IP Address:
IP address Class:
Class A: 0.0.0.0 -------------- 127.255.255.255
Class C:
Network ID:
Broadcast ID:
In case of Network ID, all the host bits are set to one’s (1’s)
Example: 192.168.1.10
192.168.1. XXXX XXXX (X = 1)
192.168.1. 1111 1111
192.168.1.255
So NID of the IP address 192.168.1.10 is 192.168.1.255
Subnet Mask:
Subnet mask
is a 32 bits value which is used to differentiate the host & network
portion of an IP address. In case of
Subnet Mask, Network is represented by 1 & Host is represented by O.
Default Subnet Mask:
Class A: 255.0.0.0
10.0.0.0 – 10.255.255.255 = Class A Private IP Addres.
127.0.0.0 – 127.255.255.255 = Loopback Address
169.254.0.0 – 169.254.255.255 = APIPA Range used Microsoft OS
172.16.0.0 – 172.31.255.255 = Class B Private IP Addres
192.168.0.0 – 192.168.255.255 = Class C Private IP Addres
224.0.0.0 – 239.255.255.255 = Multicast Communication
240.0.0.0 – 255.255.255.255 = Research & Development
255.255.255.255 = Limited Broadcast Address.
Subnetting:
The Process of dividing a large network into equal smaller parts is known as Subnetting. In subnetting some of host bits will be converted to network bits.
Example:
192.168.1.10; divide the given network in such a way where each subnetwork will
have minimum of 30 hosts & also find out the Network ID, Broadcast ID, and
First Valid IP Add & Last Valid IP of the given IP address.
Solution: From the above question, host is given. So we will use the formula (2h-2)≥No of host/network.
Step1: Find out the value of “h”, where “h” represent the Host bits.
Sample Network with Subnetted IP Address on PC:
Example: 17.16.3.10; divide the given network in such a way where each subnetwork will have minimum of 1000 hosts & also find out the Network ID, Broadcast ID, and First Valid IP Add & Last Valid IP of the given IP address.
Solution: From the above question, host is given. So we will use the formula
(2h-2)≥No of host/network.
Step1: Find out the value of “h”, where “h” represent the Host bits.
(2h-2)≥No of host/network
(2h-2)≥1000
(210-2)≥1000
(1024-2) ≥1000
1022≥1000
From the above calculation, h = 10.
Step2: Find out the value of “n”, where “n” represents the Network bits. So given network is 17.16.3.10, which is a class A network. So we will use
n+h=24
n+10=24
n=14
Step3: Find the total number of subnetwork:
(2n-2) ≥ No of sunetwork
(214 – 2) ≥ No of sunetwork
(16384-2) ≥ No of sunetwork
16382≥ No of sunetwork
Total no of subnetwork will be 16384.
Total number of valid subnetwork will be 16382.
Step4: Find out the Total Network Bits (TNB):
TNB = Default Network Bits + n
= 8 + 14
= 22
Step5: Find out the customize Subnet Mask (CSM):
CSM = 8 8 6 0
1111 1111 1111 1111 1111 1100 0000 0000
255 255 252 0
CSM = 255.255.252.0
Step6: Finding the increment:
Increment = 256 – Change value on the CSM.
= 256 – 252
= 4
Step7: Writing the Range of subnet:
17.0.0.0 – - - - - - - 17.0.3.255
17.0.4.0 – - - - - - - 17.0.7.255
----------------------------------
----------------------------------
17.1.0.0 – - - - - - - 17.1.3.255
17.1.4.0 – - - - - - - 17.1.7.255
-----------------------------------
-----------------------------------
17.16.0.0 – - - - - - - 17.16.3.255
17.16.4.0 – - - - - - - 17.16.7.255
17.16.3.10 is the IP address which is belongs to subnet (17.16.0.0---17.16.3.255). So
Network ID = 17.16.2.0
Broadcast ID = 17.16.3.255
First Valid IP Add = 17.16.0.1
Last Valid IP Add = 17.16.3.254
Task to Do:
Perform the Subnetting based on the following requirements & Find out the following: Network ID, Broadcast ID, and First Valid IP Add & Last Valid IP of the given IP address:
Internet
Protocol address is a 32 bits address for a host in a TCP/IP network.
