VIEWS: 18 PAGES: 7 POSTED ON: 11/24/2011
how to subnet: by John Moore CCNA The key thing with subnetting is using the shortcut methods. Also practice, practice, practice! Try subnettingquestions.com once you are able to answer the questions using the shortcuts it comes to you a lot easier. Practice converting to binary to decimal and over and over. Shortcuts: To find the range or increment number use 1 or both of the following: If you have a subnet mask of 255.255.255.192 convert the "interesting octet" this case 192 to binary 1100 0000 The last 1 in binary from LEFT to RIGHT is the increment, which is 64. Or the second shortcut is subtracting 256 from the interesting octet: 256 - 192 = 64 Now that you have the range you can start from 0 then increment by the "increment" number. 0 64 128 192 255 (broadcast for 192 subnet) Note: "Never go above the 255" Finding the range is easy once you have the increments. Start with the subnet number and add 1 0 1 Then go to the next subnet number and subtract 2 for your last usable: 62 63 64 Your broadcast is 63, and your range is 1-62 usable. If your IP's on your interfaces fall within this range you know that it is a usable IP, otherwise it is a broadcast or a subnet id. Sometimes with IP's in the high number range it is easier to convert the "interesting octet" to binary and do the AND operation against its mask. This way you find the subnet ID faster if you are working with several subnets. For example You have an ip of 192.168.200.2 mask of 255.255.252.0 If you did the shortcut you would come up with an increment of 4 in the third octet. Note always be mindful of what octet you are working in. 0 4 8 12 etc... but instead of figuring out all the increments it's easier (imo) to convert to binary and do the AND equation. We are working in the interesting octet, 3rd octet. 200 = 1100 1000 Mask = 1111 1100 AND = 1100 1000 1111 1100 1100 1000 subnet id = 192.168.200.0 We know our increment of 4 First usable 192.168.200.1 Last usable = 192.168.203.254 Broadcast = 192.168.203.255 Next subnet 192.168.204.0 ip-subnet zero Routers today have this turned on by default but if it is not turned on the question will reflect this on the exam. As a rule this is turned on by default unless otherwise stated in the question. What does it mean if it is not turned on? That the 0 subnet and the last broadcast subnet cannot be used. For example, 192.168.200.0 and 192.168.200.255 Our equation for the amount of subnets equals 2^S-2=Subnets If it is turned on then it is 2^S=Subnets S=amount of bits that are turned 1 or "on". VLSM: Variable Length Subnet Mask This is used to subnet into multiple smaller networks to allow for efficient use of IP addresses, increments will be different depending on the number of hosts needed per network. For example, you are assigned network 192.168.1.0/24 but you need a certain amount of hosts with different networks. Example: Given the above assigned network provide sufficient addresses for the proposed scenario without waisting IP's. 2 leased lines connecting 3 routers. Router A with direct link to Router B and Router C (router's b and c do not connect to each other) network with 10 hosts network with 25 hosts network with 50 hosts With the above we know that we have 2 point-to-point links between Router A and Router B and Router A and Router C. We start with network 192.168.1.0/24 We need 2 ip addresses for RouterA and RouterB We need to identify how many bits in this address we need: our mask is 255.255.255.0 for 2 addresses we need 2 bits 2^2-2 = 2 or in binary 0000 00 00 last 2 bits are needed for the hosts Now we can turn the rest of the bits into 1's to give us our mask (last octet shown in binary) 1111 1100 192.168.1.0/30 255.255.255.252 What are our usable IP's? Well let's look at our "increment". 