Time
1 hour 19 minutes
Difficulty
Beginner
CEU/CPE
2

Video Transcription

00:00
Hello and welcome back to the last and final video of Creative sudden it. Today we'll be going over submitting the I P V six networks.
00:09
I'll be your structure Trenton hero. And go ahead and move on here today we're gonna be suddenly thing that could be six networks. I promise you it is not as bad as it sounds
00:19
because there is so money I P addresses, right. You have to raise 228 bits.
00:27
There is a ton of I P addresses. If you're really bored. Look at some of comparisons people have done
00:34
about how many I p address is A p B six has
00:38
for the best way to do this is hopefully you have a familiarity from one of the videos on the previous lesson about high P V six networks. Uh, you have practice with in the past,
00:50
but so if we look at, we have do a quick run down here,
00:55
So hex is what we're going to utilize. It's a through F er
01:00
zero through f, right.
01:03
So we use a total of 16 total numbers and it requires four bits per hexi decimal number, right?
01:14
So we have total of 32 characters in an I P V six address
01:18
32 by four bits equals 128 bits was a total of 128 bits
01:26
per ivy six network.
01:27
So here,
01:29
if we're looking at this your entire 1st 4 blocks here
01:34
because it's a slash 24 64
01:37
is gonna be your network portion, right?
01:41
And after that is where you can actually start subdividing it and using their host bits. Right? But if you're 64 divided by far by 16 will give you four, right, because you have
01:56
16 bits per block is your 16.
01:59
Each HX is four bits by four x A decimal characters in each block equals 16.
02:07
16 bits per block equals
02:09
and you have four blocks goes 64 cylinders kind of roundabout way. So I'm gonna move on here.
02:17
So if we look at this one here, we start looking at Here is your entire network portion for this address on a slash 80 Because we just added 16 to the 64.
02:30
Is this 80?
02:32
So here we have the first network. We had a second network, right? We're so dividing this down
02:38
farther and farther and farther. Right here is your network on. It's the 1 12 Right. So that's only gonna leave you 16 bits for your host addressing right?
02:50
Which is gonna just be that last octet, right? You're just gonna have this last or not? Active with last block
02:55
is your host portion. Right? Because of accounts, 1234567 Right on your eighth
03:05
block is for your
03:07
never come portion.
03:09
So you think it was that? That is a lot of blocks of networking portion. The cool thing is, we still have 65,000 I p addresses. We can give out if we have availability of that last block of high piece. Right? There is a ton of space in this, right?
03:27
So you think about that. That's the same as a class B. And look at how many of those we can create.
03:32
So it's really core anyway, Quick run out of I P v six.
03:37
Um, we're gonna go ahead and move on to a little bit of practice,
03:44
so we're gonna have our basic network. I d. There.
03:46
What we need is a sub net for 65,000 hosts.
03:51
So if you remember what we just said here is we're gonna need a total of 16
03:59
bits for the portion there. Right? We're gonna need 16 bits for host portion.
04:04
Because let's say we want to actually bring out the calculator and do 10 race to 16. Okay? We'll do here because of the I P v six, because it is larger numbers. I'm not good enough to actually be able to do the math on this. So 10 raise. Excuse me. Sophie, too.
04:23
Race to 16 not 10. Apologize. We're gonna get 65,536
04:28
which would actually leave us 65,534
04:31
usable. Right?
04:33
So we know that. So if we use them and I'm gonna write out the whole thing here,
04:41
But essentially, what weaken dio is we can do this
04:45
a slash, Remember, we need a slash 1 12 Right, cause that'll give us 65,000 host.
04:53
So if we look at what we need for 200 hosts, if you do the binary math on that, we're gonna need eight bits,
05:01
so to raise today. I'm gonna move that before it messes anyone up to race to 16
05:08
be 65 5 36 This will equal
05:13
256 which will give us plenty for that. So here we need a minus 1 28 from eight
05:21
and her eight from 1 28 which would give us a total sub net of 1 20
05:28
which would give us the less there would only have to hexi decimal characters
05:33
to actually utilize for host address portion. Right.
05:40
So, out of 32 characters, 30 of them are being used for networking, and only two characters
05:46
are being used for too expensive. Governors are being used for the host portion, right,
05:53
Whereas an entire you know, 1/4 of an i p dry PV four addresses used for hosting a classy right.
06:01
So it gives you the kind of expansive, uh,
06:04
numbers you're looking like, How much actual I p addresses we're looking at.
06:10
All right. I want the screens. We get a little bit more space with this last question here.
06:15
So here's Poe's given a slash 96 here, and what we need is six sub nets with at least 20,000 I p addresses.
06:26
So let's figure out what we need to do so if we know that. To raise two
06:31
16 gives us 65,536 once divided in half. So too raised 15.
06:41
Well, give us about 32,000. I p addresses to raise 2 14
06:46
would give us about 16,000 roughly
06:48
so we know we're gonna have to do to raise 2 15 Right,
06:51
So it's 2128 minus 15.
06:56
It gives us a 1 13 Right.
07:00
So what? I'm a do 3 99 I'm gonna draw this out real fast.
07:04
All right, So here is where we get into the farther bit of submitting. So what I'm gonna use for now, just for simplicity, sick is just zeros. Instead of using the double coins, we can see what's going on here for
07:19
and here we're actually gonna borrow, remember, we're gonna use one sub net bit for our networking. Right?
07:29
So if we use
07:30
what say we use one sudden that bit out of a hex a decimal character,
07:35
what number did I give us? We have eight
07:40
networks, essentially, right.
07:42
So if we do zero
07:45
and this would give us all the way through seven FFF
07:50
because we have 01234567 which is eight total. Right.
07:57
So from here,
07:58
our next one
08:00
could be eight
08:01
000
08:03
and then I could finish off with f f f. So if you kind of think about a year's splitting this last block in half, right, so think of it almost is like a slash 25 I b b for an hour.
08:16
Um so here we have two networks and you could just follow that same trend. So the next one would be at the end,
08:24
0000
08:28
through seven. Huh?
08:31
It would still be 17 f f f
08:37
one
08:39
0000 R eight
08:43
through f f. Right.
08:46
So there's four networks and then we essentially know that the next block is gonna end with two.
08:54
Right?
08:58
So that's how we could actually so divided that network in order to get the Given I p space that we need to utilize.
09:05
So I'm gonna go ahead and finish up here.
09:09
So how many bids are left for hosts in an I P. V six address with a mask of slash 96 giving she seconds to do the math on this. Um,
09:22
already, hopefully you got 30 to remember, because if we do a 1 28 minus 96 that's gonna give us 32.
09:31
So that would give us 32 bits to work with or essentially the last two blocks of the I. P. V six network internally i v six address to utilize for host bits,
09:45
which, in case, you're curious as well. I'm gonna go ahead and bring up the calculator, see how many I P addresses we can actually utilize with a slash 32 here. So let's go ahead and do it too. Raised 2 32
09:58
it's gonna give us just over four million I p addresses.
10:01
So quite a few to utilize their.
10:07
Anyway, In today's video, we went over. It could be six submitting. I hope you have enjoyed the course this for and I look forward to teaching you and another course. Thank you.

Create a Subnet

This course will enable students to refresh on basic IPv4 and IPv6 concepts and to learn how to subnet an IP space in order to better provide network availability. We will dive into converting decimals to both binary and hexadecimal.

Instructed By

Instructor Profile Image
Trenton Darrow
Network Engineer at NCI Information Systems, Inc
Instructor