IP address Class:
Class A: 0.0.0.0 -------------- 127.255.255.255
Class B: 128.0.0.0 -------------- 191.255.255.255
Class C: 192.0.0.0 -------------- 223.255.255.255
Class D: 224.0.0.0 -------------- 239.255.255.255
Class E: 240.0.0.0 -------------- 255.255.255.255
Priority Bits:
Class
A: 0
Class
B: 10
Class C: 110
Class D: 1110
Class E: 1111
Network & Host Part:
Class A: N H H H
Class C: 110
Class D: 1110
Class E: 1111
Network & Host Part:
Class A: N H H H
Class B: N N H H
Class C: N N N H
Class C: N N N H
Usage of Class:
Class A:
Class
B: Used on LAN/MAN/WAN
Communication. Class C:
Class D: Used on Multicast Communication.
Class E: Used on Research & Development.
Network & Host Bits:
Class E: Used on Research & Development.
Network & Host Bits:
Bits
|
Class A
|
Class B
|
Class C
|
Total Network
Bits
|
8
|
16
|
24
|
Total Host Bits
|
24
|
16
|
8
|
Network ID:
In case of Network ID, all the
host bits are set to zero’s (0’s).
Example: 192.168.1.10
192.168.1. XXXX XXXX (X = 0)
192.168.1. 0000 0000
192.168.1.0
So NID of the IP address 192.168.1.10 is 192.168.1.0
192.168.1. XXXX XXXX (X = 0)
192.168.1. 0000 0000
192.168.1.0
So NID of the IP address 192.168.1.10 is 192.168.1.0
Broadcast ID:
In case of Network ID, all the host bits are set to one’s (1’s)
Example: 192.168.1.10
192.168.1. XXXX XXXX (X = 1)
192.168.1. 1111 1111
192.168.1.255
So NID of the IP address 192.168.1.10 is 192.168.1.255
Subnet Mask:
Default Subnet Mask:
Class A: 255.0.0.0
Class B: 255.255.0.0
Class C: 255.255.255.0
Private IP Address:
Class C: 255.255.255.0
Private IP Address:
Class A: 10.0.0.0 – 10.255.255.255
Class B: 172.16.0.0 – 172.31.255.255
Class C: 192.168.0.0 – 192.168.255.255
Reserve IP Address:
0.0.0.0 - 0.255.255.255
= Self Identification.Class B: 172.16.0.0 – 172.31.255.255
Class C: 192.168.0.0 – 192.168.255.255
Reserve IP Address:
10.0.0.0 – 10.255.255.255 = Class A Private IP Addres.
127.0.0.0 – 127.255.255.255 = Loopback Address
169.254.0.0 – 169.254.255.255 = APIPA Range used Microsoft OS
172.16.0.0 – 172.31.255.255 = Class B Private IP Addres
192.168.0.0 – 192.168.255.255 = Class C Private IP Addres
224.0.0.0 – 239.255.255.255 = Multicast Communication
240.0.0.0 – 255.255.255.255 = Research & Development
255.255.255.255 = Limited Broadcast Address.
Subnetting:
The Process of dividing a large network into equal smaller parts is known as Subnetting. In subnetting some of host bits will be converted to network bits.
- Reduce the wastage of IP address.
- Allow us to manage the network efficiently.
- Reduce the network threats.
- (2h-2)≥No of host/network.
- (2n-2) ≥ No of sunetwork.
Solution: From the above question, host is given. So we will use the formula (2h-2)≥No of host/network.
Step1: Find out the value of “h”, where “h” represent the Host bits.
(2h-2)≥No
of host/network
(2h-2)≥30
(25-2)≥30
(32-2) ≥30
30≥30
From the above calculation, h = 5.
Note: As we know, in subnetting, some of the host bits will be converted to network bits. So we can say that -
In class C: n+h=8
(2h-2)≥30
(25-2)≥30
(32-2) ≥30
30≥30
From the above calculation, h = 5.
