IP Address

IP Address is the address assigned to the network and network equipment that uses the TCP/IP protocol. The IP address consists of 32 bits (biary digits or double numbers) binary numbers which are divided into 4 ockets (bytes) consisting of 8 bits. Each bit represents a decimal number ranging from 0 to 255.

The types of IP addresses consist of:

1. Public IP

Highest public bit range address bit network address
class A 0 0 – 127* 8
class B 10 128 – 191 16
class C 110 192 – 223 24
class D 1110 224 – 239 28

2. Private

This Private IP can be used freely but is not recognized on the global internet network. Because it is usually used on closed networks that are not connected to the internet, such as ATM computer networks.

10.0.0.0 – 10.255.255.255
172.16.0.0 – 172.31.255.255
192.168.0.0 – 192.168.255.255

Conclusion
1.0.0.0 – 126.0.0.0 : Class A.
127.0.0.0 : Loopback network.
128.0.0.0 – 191.255.0.0 : Class B.
192.0.0.0 – 223.255.255.0 : Class C.
224.0.0.0 = 240.0.0.0 : Class E, reserved.
3. IPv6
consists of 16 octets, for example:
A524:72D3:2C80:DD02:0029:EC7A:002B:EA73

Subnetting

A Network Administrator often requires network sharing from an IP Address that has been assigned by the Internet Service Provider (ISP). This is because the supply of IP addresses is currently very limited due to the proliferation of sites on the internet. The way to divide the network is called subneting and the result of subneting is called subnetwork. The steps for subnetting are as follows:

Example 2:
A company gets an IP address from an ISP 160.100.0.0/16, the company has 30 departments in total, and wants all departments to have access to the internet. Determine the network for each department?

Solution ;
1. Determine which class the IP is in? B
2. How many networks are needed?
with the formula 2n > network needed
25 > 30
3. Convert to binary

network-portion host-portion
10100000 01100100 00000000 00000000
11111111 11111111 00000000 00000000

4. Take the host-portion bit according to the network needs, so that

network-portion host-portion
10100000 01100100 _ _ _ _ _ 000 00000000
11111111 11111111 1 1 1 1 1 000 00000000

note the third octet
_ _ _ _ _ 000
1 1 1 1 1 000

Method 1
By combining bits

00001 000 = 8
00010 000 = 16
00011 000 = 24
00100 000 = 32
00101 000 = 40
00110 000 = 48
……………
11111 000 = 248

Method 2
Reduce the subnet mask by 256

11111 000 = 248

256 – 248 = 8 then the subnetwork is a multiple of 8

No. Department of Subnetwork (255.255.248.0)
1 First 160.100.8.0
2 Second 160.100.16.0
3 Third 160.100.24.0
4 Fourth 160.100.32.0
5 Fifth 160.100.40.0
6 Sixth 160.100.48.0
7 Seventh 160.100.56.0
.. ………….
30 Thirty 160.100.248.0
Then

Network Broadcast Range-Hoat
160.100.8.0 160.100.15.255 160.100.8.1 – 160.100.15.254
160.100.16.0 160.100.23.255 160.100.16.1 – 160.100.23.254
160.100.24.0 160.100.31.255 160.100.24.1 – 160.100.31.254
160.100.32.0 160.100.39.255 160.100.31.254 160.100.32.0 160.100.39.255 160.100.32.1 – 160.100.39.254
160.100.40.0 160.100.47.255 160.100.40.1 – 160.100.47.254
160.100.48.0 160.100.55.255 160.100.48.1 – 160.100.55.254
160.100.56.0 160.100.63.255 160.100.56.1 – 160.100.63.254
160.100.64.0 160 100 .71.255 160.100.64.1 – 160.100.71.254
160.100.72.0 160.100.79.255 160.100.72.1 – 160.100.79.254
…… .. ………. ………….

160.100.248.0 160.100.255.255 160.100.248.1 – 160.100.255.254

VLSM (Variable Leg Subnet Mask)

The concept of subneting is indeed a solution in overcoming the number of IP addresses used. However, if you pay attention, there will be many subnets. More detailed explanation in the example:

Example 2:
A company that has 6 departments wants to divide its network, including:
1. Department A = 100 hosts
2. Department B = 57 hosts
3. Department C = 325 hosts
4. Department D = 9 hosts
5. Department E = 500 hosts
6. Department F = 25 hosts

IP Address given from ISP is 160.100.0.0/16

If we use ordinary subneting it will be easy to get but the results of subneting (such as example 1) will be wasted because the results of subneting are too many than the required number of hosts. Then we need VLSM calculations, namely:

