About CCNA Exam: CCNA exam tests you in the areas of simple LAN/WAN switching, Cisco IOS, and routing. Topics include TCP/IP model of internetworking, configuring, and troubleshooting some of the most widely used Cisco switches and routers. Also, CCNA is foundation exam for CCNP (Cisco Certified Networking Professional). The exam notes is a brief review of important points that help in quick review of key points.
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 pattern||Network address Range||Maximum networks||Maximum hosts/ nodes|
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 loop-back 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. The range of numbers from 126.96.36.199 to 188.8.131.52 are used for multicast packets. This is known as Class D address range.
3. Subnetting is nothing but creating networks within a network. Subnetting allows an organization with a single IP address (Class A /ClassB /ClassC) to have multiple subnetworks, thus allowing several physical networks within the organization.
4. How to maximize the number of subnets for a given number of hosts:
Let us take a network ID of 184.108.40.206, and find the maximum number of possible subnets and the corresponding subnet mask that can accommodate at least 500 hosts. The steps involved are outlined below:
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. 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.
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.
Therefore, the subnet mask required is 255.255.254.0.
5. How to maximize the number of hosts for a given number of subnets:
Determining the subnet mask that allows maximum number of hosts:
Let us consider an IP address 220.127.116.11 with four subnets and maximize the number of host for the given subnets. The steps involved are as below:
The number of subnets required are four. We need to add subnets of all ones and all zeros to this. This is because all zeros and all ones subnets belong to "this subnet" and "all subnets" broadcasts and can not be used. Therefore, the total number of subnets to be reserved is 4+2 = 6.
We want to implement maximum possible Hosts. Therefore, we need to minimize the number of subnets. This minimum number is 6 here. If we reserve 2 bits, it results in only 2^2=4 subnets which is less than 6. Therefore, we have to reserve 3 bits for implementing subnets, resulting in 2^3=8 subnets. This is now optimized for maximum number of Hosts (as we have optimized for minimum number of subnets).
Write the 3 bits available for subnetting in fourth octet in the form 11100000 (Five 0s being Host bits). The decimal equivalent is 2^7+2^6+2^5= 128 + 64 +32 = 224.
Therefore, the subnet mask required is 255.255.255.224.
6. 127.0.0.1 is the local loop back address.
7. In an internetwork, the number of distinct IPs' required are
One each per client computer
One each per server computer
One each per router interface
For example, your network has 2 servers, 26 clients machines, and 2 router interfaces the total number of IP addresses required are 30.
8. Finding the number of Hosts and subnets available for a given subnet mask: For example, let us find the number of hosts and subnets available for an IP 18.104.22.168 with a subnet mask of 7 bits.
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 (2^7-2)=128-2 = 126 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.
Hence, 126 subnets and 510 hosts are available.
9. The directed broadcast should reach all Hosts on the intended network (or subnet, if sub netted). For example, the directed broadcast address for an IP network 22.214.171.124 with default subnet mask is 126.96.36.199. This is arrived by putting all 1s for the host potion of the IP address.
10. Telnet, FTP, and TFTP:
Telnet is used for terminal emulation that runs programs remotely. Telnet uses TCP/IP protocol.
Telnet requires a username and password to access.
FTP (File Transfer Protocol) is a connection oriented protocol. It uses TCP/IP for file transfer. Compare this with TFTP (Trivial File Transfer Protocol) that uses UDP (Connectionless protocol). SNMP uses UDP over IP. Tracert, Ping use ICMP as their base protocol. FTP is used to transfer files.
Both FTP and Telnet are client-server protocols. Note that TCP/IP is a client server oriented protocol.
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