Subnetting is the most tested topic of CCNA. In this article I would show you the method of subnetting.

One network will not access the data of other network without the use of router. Thus we can reduce the amount of data remain in one network. Less data less overhead, collision, or broadcast storm.

This is a result of reduced network traffic.

It's easier to identify and isolate network problems in a group of Smaller connected networks than within one gigantic network. Facilitated spanning of large geographical distances Because WAN links are significantly slower and more expensive than LAN links, a single large network that spans long distances can create problems in every area earlier listed. Connecting multiple smaller networks makes the system more efficient.

Powers of 2 are important to understand and memorize for use with IP subnetting.

2^{1} | 2 | 2^{9} | 512 |

2^{2} | 4 | 2^{10} | 1024 |

2^{3} | 8 | 2^{11} | 2048 |

2^{4} | 16 | 2^{12} | 4096 |

2^{5} | 32 | 2^{13} | 8192 |

2^{6} | 64 | 2^{14} | 16384 |

2^{7} | 128 | 2^{15} | 32768 |

2^{8} | 256 | 2^{16} | 65536 |

**Before we go further let's get familiar with subnetting components**

A subnet mask is a 32-bit value that allows the receiver of IP packets to distinguish the network ID portion of the IP address from the host ID portion of the IP address. Every IP address is composed of a network component and a host component. The subnet mask has a single purpose: to identify which part of an IP address is the network component and which part is the host component. Subnet mask value 0 represent host ID while subnet mask value 1 to 255 represents Network ID in ip address.

This slash notation is sometimes called CIDR (Classless Inter-Domain Routing) notation. It's basically the method that ISPs (Internet service providers) use to allocate a number of Addresses to a company, a home—a customer. The slash notation is simply the number of 1s in a row in the subnet mask. The real reason to use CIDR notation is simply that it is easier to say and especially to type.

Subnetting happens when we extend the subnet mask past the default boundary for the address we are working with. So it's obvious that we first need to be sure of what the default mask is supposed to be for any given address. When faced with a subnetting question, the first thing to do is decide what class the address belongs to. And later decide what the default subnet mask is. One of the rules that Cisco devices follow is that a subnet mask must be a contiguous string of 1s followed by a contiguous string of 0s. There are no exceptions to this rule: A valid mask is always a string of 1s, followed by 0s to fill up the rest of the 32 bits. (There is no such rule in the real world, but we will stick to the Cisco rules here—it's a Cisco exam, after all.) Therefore, the only possible valid values in any given octet of a subnet mask are 0, 128, 192, 224, 240, 248, 252, 254, and 255. Any other value is invalid.

The process of subnetting creates several smaller classless subnets out of one larger classful . The spacing between these subnets, or how many IP addresses apart they are, is called the Block Size.

The first address in a network number is called the network address, or wire number. This address is used to uniquely identify one segment or broadcast domain from all the other segments in the network.

**The Broadcast ID**

The last address in the network number is called the directed broadcast address and is used to represent all hosts on this network segment. it is the common address of all hosts on that Network ID. This should not be confused with a full IP broadcast to the address of 255.255.255.255, which hits every IP host that can hear it; the Broadcast ID hits only hosts on a common subnet. A directed broadcast is similar to a local broadcast.

The main difference is that routers will not propagate local broadcasts between segments, but they will, by default, propagate directed broadcasts.

Any address between the network address and the directed broadcast address is called a host address for the segment. You assign these middle addresses to host devices on the segment, such as PCs, servers, routers, and switches.

There are several method of subnetting. Different author different approach to calculate the subnets. You should choose the method you can understand and perform subnetting easily. Whatever approach you choose need conversion of decimal to binary. Cram up this chart

2^{7} | 2^{6} | 2^{5} | 2^{4} | 2^{3} | 2^{2} | 2^{1} | 2^{0} |

128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |

To convert a decimal number into binary, you must turn on the bits (make them a 1) that would add up to that number, as follows:

187 = 10111011 = 128+32+16+8+2+1 224 = 11100000 = 128+64+32

To convert a binary number into decimal, you must add the bits that have been turned on (the 1s), as follows:

10101010 = 128+32+8+2 = 170 11110000 = 128+64+32+16 = 240

The IP address 138.101.114.250 is represented in binary as

10001010.01100101.01110010.11111010

The subnet mask of 255.255.255.224 is represented in binary as

11111111.11111111.11111111.11100000

When faced with a subnetting question, the first thing to do is decide what class the address belongs to. for examples:

192.168.1.1

The first octet is between 192 and 223 so it is a Class C address

Default mask for Class C: is 255.255.255.0

In exam default subnet mask is not subnetted. Now write down the given ip address as shown here. Write down the default side of IP as it is and reset of part where actual subnetting will perform in binary

192.168. 1 .00000001 255.255.255.00000000 (defaul maks)

**Step 1:- calculate the CIDR value**

CIDR are the on bit in subnet mask. As you can see in our example we have on bit only in default side.

255.255.255.00000000

So our CIDR value is 24 + 0 = 24

*Step 2:- calculate the Subnet mask*

To calculate the subnet mask use the binary to decimal chart given above. Add the decimal place value of on network bit.

<==H bit 255.255.255.00000000 N bit==>

In our example we are using on default mask so our subnet mask will be 255.255.255.0

*Step 3:- calculate the Total Host*

To calculate the total host count the H bit and use this formula

Total host = 2^{H}<==H bit 255.255.255.00000000Total host = 2^{8}= 256

*Step 4:- calculate the Valid Host*

Subtract 2 from Total host Every network or subnet has two reserved addresses that cannot be assigned to a host. These addresses are called the Network ID and the Broadcast ID, respectively. They are the first and last IPs in any network or subnet. We lose those two IP addresses from the group of values that could be assigned to hosts.

