Now we move away from the world of class ful I p address ing to class less addressing.
Well, because, for example,
if you pick a classy address 210 10 10 So 210 10 10
slash 24. Since this is a classy address or the mass being 2552552550 or represented as last 24
the network addresses 210 dot Tenn 210.0.0. The range of addresses that can be derived from this is 0.12 dot to 54
dot to 55 being a broadcast,
so you can get 254 addresses using a classy address base.
pick a Class B address space,
you get more addresses than 2 54 and a class C
class address range. You get a huge number
what if your company
I asked you to purchase
I p addresses for them
and they wanted 30 addresses?
Well, if we were still using Class A, B and C, you could come back with a classy range,
but you would have still bought 254 addresses. So because because you would have to buy the whole class the whole range.
So if somebody asked you to spend,
let's say these addresses cost $1000 a month. They ask you to spend $30,000 you spend $254,000. You're in trouble.
So class full I p address ING won't work if we need address is less than 254.
So let's pretend our company needs 30 addresses.
How do we do that? Let's start with the night. Be 1 51 0 on 45.0. Now this technically is a Class B addressed
because it falls within the range off between 1 28th
So this should be a slash 16 or 25525500
What happens if I give it a slash 27 mask? Well, the first thing I want to find out is what would the mass look like in dot and decimal?
This is the cider notation, the slash format. What would the Mass look like? What would it be to 55255255 something right?
24 bits. I definitely turned out
so the 1st 3 objects would simply be 255255 and 255 So the first act it is all ones, which gives me eight bits. Second octane is all ones, which gives me a total of 16 bits. Eight plus eight.
And then the third op ed is also all turned on, which gives me plus 8 24 bits.
Three extra bits are turned on
in the last doctor of the fourth Doctor.
Let's find out what that does to 345
So three bits turned on here
and the rest are off.
If I add up 1 28 64 32
So the mask and dotted decimal would end up being 255255255 to 24.
How many total hosts can I have in this network?
my network address. I look at the value off the last bit turned on in the subnet mask. Hence I put a square around it.
The value of the last bit turned on in the subnet mask is 32.
Remember this rule? This gives me my block size. What do I mean by block size? See in a second.
So always remember when solving questions with class less I P's always turned the sub net mask into binary. And you don't need to turn the whole mascot by angry because the first octet, second doctored and third doctor are all ones.
The four talk that the relevant that you're working in, you can turn it into binary
and look at the value off the last bit turned on.
So 27 bits turned on or slash 27 simply means 8883
And I only drew out the last octet with three bits on because the 1st 3 AQ taps are all ones anyways,
So when you look at the value of the last bit turned on, this gives me my block
That simply means that if this is my first street or my first network, my next street starts at 1 50
I just added 32 to 0.
My next street would start at
and I would keep adding 32. How did I get 64? I added 32 again and I got 64. If I had 32 again, if I had space,
the next network or the next street would start at 96 so on and so forth. All the way till 255
So, my first valid host for this first street, my first valid house address is dot Won. My last possible address on this street would be dot
31. However, remember the rule. The last possible address on any network
is your broadcast. So my valid range would be between one and
30. So 1 50.121 dot 45.1 would be my first valid address.
1 51 21.45 30 would be my last valid address.
Did I get 30 addresses? Yes. Did we need 30? Yes. What I'm looking for from you if I asked you to go by 30 addresses is
an I P address with a slash 27 mask or a 255255255 to 24 mask
again. The next street over.
My first valid would be 33. My last possible address would be 330.63 which is my broadcast. My last valid address, which is assign Herbal would be 62 my Rangers again total of 30 addresses. This would be my valid ridge.
that? Two to the power and minus two formula here. Of course I could. How many host bits do I have? 12345
True to the power five minus two to the power five is 32 minus two equals