256-252 = 4 Network starts with 0 and our next network is 4 192.168.1.0 = subnet id or Network 192.168.1.1 = first usable address 192.168.1.2 = last usable address 192.168.1.3 = broadcast (not usable) 192.168.1.4 = The NEXT subnet id or network So routera would have IP of 192.168.1.1 and routerb would have IP of 192.168.1.2 both with mask of 255.255.255.252 Now lets work on the point-to-point link from routera to routerc We already have our next network of 192.168.1.4 from above lets increment again by 4 192.168.1.4 = second network 192.168.1.5 = first usable address 192.168.1.6 = last usable address 192.168.1.7 = broadcast (not usable) 192.168.1.8 = The NEXT subnet ID or network Router A other interface IP will be 192.168.1.5 routerc point-to-point interface will be 192.168.1.6 both with mask of 255.255.255.252 Now that we have our point-to-point interfaces we need to figure out what are networks attached to each router LAN interfaces should be. First we should identify the largest network then go down from there. 50 is the largest amount so lets start here. We need to count how many bits do we need to achieve at least 50 hosts. We need 6 bits to achieve our desired amount of hosts: 1100 0000 Where does the last 1 read from left to right fall into? Hopefully you said the 64 place. Or if you use the other shortcut of 256-192 = 64. Now starting from our last network 192.168.1.8 we have a subnet mask of what? 192.168.1.8/ 255.255.255.??? Well we need 6 bits for the hosts so now how many bits did we have left over? we had 2 left over in which equals 255.255.255.192 or /26 so our network needing 50 hosts looks like this: 192.168.1.8 = subnet ID 192.168.1.9 = First usable address 192.168.1.70 = Last usable 192.168.1.71 = Broadcast address (not usable) 192.168.1.72 = Next Network We achieved the next network by adding the 64 increment (our increment changed as above) to the subnet ID. Our next highest number of hosts is 25 hosts. How many bits do we need for the hosts portion? 2^H-2 = hosts Answer is 5 bits 2^5-2 = 30 hosts Our next network above is 192.168.1.72 since we only took 5 bits we are left over with 3 bits for the subnet mask: 1110 0000 This equals 224 or 255.255.255.224 or /27 Our increment is 32 so our next network is? 72 + 32 = 104 192.168.1.72 = Subnet id for 25 hosts 192.168.1.73 = first usable address 192.168.1.102 = last usable address 192.168.1.103 = broadcast address (not usable) 192.168.1.104 = Next usable network Lastly we need 10 hosts. Can you figure out the rest? How many bits do we need for 10 hosts? What is our increment? What is the subnet mask for this network? 192.168.1.??? = Subnet ID for 10 hosts 192.168.1.??? = first usable address 192.168.1.??? = last usable address 192.168.1.??? = broadcast address 192.168.1.??? = next network Answer: How many bits do we need for 10 hosts? 4 What is our increment? 16 What is the subnet mask for this network? 255.255.255.240 or /28 192.168.1.104 = Subnet ID for 10 hosts 192.168.1.105 = first usable address 192.168.1.118 = last usable address 192.168.1.119 = broadcast address 192.168.1.120 = next network So what happens to the rest of the IP’s ranging from 192.168.1.120 to 192.168.1.254? They are used for future use and the above allows you to utilize all available IP addresses with no wasted addresses. Note: only routing protocols that can utilize VLSM is RIPv2 (not 1), OSPF, and EIGRP. Route Summarization Route Summarization is used to allow routers to summarize smaller networks in to larger networks to allow for less router updates and lower route tables, so less CPU and memory utilization on the router. For example let’s say we have networks: 172.16.1.0/24 172.16.2.0/24 172.16.3.0/24 All on routerb And Network 172.17.1.0/24 Network 172.17.2.0/24 Network 172.17.3.0/24 All on routerc And Network 192.168.1.0/24 Network 192.168.2.0/24 On routera To summarize we convert the IP’s into binary: Routerb 1011 0001.0001 0000.0000 0001.0000 0000 1011 0001.0001 0000.0000 0010.0000 0000 1011 0001.0001 0000.0000 0011.0000 0000 Starting from left to right we mark all the “like” bits until we find the bits that do not match. In this example we identify that in the 3rd octet we have all a match up to the 22nd bit. Now turn all of these to 1’s and everything to the right to 0’s. 1111 1111.1111 1111.1111 1100.0000 0000 Our mask is now 255.255.252.0 and all networks with 172.16.0.0/22 would match each so our advertisements to router A or C would only need to be the 172.16.0.0/22 network. See if you can figure out the other networks route summarization.
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