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xxv — Chapter 22, “Basic Troubleshooting”—Various show commands used to view the routing table; interpreting the show interface command; verifying your IP settings using different operating systems • Part VIII: Managing IP Services — Chapter 23, “Network Address Translation”—Configuring and verifying NAT and PAT — Chapter 24, “DHCP”—Configuring and verifying DHCP — Chapter 25, “IPv6”—Transitioning to IPv6; format of IPv6 addresses; configuring IPv6 (interfaces, tunneling, routing with RIPng) • Part IX: WANs — Chapter 26, “HDLC and PPP”—Configuring PPP, authentication of PPP using PAP or CHAP, compression in PPP; multilink in PPP, troubleshooting PPP, returning to HDLC encapsulation — Chapter 27, “Frame Relay”—Configuring basic Frame Relay, Frame Relay and subinterfaces, DLCIs, verifying and troubleshooting Frame Relay • Part X: Network Security — Chapter 28, “IP Access Control List Security”—Configuring standard ACLs, wildcard masking, creating extended ACLs, creating named ACLs, using sequence numbers in named ACLs, verifying and troubleshooting ACLs — Chapter 29, “Security Device Manager”—Connecting to a router using SDM, SDM user interfaces, SDM wizards, using SDM to configure a router as a DHCP server (or an interface as a DHCP client), using SDM to configure NAT • Part XI: Appendixes — Appendix A, “Binary/Hex/Decimal Conversion Chart”—A chart showing numbers 0 through 255 in the three numbering systems of binary, hexadecimal, and decimal — Appendix B, “Create Your Own Journal Here”—Some blank pages for you to add in your own specific commands that might not be in this book Did I Miss Anything? I am always interested to hear how my students, and now readers of my books, do on both certification exams and future studies. If you would like to contact me and let me know how this book helped you in your certification goals, please do so. Did I miss anything? Let me know. My e-mail address is ccnaguide@empson.ca. This page intentionally left blank PART I TCP/IP Version 4 Chapter 1 How to Subnet Chapter 2 VLSM Chapter 3 Route Summarization This page intentionally left blank CHAPTER 1 How to Subnet Class A–E Addresses Class Leading Bit Pattern First Octet in Decimal Notes A 0xxxxxxx 0–127 0 is invalid 2N 127 reserved for loopback testing Where N is equal to number of bits borrowed Formulae Number of total subnets created B 10xxxxxx 128–191 2N – 2 Number of valid subnets created C 110xxxxx 192–223 2H Number of total hosts per subnet Where H is equal to number of host bits D 1110xxxx 224–239 Reserved for multicasting E 1111xxxx 240–255 Reserved for future use/ testing Class A Address N H H H Class B Address N N H H Class C Address N N N H N = Network bits H = Host bits All 0s in host portion = Network or subnetwork address All 1s in host portion = Broadcast address Combination of 1s and 0s in host portion = Valid host address 2H – 2 Number of valid hosts per subnet 4 Subnetting a Class C Network Using Binary Converting Between Decimal Numbers and Binary In any given octet of an IP address, the 8 bits can be defined as follows: 27 26 25 24 23 22 21 20 128 64 32 16 8 4 2 1 To convert a decimal number into binary, you must turn on the bits (make them a 1) that would add up to that number, as follows: 187 = 10111011 = 128+32+16+8+2+1 224 = 11100000 = 128+64+32 To convert a binary number into decimal, you must add the bits that have been turned on (the 1s), as follows: 10101010 = 128+32+8+2 = 170 11110000 = 128+64+32+16 = 240 The IP address 138.101.114.250 is represented in binary as 10001010.01100101.01110010.11111010 The subnet mask of 255.255.255.192 is represented in binary as 11111111.11111111.11111111.11000000 Subnetting a Class C Network Using Binary You have a Class C address of 192.168.100.0 /24. You need nine subnets. What is the IP plan of network numbers, broadcast numbers, and valid host numbers? What is the subnet mask needed for this plan? You cannot use N bits, only H bits. Therefore, ignore 192.168.100. These numbers cannot change. Step 1 Determine how many H bits you need to borrow to create nine valid subnets. 