I'd like to share the methed I used which I call it as "fixed block size". Fortunately, the block size is always one of the following numbers:2,4,8,16,32,64,and 128, so this make thing easier.
Here we go:
Given the mask /22, I can conclude very quickly that the block size of each subnet will be "4" or, to be exact, "0.0.4.0" -- because 2 bits means 4 possibilities.
(Stop here and ask yourself "Do I need a caculator to derive 4 from 22?")
Now enumerate the numbers which are multiples of "4": 4,8,12,16...252. This means every subnet with mask /22 will begin with 4,8,12,...or 252.
(Stop here and ask yourself " Do I need a caculator to enumerate the multiples of "4"?)
Now look at the 22 again. It means the subnetting happen at the 3rd octect because [22/8]+1=3.
(Stop here and ask yourself " Do I need a caculator to derive 3rd from 22?")
Looking at the number show up at the 3rd octect :210 , I ask myself:" What is the number which is a multiple of 4 and is as close to (but less than) 210 as possible?" I get the number 208 very quickly, and this will be the network number that 210 resides in. Again, ask yourself "Do I need a calculator to do so?"
At last, I get the network address 172.16.208.0 no more than 5 seconds, though it takes me more than 20 minutes to write this article.
Every time you feel difficult to derive the result in each step, don't hestitate to pick up your calculator or a piece of paper to help you calculate. It's quite normal to use the calculator in the beginning. After several times of practicing, I believe you can also do subnetting very quickly without a calculator.
HTH
SSLIN
