This document provides an overview of subnetting concepts including: 1. It explains how binary works and the relationship between binary, decimal, and IP addresses. 2. It demonstrates how subnet masks work by using binary to determine the network and host portions of an IP address. 3. It introduces CIDR notation and how subnet masks can be represented using a slash notation like /24. 4. It provides an example of working through a subnetting problem including determining the number of subnets, hosts per subnet, valid subnet addresses, and which subnet contains a given host IP.

1. Subnetting:
Why learn it?
Tip
Knowing how to subnet
will set you apart from
most networking
professionals.

2. How binary works
Tip
It’s all in the ones and
zeros….

3. How binary works
2 1

4. How binary works
4 2 1

5. How binary works
? ? ? ? 8 4 2 1

6. How binary works
128 64 32 16 8 4 2 1

7. How binary works
128 64 32 16 8 4 2 1
1 1 1 1 1 1 1 1

8. How binary works
128 64 32 16 8 4 2 1
1 1 1 1 1 1 1 1
= 255

9. Binary Decimal
128
192
224
240
248
252
254
255
Todo
Fill out the binary
equivalent to the
decimal numbers.

10. Binary Decimal
10000000 128
11000000 192
11100000 224
11110000 240
11111000 248
11111100 252
11111110 254
11111111 255
Tip
Do you see a pattern?

11. Powers of two
Tip
Look at the answers.
Each time the number
doubles.
2^1= 2x1 which is 2
2^2= 2x2 which is 4
2^3= 2x2x2 which is 8
2^4= 2x2x2x2 which is 16
2^5= 2x2x2x2x2 which is 32
and so on.

12. IP addresses
Tip
To make it easier to
read, we divide IP
addresses into 4 octets.
Made up by binary numbers which are
grouped into octets.
192 168 1 1
11000000 10101000 00000001 00000001

13. Subnet masks
Tip
Computers don’t see
octets or the decimal
numbers, they just see a
string of binary
numbers.
Determine how many bits will be used for
the network (1’s). The remaining octets
identify the hosts (0’s).
255 255 255 0
11111111 11111111 11111111 00000000

14. Subnet masks
255 255 255 0
11111111 11111111 11111111 00000000
Network Network Network Host

15. 192 168 1 1
11000000 10101000 00000001 00000001
255 255 255 0
11111111 11111111 11111111 00000000

16. 192 168 1 1
11000000 10101000 00000001 00000001
Network Network Network Host
255 255 255 0
11111111 11111111 11111111 00000000
● The network is 192.168.1
● The host is 1

17. 192 168 1 1
11000000 10101000 00000001 00000001
255 255 0 0
11111111 11111111 00000000 00000000

18. 192 168 1 1
11000000 10101000 00000001 00000001
Network Network Host Host
255 255 0 0
11111111 11111111 00000000 00000000
● The network is 192.168
● The host is 1.1

19. Rule to memorize
The more bits you steal the more subnets
you get, but each of those subnets is
capable of supporting a lesser number of
hosts.

20. CIDR Notation
CIDR allows for the use of the slash
system for representing subnet masks.
/24 instead of 255.255.255.0
/26 instead of 255.255.255.192

21. CIDR Notation
255 255 255 0
11111111 11111111 11111111 00000000

22. CIDR Notation
255 255 255 0
11111111 11111111 11111111 00000000
8 8 8 0 = /24

23. CIDR Notation
255 255 255 192
11111111 11111111 11111111 11000000

24. CIDR Notation
255 255 255 192
11111111 11111111 11111111 11000000
8 8 8 2 = /26

25. You try some CIDR
255 255 255 240
11111111 11111111 11111111 11110000
= /

26. You try some CIDR
255 255 192 0
11111111 11111111 1100000 00000000
= /

27. You try some CIDR
255 252 0 0
11111111 11111100 0000000 00000000
= /

28. Rule to memorize
Logic tells us that the more host bits we
steal…
the more subnets we have…
but the less hosts we have available per
subnet.
Tip
Think of classrooms and
how many students can
fit in each classroom...

29. More to memorize
Logic tells us that the more host bits we
steal…
the more subnets we have…
but the less hosts we have available per
subnet.
Tip
Think of classrooms and
how many students can
fit in each classroom...

30. Tip
Memorize this and you
shall become the
subnetting Jedi…

31. Subnetting questions
255.255.255.192 or /26
● How many subnets do we have?
● How many hosts per subnet?
● What are the valid subnets?
● What are the valid hosts per subnet?
● Broadcast address?
Tip
We will walk through
this problem using our
newly found tool. .

32. How many subnets?
255.255.255.192 or /26
In a /26 how many bits have been stolen
from the hosts?
Tip
We will walk through
this problem using our
newly found tool. .

