# IPv4 Addressing-Part-III

## Examples on subnetting

**Example 1: **You have a system address of 196.202.56.0 with five Hosts on each subnet. You need to take into consideration the most extreme number of subnets. What is the subnet cover you have to apply?

**Clarification :**

I. The number of Hosts required is five. We have to include Hosts of every one of the ones and each of the zeros to this. This is on account of each of the zeros and every one of the one’s subnets have a place with “this Host” and “all Hosts” communicates and cannot be utilized. Hence, the aggregate number of Hosts delivers to be held is 5+2 = 7.

II. We need to execute the greatest conceivable subnets. In this way, we have to limit the number of Hosts. This base number is 7 here. In the event that we save 2 bits, it results in just 2^2=4 Hosts which is under 7. Thusly, we need to hold 3 bits for executing Hosts, bringing about 2^3=8 Host addresses. This is presently advanced for the most extreme number of Subnets (as we have improved for least number of Hosts). This additionally abandons us with 5 bits in the fourth Octet for executing subnets.

III. Compose the 5 bits accessible for subnetting in the fourth octet in the shape 11111000 (Five 1s being subnet bits). The decimal equal is 2^7+2^6+2^5+2^4+2^3= 128 + 64 +32 + 16 + 8 = 248.

IV. Presently the subnet veil required is 255.255.255.248.

**Example 2:** You have an IP address of 156.16.3.47 with an 11-bit subnet cover. What are your legitimate hosts?

156.16.3.47 is a class B IP address and is ordinarily utilized with a subnet cover of sort 255.255.x.x

11-bit subnet cover is identical to 255.224.0.0=11111111.11100000.00000000.00000000

Netmask – 255.224.0.0

Special case cover – 0.31.255.255

System address – 156.0.0.0/11

Communicate: 156.31.255.255

Host Min: 156.0.0.1

Host max: 156.31.255.254

Host/Network-2097150

**Example 3:** What is the subnet Id of the IP address 165.212.18.5/20?

The subnet Id of the IP address 165.212.18.5/20 is 165.212.16.0 The accessible host address extend is 165.212.16.1 – 165.212.31.254 Broadcast address: 165.212.31.255

**Example 4:** You have a system ID of 168.8.0.0. You have to isolate it into numerous subnets with something like 500 hosts for every each subnet. Which subnet veil would it be a good idea for you to utilize with the goal that you will have the capacity to partition the system into the most extreme number of subnets?

I. Discover the Class of the IP address, for this situation it is a class B organize. Class B arrange has the frame N.N.H.H. Along these lines, we have an aggregate of 16 bits (two octets) for allotting to interior systems and hosts. The base number of host tends to required is 500 (see the inquiry). The last octet relates to 2^8 = 256 hosts which are still under 500 Hosts. Accordingly, you need to obtain one more piece from the third octet to make it 256*2 = 512 Hosts. This leaves 7 bits in the third octet for allocating subnet addresses. This is equivalent to 2^7=128 subnets.

II. Compose the 7 bits accessible for subnetting in the third octet in the frame 11111110 (the last piece being the Host bit). What might as well be called the initial seven bits is 2^7+2^6+2^5+2^4+2^3+2^2+2^1= 128 + 64 +32 + 16 + 8 + 4 + 2 = 254.

III. Presently the subnet veil required is 255.255.254.0.

**Example 5:** You have an IP address 156.233.42.56 with a subnet veil of 7 bits. What number of hosts and subnets are conceivable?

Class B arrange has the frame N.N.H.H, the default subnet cover is 16 bits in length. There is extra subnet veil of 7 bits in length.

7 bits of subnet cover compares to (2^7)=128 subnets.

A few times, the subnet cover is determined with the bits accessible in the default subnet veil. For this situation, the bits accessible in default subnet veil is 16. Along these lines, add up to the number of bits accessible in the subnet veil are 16+7=23. On the off chance that you are given a subnet cover of 23 bits yearn for a class B address, it is comprehended that it contains the bits from the default subnet veil also.

9 bits (16-7) of host delivers compares to (2^9-2)=512-2 = 510 hosts.

Note that host IPs relating to each of the zeros and every one of the ones can’t be utilized.

**Example 6:** You have a system address of 196.202.56.0 with four subnets. You need to take into account the most extreme number of Hosts. What is the subnet veil you have to apply?

We need to actualize the greatest conceivable Hosts. In this manner, we have to limit the number of subnets. This base number is 4 here. In the event that we hold 2 bits, it results in 2^2=4 subnets. This is currently improved for the most extreme number of Hosts (as we have upgraded for least number of subnets).

Compose the 2 bits accessible for subnetting in the fourth octet in the frame 11000000 (Six 0s being Host bits). The decimal identical is 2^7+2^6= 128 + 64 = 192. , Now the subnet cover required is 255.255.255.192

**Example 7:** You have a system address of 196.202.56.0 with five Hosts on each subnet. You need to take into consideration the most extreme number of subnets. What is the subnet cover you have to apply?

I. The number of hosts required is five. We have to include Hosts of each of the ones and every one of the zeros to this. This is on the grounds that every one of the zeros and each of the one’s subnets has a place with “this Host” and “all Hosts” communicates and cannot be utilized. Along these lines, the aggregate number of Hosts delivers to be saved is 5+2 = 7.

II. We need to actualize the most extreme conceivable subnets. Along these lines, we have to limit the number of Hosts. This base number is 7 here. On the off chance that we save 2 bits, it results in just 2^2=4 Hosts which is under 7. In this manner, we need to save 3 bits for executing Hosts, bringing about 2^3=8 Host addresses. This is currently upgraded for the greatest number of Subnets (as we have streamlined for least number of Hosts). This additionally abandons us with 5 bits in the fourth Octet for executing subnets.

III. Compose the 5 bits accessible for subnetting in fourth octet in the frame 11111000 (Five 1s being subnet bits). The decimal equal is 2^7+2^6+2^5+2^4+2^3= 128 + 64 +32 + 16 + 8 = 248.

IV. Presently the subnet cover required is 255.255.255.248.

**Example 8:** You have a switch with two ports, each port with an unmistakable IP address. The two ports are associated with two particular subnets (sections). The main subnet has 12 customers and a server. The second subnet has 14 customers and a server. What is the aggregate number of IP tends to required?

The number of particular IPs required is

1) One each per customer PC

2) One each per server PC

3) One each per switch interface.

There are 2 servers + 26 customers + 2 switch interfaces. Accordingly, the aggregate number of IP tends to required are 30.