Computer Networking: IP Addressing and Subnetting
IP Addressing
Common Computer Numbering Systems
 Binary (base 2) 
0
to1
 Octal (base 8) 
0
to7
 Decimal (base 10) 
0
to9
 Hexadecimal (base 16) 
0
toF
BASE 10 (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)

When you see the number "4,529", you instinctively think...
Four thousand five hundred and twentynine
(at least if you speak English anyway)
From birth, we as humans are taught to count in base 10
10 is a nice even number, working in multiples of 10 is easy
0 0 0 0 4 5 2 9
   ___ 1    ___ 1 x 9 = 9
  _____ 10   _____ 10 x 2 = 20
 _______ 100  _______ 100 x 5 = 500
_________ 1000 _________ 1000 x 4 = 4000

4529
BASE 2 (0, 1)

Computer chips use base 2 or binary because of the transistor
The transistor can either be in one of two states:
 On (1)
 Off (0)
Processors can have billions of transistors
Allowing for computations of a massive range of values
Depending on the combinations of on/off states of transistors
0 0 0 0 1 0 1 0
   ___ 1    ___ 1 x 0 = 0
  _____ 2   _____ 2 x 1 = 2
 _______ 4  _______ 4 x 0 = 0
_________ 8 _________ 8 x 1 = 8

10 (base 10)
0A (base 16)
BASE 16 (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A , B , C , D , E , F)
10, 11, 12, 13, 14, 15

Hexadecimal is widely used in computing, because:
 It's more compact
 Storing hexadecimal values consumes less resources
 Converting between binary and hexadecimal is computationally easy
0 0 0 0 F F E D
   ___ 1    ___ 1 x 13 = 13
  _____ 16   _____ 16 x 14 = 224
 _______ 256  _______ 256 x 15 = 3840
_________ 4096 _________ 4096 x 15 = 61440

65517 (base 10)
Decimal to Binary and Back Again
Binary to Decimal Conversion Examples
CONVERT 0 BETWEEN DECIMAL AND BINARY

0 0 0 0 0 0 0 0
128 64 32 16 8 4 2 1
       _ 0 x 1 = 0 +
      
      _ 0 x 2 = 0 +
     
     _ 0 x 4 = 0 +
    
    _ 0 x 8 = 0 +
   
   _ 0 x 16 = 0 +
  
  _ 0 x 32 = 0 +
 
 _0 x 64 = 0 +

_ 0 x 128 = 0
___
0 (SUM)
CONVERT 255 BETWEEN DECIMAL AND BINARY

1 1 1 1 1 1 1 1
128 64 32 16 8 4 2 1
       _ 1 x 1 = 1 +
      
      _ 1 x 2 = 2 +
     
     _ 1 x 4 = 4 +
    
    _ 1 x 8 = 8 +
   
   _ 1 x 16 = 16 +
  
  _ 1 x 32 = 32 +
 
 _ 1 x 64 = 64 +

_ 1 x 128 = 128
___
255 (SUM)
You'll see that the sum of 8 bits is 255, which may lead you to wonder why you'll see a byte denoted as 256. This because in computing, 0
is also a possible binary value. If a byte can be denoted as 2^{8}, then we're calculating all the possible permutations of 8 bits  including 0
.
Effectively, if a byte is 8 bits, and if a bit can either be a 0
or a 1
, then there are 256
possible values.
CONVERT 192 BETWEEN DECIMAL AND BINARY

