Subnetting:
Logical division of IP
Network within network
Generally, Subnetting can be done in two ways
I. According to host
II. According to network
1. Given IP: 197.10.10.0, find the Subnet ID for each
department with their respective Broadcast ID.
Depart A: 5 host
Depart B: 10 host
Depart C: 20 host
Depart D: 30 host
Depart E: 40 host
Solution:
Step 1: Finding the types of class of given IP and their
host octet and network octet.
Given IP 197.10.10.0
The range of class C IP is (192.0.1.1) to
(223.255.254.254).
(Note: always check first octet to
find class )
Hence the given IP belongs Class C and it has 3 octet
for Network and 1 octet for host. i.e.
N: N: N: H
197.10.10.0
Step 2: Converting only host octet into binary system
197.10.10.00000000
(Note: Each octet is represented by 8-bits, so that IPv4 is
of 32-bits)
Step 3: Finding the values of n
We have,
2n -2 ≥ maximum no. of host
Where n represents no. of bits
2n -2 ≥ 40
At n=1,
21 -2≥ 40
0≥40, not satisfied
At n=2,
22-2≥ 40
2≥ 40, not satisfied
:
:
:
At n=6,
26-2≥ 40
64-2≥ 40
62≥ 40, satisfied
Hence n=6
Step4: Borrowing host bit to Network bit according
n=6 (from right to left <-------)
197.10.10.00000000
197.10.10.11000000
(Note: representation as
Network bit: 1
host bit : 0 )
Now converting into decimal, we have
197.10.10.192
(It is the LAST IP of given IP)
(The changes occur in first octet from right to left by 64)
Step 5: As we know, the default subnet mask for class
C IP is given by:
255.255.255.0
Hence, the subnet mask for given IP will be
255.255.255.192
Step 6: CIDR =?
(Class Inter Domain Routing)
CIDR = 32-n
CIDR = 32-6
Hence, CIDR= 26
Step 7: Subnet IDs =?
Subnet ID 1: 197.10.10.0/26
Broadcast ID: 197.10.10.63/26
Subnet ID 2: 197.10.10.64/26
Broadcast ID: 197.10.10.127/26
Subnet ID 3: 197.10.10.128/26
Broadcast ID: 197.10.10.191/26
Subnet ID 4: 197.10.10.192/26
Broadcast ID: 197.10.10.255/26
Subnet ID 5: 197.10.10.256/26; Not valid
Hence, we can derive only four subnet ID.
Step 8: Total number of Host per subnet?
The total number of Host bit (n) present in given IP:
n=6
Now, Total number of Host per subnet = 2n
26= 64
Hence, total number of Host per subnet = 64
Step 9: Usable number of Host per subnet?
Usable number of Host per subnet = 2n - 2
Hence, usable number of Host per subnet =62
2. Given IP: 12.0.0.0, find the Subnet ID for each
department with their respective Broadcast ID.
Depart A: 5 host
Depart B: 10 host
Depart C: 20 host
Depart D: 60 host
Depart E: 40 host
Solution:
Step 1: Finding the types of class of given IP and
their host octet and network octet.
Given IP 12.0.0.0
The range of class A IP is (1.0.0.0) to (126.255.255.255)
(Note: always check first octet to find class)
Hence the given IP belongs Class A and it has 3 octet
for Network and 1 octet for host. i.e.
N: H: H: H
12. 0. 0 .0
Step 2: Converting only host octet into binary system
12.00000000. 00000000.00000000
(Note: Each octet is represented by 8-bits, so that IPv4 is
of 32-bits)
Step 3: Finding the values of n
We have,
2n -2 ≥ maximum no. of host
Where n represents no. of bits
At n=6,
26-2≥60
64-2 ≥ 60
62 ≥ 60, satisfied the
condition Hence, n=6
Step 4: Borrowing host bit to Network bit according n=6
(from right to left <-------)
12.00000000. 00000000.00000000
12. 11111111.11111111.11000000
(Note: representation as
Network bit: 1
host bit : 0 )
Now converting into decimal, we have
12.255.255.192(last IP)
(The changes occur in first octet from right to left by 64)
Step 5: As we know, the default subnet mask for class A IP
is given by:
255.0.0.0
Hence, the subnet mask for given IP will be
255.255.255.192
Step 6: CIDR =?
(Class Inter Domain Routing)
CIDR = 32-n
CIDR = 32-6
Hence, CIDR = 26
Step 7: Subnet IDs =?
Subnet ID 1: 12.0.0.0/26
Broadcast ID: 12.0.0.63/26
Subnet ID 2: 12.0.0.64/26
Broadcast ID: 12.0.0.127/26
Subnet ID 3: 12.0.0.128/26
Broadcast ID: 12.0.0.191/26
Subnet ID 4: 12.0.0.192/26
Broadcast ID: 12.0.0.255/26
Subnet ID 5: 12.0.1.0/26
Broadcast ID: 12.0.1.63/26
:
:
:
Subnet ID 8: 12.0.1.192/26
Broadcast ID: 12.0.1.255/26
Subnet ID 9: 12.0.2.0/26
Broadcast ID: 12.0.2.63/26
3. Given IP: 157.110.88.0, find the Subnet ID for each bgd
ccccccccccdepartment with their respective Broadcast ID.
