Hi,
Jon gave most fastest way calculation done in your head to come up with answer of network ID , 1st host, last host-, hosts range , and broadcast address of specific host.
Would like to throw additional info that will definately help you understand the sequence involved in the process. First you have to understand few simple things things that may look hard at first glance in order to make this process seemless as well as done in your head, understanding it makes things easier than memorizing things I guess.
First thing is to understand the binary system involved in IPv4, which Im sure you already know that it is composed of 4 octets, each of 8 bit equaling a total of 32 bits.
so three basic components you need to know by heart and comprehend are these.
When you understand the esence of this you will be a pro in subneting any class network in your head.
The Rule number 1 is = State away from calculators, you may use it only to confirm your answer , but do the process in your head.
Second- = Understand IPv4 Class networks and their default classful mask - This will help you in indentifying what class network you are face with when presented with a host regardless of its mask, in the case of (supperneting).
1-126 Class A Default mask /8
128-191 Class B Default mask /16
192-223 Class C default mask /24
224-239 Class D Multicast reserved
240-255 Class E experimental
Third = Understand and know the binary value of a 32 bit address in IPv4.
You must understand and know the binary value of an octect which is:
128 64 32 16 8 4 2 1
so in your example of 172.27.205.91/21 you know that 172 falls under Class B network and a class B network uses default mask of /16 bit which is 255.255.0.0 but 172,27.205.91 uses a /21 bit mask, so knowing the binary value of an octet seen in 128 64 32 16 8 4 2 1, you count from left to right 5 positions from the 3rd octect of the mask, that is 255.255.0.0 is 16 bits which is default of class B then the 3rd octect binary value count 5 (128 64 32 16 8) 4 2 1, in the parenthesis you have 5 bits which will make a total of 21 bits mask, up to here the 5th position ends in 8, remembering that 8 will be the key number because that will be your increment figure of networks within 172.27.0.0. So far so good right?
Now, say that 172.27.205.91 had a /19 bit mask instead of /21, then you do the same principle as before , this is still a class B network, which you now know its default classful mask is 255.255.0.0,then on the 3rd octet of the mask you count from left to right in this case of /19 will be 3 bit positions from 3rd octect, 255.255.0.0 is 16 bit mask, 3rd octet binary value again is (128 64 32) 16 8 4 2 1, again, in
parenthesis you have counted 3 bit position from left to right which will come to a total of 19 bits including the (255.255 -16-bits), that will bring you to 255.255.224.0 , now a key to calculate the mask is easy, simply add 128 + 64 + 32 = 224
here you have /19 - in example above with binary value bit counting 3 positions it ends in (32), here the 32 will be the figure to increment to come up with networks incrementing by 32
172.27.0.0
172.27.32.0
172.27.64.0
172.27.96.0
172.27.128.0
172.27.160.0
172.27.192.0 -----> here you stop because 172.27.205.91 host falls here
172.27.224.0
etc..
so you have 172.27.192.0 as the network ID, 172.27.192.1 as the first host, for broadcast address simply substract one from the 3rd octect of the next network of 172.27.224.0 which is 172.27.223.255. So the broadcast address of 172.27.205.91/19 is 172.27.223.255, then substract one from the 255 broadcast address to come up with the last host address which is 172.27.223.254.
Hope this helps, again, Jon provided the most fastest way calculation.
Regards
Jorge Rodriguez