1 1 0 0 0 0 0 0
128 64 32 16 8 4 2 1
       _ 0 x 1 = 0 +
      
      _ 0 x 2 = 0 +
     
     _ 0 x 4 = 0 +
    
    _ 0 x 8 = 0 +
   
   _ 0 x 16 = 0 +
  
  _ 0 x 32 = 0 +
 
 _ 1 x 64 = 64 +

_ 1 x 128 = 128
___
192 (SUM)
CONVERT 172 BETWEEN DECIMAL AND BINARY

1 0 1 0 1 1 0 0
128 64 32 16 8 4 2 1
       _ 0 x 1 = 0 +
      
      _ 0 x 2 = 0 +
     
     _ 1 x 4 = 4 +
    
    _ 1 x 8 = 8 +
   
   _ 0 x 16 = 0 +
  
  _ 1 x 32 = 32 +
 
 _ 0 x 64 = 0 +

_ 1 x 128 = 128
___
172 (SUM)
IP Addressing Formats
I am not going to get into any debates about why you should be using IPv6 versus IPv4. The majority of homes and business  I imagine  are still using IPv4 networks for their private addressing needs.
IPv4 depletion is a concern for public IPv4 addresses. There are some purists out there that like to make a lot of noise about it. However, if you prefer IPv4 addressing for use internally, that's perfectly fine.
 An IP address  in its simplest form  is just a series of
1
and0
bits (binary or base 2 numbering)
 IPv4 is 32 bits  thirtytwo 1s and 0s
 IPv6 is 128 bits  one hundred and twenty eight 1s and 0s
 We can take base 2 and convert to base 8, base 10, and base 16
 Converting the IP address to decimal (base 10) makes it easier to store that information in databases
 I've worked with a handful of security tools where the IP address was stored exclusively in decimal
 Binary notation is too long, decimal is cleaner
 Converting the IP address to decimal (base 10) makes it easier to store that information in databases
 The networking stack of the operating system can seamlessly work with multiple notations
IP Address Conversion Tool: https://www.hacksparrow.com/tools/converters/ipaddress.html
More information here: https://www.hacksparrow.com/networking/manyfacesofipaddress.html
Network Masking
An IP address is meaningless without a network mask. Given only an IP address, it is impossible to be certain which network this IP address is a member of. You may be able to infer, but you cannot be certain.
192 . 168 . 1 . 1 < IP Address
11000000 . 10101000 . 00000001 . 00000001
255 . 255 . 255 . 0 < Subnet Mask
11111111 . 11111111 . 11111111 . 0
 Think of it like a zip code (or postal code)
 A postal code confines residences to a specific area
 A network mask confines hosts to a specific route
 We also use network masks to determine the size of the network
 The subnet mask is also  in its simplest form  just a series of
1
and0
bits
No More Classful Networking
 I don't like dwelling too much on the idea of Class A, Class B, or Class C networks
 Those days are long gone. The world now operates on Classless InterDomain Routing (CIDR)
 Instead, you should think of networks in terms of internetroutable and noninternetroutable
 Public (WAN) / Private (LAN) / Diagnostic / Experimental / Reserved
 RFC 1918 (noninternetroutable IPv4 addresses)
10.0.0.0/8
172.16.0.0/12
192.168.0.0/16
 RFC 4193 (noninternetroutable IPv6 addresses)
fc00::/7
More Information:
https://docs.netgate.com/pfsense/en/latest/network/addresses.html
https://en.wikipedia.org/wiki/Reserved_IP_addresses
Exploring Private IPv4 Network Masks
 A network mask can be broken up into two parts
 The network segment
 The host segment
 Private IPv4 Address Spaces
10.0.0.0/8
 Any host address between
10.0.0.1
and10.255.255.254
is valid
 Any host address between
172.16.0.0/12
 Any host address between
172.16.0.1
and172.31.255.254
is valid
 Any host address between
192.168.0.0/16
 Any host address between
192.168.0.1
and192.168.255.254
is valid
 Any host address between
The /8, /12, /16 are called CIDR notation or CIDR blocks. The number appearing after the slash ( /
) directly correlates to the sum of 1
bits in the binary notation of subnet mask. So /8 indicates that eight bits have been switched on.
10.0.0.0/8
10 . 0255 . 0255 . 0254
255 . 0 . 0 . 0 = /8
11111111 . 00000000 . 00000000 . 00000000 = 8 bits on (1 x 8)
^ ^ ^ ^
   
Network Host Host Host


____ This octet is masked by 11111111
Therefore, this octet is static at 10
172.16.0.0/12
172 . 1631 . 0255 . 0254
255 . 240 . 0 . 0 = /12
11111111 . 11110000 . 00000000 . 00000000 = 12 bits on (1 x 12)
^ ^^ ^ ^
 ___________ 
   