Note: As we know, in subnetting, some of the host bits will be converted to network bits. So we can say that -
In class C: n+h=8
In class B: n+h=16
In class A: n+h=24
Step2: Find out the value of “n”, where “n” represents the Network bits. So given network is 192.168.1.10, which is a class C network. So we will use
n+h=8
n+5=8
n=3
In class A: n+h=24
Step2: Find out the value of “n”, where “n” represents the Network bits. So given network is 192.168.1.10, which is a class C network. So we will use
n+h=8
n+5=8
n=3
Step3: Find the
total number of subnetwork:
(2n-2) ≥ No of sunetwork
(23 – 2) ≥ No of sunetwork
(8-2) ≥ No of sunetwork
6≥ No of sunetwork
Total no of subnetwork will be 8.
Total number of valid subnetwork will be 6.
Step4: Find out the Total Network Bits (TNB):
TNB = Default Network Bits + n
= 24 + 3
= 27
Step5: Find out the customize Subnet Mask (CSM):
CSM = 8 8 8 3
1111 1111 1111 1111 1111 1111 111 00000
255 255 255 224
CSM = 255.255.255.224
Point to Remember:
1 2 3 4 5 6 7 8
128 192 224 240 248 252 254 255
Step6: Finding the increment:
Increment = 256 – Change value on the CSM.
= 256 – 224
= 32
Step7: Writing the Range of subnet:
(2n-2) ≥ No of sunetwork
(23 – 2) ≥ No of sunetwork
(8-2) ≥ No of sunetwork
6≥ No of sunetwork
Total no of subnetwork will be 8.
Total number of valid subnetwork will be 6.
Step4: Find out the Total Network Bits (TNB):
TNB = Default Network Bits + n
= 24 + 3
= 27
Step5: Find out the customize Subnet Mask (CSM):
CSM = 8 8 8 3
1111 1111 1111 1111 1111 1111 111 00000
255 255 255 224
CSM = 255.255.255.224
Point to Remember:
1 2 3 4 5 6 7 8
128 192 224 240 248 252 254 255
Step6: Finding the increment:
Increment = 256 – Change value on the CSM.
= 256 – 224
= 32
Step7: Writing the Range of subnet:
192.168.1.0
– - - - - - - 192.168.1.31
192.168.1.32 – - - - - - - 192.168.1.63
192.168.1.64 – - - - - - - 192.168.1.95
192.168.1.96 – - - - - - - 192.168.1.127
192.168.1.128 – - - - - - - 192.168.1.159
192.168.1.160 – - - - - - - 192.168.1.191
192.168.1.192 – - - - - - - 192.168.1.223
192.168.1.224 – - - - - - - 192.168.1.255
192.168.1.10 is the IP address which is belongs to Subnet (192.168.1.0 --- 192.168.1.31). So
Network ID = 192.168.1.0
Broadcast ID = 192.168.1.31
First Valid IP address = 192.168.1.1
Last Valid IP Address = 192.168.1.30
Note: 192.168.1.32 – - - - - - - 192.168.1.63
192.168.1.64 – - - - - - - 192.168.1.95
192.168.1.96 – - - - - - - 192.168.1.127
192.168.1.128 – - - - - - - 192.168.1.159
192.168.1.160 – - - - - - - 192.168.1.191
192.168.1.192 – - - - - - - 192.168.1.223
192.168.1.224 – - - - - - - 192.168.1.255
192.168.1.10 is the IP address which is belongs to Subnet (192.168.1.0 --- 192.168.1.31). So
Network ID = 192.168.1.0
Broadcast ID = 192.168.1.31
First Valid IP address = 192.168.1.1
Last Valid IP Address = 192.168.1.30
(192.168.1.0 – - - - - - - 192.168.1.31) is a
subnetwork which is known as Subnet Zero. Cisco IOS prior to 12.0, the IP
Subnet Zero is restricted (i.e. Disable). IOS 12.0 to all new IOS, the IP
Subnet Zero is enabling by default.
Sample Network with Subnetted IP Address on PC:
Task to Do:
Perform the Subnetting based on the following requirements & Find out the following: Network ID, Broadcast ID, and First Valid IP Add & Last Valid IP of the given IP address:
Perform the Subnetting based on the following requirements & Find out the following: Network ID, Broadcast ID, and First Valid IP Add & Last Valid IP of the given IP address:
- 200.200.200.200; Host = 75
- 221.195.156.156; Host = 45
- 193.151.200.200; Host = 125
- 205.205.103.31; Host = 20
- 219.215.215.213; Host = 32
Subnetting for Class B:
Example: 172.16.3.10; divide the given network in such a way where each subnetwork will have minimum of 300 hosts & also find out the Network ID, Broadcast ID, and First Valid IP Add & Last Valid IP of the given IP address.
