Class B subnetting

Could someone supply a good, challenging class B subnetting problem where one is given a node and told to find which network it belongs to? I always have trouble when I have to tinker with both the third and fourth octet instead of the simple class Bs where one only has to adjust the fourth.

I looked but couldn't find a section here that discussed class B subnetting. Perhaps there isn't one? If there is a section, could someone supply the link? Thanks. I need a little practice.

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Comments

rjbarlow

OK, node 145.67.223.71/23

node 145.67.223.139/27

What the subnet and broadcast numbers?

they go well?

However I think in the net You can find a lot of these.

Tricon7

rjbarlow wrote:

OK, node 145.67.223.71/23

node 145.67.223.139/27

What the subnet and broadcast numbers?

they go well?

However I think in the net You can find a lot of these.

Ok, I believe I got the second one all right:

Subnet mask is 255.255.255.224
Block size is 32
145.67.223.139 falls within the .128 subnet, so the network is 145.67.223.128 and the broadcast would be 145.67.223.159.

The second example, however, eluded me. With a subnet mask of 255.255.254.0 the block size is only 2, so the network/host division is in the third octet. At this point I broke down (not literally) and couldn't go any further. Any help?

Pash

Well your on track. So what would be your starting network address? and what would be the last useable address on that subnet?

Tricon7

Pash wrote:

Well your on track. So what would be your starting network address? and what would be the last useable address on that subnet?

I'm tempted to say that since the block size is 2, then it would go .0, .2, .4. .6, etc for the networks in the fourth (?) octet. However, if this is so, then the first address after the network is also the broadcast, leaving no usable hosts, so this isn't right. And the fact that my subnet maks division is in the third octet, between the first and second bit, only makes this more confusing for me. Funny, but I don't have any trouble with VLSM, but this kind of problem gives me fits. It's my kryptonite, I guess.

Netstudent

As you said it is blocks of 2 and the 3rd octet is being used for subnets. You already knew that

**Small hint**

Your first subnet should be 145.67.2.0(if you are using the traditional method)

Hosts should start with 2.1, 2.2, 2.3, --->2.255, 3.0, 3.1

Second will be 145.67.4.0

So knowing that you should be able to compute the subnet for the IP 145.67.223.139

The broadcast for the first subnet should be 145.67.3.?

tech-airman

Tricon7 wrote:

Could someone supply a good, challenging class B subnetting problem where one is given a node and told to find which network it belongs to? I always have trouble when I have to tinker with both the third and fourth octet instead of the simple class Bs where one only has to adjust the fourth.

I looked but couldn't find a section here that discussed class B subnetting. Perhaps there isn't one? If there is a section, could someone supply the link? Thanks. I need a little practice.

Tricon7,

I use the method of determining the "relevant octet." That way, there's only one octet that you have to convert from decimal to binary, work in binary, then convert from binary back to decimal.

First question is to determine what the default mask given the IP address. Remember from CNAP that:

Class A addresses start with binary 0 in the first octet.
Class B addresses start with binary 10 in the first octet.
Class C addresses start with binary 110 in the first octet.

So now that you've figured out the class of the IP address, you use that information to figure out the default mask. The default masks are:

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

Remember, the purpose of the network mask or subnetwork mask is to find out the network address. The process that the network address is determined is by the logical AND process with the bits in the IP address with the bits in the sub/network mask.

Going back to my term "relevant octet," let's look at a Class C address of 192.168.1.1 . We know that 192. is a Class C address because 192. decimal is 11000000. binary. So we know the default mask for a Class C address is 255.255.255.0 . Now, at this point, we can still keep things in decimal but understand there's an underlying binary process occuring. So let's "logical AND" the decimal IP address and the decimal default mask. I will use the following acronyms: 1) IP = IP address 2) DM = Default mask 3) NA = Network address.

IP: 192.168.001.001
DM: 255.255.255.000
AND process
NA: 192.168.001.000

So for the IP address of 192.168.1.1 the network address is 192.168.1.0 . Notice the difference is in the fourth octet. So for the case of a Class C IP address, the "relevant octet" is the fourth octet. This is important because due to the limited time of the exam, it's quicker to convert a single octet from decimal to binary, work in binary, then convert from binary to decimal than the entire 32 bit IP address from decimal to binary.

