Subnetting
Made EZ
32768 16384
8192 4096 2048 1024
512
256
Binary
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1
Sub Mask
128
192
224
240
248
252
254
255
128
192
224
240
248
252
254
255
CIDR
/17
/18
/19
/20
/21
/22
/23
/24
/25
/26
/27
/28
/29
/30
/31
/32
510
254
126
62
30
14
6
2
0
0
-
-
-
-
Host
32766 16382
8190 4094 2046 1022
B Networks
0
2
6
14
30
62
126
254
510
C Networks
-
-
-
-
-
-
-
-
0
1022 2046 4094 8190 16382
2
6
14
30
62
Subnetting is a major part of your CCNA exam. The chart above
will speed the time taken when answering subnetting questions
you will encounter on your exam.
The following presentation will show you:
1. How to create the above chart.
2. How to use the chart when answer subnetting questions.
Binary
Brian
Sub Mask
Says
CIDR
Class
Host
Has
B Networks
Been
C Networks
Canceled
Your subnetting chart made EZ foundation is remembering the row
titles. The acronym I use to remember the row titles is: “Brian Says
Class Has Been Canceled”. Feel free however, to use whatever
you like.
The title rows from top to bottom are as follows: “Binary”, “Sub
Mask”, “CIDR”, “Host”, “B Networks”, & “C Networks”.
Binary
128
64
32
16
8
4
2
1
128
64
32
16
8
4
Sub Mask
CIDR
Host
B Networks
C Networks
Next complete the row titled “Binary”. The “Binary” row is going to
use the same chart we used to convert decimal numbers to binary
and vice versa.
With our subnetting chart we focus on the 3rd and 4th octets of a
TCP/IP address so we write this chart out twice.
2
1
Binary
128
64
32
16
8
4
2
1
Sub Mask
128
192
224
240
248
252
254
255
128
64
32
16
8
4
2
CIDR
Host
B Networks
C Networks
The row “Sub Mask” was abbreviated for subnet mask. To
determine the subnet mask row, start with the number “128” it is
easy to remember because it falls in line with “128” from the
“Binary” row in the 3rd octet just above it.
Then take the first column in “Sub Mask” row, which has a value of
“128”, and add it to next column in the “Binary” row. In this case,
the number is “64”. Adding these two together will determine the
value of the “Sub Mask” row in the second column. 128 + 64 =
192.
Repeat this until you reach the end of the 3rd octet numeric value
255. You should be adding values: 192 + 32 = 224, 224 + 16 = 240,
240 + 8 = 248, 248 + 4 = 252, 252 + 2 = 254, & 254 + 1 = 255.
Then the line you just created gets copied from the 3rd octet into
the 4th octet.
1
Binary
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1
Sub Mask
128
192
224
240
248
252
254
255
128
192
224
240
248
252
254
255
CIDR
/17
/18
/19
/20
/21
/22
/23
/24
/25
/26
/27
/28
/29
/30
/31
/32
Host
B Networks
C Networks
Moving on to the row titled “CIDR”. It is always important to
remember our TCP/IP addresses are 32 bits long, and with our
chart we concentrate on the last 16 bits.
/1 /2 /3 /4 /5 /6 /7 /8 /9 /10 /11 /12 /13 /14 /15 /16 /17 /18 /19 /20 /21 /22 /23 /24 /25 /26 /27 /28 /29 /30 /31 /32
“CIDR” row is simple count from 1 through 32. We drop 1 through
16 because these numbers are found in the first 2 octets. Then
take 17 through 32 counting by 1 left to right.
32768 16384
8192 4096 2048 1024
512
256
Binary
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1
Sub Mask
128
192
224
240
248
252
254
255
128
192
224
240
248
252
254
255
CIDR
/17
/18
/19
/20
/21
/22
/23
/24
/25
/26
/27
/28
/29
/30
/31
/32
510
254
126
62
30
14
6
2
0
0
32766 16382
Host
8190 4094 2046 1022
B Networks
C Networks
“Host” row is short for Host per Network. To start we need to add a
row above the binary line in the 3rd octet. To create this, just double
the previous number.