Solution: From the above question, host is given. So we will use the formula
(2h-2)≥No of host/network.
Step1: Find out the value of “h”, where “h” represent the Host bits.
(2h-2)≥No of host/network
(2h-2)≥300
(29-2)≥300
(512-2) ≥300
510≥300
From the above calculation, h = 9.
Step2: Find out the value of “n”, where “n” represents the Network bits
So given network is 172.16.3.10, which is a class B network. So we will use
n+h=16
n+9=16
n=7
Step3: Find the total number of subnetwork:
(2n-2) ≥ No of sunetwork
(27 – 2) ≥ No of sunetwork
(128-2) ≥ No of sunetwork
126≥ No of sunetwork
Total no of subnetwork will be 128.
Total number of valid subnetwork will be 126
Step4: Find out the Total Network Bits (TNB):
TNB = Default Network Bits + n
= 16 + 7
= 23
Step5: Find out the customize Subnet Mask (CSM):
CSM = 8 8 7 0
1111 1111 1111 1111 1111 1110 0000 0000
255 255 254 0
CSM = 255.255.254.0
Step6: Finding the increment:
Increment = 256 – Change value on the CSM.
= 256 – 254
= 2
Step7: Writing the Range of subnet:
172.16.0.0 – - - - - - - 172.16.1.255
172.16.2.0 – - - - - - - 172.16.3.255
172.16.4.0 – - - - - - - 172.16.5.255
-----------------------------------------
-----------------------------------------
172.16.254.0 – - - - - - - 172.16.255.255
172.16.3.10 is the IP address which is belongs to subnet (172.16.2.0 --- 172.16.3.255). So
Network ID = 172.16.2.0
Broadcast ID = 172.16.3.255
First Valid IP Add = 172.16.2.1
Last Valid IP Add = 172.16.3.254
Example: 172.16.3.10; divide the given network in such a way where each subnetwork will have minimum of 300 hosts & also find out the Network ID, Broadcast ID, and First Valid IP Add & Last Valid IP of the given IP address.
Solution: From the above question, host is given. So we will use the formula
(2h-2)≥No of host/network.
Step1: Find out the value of “h”, where “h” represent the Host bits.
(2h-2)≥No of host/network
(2h-2)≥300
(29-2)≥300
(512-2) ≥300
510≥300
From the above calculation, h = 9.
Step2: Find out the value of “n”, where “n” represents the Network bits
So given network is 172.16.3.10, which is a class B network. So we will use
n+h=16
n+9=16
n=7
Step3: Find the total number of subnetwork:
(2n-2) ≥ No of sunetwork
(27 – 2) ≥ No of sunetwork
(128-2) ≥ No of sunetwork
126≥ No of sunetwork
Total no of subnetwork will be 128.
Total number of valid subnetwork will be 126
Step4: Find out the Total Network Bits (TNB):
TNB = Default Network Bits + n
= 16 + 7
= 23
Step5: Find out the customize Subnet Mask (CSM):
CSM = 8 8 7 0
1111 1111 1111 1111 1111 1110 0000 0000
255 255 254 0
CSM = 255.255.254.0
Step6: Finding the increment:
Increment = 256 – Change value on the CSM.
= 256 – 254
= 2
Step7: Writing the Range of subnet:
172.16.0.0 – - - - - - - 172.16.1.255
172.16.2.0 – - - - - - - 172.16.3.255
172.16.4.0 – - - - - - - 172.16.5.255
-----------------------------------------
-----------------------------------------
172.16.254.0 – - - - - - - 172.16.255.255
172.16.3.10 is the IP address which is belongs to subnet (172.16.2.0 --- 172.16.3.255). So
Network ID = 172.16.2.0
Broadcast ID = 172.16.3.255
First Valid IP Add = 172.16.2.1
Last Valid IP Add = 172.16.3.254
Task to Do:
Perform the Subnetting based on the following requirements & Find out the following: Network ID, Broadcast ID, and First Valid IP Add & Last Valid IP of the given IP address:
Perform the Subnetting based on the following requirements & Find out the following: Network ID, Broadcast ID, and First Valid IP Add & Last Valid IP of the given IP address:
- 151.163.20.3; Host = 500
- 191.164.12.134; Host = 750
- 142.158.26.30; Host = 128
- 158.156.29.32; Host = 817
- 183.169.176.168; Host = 150
Example: 17.16.3.10; divide the given network in such a way where each subnetwork will have minimum of 1000 hosts & also find out the Network ID, Broadcast ID, and First Valid IP Add & Last Valid IP of the given IP address.
