LEARN SUBNETTING BEAST IN 5 MINUTES!!!!!
Sunday, August 31, 2008
The credit for this courses goes to MR.spacyfreak He is currently pursing his ccsa and done with following courses CCNA / MCSE / MCSA / MCP / LPI 70-270 / 70-219 / 70-216 / 70-218 / 70-224 / 70-215 / 70-217 / 70-292 / LPI 101 / LPI 102 / 640-801 / 642-502 / 642-551 You think Subnetting is a beast? You think you have to be Superbrain to understand it? You are wrong! Here the step-by-step course. After reading and some self-training, you should be able to fix Subnetting-Questions in CCNA Exam without any problems in a snatch. Relax! What is a Subnetmask? With Subnetmasks, we can divide an IP-Address in network-part and in host-part.A given IP-Network can be divided in smaller parts. Each of this smaller parts is called a "Subnet". If we for example have the network 192.168.10.0 255.255.255.0 We have here ONE Class C - network, with 253 useable IPs for Client-PCs. The useable IP Range of this network is 192.168.10.1 - 192.168.10.254 The very last IP of each Subnet is called Broadcast-Address. This address is in that example 192.168.10.255 and its NOT useable for host-pcs. If we want to divide this network in two parts, we must use subnetting. With Subnetmask 255.255.255.128 we would divide the network in two parts. 192.168.10.1 - 192.168.10.127 192.168.10.128 - 192.168.10.255 So in this example, BEFORE we had one big Network. With the change of the Subnetmask we did divide it in two smaller networks.  First with Subnetmask 255.255.255.0 we had THIS network:192.168.10.0 >>> This is the "Network-IP" which is NOT useable for Host-PCs 192.168.10.1 192.168.10.2 192.168.10.3 192.168.10.4 192.168.10.5 ... ... ... 192.168.10.253 192.168.10.254 192.168.10.255 >>>This is the Broadcast-IP, which is NOT useable for Host-PCs Now with Subnetmask 255.255.255.128 we have THIS two networks: First Subnet: 192.168.10.0 >>> This is the "Network-IP" which is NOT useable for Host-PCs 192.168.10.1 192.168.10.2 192.168.10.3 192.168.10.4 192.168.10.5 ... ... ... 192.168.10.125 192.168.10.126 192.168.10.127 >>>This is the Broadcast-IP, which is NOT useable for Host-PCs Second Subnet: 192.168.10.128 >>> This is the "Network-IP" which is NOT useable for Host-PCs 192.168.10.129 192.168.10.130 192.168.10.131 192.168.10..132 192.168.10.133 ... ... ... 192.168.10.253 192.168.10.254 192.168.10.255 >>>This is the Broadcast-IP, which is NOT useable for Host-PCs The Subnetmask defines how big the subnet is. That means - how many Client-PCs will have place in that subnetwork. A Subnetmask of 255.255.255.0 means in binary 11111111.11111111.11111111.00000000 So, what do we see? 4 Blocks, divided with a ".". Each of these blocks is also called "octett".Because - each Block has 8 bits. To be able to do subnet-calculation, we first must understand binary calculation. Lets take the first block. The first "1" stands for a 128. The second "1" stands for a 64. The third "1" stands for a 32. The fourth "1" stands for a 16. The fifth "1" stands for a 8. And so on. That means: 11111111=255 11110000=240 11100000=224 If we see something like "/24", that means that 24 bits are set to "1", from the left side. Examples: /16 = 255.255.0.0 = 11111111.11111111.00000000.00000000 /20 = 255.255.240.0 = 11111111.11111111.11110000 If we would take a subnetmask of 255.255.255.255 that would be 128+64+32+16+8+4+2+1.128+64+32+16+8+4+2+1. 128+64+32+16+8+4+2+1. 128+64+32+16+8+4+2+1 and in binary it would be 11111111.11111111.11111111.11111111 Calculation of Subnetmask big enough for a specified number of Hosts If they ask.. "create a subnet with minimum 10 host IPs" than 1. calculate a power of two, that is minimum 10 2^3=8. That is not enough 2^4=16 That is higher than 10. Good. 2. Now put the LAST 4 Bits of your subnetmask to 0. 11111111.11111111.11111111.11110000 That is in decimal 255.255.255.240 With THIS Subnetmask, you have minimum 10 Host-Ips in the Subnet,without wasting to much IP-Addresses. ------------------------------------------------------------ Other example If they ask Create a subnet with minimum 70 Host-IPs 1. Calculate a Power of 2 that is MINIMUM 70 2^6=64. Not enough. 2^7=128. Thats higher than 70. Good. 2. Put the LAST 7 Bits of your Subnetmask to 0. 