1. Sort by required hosts
   1. Department E = 500 hosts
   2. Department C = 325 hosts
   3. Department A = 100 hosts
   4. Department B = 57 hosts
   5. Department F = 25 hosts
   6. Department D = 9 hosts
2. Convert to binary

network-portion host-portion
10100000 01100100 00000000 00000000
11111111 11111111 00000000 00000000
If the subneting is taken from the network then the VLSM is taken from the host

l For 500 hosts
network-portion host-portion
10100000 01100100 00000000 00000000
11111111 11111111 00000000 00000000

For 500 hosts, 9 bits are taken from the host-portion because
2n-2 > number of hosts

The result is 160.100.0.0/23

Network Broadcast Range-Hoat
160.100.0.0/23 160.100.0.255 160.100.0.1 – 160.100.1.254
160.100.2.0/23 160.100.2.255 160.100.2.1 – 160.100.3.254
160.100.4.0/23 160.100.4.255 160.100.4.1 – 160.100.5.254
160.100.6.0/23 160.100.6.255 160.100.6.1 – 160.100.7.254
160.100.8.0/23 160.100.8.255 160.100.8.1 – 160.100.9.254
…….. ………. ………….
160.100.254.0/23 160.100.254.255 160.100.254.1 – 160.100.255.254

l For 325 hosts we can still use a subnet of 500 hosts because it is still in arena 29 and choose an unused subnet.
l For 100 hosts use 28 > 100 and take one of the previously unused subnets.
e.g. 160.100.2.0/24

network-portion host-portion
10100000 01100100 00000010 00000000
11111111 11111111 00000010 00000000

then
Network Broadcast Range-Hoat
160.100.2.0/24 160.100.2.255 160.100.2.1 – 160.100.2.254
160.100.3.0/24 160.100.3.255 160.100.3.1 – 160.100.3.254

l For 57 hosts use 26 >57 and take one of the previously unused subnets.
e.g. 160.100.3.0/24

network-portion host-portion
10100000 01100100 00000010 00000000
11111111 11111111 00000011 00000000

then
Network Broadcast Range-Hoat
160.100.3.0/26 160.100.3.91 160.100.3.1 – 160.100.3.90
160.100.3.64/26 160.100.3.63 160.100.3.65 – 160.100.3.126
160.100.3.128/26 160.100.3.127 160.100.3.129 – 160 100. 3.190
160.100.3.192/26 160.100.3.191 160.100.3.193 – 160.100.3.254

l For 25 hosts use 25 > 25 and take one of the previously unused subnets.
e.g. 160.100.3.192/25

network-portion host-portion
10100000 01100100 00000010 00000000
11111111 11111111 00000011 00000000
then

Network Broadcast Range-Hoat
160.100.3.192/27 160.100.3.223 160.100.3.193 – 160.100.3.222
160.100.3.224/27 160.100.3.255 160.100.3.225 – 160.100.3.254

l For 9 hosts use 24 > 16 and take one of the previously unused subnets.
e.g. 160.100.3.224/25

network-portion host-portion
10100000 01100100 00000010 00000000
11111111 11111111 00000011 00000000

then
Network Broadcast Range-Hoat
160.100.3.224/28 160.100.3.239 160.100.3.225 – 160.100.3.227
160.100.3.240/28 160.100.3.255 160.100.3.241 – 160.100.3.254

SUBNETTING ON IP ADDRESS CLASS B

First, the subnet mask that can be used for subnetting class B is as below. I deliberately separated it into two, the left and right blocks because each has a different technique, especially for the octet that is “played” based on the subnet block. The CIDR /17 to /24 method is exactly the same as the Class C subnetting, only the subnet blocks are inserted directly into the third octet, not like Class C is “played” in the fourth octet. While the CIDR /25 to /30 (multiple) of the subnet block we “play” in the fourth octet, but after the third octet is finished, we move forward (coeunter) from 0, 1, 2, 3, and so on.

Now let’s try two questions for both subnetting techniques for Class B. We start from using a subnetmask with a CIDR of /17 to /24. Example network address 172.16.0.0/18.

Analysis: 172.16.0.0 means class B, with Subnet Mask /18 means 11111111.11111111.110000000000000 (255.255.192.0).

Calculation:

• Number of Subnets = 2x, where x is the number of binaries 1 in the last 2 octets. So the number of subnets is 22 = 4 subnets
• Number of Hosts per Subnet = 2y – 2, where y is the reciprocal of x i.e. the number of 0 binaries in the last 2 octets. So the number of hosts per subnet is 214 – 2 = 16,382 hosts
• Block Subnet = 256 – 192 = 64. The next subnets are 64 + 64 = 128, and 128+64=192. So the complete subnets are 0, 64, 128, 192.
• Valid host and broadcast addresses?