Solution: From the above question, host is given. So we will use the formula
(2h-2)≥No of host/network.
Step1: Find out the value of “h”, where “h” represent the Host bits.
(2h-2)≥No of host/network
(2h-2)≥1000
(210-2)≥1000
(1024-2) ≥1000
1022≥1000
From the above calculation, h = 10.
Step2: Find out the value of “n”, where “n” represents the Network bits. So given network is 17.16.3.10, which is a class A network. So we will use
n+h=24
n+10=24
n=14
Step3: Find the total number of subnetwork:
(2n-2) ≥ No of sunetwork
(214 – 2) ≥ No of sunetwork
(16384-2) ≥ No of sunetwork
16382≥ No of sunetwork
Total no of subnetwork will be 16384.
Total number of valid subnetwork will be 16382.
Step4: Find out the Total Network Bits (TNB):
TNB = Default Network Bits + n
= 8 + 14
= 22
Step5: Find out the customize Subnet Mask (CSM):
CSM = 8 8 6 0
1111 1111 1111 1111 1111 1100 0000 0000
255 255 252 0
CSM = 255.255.252.0
Step6: Finding the increment:
Increment = 256 – Change value on the CSM.
= 256 – 252
= 4
Step7: Writing the Range of subnet:
17.0.0.0 – - - - - - - 17.0.3.255
17.0.4.0 – - - - - - - 17.0.7.255
----------------------------------
----------------------------------
17.1.0.0 – - - - - - - 17.1.3.255
17.1.4.0 – - - - - - - 17.1.7.255
-----------------------------------
-----------------------------------
17.16.0.0 – - - - - - - 17.16.3.255
17.16.4.0 – - - - - - - 17.16.7.255
17.16.3.10 is the IP address which is belongs to subnet (17.16.0.0---17.16.3.255). So
Network ID = 17.16.2.0
Broadcast ID = 17.16.3.255
First Valid IP Add = 17.16.0.1
Last Valid IP Add = 17.16.3.254
Task to Do:
Perform the Subnetting based on the following requirements & Find out the following: Network ID, Broadcast ID, and First Valid IP Add & Last Valid IP of the given IP address:
- 10.1.15.10; Host = 600
- 115.100.15.100; Host = 1000
- 111.125.136.124; Host = 455
- 56.125.24.29; Host = 799
- 76.75.82.86; Host = 1500
CIDR (Classless Inter
Domain Routing):
Classless Inter-Domain Routing (CIDR) is a method for allocating IP addresses and routing Internet Protocol packets.
CIDR was developed in the 1990s as a standard scheme for routing network traffic across the Internet. Before CIDR technology was developed, Internet routers managed network traffic based on the class of IP addresses. In this system, the value of an IP address determines its Subnetwork for the purposes of routing.
CIDR Notation:
Classless Inter-Domain Routing (CIDR) is a method for allocating IP addresses and routing Internet Protocol packets.
CIDR was developed in the 1990s as a standard scheme for routing network traffic across the Internet. Before CIDR technology was developed, Internet routers managed network traffic based on the class of IP addresses. In this system, the value of an IP address determines its Subnetwork for the purposes of routing.
CIDR Notation:
CIDR specifies an IP address
range using a combination of an IP address and its associated network mask. IDR
notation uses the following format –
XXX.XXX.XXX.XXX/n
Where n is the number of (leftmost) '1' bits in the mask. For example, 192.168.12.0/23.
Example: 192.168.1.100/28; Find out the NID, BID, FVIP, LVIP, Total No of Subnets & Total no of host per subnet.
Solution: Here in this question Total Network Bits is given, which is 28.
Step1: Finding the Customized Subnet Mask from the Total Network Bits (TNB):
Where n is the number of (leftmost) '1' bits in the mask. For example, 192.168.12.0/23.