Total host - 2 256 -2 = 254

*Step 5:- calculate the Network*

To calculate the Network count the N bit and use this formula

Network = 2^{N}255.255.255.00000000N bit==>Network = 2^{0}= 1

*Step 6:- Find out the block Size*

Finding block size is very easy just subtract the subnet mask from 256

256 – Subnet mask (only the last octal, don't include the default subnet mask) 256 - 0 = 256

*Step 7:- Write down the subnet chart*

Network 1

CIDR Value /24 | IP | Sunetmask |

Net ID | 192.168.1.0 | 255.255.255.0 |

First Valid Host | 192.168.1.1 | 255.255.255.0 |

Last Valid Host | 192.168.1.254 | 255.255.255.0 |

Broadcast ID | 192.168.1.255 | 255.255.255.0 |

Now do the subnetting of CIDR /25 using same method

** Step 1:- calculate the CIDR value** CIDR = sum of all on bit in subnet mask

255.255.255.10000000

So our CIDR value is 24 + 1 = 25

*Step 2:- calculate the Subnet mask*

Add the decimal place value of on network bit.

<==H bit 255.255.255.10000000 N bit==>

In our example we have one on bit and as you can see in decimal chart the place value of 1000000 is 128 so our subnet mask will be 255.255.255.128

*Step 3:- calculate the Total Host*

Total host = 2^{H}<==H bit 255.255.255.10000000Total host = 2^{7}= 128

*Step 4:- calculate the Valid Host*

Subtract 2 from Total host

Total host - 2 128 -2 = 126

*Step 5:- calculate the Network*

To calculate the Network count the N bit and use this formula

Network = 2^{1}255.255.255.10000000 N bit==>Network = 2^{1}= 2

*Step 6:- Find out the block Size*

256 – Subnet mask (only the last octal, don't include the default subnet mask) 256 - 128 = 128

With help of block size you can easy find out the network ID and broadcast ID of all possible networks as we have 8 bits in one octal those can give maximum of **2 ^{8} = 256 decimal number**

We start from 0 so it will end up on 255 (Do not get confuse because we are counting from 0 not from 1 so the last digit will be 255 not 256. It will 256 only when you count from 1 ). All subnetting will perform between these two numbers.

Create a table of x Columns where x is the number of your network

First ip of first network will always be 0 and last ip of last network will be 255 fill its in chart

Now you have network ID of first network and broadcast ID of last network.

Now add block size in the first ip of first network to get the network ID of second network and so on till we get the network id of last network

First network ID 0 Second Network ID 0 +128 = 128

Fill this in Chart.

As you can see from 128 next network is started so the last IP of first network will be 127 fill it in chart. With this method you can fill the last ip of all networks.

Now you have first ip ( network ID ) of all networks and the last ip (Broadcast ID) of all networks. At this point you can easily fill the valid ip in each network. As valid hosts are all ip address those fall between network ip and host ip.

*Step 7:- Write down the subnet chart *

CIDR /25 | Network 1 | Network 2 |

Net ID | 192.168.1.0 | 192.168.1.128 |

First Valid Host | 192.168.1.1 | 192.168.1.129 |

Last Valid Host | 192.168.1.126 | 192.168.1.254 |

Broadcast ID | 192.168.1.127 | 192.168.1.255 |

Binary ANDing is the process of performing multiplication to two binary numbers. In the decimal numbering system, ANDing is addition: 2 and 3 equals 5. In decimal, there are an countless number of answers when ANDing two numbers together. However, in the binary numbering system, the AND function give up only two possible outcomes, based on four different combinations. These answers, can be displayed as a truth table:

0 and 0 = 0 1 and 0 = 0 0 and 1 = 0 1 and 1 = 1

You use ANDing most often when comparing an IP address to its subnet mask. The end result of ANDing these two numbers together is to give up the network number of that address.

What is the network number of the IP address 192.168.100.115 if it has a subnet mask of 255.255.255.240?

Answer

Step 1 Convert both the IP address and the subnet mask to binary:

192.168.100.115 = 11000000.10101000.01100100.01110011 255.255.255.240 = 11111111.11111111.11111111.11110000

Step 2 Perform the AND operation to each pair of bits—1 bit from the address ANDed to the corresponding bit in the subnet mask. Refer to the truth table for the possible outcomes:

192.168.100.115 = 11000000.10101000.01100100.01110011 255.255.255.240 = 11111111.11111111.11111111.11110000 ANDed result = 11000000.10101000.01100100.01110000

Step 3 Convert the answer back into decimal:

11000000.10101000.01100100.01110000 = 192.168.100.112

The IP address 192.168.100.115 belongs to the 192.168.100.112 network when a mask of 255.255.255.240 is used.

Conversion of decimal to binary and vice versa to get network ID is too time consuming process in exam. So I found this easy method.

Step 1:- Decide from which class this IP belongs and what's its default subnet mask

As given IP have 192 in its first octal so it's a class C IP. And default subnet mask of class C is 255.255.255.0

Step2:- Find out the block size. ( As we describe above)

256 -240 = 16

Step3:- Write down all possible network using block size till we do not get our host partition in middle of two network

0,16,32,48,64,80,96,112,128,

As our host number is 115 which fall in the network of 112 so our network ID is

192.168.1.112

And our host's broad cast ID is 192.168.1.127 as from 128 onward next network will start. Easy as I promise

**Click the [New Problem] button to start**

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