Now, we have to take the numbers we just created in the 3rd octet,
the numbers in the “Binary” row in the 4th octet, and subtracting 2
from each to calculate the number of host per network.
32768 16384
8192 4096 2048 1024
512
256
Binary
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1
Sub Mask
128
192
224
240
248
252
254
255
128
192
224
240
248
252
254
255
CIDR
/17
/18
/19
/20
/21
/22
/23
/24
/25
/26
/27
/28
/29
/30
/31
/32
510
254
126
62
30
14
6
2
0
0
126
254
510
-
-
32766 16382
Host
B Networks
0
2
8190 4094 2046 1022
6
14
30
62
1022 2046 4094 8190 16382
C Networks
“B Networks” row is to answer questions when subnetting a Class
B network address to determine how many networks are created.
To create the “B Networks” row, take the numbers right to left from
the “Host” row after dropping the first zero. Copy these numbers
into the “B Networks” left to right.
Check yourself 254 in Host is above 254 in B Networks
You will not need to calculate the last 2 positions in this row
because neither are capable of holding any host as you can see in
the row above.
32768 16384
8192 4096 2048 1024
512
256
Binary
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1
Sub Mask
128
192
224
240
248
252
254
255
128
192
224
240
248
252
254
255
CIDR
/17
/18
/19
/20
/21
/22
/23
/24
/25
/26
/27
/28
/29
/30
/31
/32
510
254
126
62
30
14
6
2
0
0
-
-
-
-
32766 16382
Host
8190 4094 2046 1022
B Networks
0
2
6
14
30
62
126
254
510
C Networks
-
-
-
-
-
-
-
-
0
1022 2046 4094 8190 16382
2
6
14
30
62
“C Networks” row is to answer questions when subnetting a Class
C network address to determine how many networks are created.
To create the “C Networks” row, take the “B Networks” section in
the 3rd octet and copy it to the 4th octet in “C Networks”.
You will not need to calculate the last 2 positions in this row
because neither are capable of holding any host.
In the “C Networks” row under the third octet will be blank because
a Class C address will begin to be subnetted in the 4th octet.
32768 16384
8192 4096 2048 1024
512
256
Binary
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1
Sub Mask
128
192
224
240
248
252
254
255
128
192
224
240
248
252
254
255
CIDR
/17
/18
/19
/20
/21
/22
/23
/24
/25
/26
/27
/28
/29
/30
/31
/32
510
254
126
62
30
14
6
2
0
0
-
-
-
-
32766 16382
Host
8190 4094 2046 1022
B Networks
0
2
6
14
30
62
126
254
510
C Networks
-
-
-
-
-
-
-
-
0
1022 2046 4094 8190 16382
2
6
14
30
One last topic to understand with this chart is what to do when IP
Subnet Zero is ENABLED. When IP Subnet Zero is ENABLED
be sure to add 2 to the number of networks for both rows “B
Networks” & “C Networks”
62
Net ID
0
16
32
48
64
80
96
112
128
144
160
176
192
208
224
240
Broadcast
15
31
47
63
79
95
111
127
143
159
175
191
207
223
239
255
Another chart I highly recommend creating is “Count by 16”. This
will prove very helpful when determining Valid Hosts, Network IDs,
and Broadcast Addresses.
Created by counting by 16 starting with 0, from left to right.
Then by taking the next Network ID subtract 1 and this will give you
the Broadcast address for the previous network.
Now any IP address between the Network ID and Broadcast ID is a
Valid Host.
256
32768 16384
8192 4096 2048 1024
512
256
Binary
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1
Sub Mask
128
192
224
240
248
252
254
255
128
192
224
240
248
252
254
255
CIDR
/17
/18
/19
/20
/21
/22
/23
/24
/25
/26
/27
/28
/29
/30
/31
/32
510
254
126
62
30
14
6
2
0
0
-
-
-
-
32766 16382
Host
8190 4094 2046 1022
B Networks
0
2
6
14
30
62
126
254
510
C Networks
-
-
-
-
-
-
-
-
0
1022 2046 4094 8190 16382
2
6
14
30
62
1. How many host will I have per network using a /30 address? IP
subnet zero has been configured.
