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# Subnetting Basics Tutorial

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How to Identify Subnet & Host bits

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### Subnetting Basics Tutorial

1. 1. Subnetting Basics ? ! ? NO ! 1
2. 2. Subnetting Basics Made Easy By: Michael D. Mann © Copyright 2011 by : Michael D. Mann 2
3. 3. Subnetting Basics Made Easy Part 1: Prerequisites 3
4. 4. Subnetting Basics Prerequisite Knowledge Class 1st Octet Valid Network Total Number of Range Numbers Networks hosts per for this network class 1 -126 1.0.0.0 to 126 224 – 2 A 126.0.0.0 16,777,214N.H.H.H 128 – 191 128.0.0.0 to 16,384 216 – 2 B 191.255.0.0 65,534N.N.H.H 192 – 223 192.0.0.0 - 2,097,152 28 – 2 C 223.255.255.0 254N.N.N.H N = network octet H = host octet 4
5. 5. Subnetting Basics Prerequisite Knowledge DefinitionsNetwork ID : all host bits are set to 0 192.168.32.0000 0000 Ex : 192.168.32.0 /24Broadcast ID : all host bits are set to 1 192.168.32.1111 1111 Ex : 192.168.32.255 /24 IP Address : any value in between 192.168.32.0000 0001 – 1111 1110 Ex : 192.168.32.1 - .254 /24 5
6. 6. Subnetting Basics Prerequisite Knowledge Prefix Length Notation There are 2 ways to express a subnet mask:Dotted Decimal Notation: Prefix Length Notation:Class C mask: 255.255.255.0 Class C mask: /24 255.255.255.0 = 1111 1111. 1111 1111. 1111 1111. 0000 0000 Notice that there are 24 blue (network) bits. The PL number (/24)indicates how many bits are being used to represent the network ID. Notice how the host bits are always 0 in a subnet mask 6
7. 7. Subnetting Basics Prerequisite Knowledge The default Class A, B, & C subnet masks Class A default subnet mask : 255.0.0.0 ( /8 )Binary: 1111 1111.0000 0000.0000 0000.0000 0000Class B default subnet mask : 255.255.0.0 ( /16 )Binary: 1111 1111.1111 1111.0000 0000.0000 0000Class C default subnet mask : 255.255.255.0 ( /24 )Binary: 1111 1111.1111 1111.1111 1111.0000 0000 7
8. 8. Subnetting Basics Prerequisite Knowledge Binary Number System 1 BYTE = 8 BITS : 1010 0001Each bit position has an associated PLACE VALUE as indicated : 27 26 25 24 23 22 21 20 1 0 1 0 0 0 0 1 128 + 32 + 1 = 16110 Add up the place values wherever you have a “1” bit This is how you convert from binary to decimal 8
9. 9. Subnetting Basics Made EasyPart 2: Identifying Subnet and Host Bits 9
10. 10. Subnetting Basics Identifying Subnet Bits in a Class C IP AddressExample (unsubnetted or classful): 192.168.32.158 /24Example (subnetted or classless): 192.168.32.158 /28Step 1: Notice that the prefix length (PL) indicator of thesubnetted example is /28. This number is greater than thedefault PL indicator of /24. This tells you that the IP address issubnetted. 10
11. 11. Subnetting Basics Identifying Subnet Bits in a Class C IP AddressExample: 192.168.32. 158 /2815810 = 1001 11102Step 2: Examine the host octet by converting it to a binarynumber. 11
12. 12. Subnetting Basics Identifying Subnet Bits in a Class C IP AddressExample: 192.168.32.158 /28 15810 = 1001 11102Step 3a Subnet Bits Rule: Subnet bits are “assigned” startingwith the high order bits in the host octet.Step 3b: Subtract the default class C PL indicator from thecurrent PL indicator to get the number of bits to “assign”./28 - /24 = 4 bits must be “borrowed” for use as subnet bits. 12
13. 13. Subnetting Basics Identifying Subnet Bits in a Class C IP AddressExample: 192.168.32.158 /28 15810 = 1001 11102Step 4: 4 Subnet bits are “borrowed” starting with the highorder bits in the host octet. Color code these bits in blue usingan italicized font and color code the remaining host bits in red:15810 = 1001 11102 13
14. 14. Subnetting BasicsPart 3: Calculating The Number of Subnets 14
15. 15. Subnetting Basics Calculating The Number of SubnetsExample: 192.168.32.158 /28 15810 = 1001 11102Procedure: Use the equation 2X where X = the number ofsubnet bits. There are four subnet (i.e. blue, italicized ) bits.Therefore: 24 = 16, so there are 16 subnets available. 15
16. 16. Subnetting Basics Calculating The Number of SubnetsExample: 192.168.32.158 /28 15810 = 1001 11102The 16 subnets are indicated by all possible combinations of the four subnetbits. Note that the subnet ID numbers are shown in parentheses next to thebinary.0000 (ID 0) 1000 (ID 128)0001 (ID 16) 1001 (ID 144)0010 (ID 32) 1010 (ID 160)0011 (ID 48) 1011 (ID 176)0100 (ID 64) 1100 (ID 192)0101 (ID 80) 1101 (ID 208)0110 (ID 96) 1110 (ID 224)0111 (ID 112) 1111 (ID 240) 16
17. 17. Subnetting Basics Calculating The Number of Subnets Example: 192.168.32.158 /28 15810 = 1001 11102The subnet ID numbers shown in parentheses in the previousslide are the result of binary to decimal conversion using themethod as demonstrated in slide 8. The binary place values arerepeated below for your convenience:27 = 128 26 = 64 25 = 32 24 = 16 23 = 8 22 = 4 21 = 2 20 = 1 17
18. 18. Subnetting BasicsPart 4: Calculating The Number of Hosts 18
19. 19. Subnetting Basics Calculating The Number of HostsExample: 192.168.32.158 /28 15810 = 1001 11102Procedure: Use the equation 2X - 2 where X = the number ofhost bits. There are four host (i.e. red) bits.Therefore: 24 – 2 = 14, so there are 14 hosts available. 19
20. 20. Subnetting Basics Calculating The Number of HostsExample: 192.168.32.158 /28 15810 = 1001 11102The host ID numbers are given by all possible combinations ofthe four host bits:0000 (ID 0) 1000 (ID 8)0001 (ID 1) 1001 (ID 9)0010 (ID 2) 1010 (ID 10)0011 (ID 3) 1011 (ID 11)0100 (ID 4) 1100 (ID 12)0101 (ID 5) 1101 (ID 13)0110 (ID 6) 1110 (ID 14) 200111 (ID 7) 1111 (ID 15)
21. 21. Subnetting Basics Calculating The Number of HostsExample: 192.168.32.158 /28 15810 = 1001 11102Why are the first and last hosts grayed out?Hint: See slide 5.0000 (ID 0) 1000 (ID 8)0001 (ID 1) 1001 (ID 9)0010 (ID 2) 1010 (ID 10)0011 (ID 3) 1011 (ID 11)0100 (ID 4) 1100 (ID 12)0101 (ID 5) 1101 (ID 13)0110 (ID 6) 1110 (ID 14) 210111 (ID 7) 1111 (ID 15)
22. 22. Subnetting Basics Putting It All Together Example: 192.168.32.158 /28 15810 = 1001 11102 144 14 The host octet in this class C private IP addressidentifies subnet ID #144, which contains host #14. 22
23. 23. Subnetting BasicsBe sure to watch the complete Subnetting Basics presentation found in the Course Documents section of your Bb course shell. 23