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?