I went step by step, so it looks longer than it really is in fact this is
exactly what you computer does in order to know if a destination IP is part
of its own network or if it needs to use the Default Gateway to route it.
if you practice binary convertions you will have a very good chance at your
exam.
"Kevin Goad" <None@none.com> wrote in message
news:8dCdncAvx6PZZ7SiU-KYgw@comcast.com...
> JasonW wrote:
> > Forgive me if I'm missing the point of your question, but here is how I
> > approach it.
> > If you use the (2^n)-2 formula, then that precludes the use of the zero
> > subnet and the broadcast (hence the minus two).
> > I find that the easiest way for me to figure subnets is to draw out the
> > octets in binary starting with the first divided octet, like:
> > 192.168.---|-----.--------
> >
> > There are three slots to the left of the interval, so (2^3)-2 is the
number
> > of useable subnets (6). There are 13 slots to the right of the interval,
so
> > (2^13)-2 is the number of useable hosts per subnet (8190).
> >
> > The interval is just to the left of the division, in this case 32. So
the
> > first useable subnet, if zero subnet is not enabled, is 192.168.32.0
(the
> > first host would be 192.168.32.1). The range of the first subnet will
run
> > from 192.168.32.1 to 192.168.63.254 or graphically:
> > 192.168.001|00000.00000001 to 192.168.001|11111.11111110
> >
> > The next subnet (64) changes the bits on the left of the interval:
> > 192.168.010|00000.00000000
> > (or 192.168.64.0, the first host would be 192.168.64.1)
> >
> > Does that help?
> >
> > -JasonW
> >
> >
>
> Yes, that helps my understanding of it. Now im just working on speed,
> and the quickest possible way to do it.
>
> BTW, does anyone have a formula/quick way to find out which network an
> address is on, given a mask and Address?
>
> Thanks.
>
|
|
|
|
|