Lesson 30 - IPv4 Subnetting - Practice

In the previous post, I showed you three major rules used in calculating subnets. This knowledge can only be verified in practice though. Let me show you a few examples related to subnet calculations. I hope that looking at this topic from different angles is going to help you understand the concept better and feel confident when planning your IP addressing scheme. The first four questions are merely appetizers for a bigger dish: VLSM.

I am going to refer to my previous post's rules while answering the questions (rule 1, rule 2 and rule 3).

If you still do not remember the weights of all bits, you may consider using this little aid presented below (pic. 1) while calculating subnets, and converting binary network masks into decimal values.

Pic. 1- Subnet Calculation Aid.

This tool is useful before you remember all the weights from left to right and right to left.

Pic. 2 - Example of Subnet Binary-to-Decimal Conversion.

Question 1

Given the prefix 192.168.1.0/24, what should be the length of subnet mask allowing up to 9 subnets?

Answer 1

The address belongs to the class C and uses its default network mask. That leaves us with 8 bits to play with (the last byte). Before we change anything, our address and network mask converted into the binary notation look like shown below (pic. 3).

Pic. 3 - 192.168.1.0/24 in Binary.

In order to create 9 subnets we must extend the existing length of the network mask by 4 bits which allows up to 16 subnets (use calculation aid in pic. 1). If I tried to extend it by 3 bits only, the maximum subnets allowed would be only 8 subnets (rule 2 in lesson 29). So, I must use 4 bits and the result is: 192.168.1.0/28 (192.168.1.0 255.255.255.240).

Pic. 4 - The Answer to Question 1

Question 2

Given the host address 192.168.1.177/29, what are the subnet and broadcast addresses?

Answer 2

In order to determine the subnet and broadcast address of the subnet of this host address, we must look at the length of the network mask first. It is 29 bits (24+5). This tells us that the last byte of the address has 5 bits masked (subnet bits) and 3 bits unmasked (host bits). It is a good idea to look at the the last byte of the address (177) with its network mask using binary notation. Pic. 5 below shows you this clearly.

Pic. 5 - 192.168.1.177/29 in Binary.

Since we must determine the the subnet in which the host resides (177 = 10110001), the host portion of the prefix (host bits reside in the last byte) must all be set to '0'. The byte value with the host zeroed is the address of the subnet (rule 1 pkt.1 in lesson 29). This is the result:

Pic. 6 - Host Bits Zeroed = Subnet Address.

The second part of the question relates to the broadcast address of the subnet. As you remember, in order to obtain the broadcast address, you must put '1' on all host bits of the subnet/network. The subnet has already been determined (pic. 6), so let's put '1' on all bits of the host portion:

.10110111
.10110000 = 176 <- subnet address
..00000111 = 7 <- host bits set to '1'

In decimal it is: 176 + 7 = 183
The broadcast address is: 183.

The below picture illustrates it using binary numbers.

Pic. 7 - Host Bits Set to '1' = Broadcast Address.

Question 3

Given the  prefix 172.16.0.0/17, how many subnets can you create?

Answer3

This is a bit tricky isn't it? In order to answer this question, you don't need any calculator, paper or pen. You must trust the rule 2 in lesson 29. The address and its network mask (called prefix) converted into binary look like presented below:

Pic. 8 - The Number of Subnets for 172.16.0.0/17

As you see the number of bits we have extended the class B address is: 1. So, the number of subnets we can create with it is: 2 subnets, since this subnet bit can be either 1 or 0.

Pic. 9 - Questions 3 Answer

Question 4

What length of network mask would be the most optimal for router's point-to-point connection?

Answer 4

The key to this question is to understand that point-to-point connection needs only 2 host addresses (two points that are connected together). Knowing this, the rest is a piece of cake. We use rule 3 inlesson 29 to determine the length of the network mask that allows 2 host addresses. Check out the picture 10.

Pic. 10 - Calculating Point-to-Point Connection Host Addresses.

If you count ones above the optimal network mask for point-to-point connection is /30. The decimal value is: 255.255.255.252.

Question 5 - Variable Length Subnet Masking (VLSM)

It's time for a big one. Given the topology (pic. 11), calculate IP addresses for each subnet trying to optimize them according the host address requirements. The IP address you should use to create subnets is: 192.168.1.0/24. The number of host addresses in the subnets are as follows:

Subnet 1 = 46 host addresses

Subnet 2 = 16 host addresses

Subnet 3 = 10 host addresses

Subnet 4 = 2 host addresses

Subnet 5 = 2 host addresses

Pic. 11 - VLSM Topology.

Icons designed by: Andrzej Szoblik - http://www.newo.pl

As always, if you know the rules and the method, it is going to be easy thing to do. The rules have been discussed in lesson 29, so let me go about this kind of task now.