11111111.11111111.11111111.10000000 That is in decimal 255.255.255.128 You have a Subnetmask, with more than 70 Host-IPs. ------------------------------------------------------------------------- Calculation what is the Broadcast-IP of a Subnet When they ask "There is subnet 172.16.64.0/20. What is the BROADCAST ADDRESS of that Subnet ,dude?" 1. Step /20 meens 255.255.240.0 2. Step Now analyze the Subnet Oktett to find out the "network-jumps" 240 means 11110000 The LAST of the 1s is under decimal 16. That are our "network jumps" (128/64/32/16/8/4/2/1) 3. Step Write down the network-jumps 172.16.64.0 - 172.16.79.255 +16 172.16.80.0 - 172.16.95.255 +16 172.16.96.0 - 172.16.111.255 +16 172.16.112.0 - 172.16.127.255 Because the NEXT Subnet in the example is 172.16.80.0, the broadcast must be 172.16.79.255, cause THAT is the IP BEFORE the next Subnet starts = the BroadcastAddress. ------------------------ Other example of Broadcast-IP calculation: If it would be 172.16.64.0 /26 Same procedure /26 means 255.255.255.192 192 is binary 11000000 The LAST 1 stands under the 64. That are in that example our "net-jumps". 172.16.64.0 - 172.16.64.63 << 172.16.64.128 - 172.16.64.191 --------------------------------------------------------------------- Calculating first and last possible IP for a Host You have Network 192.168.20.32 /27 The very first IP is reserved for Default Gateway! What is the first and last valid IP for a Host-PC? 1.Step /27 is 255.255.255.224 2.Step 224 means 11100000 The LAST 1 is under the 32. That are our "network-jumps" in this example (128/64/32/16/8/4/2/1) Valid IPs in that subnet: 192.168.20.33 - 192.168.20.62 (192.168.20.63 is NOT useable, this is the very last IP and so the BROADCAST-IP). 192.168.20.64 << reserved for Default Gateway, our first Host-PC IP would be 192.168.20.34 The very last Host-PC IP would be 192.168.20.62 ################################################################### Here some examples of real exam questions, and step by step solutions: Given that you have a class B IP address network range, which of the subnet masks below will allow for 100 subnets with 500 usable host addresses per subnet? A. 255.255.0.0 B. 255.255.224.0 C. 255.255.254.0 D. 255.255.255.0 E. 255.255.255.224 Solution: Allways the same game... Like in 5 minute course.. Power of 2 that is minimum 500? 2^7=128 2^8=256 2^9=512 >>voila! Now - put the last 9 Bits of your Subnetmask to "0" 11111111.11111111.11111110.00000000 That is in decimal 255.255.254.0 ===================================================================== If a host on a network has the address 172.16.45.14/30, what is the address of the subnetwork to which this host belongs? A. 172.16.45.0 B. 172.16.45.4 C. 172.16.45.8 D. 172.16.45.12 E. 172.16.45.18 Solution: 172.16.45.14/30 /30 means 11111111.11111111.11111111.11111100 The last of the ones stands under the "4". That is our increment or network jump. 172.16.45.0 - 172.16.45.3 172.16.45.4 - 172.16.45.7 172.16.45.8 - 172.16.45.11 172.16.45.12 - 172.16.45.15 172.16.45.16 - 172.16.45.19 As we see, the Ip is in the Range of 172.16.45.12 - 172.16.45.15. So the network Address is 172.16.45.12 ================================================= QUESTION NO: 9 Which two of the addresses below are available for host addresses on the subnet 192.168.15.19/28? (Select two answer choices) A. 192.168.15.17 B. 192.168.15.14 C. 192.168.15.29 D. 192.168.15.16 E. 192.168.15.31 F. None of the above Solution: /28 means 11111111.11111111.11111111.11110000 The last 1 stands under the 16. This is increment or network-jumps. 192.168.15.0 - 192.168.15.15 192.168.15.16 - 192.168.15.31 192.168.15.32 - 192.168.15.47 Only A and C are IPs in the right range. Also E is in the right range. But - this is not useable for hosts, cause its broadcastaddress. So answer is A and C. ################################################################ ------------------------------------------------------------------------------------------------------------------------- Calculation of Wildcard-Masks (Needed for Access Lists and OSPF Configuration) You have Network 192.168.32.0 /28 Only THIS network should be denied of accessing a network or server. 1. Step calculate the wildcard mask /28 means 255.255.255.240 binary this is 11111111.11111111.11111111.11110000 For wildcard-mask only the ZEROS are interesting. 11110000 Make a addition of all the fields, that are set to zero 128/64/32/16/8/4/2/1 That is 8+4+2+1=15 So the wildcard-mask will be 0.0.0.15 access-list will be access-list 1 deny 192.168.32.0 0.0.0.15 access-list 1 permit ip any any now, we have to bind that access-list to a routerinterface. In the example, this is e0. interface e0 ip access-group 1 out (or in!) exit ------------------------------------------------------------------------------------------------------------------------- PS. Its good to write on a BIG paper the powers of 2 2^2=4 2^3=8 2^4=16 2^5=32 2^6=64 2^7=128 2^8=256 2^9=512 2^10=1024 2^11=2048 2^12=4096 And write on that paper the numbers 128 192 224 240 248 252 254 Cause this are the Numbers, you will allways need in calculating Subnets. Burn them in your mind! Hang the paper in front of your eyes to never forget them.Then you will be able to calculate Subnets in your head in a half second! Isnt live easy? | |
| | |