Answer: 2
• Remember a /30 mask allows for 2 hosts per network,
commonly used for serial connections. (Point to Point)
32768 16384
8192 4096 2048 1024
512
256
Binary
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1
Sub Mask
128
192
224
240
248
252
254
255
128
192
224
240
248
252
254
255
CIDR
/17
/18
/19
/20
/21
/22
/23
/24
/25
/26
/27
/28
/29
/30
/31
/32
510
254
126
62
30
14
6
2
0
0
-
-
-
-
32766 16382
Host
8190 4094 2046 1022
B Networks
0
2
6
14
30
62
126
254
510
C Networks
-
-
-
-
-
-
-
-
0
1022 2046 4094 8190 16382
2
6
14
30
How many networks are created using a 192.168.0.0 with a
subnet mask 255.255.255.224? IP Subnet Zero is applied.
Answer: 8
62
32768 16384
8192 4096 2048 1024
512
256
Binary
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1
Sub Mask
128
192
224
240
248
252
254
255
128
192
224
240
248
252
254
255
CIDR
/17
/18
/19
/20
/21
/22
/23
/24
/25
/26
/27
/28
/29
/30
/31
/32
510
254
126
62
30
14
6
2
0
0
-
-
-
-
32766 16382
Host
8190 4094 2046 1022
B Networks
0
2
6
14
30
62
126
254
510
C Networks
-
-
-
-
-
-
-
-
0
1022 2046 4094 8190 16382
2
6
14
30
62
You need to subnet 160.200.3.0 to support 30 site locations and as
many hosts supported per subnet as possible. What CIDR would
you recommend to use? (IP subnet zero is not applied.)
Answer: /21
32768 16384
8192 4096 2048 1024
512
256
Binary
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1
Sub Mask
128
192
224
240
248
252
254
255
128
192
224
240
248
252
254
255
CIDR
/17
/18
/19
/20
/21
/22
/23
/24
/25
/26
/27
/28
/29
/30
/31
/32
510
254
126
62
30
14
6
2
0
0
-
-
-
-
32766 16382
Host
8190 4094 2046 1022
B Networks
0
2
6
14
30
62
126
254
510
C Networks
-
-
-
-
-
-
-
-
0
1022 2046 4094 8190 16382
2
6
14
30
62
You have been assigned the network of 191.5.5.0. How many hosts do
you have using a mask of 255.255.240.0? IP subnet zero is applied.
Answer: 4094
32768 16384
8192 4096 2048 1024
512
256
Binary
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1
Sub Mask
128
192
224
240
248
252
254
255
128
192
224
240
248
252
254
255
CIDR
/17
/18
/19
/20
/21
/22
/23
/24
/25
/26
/27
/28
/29
/30
/31
/32
510
254
126
62
30
14
6
2
0
0
-
-
-
-
32766 16382
Host
8190 4094 2046 1022
B Networks
0
2
6
14
30
62
126
254
510
C Networks
-
-
-
-
-
-
-
-
0
1022 2046 4094 8190 16382
2
6
14
30
62
Your company has a class C network address. The company requires 5 usable
subnets. Each subnet mask must accommodate at least 18 hosts. Which subnet
mask should you use? (IP subnet zero is not applied)
Answer: 255.255.255.224
32768 16384
8192 4096 2048 1024
512
256
Binary
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1
Sub Mask
128
192
224
240
248
252
254
255
128
192
224
240
248
252
254
255
CIDR
/17
/18
/19
/20
/21
/22
/23
/24
/25
/26
/27
/28
/29
/30
/31
/32
510
254
126
62
30
14
6
2
0
0
-
-
-
-
32766 16382
Host
8190 4094 2046 1022
B Networks
0
2
6
14
30
62
126
254
510
C Networks
-
-
-
-
-
-
-
-
0
1022 2046 4094 8190 16382
2
6
14
30
62
Which of the following addresses is representative of a unicast address?