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NOTICE!
If your design looks similar to mine (optimizing addresses to the number of host required) you muststart the calculation with the largest number of host addresses requirement and work your way down to the least number of host addresses.

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This is one of the many methods available. It helps quickly calculate all subnet ranges without using calculator (pen and a piece of paper should do).

Step 1
Determine the length of the network mask for each subnet in question. Keep in mind we focus in on the last byte of IP address 192.168.1.0 (8 bits).
The first three bytes do not change!

Subnet 1 = 46 Host Addresses

In order to allocate 46 addresses we must use 6 host bits. Why? 5 bits will not be enough as 2 raised to the power of 5 is 32. Also, we must decrement two addresses for subnet and broadcast addresses. So using 5 bits would give you only 30 host addresses. Here we go with 6 bits then:

Pic. 12 - Subnet 1 in Binary.

Subnet 2 = 16 Host Addresses

We must repeat the same math for the remaining subnets.  How many host bits to allocate for 16 hosts (subnet 2)? We must use 5 bits. In case we wanted to use only 4 host bits, the maximum number of hosts is 14 (16 - 2).

Pic. 13 - Subnet 2 in Binary

Subnet 3 = 10 Host Addresses

We continue using the same logic.

Pic. 14 - Subnet 3 in Binary.

 Subnet 4 and 5 = 2 Host Addresses Each

On point-to-point links only 2 host addresses area needed. The most optimal network mask is /30(30 bits).

Pic. 15 - Subnet 4 and 5 in Binary.

Step 2

Now, that we know the length of network mask for each subnet, we can start calculating the IP address ranges. 

The subnet 1 address is: 192.168.1.0/26.

The value of the lowest bit in the network mask is going to be our increment used to calculate the next available subnet address. With /26 the increment value is 64 (pic. 16).

So, if we add the increment to the last byte, we get the number of our next available subnet address:

192.168.1.0 + 64 = 192.168.1.64.

From there, this next subnet address (value) - 1 is the broadcast of our current subnet:

192.168.1.64 - 1 = 192.168.1.63 (current broadcast address)

Current subnet value + 1 = the first host address:

192.168.1.0 + 1 = 192.168.1.1 (first host address of current subnet)

Current broadcast address - 1 = the last host's address:

192.168.1.63 - 1 = 192.168.1.62 (last host address of current subnet).

Look at the below pictures which illustrate this method.

Pic. 16 - Subnet 1 - IP addresses

Pic. 17 - Subnet 2 - IP addresses

Pic. 18 - Subnet 3 - IP addresses

 Pic. 19 - Subnet 4 - IP addresses

Pic. 20 - Subnet 5 - IP addresses

Now, we're ready to start talking about routing. In my next post, I will talk about a router, its functions,and  basic operation. From there, we'll start exploring routing protocols.

Lesson 26 - Binary World

Now this is a real fun part, which I love and am never tired of explaining :) 

The routing world awaits. But before we plunge into its depths, it is imperative that we become fluent with conversions between decimal and binary numeral systems. Why? Because all electronic devices such as our beloved computers and routers use binary numbers. On the other hand, we humans tend to use decimal notations almost all the time.

Believe it or not but the binary numeral system is easier to understand than the decimal one. But our problem is that we have been using the decimal notation for so long that looking at anything different feels very awkward.

It is good to know that all numeral systems (binary, octal, decimal, hexadecimal, the-one-you-may-want-to-create) are based on the same foundations. There are two major components of a numeral system: the base and the exponent.

Pic. 1 - Numeral System Components - The Formula.

I almost hear your thought: 'Eh... what?' Fear not. It's easy to understand if you remember the math at the elementary level.

The base in the above formula is determined based on how many digits (or characters) we use to denote the number. Consider the decimal system. We have exactly 10 digits to express the volume:0,1,2,3,4,5,6,7,8,9. So, the in decimal numeral system the base = 10 (ten digits available). What happens when we have more of something than 9?

It is simple we engage a new column of numbers on the left and zero the number on the right: 10. This way we get ten. The first column denotes the number of 10s and the next column, denotes the units. As soon as we run out of numbers in the unit columns, we increase the number in the columns of tens. For instance: 17, 18, 19... 20.

What happens when we run out of the number of the column of tens? We add one more column on the left, which is the columns of 100s and zero all other columns on the right. For instance: 97, 98,99... 100.

We know that in the four column number: 1048 the first column (1) on the left denotes thousands, the next to the right (0) denotes hundreds, the one next to it (4) denotes tens, and the last one (8) describes units. The column in each and every numeral system has the weight.

Check the picture below to see this in more detail.