Now, the main question is how to subnet a Class B address. Let's go back to default masks. The default mask for a Class B address is 255.255.0.0 . That means from the get go, the potential "relevant octet" could be either the third or fourth. Then you compare the subnet mask with the default mask to determine the "relevant octet." So you "read" the subnet mask by...

SM: 255.255.255.192

Notice how in this example, there's a number less than 255 in the fourth octet? Guess what? That means the relevant octet for this Class B example is just like the previous Class C example. So let's say you've got an IP address of 172.16.144.25 with the subnet mask of 255.255.225.192 . So let's write that down...

IP: 172.016.144.144
SM: 225.255.255.192

Now, let's convert ONLY the fourth octet from decimal to binary.

IP: 172.016.144.10010000
SM: 255.255.255.11000000

Now, let's apply the AND process ONLY to the fourth octet.

IP: 172.016.144.10010000
SM: 255.255.255.11000000
AND Process
SA: 172.016.144.10000000

The SA represents the "Subnetwork Address" which is what you're trying to find. Now, let's finish up and convert ONLY the fourth octet from binary to decimal.

IP: 172.016.144.10010000
SM: 255.255.11000000
AND Process
SA: 172.016.144.10000000 (binary)
SA: 172.016.144.128 (decimal)

So, for a subnet mask of 255.255.255.192 used on an IP address of 172.16.144.144, the subnetwork address is 172.16.144.128.

Now, let's figure one out for the third octet. Let's use the same IP address of 172.16.144.144, but instead of a subnet mask of 255.255.255.192, use a subnet mask of 255.255.192.0. So like previously, let's write those down...

IP: 172.016.144.144
SM: 255.255.192.000

We can verify that the third octet is the relevant octet because in the subnet mask, the third octet contains a number less than 255. So let's convert ONLY the third octet from decimal to binary.

IP: 172.016.10010000.144
SM: 255.255.11000000.000

Now apply the AND process....

IP: 172.016.10010000.144
SM: 255.255.11000000.000
AND process
SA: 172.016.10000000.000

Let's convert ONLY the third octet from binary to decimal....

IP: 172.016.10010000.144
SM: 255.255.11000000.000
AND process
SA: 172.016.10000000.000 (binary)
SA: 172.016.128.000 (decimal)

So, for a subnet mask of 255.255.192.0 used on an IP address of 172.16.144.144, the subnetwork address is 172.16.128.0 .

So in summary, the steps used were:

What is the class of the IP address?
What is the default mask for the IP address?
What is the subnet number of the IP address?
What is the first valid IP address for the IP address?
What is the last valid IP address for the IP address?
What is the broadcast address for the IP address?

The last three steps are extensions of finding out the subnetwork address to help determine other configuration issues such as IP address misconfiguration issues.

I hope this helps.

Netstudent

very nice and explanatory airman. I think everyone should learn the binary method first so that they know exactly what it all means and know how the machine is going to inerpret the logic.

Tricon7

Great explanation, airman. Maybe I should print out your explanation and keep it in my back pocket to glance at between now and the test. I hear that subnetting is heavy on the INTRO, so I want to make sure I nail it.

Btw, does or does not Cisco use subnet zero in their subnetting questions on their tests?

Netstudent

I have not taken the test yet, but I know that it will specify when they want you to use it.
But I find it hard to assume that they would put both answers for each method in the same multiple choice question and not explicitly specify. You know what I mean?

rjbarlow

Great explanation tech-airman, compliments! Tricon, I think You need first at all to understand the real effective binary computation of the subnet number before use a decimal trick.
That because in a while, mentally, You need to "put down all remaining 0 bits" to find the subnet number and all 1 bits to find the broadcast number.
Focuse always on the interesting octet (in the address and in the mask), that is the ultimate octet on where are present subnet bits in the mask and, secondarily, after, on the next octet (if present).
When You has understood the binary computation I'm sure You will eat these type of question like bread. Ciao.

