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I have a hard time trying to find subnets and hosts for each subnet.
the problem is as follows:
"Using this network ID 192.168.128.0 and assume that you need 8 subnets with 30 hosts in each subnet. Calculate the subnet mask require".Please I need help. I know how to do the binary what then I get stuck.
I would really apreciate help.
thanks
GG

From: http://computing.net/networking/wwwboard/forum/24108.html
Response Number 6
Name: Bryco
Date: February 11, 2005 at 04:25:25 Pacific
Subject: help with subnetting
Reply:
Eight bits binary shown as follows:
1-----1---1---1--1---1---1---1
128--64--32--16--8---4---2---1
Borrowing 1 bit is 128 leaving 7 bits left over.
Adding them from the right leaves:
1+2+4+8+16+32+64=127 less the network ID making 126 usable addresses in each subnet mask. Host ID being 1 and 128 as broadcast along with 129 as the second host ID and 255 as the broadcast for the second.If we borrow no bits then that equates to two to the the zero power equaling 1 subnet and in the above case we borrowed one bit or two to the first power = 2 or two subnets.
Borrowing 2 bits (adding from the left) is 128+64=192 leaving 6 bits left over.
Adding them from the right leaves:
1+2+4+8+16+32=63 available addresses in each subnet less the host ID making 62 usable addresses in each subnet mask for a total of 4 subnets or since we borrowed two bits we can say 2 to the 2nd power being 4.Borrowing 3 bits (adding from the left) is 128+64+32=224 leaving 5 bits left over.
Adding them from the right leaves:
1+2+4+8+16=31 available addreses in each submask less the host ID making 30 usable addresses in each subnet mask.128+64+32+16+8+4+2+1=255 where 255 is broadcast and 0 being the host ID making a total of 256.
Get it?
Bryan

Jimi,
You still need to understand conceptionally how it works. That's like saying the best way to learn math is use a calculator.
"...but in my defense, it was dark, I was drunk, and it was delicious!"

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