33. How many subnets?
255.255.255.192 or /26
In a /26 how many bits have been stolen
from the hosts?
128 64 32 16 8 4 2 1
1 1 0 0 0 0 0 0

34. How many hosts per subnet?
255.255.255.192 or /26
How many bits remain for the hosts? Tip
We will walk through
this problem using our
newly found tool. .

35. How many hosts per subnet?
255.255.255.192 or /26
How many bits remain for the hosts?
128 64 32 16 8 4 2 1
1 1 0 0 0 0 0 0

36. What are the valid subnets?
255.255.255.192 or /26
Just like we ticked two places on the top
Subnet column to get to 192, we will tick
two places on the Bits row…
What did you get?
Tip
We will walk through
this problem using our
newly found tool. .

37. What are the valid subnets?
Tip
We will walk through
this problem using our
newly found tool. .

38. What are the valid subnets?
255.255.255.192 or /26
x.x.x.0 (then +64)
x.x.x.64 (then +64)
x.x.x.128 (then +64)
x.x.x.192
Tip
We will walk through
this problem using our
newly found tool. .

39. What are the valid hosts?
255.255.255.192 or /26
In this example we don’t know.
We don’t have enough information.

40. What are the valid hosts?
255.255.255.192 or /26
In what subnet can you find 192.16.150.76 /26
Subnet 1 = 192.16.150.0 hosts 1 - 62 (63 is the broadcast).
Subnet 2 = 192.16.150.64 hosts 65 - 126 (127 is the broadcast).
Subnet 3 = 192.16.150.128 hosts 129 - 190 (191 is the broadcast).
Subnet 4 = 192.16.150.192 hosts 193 - 254 (255 is the broadcast).

41. What are the valid hosts?
255.255.255.192 or /26
In what subnet can you find 192.16.150.76 /26
Subnet 1 = 192.16.150.0 hosts 1 - 62 (63 is the broadcast).
Subnet 2 = 192.16.150.64 hosts 65 - 126 (127 is the broadcast)
Subnet 3 = 192.16.150.128 hosts 129 - 190 (191 is the broadcast).
Subnet 4 = 192.16.150.192 hosts 193 - 254 (255 is the broadcast).

42. Next question...
Which subnet is host 200.100.206.99 /27?
Tip
We will walk through
this problem using our
newly found tool. .

43. How many subnets?
Which subnet is host 200.100.206.99 /27?
In a /27 we have taken an additional bit from the hosts!!
128 64 32 16 8 4 2 1
1 1 1 0 0 0 0 0

44. Let’s figure out the subnet mask
11111111 11111111 11111111 11100000
8 8 8 3 = /27
Which subnet is host 200.100.206.99 /27?

45. Let’s figure out the subnet mask
Which subnet is host 200.100.206.99 /27?

46. Let’s figure out the subnet mask
255 255 255 224
11111111 11111111 11111111 11100000
8 8 8 3 = /27
Which subnet is host 200.100.206.99 /27?

47. What do we know so far?
Which subnet is host 200.100.206.99 /27?
We know the subnet mask is 255.255.255.224

48. How many hosts per subnet?
255.255.255.192 or /26
How many bits remain for the hosts?
128 64 32 16 8 4 2 1
1 1 0 0 0 0 0 0

49. What are the valid subnets?
255.255.255.192 or /26
Just like we ticked two places on the top
Subnet column to get to 192, we will tick
two places on the Bits row…
What did you get?
Tip
We will walk through
this problem using our
newly found tool. .

50. What are the valid subnets?
Tip
We will walk through
this problem using our
newly found tool. .

51. What are the valid subnets?
255.255.255.192 or /26
x.x.x.0 (then +64)
x.x.x.64 (then +64)
x.x.x.128 (then +64)
x.x.x.192
Tip
We will walk through
this problem using our
newly found tool. .

52. What are the valid hosts?
255.255.255.192 or /26
In this example we don’t know.
We don’t have enough information.

53. What are the valid hosts?
255.255.255.192 or /26
In what subnet can you find 192.16.150.76 /26
Subnet 1 = 192.16.150.0 hosts 1 - 62 (63 is the broadcast).
Subnet 2 = 192.16.150.64 hosts 65 - 126 (127 is the broadcast).
Subnet 3 = 192.16.150.128 hosts 129 - 190 (191 is the broadcast).
Subnet 4 = 192.16.150.192 hosts 193 - 254 (255 is the broadcast).

54. What are the valid hosts?
255.255.255.192 or /26
In what subnet can you find 192.16.150.76 /26
Subnet 1 = 192.16.150.0 hosts 1 - 62 (63 is the broadcast).
Subnet 2 = 192.16.150.64 hosts 65 - 126 (127 is the broadcast)
Subnet 3 = 192.16.150.128 hosts 129 - 190 (191 is the broadcast).
Subnet 4 = 192.16.150.192 hosts 193 - 254 (255 is the broadcast).

55. Class dismissed