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CCENT™ (Cisco Certified Entry Network Technician) Certification Exam Notes


C. IP Addressing and Subnetting

1. IP addresses are written using decimal numbers separated by decimal points. This is called dotted decimal notation of expressing IP addresses.
The different classes of IP addresses is as below:
Class Format Leading Bit Network address Maximum Maximum hosts pattern Range networks / nodes
A N.H.H.H 0 0 - 126 127 16,777,214
B N.N.H.H 10 128 - 191 16,384 65,534
C N.N.N.H 110 192 - 223 2,097,152 254
Network address of all zeros means "This network or segment".
Network address of all 1s means "all networks", same as hexadecimal of all Fs.
Network number 127 is reserved for loopback tests.
Host (Node) address of all zeros mean "This Host (Node)".
Host (Node) address of all 1s mean "all Hosts (Nodes)" on the specified network.

2. Class B network has the form N.N.H.H, the default subnet mask is 16 bits long. There is additional subnet mask of 7 bits long.
7 bits of subnet mask corresponds to (2^7)=128 subnets.
9 bits (16-7) of host addresses corresponds to (2^9-2)=512-2 = 510 hosts.
Some times, the subnet mask is specified with the bits available in the default subnet mask. In this case the bits available in default subnet mask is 16. Therefore, total number of bits available in the subnet mask are 16+7=23. If you are given a subnet mask of 23 bits long for a class B address, it is understood that it contains the bits from the default subnet mask as well.
Class C IP address has the form N.N.N.H. You have 8 bits available for host portion, of which 4 bits are used for subnetting. This means, you have 2^4 = 16 subnets. Similarly, you have 4 bit available for host Ids. This equals 2^4 = 16 host Ids. Out of this, host Ids of all 1s' and all 0s' are reserved. This leaves us with 14 host Ids per subnet.

3. The address range and the corresponding binary values for the first octet of IP classes are as given below:
Class A: 0-127 binary: 00000000-01111111
Class B: 128-191 binary: 10000000 – 10111111
Class C: 192-223 binary: 11000000 – 11011111
Class D: 224-239 binary: 11100000 – 11101111
Note that the IP address 172.54.17.6 belongs to class B, and hence the applicable binary address range is 10000000 – 10111111.

4. The following are true about the IP address 132.145.12.5/25
a. Subnet mask: 255.255.255.128
b. Subnet address: 132.145.12.0
c. Broadcast address: 132.145.12.127
d. Lowest valid host address in the given subnet: 132.145.12.1 (A host address of all zeros cannot be used)
e. Highest valid host address in the given subnet: 132.145.12.126 (A host address of all ones cannot be used)
f. Valid host address range: 192.145.12.1 - 192.145.12.126

5. The number of distinct IP’s required in an IP network are:
1. One each per client computer
2. One each per server computer
3. One each per router interface.

6. The range of numbers from 224.0.0.0 to 239.255.255.255 are used for multicast packets. This is known as Class D address range.

7. Subnetting:
Subnetting is nothing but creating networks within a network. Subnetting allows an organization with a single IP address (Class A /Class B /Class C) to have multiple subnetworks, thus allowing several physical networks with in the organization.
The subnet mask is computed as below:
1. Find the Class of the IP address, in this case it is a class B network. Class B network has the form N.N.H.H. Therefore, we have a total of 16 bits (two octets) for assigning to internal networks and hosts. The minimum number of host addresses required is 500 (see the question). The last octet corresponds to 2^8 = 256 hosts which is still less than 500 Hosts.. Therefore, you have to borrow one more bit from the third octet to make it 256*2 = 512 Hosts. This leaves 7 bits in the third octet for assigning subnet addresses. This is equal to 2^7=128 subnets.
2. Write the 7 bits available for subnetting in third octet in the form 11111110 (last bit being the Host bit). The decimal equivalent of the first 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.
3. Now the subnet mask required is 255.255.254.0.

8. Consider a host IP 156.16.3.47: The available hosts with a 11 bit subnet mask, may be computed as below: 
11 bit subnet mask is equivalent to 255.255.255.224; It will accommodate 2,046 subnets, each with 30 hosts. The host address of 156.16.3.47 is in the 32 subnet range. The valid hosts for subnet 32 are 33-62. The broadcast address is 63.

9. To find the subnet that a host 56.16.3.52 with 12-bit subnet mask belongs to:
A 12-bit subnet mask gives us 255.255.255.240; 4094 subnets, each with 14 hosts. (Host addresses of all zeros and all 1s can't be assigned). The 156.16.3.52 is in the 48 subnet range. The valid range is through 49 - 62. 63 is a broadcast address.
Here is how you get the subnet range:
1. Find the subnet mask. In this case, default subnet mask for Class B address is 255.255.0.0. There are additional; 12 bits in the subnet mask. Now the subnet mask looks like:
11111111.11111111.11111111.11110000. This is equal to 255.255.255.240.0.
2. Now, deduct the lowest value octet (Which is non zero), from 256. Here, (256-240)=16.
This is the value that the subnets are incremented. Therefore, you will have hosts with values from:.
156.16.3.1 to 156.16.3.14 (All 0s and all 1s host addresses can not be used)
156.16.3.17 to 156.16.3.30
156.16.3.33 to 156.16.3.46
156.16.3.49 to 156.16.3.62 and so on.
It is important to know that subnets are incrementing by a factor of 16.

10. The subnet Id of a host with IP address 165.212.18.5/20 is 165.212.16.0
The available host address range is 165.212.16.1 - 165.212.31.254
Broadcast address: 165.212.31.255

11. Variable-Length Subnet Masking:
Variable-length subnet masking (VLSM) means implementing more than one mask on the same major class of a network. It allows for a more efficient use of IP address space both in terms of hosts and subnets, and can be essential on a network that does not have an abundance of IP address space.
The following routing protocols support VLSM:
1. RIP v2
2. EIGRP
3. IS-IS
4. OSPF
5. BGP

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