Still confused on how to get block size

pham0329

IMO, rather than looking at what class the address falls in to find your "interesting" octet, I look at the subnet mask. For example, if we have 192.168.0.0/18, you wouldn't want to start in the 4th octet just because 192 is a class C

JockVSJock

pham0329 wrote: »

rather than looking at what class the address falls in to find your "interesting" octet

I keep seeing this, however I don't know what it means. What does interesting octet mean?

Ltat42a

When you subnet, the interesting octet is one of the octets that is something other than the default for each class of address.

Default Class A = 255.0.0.0
Default Class B = 255.255.0.0
Default Class C = 255.255.255.0

In your comment above...192.168.4.0/28 =
192.168.4.0
255.255.255.240

The interesting octet here is the 4th (.240) octet because the default mask for Class C is 255.255.255.0

If you have something like 172.18.2.20/20, this translates to:
172.18.2.20
255.255.240.0

The interesting octet here is the 3rd one. Your subnetting will start in the 3rd octet.
Starting with subnet zero, the 172.18.2.20 host is in the 172.18.0.0

1st 172.18.0.0
2nd 172.18.16.0
3rd 172.18.32.0
etc
etc
etc

For class A addresses, you can subnet using the 2nd, 3rd, & 4th octet
For class B addresses, you can subnet with the 3rd & 4th octet
For class C addresses, you can only subnet with the 4th octet

does that help??

Excellent1

JockVSJock wrote: »

I keep seeing this, however I don't know what it means. What does interesting octet mean?

The interesting octet is the one that's not 255 or 0. Basically, it's the octet you're borrowing host bits from for your subnetworks. In a classful mask with the corresponding classful network (class A network with a /8, for example) you have no subnetting; just the standard 8 network and 24 host bits.

thedrama

JockVSJock wrote: »

Ok, had another problem (I wish there was a way to select only these problems and do these so I can get better at them).

I didn't get this correct



I though the block size was 16 (256 - 240) and got the following for the network address:

Network - 192.168.4.240
1st host - 192.168.26.240

192.168.4.0/28 28-24= 4 bits borrowed. This means 2^4 = 16 subnets

if 4 bits are borrowed, as the default subnet mask of Class C is 255.255.255.0 , now it becomes 255.255.255.11110000 equals to 255.255.255.(128+64+32+16=> 255.255.255.240)

This leads us to 256-240 = 16 = block size

Now, if 192.168.4.0 is the first subnet
192.168.4.16 is the second subnet

Now valid hosts between the second subnet and the third are 192.168.4.17- 192.168.4.30 due to 192.168.4.31 is the broadcast address, it can
not be assigned to hosts.

instant000

JockVSJock wrote: »

Ok, had another problem (I wish there was a way to select only these problems and do these so I can get better at them).

I didn't get this correct

Question: Your router needs to be assigned the first valid host address of the 2nd subnet on network 192.168.4.0/28. What address would you assign?
Answer: 192.168.4.17

I though the block size was 16 (256 - 240) and got the following for the network address:

Network - 192.168.4.240
1st host - 192.168.26.240

Problem:
Given: Network: 192.168.4.0/28
Find: First Valid host on 2nd subnet

In this case, it's just asking you to find the first valid host in the 2nd subnet. In order to do that, you'd have to know where the subnets were. In order to do that, you'd need to know your block sizes.

Example here:

Step 1:
Find the subnet mask
In this case, a /28 is 255.255.255.240

Step 2:
Find the interesting octed:
In this case, it is the 4th octet
(As other posters have said, it is the octet that's not 255 or 0 in value.)

Step 3:
Find your block size.
Your block size will be 256 minus the interesting octet.
In this case, 256-240 = 16

Step 4:
Increment via the block size, to find your subnets
0,16,32,48 ...

1st Network: 192.168.4.0
2nd Network: 192.168.4.16
3rd Network: 192.168.4.32
4th Network: 192.168.4.48
.... (and so on)

Step 5: Find broadcast addresses, and first and last hosts

...                   Network       | First  - Last host                 | Broadcast
1st Network:  192.168.4.0   | 192.168.4.1-192.168.4.14   |  192.168.4.15
2nd Network: 192.168.4.16 | 192.168.4.17-192.168.4.30  | 192.168.4.31
3rd Network: 192.168.4.32  | 192.168.4.33-192.168.4.46  | 192.168.4.37
4th Network:  192.168.4.48 | 192.168.4.49-192.168.4.62  | 192.168.4.63

Answer to your question:
192.168.4.17 is the first valid host in the 2nd subnet of 192.168.4.0/28


The INTERESTING octet is important, as it refers to the octet that you're concentrating on.

For example, if the mask is a /19, then the interesting octet would be the third octet.

############################################

I Will go through one right quick, with two questions:
1 - 192.168.4.0/20 Falls into which network?
2 - What would be the first valid host of the third subnet of that network?

192.168.4.0/20

/20 CIDR translates to a mask of 255.255.240.0

Third octet is the interesting octet

256-240 = 16

16 is the block size

Increments: 0, 16, 32, 48, 64 ...

Subnets:

192.168.0.0 < --- 192.168.4.0 falls between this number and the next network
192.168.16.0
192.168.32.0 <--- third network
192.168.48.0
192.168.64.0

Answers:
1. 192.168.4.0/20 is in the 192.168.0.0 network
2. First valid host in the third network: 192.168.32.1

#######################################

And another example:
1. 172.20.5.17/12 falls into which network?
2. What is the last valid host in that network?

/12 means a mask of 255.240.0.0

256-240 = 16

Interesting octet is the 2nd octet, per the mask above
Counting Networks ...
0,16,32,48,64


First Network: 172.0.0.0
Second Network: 172.16.0.0 <<<

---

172.20.5.17 falls right in here!
Third Network: 172.32.0.0
Fourth Network: 172.48.0.0
Fifth Network: 172.64.0.0

1. 172.20.5.17/12 Falls into the 172.16.0.0 Network
2. 172.31.255.254 is the last valid host in that network

###########################

Note: Since we're dealing with binary, the same masks will always have the same block sizes. Once you work with this enough, you'll begin to see the patterns in the math.

You'll go from looking at these and wondering what you're supposed to be doing, to knowing that when someone wants 4 class C's, to give them a /22 mask, without a second thought.

Just keep practicing the subnetting problems!

Hope this helps!

As others have said: Find a method, and stick to it!

We probably take shortcuts now, and may not explain every step properly, since once you do it enough, you can look at the numbers and know the answer. Find some reputable person and method to subscribe to, and stick with it. Then, whatever method you're using, we can assist you with that method.

Ltat42a

Here's a nice VLSM example. Check out his website & Youtube channel, lots of good stuff.

Solve a VLSM subnet problem - Cisco CCNA - YouTube

JockVSJock

I've finally got the hang of this, thanks everyone for their help.

Its too bad, you can't configure subnetting.org to just do these types of problems. Would love to be able to spend 25 minutes a day doing these.

subnetting.org - Free Subnetting Questions and Answers Randomly Generated Online

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