Network Network Host Host
 
 
 ____ This octet is masked by 11110000
 Therefore, this octet has some variable range

____ This octet is masked by 11111111
Therefore, it is static at 172
This network is between /8
and /16
, which means that it was subnetted from a /8
network. So, we took a 172.16.0.0/8
and made it 172.16.0.0/12
. RFC 1918 has allocated one of 16 networks to be used as private, noninternetroutable address space.
172.16.0.0/12 highlighted in red
This means that the remainder of networks in the 172.0.0.0/12
address space are internetroutable address blocks.
If you haven't noticed yet, the network address in the picture above increases in multiples of 16  0, 16, 32, 48, 64, and so on. This is because our network mask is 11110000
Given a subnet mask, you can calculate the total number of possible networks the following ways:
 Find the partial network mask
 A fully masked network ID is 8 bits 
11111111
/8
,/16
,/24
,/32
represent CIDR masks where network ID octets are fully masked11111111.00000000.00000000.00000000
=/8
11111111.11111111.00000000.00000000
=/16
11111111.11111111.11111111.00000000
=/24
/12
is between/8
and/16
11111111.11110000.00000000.0000000
full partial host host
 A fully masked network ID is 8 bits 
 Then...
 Use the decimal value of the network mask
 Or, use the number of host bits as an exponent
Decimal Value
11110000
(base 2) =240
(base 10) Total of 8 bits in a binary octet, with
256
possible values in decimal (as discussed earlier) 256
240
=16
Host Bit Exponent
 There are 4 bits in the host portion of the subnet mask
 Binary is base 2, so 2^{4 }= 16
 Again, 4 is the exponent here, as that's the number of host bits
 The total possible values in binary, you'll recall, are
1
or0
, or base 2.  2^{4} = 16, hence the network ID increases in multiples of 16
172.0.x.x
to 172.15.x.x
 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15  16 possible values (including zero)172.16.x.x
to 172.31.x.x
 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31  16 possible values
Etc...
We'll get more practice with this a bit more down the page
Conversely, any octet where the network mask is 11111111
is fully masked, and the value can only be one value from 0
to 255
192.168.0.0/16
Technically speaking, if there's a 192.168.0.0/16
, there is a 192.0.0.0/8
, as demonstrated in the screenshot below.
https://en.wikipedia.org/wiki/List_of_assigned_/8_IPv4_address_blocks
Effectively, RFC 1918 has shrunk the /8
to /16
and said, that any address falling in 192.168.0.0/16
should be considered noninternetroutable (private).
192 . 168 . 0255 . 0254
255 . 255 . 0 . 0 = /16
11111111 . 11111111 . 00000000 . 00000000 = 16 bits on (1 x 16)
^ ^ ^ ^
   
Network Network Host Host
 
__________


____ Both of these octets are masked by 11111111
Therefore, both of these octets are static
at 192 and 168 respectively
All of this is important, because every device with a network interface card (NIC) is going to have a routing table:
 Hosts need to know what their Local Area Network (LAN) and what constitutes a foreign network
 Can the host put the packet on the wire and let the switch take care of it?
 Or, does it need to send it to the default gateway to be routed somewhere elese?
 Routers need to be able to know how to move packets
Subnetting
DON'T OVERTHINK IT!
Folks spend a lot of time upskilling in the HOW of subnetting, but I don't think training materials really spend enough time on the WHY of subnetting.
Why Do We Subnet?
A) To shrink the size of a network
B) To make multiple networks from a single network block
Scenario A: Shrink the 192.168.0.0/16
network
 RFC 1918 has allocated the
192.168.0.0/16
address block for use in private computer networking  A
/16
network will yield 65,534 possible addresses.  There's no way as residential users that we're going to use all of these addresses internally
192 . 168 . 0 . 0
255 . 255 . 0 . 0
11111111 . 11111111 . 00000000 . 00000000 = /16
^ ^ ^ ^
   
Network Network Host Host
Before
 We can "borrow bits" from the host segment and switch them to
1
to make them a network segment  In this example, since we flipped all of the bits to
11111111
, we can use any value between0
and255
 We've add eight more bits to the network mask, making it a
/24
network 192.168.1.0/24
and192.168.0.0/24
are very common subnets shipped on consumergrade routers.
192 . 168 . 0255 . 1254
255 . 255 . 255 . 0
11111111 . 11111111 . 11111111 . 00000000 = /24
^ ^ ^ ^
   