Depart A: 5 host
Depart B: 10 host
Depart C: 20 host
Depart D: 60 host
Depart E: 40 host
Solution:
Step 1: Finding the types of class of given IP and their
host octet and network octet.
Given IP 157.110.88.0
Hence, class B------ N:N:H:H
Step 2: Converting only host octet into binary system
157.110.01011000.00000000
Step 3: Finding the values of n
We have,
2n -2 ≥ maximum no. of host
Where n represents no. of bits
At n=6,
26-2≥60
64-2 ≥ 60
62 ≥ 60, satisfied the
condition Hence, n=6
Step4: Borrowing host bit to Network bit according n=6
(from right to left <-------)
157.110.01011000.00000000
After borrowing, we have,
157.110.11111111.11000000
(The all remaining part of the host is replaced by 1 followed by
0 as n=6)
Now converting into decimal, we
have 157.110.255 .192 (Last IP)
Step 5: As we know, the default subnet mask for class B
IP is given by:
255.255.0.0
Hence, the subnet mask for given IP will be
255.255.255.192
Step 6: CIDR =?
(Class Inter Domain Routing)
CIDR = 32-n
CIDR = 32-6
Therefore, CIDR = 26
Step 7: Subnet IDs =?
Subnet ID 1: 157.110.88.0/26
Broadcast ID: 157.110.88.64/26
Subnet ID 2: 157.110.88.64/26
Broadcast ID: 157.110.88.127/26
Subnet ID 3: 157.110.88.128/26
Broadcast ID: 157.110.88.191/26
Subnet ID 4: 157.110.88.192/26
Broadcast ID: 157.110.88.255/26
Subnet ID 4: 157.110.89.0/26
Broadcast ID: 157.110.89.63/26
4. Given IP: 12.0.0.0, find the Subnet ID for each
department with their respective Broadcast ID.
Depart A: 16 host
Depart B: 32 host
Depart C: 64 host
Depart D: 1024 host
Depart E: 512 host
Solution:
Step 1: Finding the types of class of given IP and their
host octet and network octet.
Given IP 12.0.0.0
Hence, class A------ N: H: H: H
Step 2: Converting only host octet into binary system
12.0.0.0
12.00000000.00000000.00000000
Step 3: Finding the values of n
We have,
2n -2 ≥ maximum no. of host
Where n represents no. of bits
At n= 10,
210 - 2 ≥ 1024
1024 - 2 ≥ 1024
1022 ≥ 1024, not satisfied
At n= 11,
211 - 2 ≥ 1024
2048 -2 ≥ 1024
2046 ≥ 1024, satisfied
Hence n=11
Step4: Borrowing host bit to Network bit according
n=11(from right to left <-------)
12.00000000. 00000000.00000000
12.11111111. 11111000.00000000
(Note: The changes occur in second octet from right to
left by 8)
Now converting into decimal, we
have 12.255.248.0 (Last IP)
Step 5: As we know, the default subnet mask for class A
IP is given by: 255.0.0.0
Hence, the subnet mask for given IP will be
255.255.248.0
Step 6: CIDR =?
(Class Inter Domain Routing)
CIDR = 32-n
CIDR = 32-11
CIDR = 21
Step 7: Subnet IDs =?
Subnet ID 1: 12.0.0.0/21
Broadcast ID 1: 12.0.7.255/21
Subnet ID 2: 12.0.8.0/21
Broadcast ID 2: 12.0.15.255/21
Subnet ID 3: 12.0.16.0/21
Broadcast ID 3: 12.0.23.255/21
Subnet ID 4: 12.0.24.0/21
Broadcast ID 3: 12.0.31.255/21
:
:
:
Subnet ID : 12.0.248.0/21
Broadcast ID: 12.0.255.255/21
Subnet ID : 12.1.0.0/21
Broadcast ID : 12.7.255.255/21
2. According to Network
5. Consider the IP address 157.110.88.0, we need 30
subnet in such a way that each subnet may host upto 64
host.
Step 1: The IP 157.110.88.0 is of class B.
N: N: H: H
Step 2: Only representing network octet i.e. Network
ID
157.110.00000000.00000000
Step 3: Finding the value of n
2n ≥ maximum no. of subnet required
2n ≥ 30
Therefore, n= 5
Step 4: Borrowing host bits to network bits from
(LEFT to RIGHT )
157.110.00000000.00000000
157.110.11111000.00000000
Converting into decimal
157.110.248.0 (Last IP)
(NOTE: The changes occur in second octet from right to
left by 8)
Step 5: Subnet mask?
255.255.248.0
Step 6: CIDR =?
CIDR = Total no. of network bits + n
CIDR = 16+ 5
CIDR = 21
Step 7: Subnet ID?
Subnet ID 1: 157.110.0.0/21 05:00 05:59 06:00
Broadcast ID 1: 157.110.7.255/21
Subnet ID 2: 157.110.8.0/21
Broadcast ID 2: 157.110.15.255/21
Subnet ID 3: 157.110.16.0/21
Broadcast ID 3: 157.110.23.255/21 and so on.
