Classless IP Addressing (Part 2)
Video Activity
Classless IP Addressing (part 2) In this lesson we further break down subnet masking by examining what happens when we change the last octet of our IP address to create deeper layers further that our subnet mask when expanding the network. We'll diagram and breakdown the calculation to demonstrate who you calculate out in bits the subnet mask when ...
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Time
27 hours 46 minutes
Difficulty
Intermediate
CEU/CPE
31
Video Description
Classless IP Addressing (part 2) In this lesson we further break down subnet masking by examining what happens when we change the last octet of our IP address to create deeper layers further that our subnet mask when expanding the network. We'll diagram and breakdown the calculation to demonstrate who you calculate out in bits the subnet mask when the last octet is already a subnet masked IP address.
Video Transcription
00:04
next, I'm going to keep my I P address the same, but I'm going to change my mass to a slash 30.
00:12
So if I were to ask you,
00:15
what would the Mass look like? Indicted decimal course Last 30 is more than 24 bits. So you know the 1st 3 octet CE, which is eight bits, eight bits and eight bits. The mask is going to be 255 not
00:30
255 dot
00:33
255
00:36
That's something.
00:38
So I'm going to draw out the last octet slashed 31 2345667812
00:49
for 8 16 32
00:52
64 on 1 28
00:56
So the 1st 2nd and third akademik 24 bits eight plus eight plus eight is 24.
01:02
I have to get to the number 30
01:04
25 26 27 28 29 30
01:10
1 28 plus 64 plus 32 plus 16 plus eight before gives me 2 52 so the mass would become 255255255.25
01:26
Shoot. Now, if you notice the value of the last bit turned on here
01:30
is four.
01:33
So
01:34
started 1 51 01 45.0 slash 30
01:42
My next street over or my next network starts at
01:46
1 51 01 45.4 its last 30. My next one will start at 1 51 01 45 dot Add 4 to 4. You get eight. I'm just making jumps of four.
02:04
My next one would start at 1 51 01 45.12 slash 30.
02:12
The valid addresses over here and the first network would be 0.1 and not two
02:17
0.3 would be my broadcast
02:22
again. In the next street over
02:23
my house numbers would be 0.5 and 0.6
02:29
0.7 again would be broadcast
02:35
my next street over. My valid addresses are gonna be 0.9 and dot Tenn 0.0.11 would be my broadcast.
02:45
The point I'm getting at is that with a slashed 30 mask,
02:50
you only get two addresses.
02:52
So a slashed 30 mask or 255255255 to 5 to mask should always be used
03:00
over point to point links.
03:02
So if you have rather one on this side and router to on this side.
03:07
One address goes here on one address goes here so over point to point lengths that could be on the exam. You always use this last 30 or a 2 52 masked. Otherwise you're wasting addresses.
03:20
You only need to address his one on each side. Any other mass like a slash 29 mask would yield six addresses and you're wasting for.
03:29
And as you guys notice, I'm not even playing with I p. I'm simply playing with the mask.
03:35
Let's make this
03:36
slash 19
03:38
and let's make this
03:42
64 God zero. My question to you guys are question could be What is the valid host range for this network?
03:51
Well, let's figure it out
03:53
again. I'm going to take the sub net mass, which is slash 19 and break it down into binary.
04:00
So this time I'll do the whole thing.
04:03
12345678.12345678
04:15
12345678 Dust
04:20
12345678
04:26
So are we working in the first octet? Second octet, 3rd October four talked yet. Let's find out. 19 bits turned on
04:32
123456789 10 11 12 13 14 15 16 17 18 19. I'm working in the third octet. The rest of the bits
04:44
are off.
04:46
What is the value of the last bit turned on? Let's find out.
04:50
1248 16 32 64 1
04:58
1 28
05:00
So the value of the last bit turned on
05:02
is
05:03
32.
05:06
So we have found out that the value of the last bit turned on is 32
05:11
1 28 plus 64 plus 32 is to 24. So this is to 24
05:17
the first octet. Everything's on. So it is 255
05:21
Second doctor is 255 3rd octet is to 24 the last octet. Everything is off, so it's simply zero.
05:30
So I'm gonna go ahead and erase this because they need the space.
05:33
Just remember that the value of the last bit turned on was 32. So the mask will be
05:40
255
05:42
255 indicted decimal. It would be to 55255 to 24.0 You start at 1 50
05:51
1010
05:55
0.0
05:58
slash 19.
06:00
What was the value of the last bit turned on?
06:03
It was 32
06:05
and I'm looking for the 64 network. So my first street started at 1 51 a 10.0
06:14
My next street will start at
06:16
I should write this a little bit higher. So my first restarted at 1 51 a 10.0 slash 19
06:27
My next network or street
06:29
is we're working in the terror doctor, remember?
06:32
1 51 01 And I simply add 32. Since the value off the last bit turned on was 32
06:41
32.0 slash 19
06:46
my next network would be
06:48
1 50
06:50
101 64 dot
06:55
zero slash
06:58
19.
06:59
My next network would be 1 50
07:02
101 96. I just simply at 32 to 64 to yield 96
07:10
not zero slash 19.
07:14
So I had asked, What is the valid host range for this network?
07:18
Well,
07:19
the first address would be
07:23
64.1.
07:28
The last possible address
07:30
for this street would be one before the next street. Since the next street is starts at 96 it would be 95 dot
07:40
255
07:43
95
07:46
dot to 54
07:48
would be my last valid address. 95 255 as always, being my
07:56
broadcast. So my valid range would be 64.1
08:00
through 95.254 Not hard, guys. Uh, as long as you remember the rule,
08:07
you look at the value of the last of it turned on.
08:11
And then that gives you the block size. Then you start a zero in that op ed and keep adding that number until you get to your element network that you're trying to find the range for.
08:20
Okay, I'm gonna go ahead and raise the board. We will do one more
08:24
example question, and then we will move on
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