Network Network Network Host
 
 ______ This is the host octet now
 You can address any host from 1 through 254


______ Choose a number between 0 through 255
However, this now becomes part of the network ID
So, this number is static, whichever you choose
After
Scenario B: Increase the number of networks
Context
 Think about this from a business perspective
 You are working at an office branch and you are the network administrator
 The network engineer at HQ gave you a
10.10.0.0/16
address space to work with On its own, this address space yields:
 1 possible network
 65,534 possible hosts
 You're required to work with this space and have no control over additional networks
 You don't want to create routing conflicts with other branches in the organization
 You control
10.10.0.0/16
and any subvariant of this address space
 On its own, this address space yields:
The Problem
 You need to break this up into smaller chunks for multiple departments:
 Sales
 HR
 Engineering
 Once you've broken it up, you'll give this information to the network engineer, so that they can set up routes and firewall rules accordingly
10 . 10 . 0 . 0
255 . 255 . 0 . 0
11111111 . 11111111 . 00000000 . 00000000 = /16
^ ^ ^ ^
   
Network Network Host Host
The Solution
 We know we need 3 networks
 We need to calculate the number of host bits to borrow for the network mask.
 The formula to check this is 2^{n}, where
n
is the number of bits borrowed 2^{1} = 2, meaning 2 subnets are possible if we borrow 1 bit; not enough
 2^{2} = 4, meaning 4 subnets are possible if we borrow 2 bits; more than we need, but the best possible choice
10 . 10 . 0 . 0
255 . 255 . 192 . 0
11111111 . 11111111 . 110000000 . 00000000 = /18
^ ^ ^ ^
   
Network Network Host Host
More Subnetting Practice
Shrink a Network Block
Scenario 1: Only 254 hosts per network
Problem
 You're setting up a lab network and you've decided to use the
10.0.0.0/8
RFC 1918 spaces  In it's current state, you've got the following network mask
10 . 0 . 0 . 0
11111111 . 00000000 . 00000000 . 00000000
Network Host Host Host
8 network bits
24 host bits
 2^{24} = 16,777,216 = total addresses
 (2^{24})  2 = 16,777,214 = total usable addresses (minus network and broadcast)
 You don't need this many addresses in your lab environment
Calculate the Number of Usable Host Bits
 (2^{8})  2 = 254 = total usable addresses = only 8 host bits
 We know there's a total of 32 bits in the network mask
 32 total bits  8 host bits = 24 = number of network bits
 Yielding network mask of
11111111.11111111.11111111.00000000
Number of Usable Networks
10 . 0255 . 0255 . 1254
11111111 . 11111111 . 11111111 . 00000000
Network . Network . Network . Hosts
  
  ____ Fully masked by 11111111
  Which ever decimal value we set it to is static
  Can choose any decimal value between 0255
  This will be the network ID
 
 ____ Fully masked by 11111111
 Which ever decimal value we set it to is static
 Can choose any decimal value between 0255
 This will be the network ID

____ Fully masked by 11111111
This value cannot be anything other than 10
The 10.0.0.0/8 network is the only "10" IP address block
Reserved by RFC 1918 for private networking
10
cannot be changed The second octet is variable and open to our choosing  8 bits
 The third octet is variable and open to our choosing  8 bits
 8 bits + 8 bits = 16 bits
 2^{16 }= 65,536 possible networks, with each network capable of having 254 hosts
Example
10.1.2.0/24
10 . 1 . 2 . 1254
255 . 255 . 255 . 0
11111111 . 11111111 . 11111111 . 00000000 = /24
Network Network Network Hosts
10.1.2  are the network ID and are static
The last octet can be any value between 1254 as a usable address
Scenario 2: Only 30 hosts per network
Problem
 You're setting up a lab network and you've decided to use the
172.16.0.0/12
RFC 1918 space  In it's current state, you've got the following network mask
172 . 1631 . 0 . 0
11111111 . 11110000 . 00000000 . 00000000
Network '''' Host Host Host
Net Host
12 network bits
20 host bits
 2^{20} = 1,048,576 = total addresses
 (2^{20})  2 = 1,048,574 = total usable addresses (minus network and broadcast)
 You don't need this many addresses in your lab environment
Calculate the Number of Usable Host Bits
 (2^{5})  2 = 30 = total usable addresses = only 5 host bits
 We know there's a total of 32 bits in the network mask
 32 total bits  5 host bits = 27 = number of network bits
 Yielding network mask of
11111111.11111111.11111111.11100000
Number of Usable Networks
172 . 1631 . 0255 . 1254
11111111 . 11111111 . 11111111 . 11100000
Network . Network . Network . ''''
    