CCNA Q: The given IP 192.168.15.240/28, find out
that it is which part of the subnet?
Solution:
The given IP is of class C and default subnet of class C is 24
But, subnet: 24+4=28
Now,
192.168.15.11110000
(The changes occurs in first octet by 16 from left to right i.e.
each subnet has difference of 16)
Subnte ID 1192.168.15.0/28
192.168.15.16/28
192.168.15.32/28
192.168.15.48/28
192.168.15.64/28
192.168.15.80/28
192.168.15.96/28
192.168.15.112/28
Subnet ID 9 -----> 192.168.15.128/28
192.168.15.144/28
192.168.15.160/28
192.168.15.176/28
192.168.15.192/28
192.168.15.208/28
192.168.15.224/28
Subnet ID 16 -----> 192.168.15.240/28
Hence, the given IP 192.168.15. 240/28 is 16th part of subnet
#VLSM (Variable Length Subnet Mask)
6. Given IP: 197.10.10.0, find the VSLM each department with
their respective Broadcast ID and subnet mask.
Depart A: 4 host
Depart B: 10 host
Depart C: 20 host
Depart D: 30 host
Depart E: 2 host
Solution:
Step 1: The given IP 197.10.10.0 is of class C and it has:
N:N:N:H
Step 2: Converting only host octet into binary system.
197.10.10.00000000
Step 3: n =?
2n – 2 >= max no. of host
25- 2>= 30
30>=30, satisfied.
Hence, n = 5
Step 4: Borrowing host bit to Network bit
according n = 5
197.10.10.00000000
197.10.10.11100000
Converting into decimal
197.10.10.224
Subnet mask: 255.255.255.224
Step 5: CIDR =?
CIDR = 32- 5
CIDR = 27
Step 6:
Subnet ID 1: 197.10.10.0/27
Broadcast ID: 197.10.10.31/27
Subnet mask: 255.255.255.224
Subnet ID2: 197.10.10.32
2^n - 2 >= 20
n =5
Borrowing host bit to n/w bit according n=5
197.10.10.00000000
197.10.10.11100000
Converting to decimal
197.10.10.224
(The changes occur in first octet from right to left by 32)
Step5: CIDR=?
CIDR= 32-n
CIDR= 32-5=27
Subnet ID 2 : 197.10.10.32/27
Broadcast ID : 197.10.10.63/27
subnet mask : 255.255.225.224
Subnet ID 3: 197.10.10.64
2^n - 2 >= 10
n=4
Borrowing host bit to n/w bit according n=4
197.10.10.00000000
197.10.10.11110000
Converting to decimal
197.10.10.240
(The changes occur in first octet from right to left by
16)
CIDR= 32-4 => 28
Subnet ID 3: 197.10.10.64/28
Broadcast ID: 197.10. 10.79/28
Subnet mask: 255.225.225.240
Subnet ID 4: 197.10.10.80
2^n - 2 >= 4
n=3
197.10.10.00000000
197.10.10.11111000
Converting to decimal
197.10.10.248
(The changes occur in first octet from right to left by
8)
CIDR= 32-3 => 29
Subnet ID 4: 197.10.10.80/29
Broadcast ID: 197.10.10.87/29
Subnet mask: 255.255.255.248
Subnet ID5: 197.10.10.88
2^n - 2 >= 2
n=2
197.10.10.00000000
197.10.10.11111100
Converting to decimal
197.10.10.252
(The changes occur in first octet from right to left by 4)
CIDR= 32-2 => 30
Subnet ID5: 197.10.10.88/30
Broadcast ID: 197.10.10.91/30
Subnet mask: 255.255.255.252
Q: Given IP : 197.10.72. 32, find the
VLSM(Variable Length Subnet Mask) for
each depart
B: 10 host
C: 34 host
D: 44 host
E : 2 host
How to find the Network ID in Classless Routing?
Example 1 : Given IP : 192.0.30.1/26, find its
Network ID.
n= 32 - CIDR
n = 32 - 26
n= 6
Converting the given IP into Binary system.
11000000.00000000.00011110.00000001
Now, making zero as n=6 from right to
left. we have
11000000.00000000.00011110.00000000
Converting into decimal
192.0.30.0/26
Hence, the Network ID of given IP is 192.0.30.0/26
Example 1: Given IP: 192.0.30.1/22, find its Network ID.
n= 32 - CIDR
n = 32 - 22
n= 10
Converting the given IP into Binary system.
11000000.00000000.00011110.00000001
Now, making zero as n=10 from right to
left. We have
11000000.00000000.00011100.00000000
Converting into decimal
192.0.28.0/22
Hence, the Network ID of given IP is 192.0.28.0/22
Subnet ID 1: 157.110.88.0/27
Host bit= 32-27= 5 bit
Total no of host= 25 =32
157.110.88.0/26 netwrok ID
157.110.88.1/26
157.110.88.2/26
157.110.88.3
157.110.88.4
157.110.88.5
157.110.88.6
157.110.88.7
157.110.88.30/26
:
:
157.110.88.31/26 Broadcast ID