Example: 192.168.1.100/28; Find out the NID, BID, FVIP, LVIP, Total No of Subnets & Total no of host per subnet.
Solution: Here in this question Total Network Bits is given, which is 28.
Step1: Finding the Customized Subnet Mask from the Total Network Bits (TNB):
8 8 8 4
255 255 255 240
Step2: Finding the value of “n”:
TNB = Default Network Bits + n
28 = 24 + n
n = 4
Step3: Find the value of ‘h”:
So given network is 192.168.1.100, which is a class C network. So we will use
n+h=8
h+4=8
h=4
Step4: Finding the increment:
Increment = 256 – Change value on the CSM.
= 256 – 240
= 16
255 255 255 240
Step2: Finding the value of “n”:
TNB = Default Network Bits + n
28 = 24 + n
n = 4
Step3: Find the value of ‘h”:
So given network is 192.168.1.100, which is a class C network. So we will use
n+h=8
h+4=8
h=4
Step4: Finding the increment:
Increment = 256 – Change value on the CSM.
= 256 – 240
= 16
Step5: Writing the
Range:
192.168.1.0 – - - - - - - 192.168.1.15
192.168.1.16 – - - - - - - 192.168.1.31
192.168.1.32 – - - - - - - 192.168.1.47
192.168.1.48 – - - - - - - 192.168.1.63
192.168.1.64 – - - - - - - 192.168.1.79
192.168.1.80 – - - - - - - 192.168.1.95
192.168.1.96 – - - - - - - 192.168.1.111
---------------------------------------------
---------------------------------------------
192.168.1.240 – - - - - - - 192.168.1.255
192.168.1.100 is the IP address which is belongs to subnet (192.168.1.96 – 192.168.1.111). So
Network ID = 192.168.1.96
Broadcast ID = 192.168.1.111
First Valid IP Add = 192.168.1.97
Last Valid IP Add = 192.168.1.110
Finding the Total number of Subnets:
(2n-2) ≥ No of sunetwork
(24 – 2) ≥ No of sunetwork
(16-2) ≥ No of sunetwork
14≥ No of sunetwork
Total no of subnetwork will be 16.
Total number of valid subnetwork will be 14.
Finding the Total number of host per subnets:
(2h-2) ≥ No of host/sunetwork
(24 – 2) ≥ No of host/sunetwork
(16-2) ≥ No of host/sunetwork
14≥ No of host/sunetwork
Total number of Host/Subnetwork is 14
192.168.1.0 – - - - - - - 192.168.1.15
192.168.1.16 – - - - - - - 192.168.1.31
192.168.1.32 – - - - - - - 192.168.1.47
192.168.1.48 – - - - - - - 192.168.1.63
192.168.1.64 – - - - - - - 192.168.1.79
192.168.1.80 – - - - - - - 192.168.1.95
192.168.1.96 – - - - - - - 192.168.1.111
---------------------------------------------
---------------------------------------------
192.168.1.240 – - - - - - - 192.168.1.255
192.168.1.100 is the IP address which is belongs to subnet (192.168.1.96 – 192.168.1.111). So
Network ID = 192.168.1.96
Broadcast ID = 192.168.1.111
First Valid IP Add = 192.168.1.97
Last Valid IP Add = 192.168.1.110
Finding the Total number of Subnets:
(2n-2) ≥ No of sunetwork
(24 – 2) ≥ No of sunetwork
(16-2) ≥ No of sunetwork
14≥ No of sunetwork
Total no of subnetwork will be 16.
Total number of valid subnetwork will be 14.
Finding the Total number of host per subnets:
(2h-2) ≥ No of host/sunetwork
(24 – 2) ≥ No of host/sunetwork
(16-2) ≥ No of host/sunetwork
14≥ No of host/sunetwork
Total number of Host/Subnetwork is 14
Task to do:
Find the NID, BID, FVIP, LVIP, and Total No of Subnets & Total no of host per subnet based on the following:
Find the NID, BID, FVIP, LVIP, and Total No of Subnets & Total no of host per subnet based on the following:
- 195.168.1.15/25
- 174.16.100.100/24
- 115.10.10.152/16
- 142.158.158.120/23
- 182.85.168.200/22
- 199.199.199.100/30
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