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subnetting question

 
 
anonymous
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Posts: n/a
 
      02-13-2004
I was using a class C subnet mask (255.255.255.0) scope
for my private network. I was going to increase it to
255.255.252.0 which would give me 1023 addresses less 2 I
believe? However, now I'm not sure how the range
would/will be in third octet. It was easy with the
255.255.255.0 range because I just went from 146.157.30.0
to 146.157.30.254. Now what sort of ranges to I have in
the third octet? Will the scope go from 146.157.252.0 to
146.157.255.0 giving me 1020 addresses?
 
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Andy Ruth [MS]
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      02-13-2004
With a subnet mask of 255.255.252.0, you have 10 bits for host ID use and 10
bits is 1+2+4+8+16+32+64+128+256+512 for 1023. The subtracting two would
refer to not using the all zeros combination and not using the all ones
combination. 146.157.30.0 = 146.157.00011110.00000000 with a mask of
255.255.11111100.00000000, so your first unique host ID (assuming all zeros
is not used) would be 146.157.00011110.00000001, which is 146.157.30.1 and
the end of that range would be 146.157.00011111.11111110, which is equal to
146.157.31.254, assuming you did not count all zeros or all ones as unique
identifiers. That would give you 1022 addresses. All of this is assuming I
have done my math correctly without the help of coffee.

The shortcut to figuring roughly how many addresses you would have is to
figure out the binary worth of the last bit of your subnet mask and subtract
two. For instance, 255.255.252.0 = 255.255.11111100.00000000, so the last
bit is the 11th bit from the least significant bit/end
(1/2/4/8/16/32/64/128/256/512/1024 = 1024) and I would subtract two for
1022.

I hope that makes sense. It is easier to explain in person.

--
Andy Ruth
Microsoft Learning

This posting is provided "AS IS" with no warranties, and confers no
rights.

"anonymous" <(E-Mail Removed)> wrote in message
news:1014001c3f24d$c0a099a0$(E-Mail Removed)...
> I was using a class C subnet mask (255.255.255.0) scope
> for my private network. I was going to increase it to
> 255.255.252.0 which would give me 1023 addresses less 2 I
> believe? However, now I'm not sure how the range
> would/will be in third octet. It was easy with the
> 255.255.255.0 range because I just went from 146.157.30.0
> to 146.157.30.254. Now what sort of ranges to I have in
> the third octet? Will the scope go from 146.157.252.0 to
> 146.157.255.0 giving me 1020 addresses?



 
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James Martin
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Posts: n/a
 
      02-13-2004
http://www.learntosubnet.com unless you don't really want to know, then
google for a free subnet calculator.


"anonymous" <(E-Mail Removed)> wrote in message
news:1014001c3f24d$c0a099a0$(E-Mail Removed)...
> I was using a class C subnet mask (255.255.255.0) scope
> for my private network. I was going to increase it to
> 255.255.252.0 which would give me 1023 addresses less 2 I
> believe? However, now I'm not sure how the range
> would/will be in third octet. It was easy with the
> 255.255.255.0 range because I just went from 146.157.30.0
> to 146.157.30.254. Now what sort of ranges to I have in
> the third octet? Will the scope go from 146.157.252.0 to
> 146.157.255.0 giving me 1020 addresses?



 
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Guest
Posts: n/a
 
      02-16-2004
here is a quick guide that will never stear you wrong (of
the top of my head of course)

fisrst to calculate the number of host, use the formula:
2 (to the power of number of host bits) - 2

in your case 2 to the power of 10 -2 = 1022 host

as for the range use this chart

in the third octet, find you low order bit (last bit set
to one)
128 64 32 16 8 4 2 1
1 1 1 1 1 1 0 0 = 252

last bit that is set to 1 is slot 4. So your subnet will
be broken up into blocks of 4.

i.e.
146.157.0.1 - 146.157.3.254
146.157.4.1 - 146.157.7.254
146.157.8.1 - 146.157.11.254

and so on

>-----Original Message-----
>I was using a class C subnet mask (255.255.255.0) scope
>for my private network. I was going to increase it to
>255.255.252.0 which would give me 1023 addresses less 2 I
>believe? However, now I'm not sure how the range
>would/will be in third octet. It was easy with the
>255.255.255.0 range because I just went from 146.157.30.0
>to 146.157.30.254. Now what sort of ranges to I have in
>the third octet? Will the scope go from 146.157.252.0 to
>146.157.255.0 giving me 1020 addresses?
>.


 
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Samantha Hyatt
Guest
Posts: n/a
 
      02-17-2004
He (or she ) is right. That's exactly how I would have explained it. It's a quick easy way to figure out subnetting.... especially if you don't like dealing in binary.

--
Samantha Hyatt
http://www.velocityreviews.com/forums/(E-Mail Removed)
(E-Mail Removed)
410-371-2645

"What lies behind us and what lies before us are small matters compared to what lies within us."
-- Ralph Waldo Emerson
<(E-Mail Removed)> wrote in message news:119e901c3f4e3$f7ddaf00$(E-Mail Removed)...
here is a quick guide that will never stear you wrong (of
the top of my head of course)

fisrst to calculate the number of host, use the formula:
2 (to the power of number of host bits) - 2

in your case 2 to the power of 10 -2 = 1022 host

as for the range use this chart

in the third octet, find you low order bit (last bit set
to one)
128 64 32 16 8 4 2 1
1 1 1 1 1 1 0 0 = 252

last bit that is set to 1 is slot 4. So your subnet will
be broken up into blocks of 4.

i.e.
146.157.0.1 - 146.157.3.254
146.157.4.1 - 146.157.7.254
146.157.8.1 - 146.157.11.254

and so on

>-----Original Message-----
>I was using a class C subnet mask (255.255.255.0) scope
>for my private network. I was going to increase it to
>255.255.252.0 which would give me 1023 addresses less 2 I
>believe? However, now I'm not sure how the range
>would/will be in third octet. It was easy with the
>255.255.255.0 range because I just went from 146.157.30.0
>to 146.157.30.254. Now what sort of ranges to I have in
>the third octet? Will the scope go from 146.157.252.0 to
>146.157.255.0 giving me 1020 addresses?
>.


 
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