Slowhand

You could check out Learn to Subnet and Steve Kehlet's Subnetting Quiz if you're looking for practice. Also the CCNA Prep-Center has some resources for learning subnetting.

tech-airman

Tricon7 wrote:

Great explanation, airman. Maybe I should print out your explanation and keep it in my back pocket to glance at between now and the test. I hear that subnetting is heavy on the INTRO, so I want to make sure I nail it.

Btw, does or does not Cisco use subnet zero in their subnetting questions on their tests?

Due to my acceptance of the Cisco Career Certification Confidentiality Agreement, I cannot confirm nor deny your question as far as how subnetting questions are phrased in a Cisco exam.

However, as a general rule for me is that "subnetting begins with the routing protocol." Currently, I understand that routing protocols live in a sublater of the the Network OSI model layer or the Internet TCP/IP model layer. Since the 640-821 INTRO exam has a "DESIGN and Support" section and I still believe that Designing is a process that goes from the Application layer down to the Physical layer, by understanding the characteristics of routing protocols you should be able to "start subnetting" by deciding if the routing protocol supports subnet zero or not then finish up the subnetting process.

As a practical extension of my subnetting process, since calculators are not permitted in the actual Cisco exam room, use the calculator you've got in your fingers. Here's how the fingers can be used as a calculator.

 Left hand     Right hand

 | | | |        | | | |
 | | | |        | | | |

Now, you've got 8 fingers, and how many bits are in an octet? You got it, 8 bits.

So, to represent a '0', I hover my fingers above the table edge like below.

 Left hand   Right hand

 | | | |     | | | |
 | | | |     | | | |
------------------------- [table edge]

In binary the above would be 00000000.

Now, to represent a '1', bend your finger down until it touches the table surface like...

 Left hand   Right hand

  / / / /  \ \ \ \  
  | | | |   | | | |
----------------------- [table edge]

In binary the above would be 11111111.

Now, for the purposes of subnetting, there's many ways to use your "finger calculator" to help. The first case is to determine the number of subnets required. Let's say you need 2 subnets. Let's say that subnet zero CAN'T be used. So you're going to need to allow for 4 subnets, two of them you're not going to use. Since subnetting "borrows bits" from the host bits, then you're going to start your counting from the left. So here goes...

  Left hand    Right hand
   2+2=4 subnets 
  / / | |    | | | |
  | | | |    | | | |

Now that you've "subnetted" the relevant octet, now what? You need to figure out the subnet mask that will give you those subnets. This is a very important sequence to remember. Sing this during dinner tonight, sing this in your sleep, sing this in the morning, sing this in the car...

from left to right
1 finger down = 128
2 fingers down = 128+64 = 192
3 fingers down = 128+64+32 = 224
4 fingers down = 128+64+32+16 = 240
5 fingers down = 128+64+32+16+8 = 248
6 fingers down = 128+64+32+16+8+4 = 252
7 fingers down = 128+64+32+16+8+4+2 = 254
8 fingers down = 128+64+32+16+8+4+2+1 = 255

o/~ 128, 192, 224, 240, 248, 252, 254, 255 o/~

That will help convert the relevant octet of the subnet mask from binary to decimal. In the example of 4 subnets, that would be 192. So if this was a Class C case, then the subnet mask would be 255.255.255.192 . Depending on the "relevant octet" for a Class B address, it's either 255.255.192.0 or 255.255.255.192 . Are you at least still humming 0/~ 128, 192, 224, 240, 248, 252, 254, 255 o/~ ?

Finally, good luck tomorrow.

happy420golucky

Could someone supply a good, challenging class B subnetting problem where one is given a node and told to find which network it belongs to? I always have trouble when I have to tinker with both the third and fourth octet instead of the simple class Bs where one only has to adjust the fourth.

I looked but couldn't find a section here that discussed class B subnetting. Perhaps there isn't one? If there is a section, could someone supply the link? Thanks. I need a little practice.

Here is a good site to practice subnetting dude!!!!!!
http://www.subnettingquestions.com/default_int.asp

Sulblk27

Thank you tech-airman