   Net Hosts
  
  ____ Fully masked by 11111111
  Which ever decimal value we set it to is static
  Can choose any decimal value between 0255
  This will be the network ID
 
 ____ Per RFC 1918, this octet must be
 Any deciaml value of 16 through 31

____ Per RFC 1918, this octet may only be 172
172
cannot be changed The second octet may be between
16
and31
per RFC 1918  The third octet is variable and open to our choosing  8 bits
 We've got a total of 27 bits in the network mask 
/27
 The original network mask is 12 bits 
/12
 27  12 = 15
 2^{15 }= 32,768 possible networks, with each network capable of having 30 hosts
Example
172.20.234.0/27
Break up a Network Block
Scenario 1: Yield 3 Networks
Problem
 You've been given a
192.168.1.0/24
network to work with  You need to divide this network up into three subnets
Convert the Subnet Mask to Binary
/24
means 24 bits are flipped on in the subnet mask11111111.11111111.11111111.00000000
(1 x 24 = 24)
Network Network Network Host
Calculate the Number of Host Bits to Borrow
 2^{n }formula where
n
is the number of bits to borrow  2^{1 }= 2 networks, which does not yield enough networks for us, as we need 3
 2^{2 }= 4 networks, which is more than we need, but the best we can do given the circumstances
Network Mask
255 . 255 . 255 . 192
11111111 . 11111111 . 11100000 . 11000000 = /26 (1 x 26)
Network . Network . Network . ''
 
Network Host
 Our new CIDR notation is now
/26
(1 x 26) 192.168.1.0/26
Calculate the Network Size
 32 bits in a network mask
 32  26 = 6 host bits
 2^{6 }= 64 hosts per network
Calculate the Valid Decimal Values
192 . 168 . 1 . 0
255 . 255 . 255 . 192
11111111 . 11111111 . 11111111 . 11000000 = /26 (1 x 26)
Network . Network . Network . ''
    
   Network Host
   
   ____ 6 host bits = 2^6 = 64
   64 total addresses per network
   Increments of 64
  
  
  ____ Fully masked by 11111111
  This octet is static at
  1
 
 _____ Fully masked by 11111111
 This octet is static at
 168

____ Fully masked by 11111111
This octet is static at
192
Note that the network ID address increases by increments of 64
Scenario 2: Yield 8 Networks
Problem
 You've been given a
10.12.0.0/16
network to work with  You need to divide this network up into eight subnets
Convert the Subnet Mask to Binary
/16
means 16 bits are flipped on in the subnet mask11111111.11111111.00000000.00000000
(1 x 16 = 16)
Network Network Host Host
Calculate the Number of Host Bits to Borrow
 2^{n }formula where
n
is the number of bits to borrow  2^{3 }= 8, therefore we need to borrow 3 bits from the host portion of the subnet mask
Network Mask
255 . 255 . 224 . 0
11111111 . 11111111 . 11100000 . 00000000 = /19 (1 x 19)
Network . Network . '''' . Host
 
Network Host
 Our new CIDR notation is now
/19
(1 x 19) 10.12.0.0/19
Calculate the Network Size
 32 bits in a network mask
 32  19 = 13 host bits
 2^{13 }= 8,192 hosts per network
Calculate the Valid Decimal Values
10 . 12 . 0 . 0
255 . 255 . 224 . 0
11111111 . 11111111 . 11100000 . 00000000 = /19 (1 x 19)
Network . Network . '''' . Host
  
Network Host ____ 8 host bits
 2^8 = 0255


_____ 5 host bits
2^5 = 031
Increments of 32
Note that the network ID address increases by increments of 32