Next we try another one for Class B, especially for those using the CIDR /25 to /30 subnetmask. Example network address 172.16.0.0/25.

Analysis: 172.16.0.0 means class B, with Subnet Mask /25 means 11111111.11111111.11111111.10000000 (255.255.255.128).

Calculation:

• Number of Subnets = 29 = 512 subnets
• Number of Hosts per Subnet = 27 – 2 = 126 hosts
• Subnet block = 256 – 128 = 128. So the complete is (0, 128)
• Valid host and broadcast addresses?

SUBNETTING ON IP ADDRESS CLASS A

If it is solid and understands correctly, we will continue to Class A. The concepts are all the same. The difference is in which OCTET we play the subnet blocks. If Class C is in the 4th (last) octet, class B is in the 3rd and 4th octet (last 2 octet), if Class A is in the 2nd, 3rd and 4th octet (last 3 octet). Then the subnet masks that can be used for subnetting class A are all subnet masks from CIDR /8 to /30.

We try to practice for the network address 10.0.0.0/16.

Analysis: 10.0.0.0 means class A, with Subnet Mask /16 means 111111111.11111111.000000000000000 (255.255.0.0).

Calculation:

• Number of Subnets = 28 = 256 subnets
• Number of Hosts per Subnet = 216 – 2 = 65534 hosts
• Block Subnet = 256 – 255 = 1. So the complete subnet is: 0.1,2,3,4, and so on.
• Valid host and broadcast addresses?

Note: All subnet calculations above assume that IP Subnet-Zeroes (and IP Subnet-Ones) are calculated by default. Todd Lamle’s latest version of the book as well as CCNA after 2005 have accommodated this IP Subnet-Zeroes (and IP Subnet-Ones) problem. CCNA pre-2005 does not include it by default (though in fact we can activate it with the command ip subnet-zeroes), so maybe in some books about CCNA and CNAP test questions, you still find the formula for calculating the number of subnets = 2x – 2

IP Address

IP Address is the address assigned to the network and network equipment that uses the TCP / IP protocol. IP addresses consist of 32 bit binary numbers which can be written as four decimal places separated by periods such as 192.16.10.01 or for example in wxyz format. IP addresses are the most widely used protocols for forwarding (routing) information on the network.

IP addresses have classes as in table 2.4.

Table 2.4. IP address
classes Class Range Network ID Host ID Default Subnet Mask
A 1-126 w xyz 255.0.0.0
B 128-191 wx yz 255.255.0.0
C 192-223 wxy z 255.255.255.0

note: there is still class D that is rarely used, and there is IPV6 that will be used if this IPV4 is not sufficient.

For example, there is an IP 192.168.0.100, so it includes a Class C IP Address

Subnetting

If an owner of a class B IP address, for example, requires more than one network ID, he must apply to Internic to get a new IP address. However, the supply of IP addresses is very limited due to the proliferation of sites on the internet.

To overcome this, a technique emerged to multiply the network ID from an existing network. This is called subnetting, in which a portion of the host ID is sacrificed for use in creating additional network IDs.

For example, in class B, the network ID is 130.200.0.0 with a subnet mask of 255.255.224.0 where the third octet is enclosed by 224. It can be calculated using the formula 256-224=32. then the subnet groups that can be used are multiples of 32, 64, 128, 160, and 192. Thus, the IP address groups that can be used are:

130.200.32.1 to 130.200.63.254
130.200.64.1 to 130.200.95.254
130.200.96.1 to 130.200.127.254
130.200.128.1 to 130.200.159.254
130.200.160.1 to 130.200.191.254
130.200.192.1 to 130.200.223.254

Or it will be easier with a good formulation in determining the subnet and the number of hosts per subnet. The number of subnets = 2n-2, n = the number of hidden bits

Number of hosts per subnet = 2N-2, N = number of bits not hidden

For example, suppose a subnet has a network address of 193.20.32.0 with a subnet mask of 255.255.255.224. So: The
number of subnets is 6, because from the network address 193.20.32.0 by paying attention to the number from the first octet, which is 193, it can be seen that it is in class C. By observing the subnetmask 255.255.255.224 or 11111111.11111111.1111111. 11100000 can be seen that the three bits of the host ID are shrouded, so we get n = 3 and get: number of subnets = 23-2 = 6.

As for the number of hosts per subnet is 30, this is obtained from 5 bits that are not hidden, then N = 5 and will be obtained: the number of hosts per subnet = 25-2 = 30.

The hidden bit is the bit that is represented by the number 1, while the bit that is not hidden is the bit that is represented by the number 0.