A. 224.1.5.2
B. 255.255.255.255
C. FFFF.FFFF.FFFF
D. 192.168.24.59/30
E. 172.31.128.255/18
We will need to count by 64s in the 3rd octet. Don’t forget the fourth octet, a 0 in
224.0.0.0
255.255.255.255
This
We
will
is aneed
MAC
through
to
Address
count
is239.255.255.255
TCP/IP
that
by 4s.
Broadcast
is aShortcut
Broadcast
areAddress.
Class
use
asawell.
Dmultiple
Address,of 4 to get
Network ID, and a 255 as a Broadcast Address. Example 64.0 minus 1 from both
designated
you
close
fortarget
multicast.
the third
and to
forth
octetsnumber!
to calculate the Broadcast is 63.255!!!!!
Net ID
0.0
Broadcast
63.255
Net ID
64.0
40
128.0 44
48
52 256.0
56
192.0
127.255
43
47 255.255
51
55
Broadcast 191.255
59
60
32768 16384
8192 4096 2048 1024
512
256
Binary
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1
Sub Mask
128
192
224
240
248
252
254
255
128
192
224
240
248
252
254
255
CIDR
/17
/18
/19
/20
/21
/22
/23
/24
/25
/26
/27
/28
/29
/30
/31
/32
510
254
126
62
30
14
6
2
0
0
-
-
-
-
32766 16382
Host
8190 4094 2046 1022
B Networks
0
2
6
14
30
62
126
254
510
C Networks
-
-
-
-
-
-
-
-
0
1022 2046 4094 8190 16382
2
6
14
30
62
Consider the address 192.168.15.19/28, which of the following addresses are valid host
addresses on this subnet? (Select two options.)
A. 192.168.15.17
B. 192.168.15.0
C. 192.168.15.29
F. 192.168.15.35
D. 192.168.15.16
E. 192.168.15.31
We will
to count
by 16 to determine
Network
Broadcast
IDs, to
Next
weneed
will have
to determine
the network
ID and IDs
the &
broadcast
address
help
us determine valid hosts
for 192.168.15.19/28.
Net ID
0
16
32
48
64
80
96
112
128
144
160
176
192
208
224
240
Broadcast
15
31
47
63
79
95
111
127
143
159
175
191
207
223
239
255
256
32768 16384
8192 4096 2048 1024
512
256
Binary
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1
Sub Mask
128
192
224
240
248
252
254
255
128
192
224
240
248
252
254
255
CIDR
/17
/18
/19
/20
/21
/22
/23
/24
/25
/26
/27
/28
/29
/30
/31
/32
510
254
126
62
30
14
6
2
0
0
-
-
-
-
Host
32766 16382
8190 4094 2046 1022
B Networks
0
2
6
14
30
62
126
254
510
C Networks
-
-
-
-
-
-
-
-
0
1022 2046 4094 8190 16382
2
6
14
30
62
What is the subnet work number of the IP address
159.56.219.56/21?
159.56.219.56/21
A. 159.56.0.0
D. 159.56.52.0
B. 159.56.219.0
E. 159.56.216.0
C. 159.56.8.0
F. 159.56.199.0
We will need to count by 8s in the 3rd octet. Don’t forget the fourth
octet, a 0 in Network ID, and a 255 as a Broadcast Address. Example
224.0 minus 1 from the 3rd & 4th octets to calculate the Broadcast is
223.255!!!