Pic. 2 - Decimal Numbers - Formula.

Using the formula you calculate the numbers in decimal as shown in the next picture (pic. 3). Each number in the column must be multiplied by the corresponding weight of the column. These in turn, must be added together to give you the number. Pic. 3 shows that in more detail. Make sure that you understand it before you proceed with your reading.

Pic. 3 - Example of Decimal Numbers According to the Formula.

In exact same way we create others such as the binary numeral system. Here, our base is going to use only two digits: 1 or 0 called bits. Thus, our base = 2 (only two digits available). Since, we most often cluster 8 bits together to form a byte, our colums and weights are going to look as in the picture below (pic. 4).

Pic. 4 - Binary Numbers - Formula.

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NOTICE!
Cisco expects their engineers to be able to do the binary-to-decimal and decimal-to-binary conversions WITHOUT using a calculator (using a paper and pen only).

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Binary-to-Decimal Conversion
In order to convert the binary number to its decimal equivalent, you must follow the exact same rules explained for decimal system. You multiply the number in the column by its weight in every columns. The products of these multiplications must be added together.Take a look at the example in the pic. 5.

Pic. 5 - Examples of Binary-to-Decimal Conversion.

In order to practice those try to quickly convert the following binary to decimal (DO NOT USE A CALCULATOR):
Ex1: 11011010 (calculated above)
Ex2: 10011101 (calculated above)
Ex3: 11001011
Ex4: 00100101
Ex5: 01010101
Ex6: 11111010
Ex7: 01101101.

It should not take you more than a few seconds per example. Check your results using a calculator.

When it comes to opposite conversion it is a bit more difficult, but still at the elementary math level.

Decimal-to-Binary Conversion
The best way to learn how to do the conversion is actually do it and analyse every step of the way.

Example1: the decimal value of 172, to be converted into the binary.

The following table is going to come in handy.

Pic. 6 - Decimal-to-Binary Conversion.

Step 1
Find the closest weight value in the binary table that does NOT exceed the value you convert.

In the example1, the closest weight is: 128. The weight of the next column (here, non-existent column 9) would exceed the value of 172, since the weight of it is 256.


Step 2
In our handy table, write in ‘1’ in the column you have chosen.

Pic. 7 - Decimal-to-Binary Conversion.

Step 3
From the initial decimal number take away the weight you have used (the ‘1’ you put in the table takes away the weight from the number you are converting).

In the example1, the decimal number we have is: 172 and the weight we just used by putting a ‘1’ in our table is: 128. So, our calculation looks like the one below:

172 – 128 = 44

Step 4
Check if the number you have left (in our example the value left is: 44) is larger or smaller than the weight of the next column on the right-hand side in our table.


a) If the number you have left is larger than the weight in the next column to the right, put in ‘1’ in the next column and subtract its weight from the number you have left.

b) If the number you have left is smaller than the weight in the next column to the right, put in ‘0’ in the next column and perform the same check against the next column (step 4a). You have to do it until the check in step 4a is true.

In the example1, the value we have left is: 44. We check it against the weight in the next column on the right. The number (44) is smaller than the weight in the next column to the right (64). Since 44 < 64, we follow the instructions in the step 4b.

Pic. 8 - Decimal-to-Binary Conversion.

The next check against step 4a is true! Since 44 > 32, we put in a ‘1’ in the next column and take away the weight (32) from the number we have left (44).

Pic. 9 - Decimal-to-Binary Conversion.

44 – 32 = 12

We go back to the step 4.

The number we are left with (12) is smaller than the weight of the next column to the right (16). Since12 < 16, we proceed to the step 4b.

Pic. 10 - Decimal-to-Binary Conversion.

The next check against step 4a is true! Because 12 > 8 we put in a ‘1’ in the next available column on the right and take away its weight (8) from the number we are left with now (12).

Pic. 11 - Decimal-to-Binary Conversion.

12 - 8 = 4

The difference is: 4. I hope you already know what is going to happen. Step 4a is true, so we put in a ‘1’ in the column and take away the weight from the number.

Pic. 12 - Decimal-to-Binary Conversion.

4 - 4 = 0 

The difference now reaches 0, so we fill in the remaining columns with 0.

Pic. 13 - Decimal-to-Binary Conversion.

In order to practice those try to quickly convert the following binary to decimal (DO NOT USE A CALCULATOR):

Ex1: 172 (calculated above)
Ex2: 67
Ex3: 124
Ex4: 168
Ex5: 215
Ex6: 237

It should not take you more than a few seconds per example. Check your results using a calculator.

Understanding and fluency with the conversion is the pre-requisite to calculating IP subnets, maximizing the addresses per subnets etc.

In my next post, we will take a look at IP address and play with subnet calculations.