200.0
208.0
216.0
224.0
Net ID
Broadcast
207.255
215.255
223.255
32768 16384
8192 4096 2048 1024
512
256
Binary
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1
Sub Mask
128
192
224
240
248
252
254
255
128
192
224
240
248
252
254
255
CIDR
/17
/18
/19
/20
/21
/22
/23
/24
/25
/26
/27
/28
/29
/30
/31
/32
510
254
126
62
30
14
6
2
0
0
-
-
-
-
Host
32766 16382
8190 4094 2046 1022
B Networks
0
2
6
14
30
62
126
254
510
C Networks
-
-
-
-
-
-
-
-
0
1022 2046 4094 8190 16382
2
6
14
30
62
Which of the following are valid host addresses given the IP address
135.205.0.0/19? IP
IP subnet
subnetzero
zeroisisnot
notactive.
active. (Choose 3)
A. 135.205.20.100
B. 135.205.40.0
C. 135.205.95.255
F. 135.205.225.150
D. 135.205.96.255
E. 135.205.80.100
We will need to count by 32s in the 3rd octet. Don’t forget the fourth octet, a 0 in
Network ID, and a 255 as a Broadcast Address. Example 224.0 minus 1 from the
3rd & 4th octets to calculate the Broadcast is 223.255!!!
Net ID
0.0
32.0
64.0
96.0
128.0
160.0
192.0
224.0
Broadcast
31.255
63.255
95.255
127.255
159.255
191.255
223.255
255.255
256.0
32768 16384
8192 4096 2048 1024
512
256
Binary
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1
Sub Mask
128
192
224
240
248
252
254
255
128
192
224
240
248
252
254
255
CIDR
/17
/18
/19
/20
/21
/22
/23
/24
/25
/26
/27
/28
/29
/30
/31
/32
510
254
126
62
30
14
6
2
0
0
-
-
-
-
Host
32766 16382
8190 4094 2046 1022
B Networks
0
2
6
14
30
62
126
254
510
C Networks
-
-
-
-
-
-
-
-
0
1022 2046 4094 8190 16382
2
6
14
30
62
Given the following list of available IP addresses you need to find which are
valid on the 4 network given the network 159.56.0.0. This address space has
already been divided by Brian into 16 equal subnets, because he said it would
make it harder to figure out. Choose 3 IP addresses that meet the defined
requirements. IP subnet
has been
applied.
IP subnet
has been
applied.
A. 159.56.50.255
D. 159.56.52.0
We will need to count by 16s in the 3rd
octet. Don’t forget the fourth octet, a 0
in Network ID, and a 255 as a
Broadcast Address. Example 16.0
minus 1 from the 3rd & 4th octets to
calculate the Broadcast is 15.255!!!
B. 159.56.219.45
E. 159.56.216.50
C. 159.56.8.0
F. 159.56.63.1
Net ID
0.0
16.0
32.0
48.0
Broadcast
15.255
31.255
47.255
63.255
64.0
32768 16384
8192 4096 2048 1024
512
256
Binary
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1
Sub Mask
128
192
224
240
248
252
254
255
128
192
224
240
248
252
254
255
CIDR
/17
/18
/19
/20
/21
/22
/23
/24
/25
/26
/27
/28
/29
/30
/31
/32
510
254
126
62
30
14
6
2
0
0
-
-
-
-
32766 16382
Host
8190 4094 2046 1022
B Networks
0
2
6
14
30
62
126
254
510
C Networks
-
-
-
-
-
-
-
-
0
1022 2046 4094 8190 16382
2
6
14
30
62
You have been assigned the task of configuring a serial interface of a router.
You were given the IP addresses listed below, use the eighth IP address on the
12 network. 192.168.5.0/24 is the network used by your organization. They have
chosen to subnet that network allowing 14 hosts per subnet. IP subnet zero is
not
active.
IP
subnet
zero is not active.
A. 192.168.5.184
B. 192.168.5.232
C. 192.168.5.201
F. 192.168.5.162
D. 192.168.5.200
E. 192.168.5.8
We will need to count by 16s in fourth octet. Using the value in the binary row the
192 + 8 = 200
increment.
Net ID
0
16
32
48
64
80
96
112
128
144
160
176
192
Broadcast
15
31
47
63
79
95
111
127
143
159
175
191
223
224
Congratulations

Practice

Watch the octet subnetting occurs

Remember if trying to find Network ID, Broadcast
Address, or Valid Host you need to find your
increment (the number in binary row)