2N – 2 ≥ 9 N = 4, so you need to borrow 4 H bits and turn them into N bits. Start with 8 H bits HHHHHHHH Borrow 4 bits NNNNHHHH Subnetting a Class C Network Using Binary Step 2 Determine the first valid subnet in binary. 0001HHHH Cannot use subnet 0000 because it is invalid. Therefore, you must start with the bit pattern of 0001 00010000 All 0s in host portion = subnetwork number 00010001 First valid host number . . . 00011110 Last valid host number 00011111 All 1s in host portion = broadcast number Step 3 Convert binary to decimal. 00010000 = 16 Subnetwork number 00010001 = 17 First valid host number . . . 00011110 = 30 Last valid host number 00011111 = 31 All 1s in host portion = broadcast number Step 4 Determine the second valid subnet in binary. 0010HHHH 0010 = 2 in binary = second valid subnet 00100000 All 0s in host portion = subnetwork number 00100001 First valid host number . . . 00101110 Last valid host number 00101111 All 1s in host portion = broadcast number 5 6 Subnetting a Class C Network Using Binary Step 5 Convert binary to decimal. 00100000 = 32 Subnetwork number 00100001 = 33 First valid host number . . . 00101110 = 46 Last valid host number 00101111 = 47 All 1s in host portion = broadcast number Step 6 Create an IP plan table. Valid Subnet Network Number Range of Valid Hosts Broadcast Number 1 16 17–30 31 2 32 33–46 47 3 48 49–62 63 Notice a pattern? Counting by 16. Step 7 Verify the pattern in binary. (The third valid subnet in binary is used here.) 0011HHHH Third valid subnet 00110000 = 48 Subnetwork number 00110001 = 49 First valid host number . . . 00111110 = 62 Last valid host number 00111111 = 63 Broadcast number Subnetting a Class C Network Using Binary Step 8 Finish the IP plan table. Network Address (0000) Range of Valid Hosts (0001–1110) Broadcast Address (1111) 0 (0000) invalid 192.168.100.0 192.168.100.1– 192.168.100.14 192.168.100.15 1 (0001) 192.168.100.16 192.168.100.17– 192.168.100.30 192.168.100.31 2 (0010) 192.168.100.32 192.168.100.33– 192.168.100.46 192.168.100.47 3 (0011) 192.168.100.48 192.168.100.49– 192.168.100.62 192.168.100.63 4 (0100) 192.168.100.64 192.168.100.65– 192.168.100.78 192.168.100.79 5 (0101) 192.168.100.80 192.168.100.81– 192.168.100.94 192.168.100.95 6 (0110) 192.168.100.96 192.168.100.97– 192.168.100.110 192.168.100.111 7 (0111) 192.168.100.112 192.168.100.113– 192.168.100.126 192.168.100.127 8 (1000) 192.168.100.128 192.168.100.129– 192.168.100.142 192.168.100.143 9 (1001) 192.168.100.144 192.168.100.145– 192.168.100.158 192.168.100.159 10 (1010) 192.168.100.160 192.168.100.161– 192.168.100.174 192.168.100.175 11 (1011) 192.168.100.176 192.168.100.177– 192.168.100.190 192.168.100.191 12 (1100) 192.168.100.192 192.168.100.193– 192.168.100.206 192.168.100.207 13 (1101) 192.168.100.208 192.168.100.209– 192.168.100.222 192.168.100.223 14 (1110) 192.168.100.224 192.168.100.225– 192.168.100.238 192.168.100.239 Subnet 7 8 Subnetting a Class B Network Using Binary 15 (1111) invalid 192.168.100.240 192.168.100.241– 192.168.100.254 192.168.100.255 Quick Check Always an even number First valid host is always an odd # Always an odd number Last valid host is always an even # Use any nine subnets—the rest are for future growth. Step 9 Calculate the subnet mask. The default subnet mask for a Class C network is as follows: Decimal Binary 255.255.255.0 11111111.11111111.11111111.00000000 1 = Network or subnetwork bit 0 = Host bit You borrowed 4 bits; therefore, the new subnet mask is the following: 11111111.11111111.11111111.11110000 255.255.255.240 NOTE: You subnet a Class B or a Class A network with exactly the same steps as for a Class C network; the only difference is that you start with more H bits. Subnetting a Class B Network Using Binary You have a Class B address of 172.16.0.0 /16. You need nine subnets. What is the IP plan of network numbers, broadcast numbers, and valid host numbers? What is the subnet mask needed for this plan? You cannot use N bits, only H bits. Therefore, ignore 172.16. These numbers cannot change. Step 1 Determine how many H bits you need to borrow to create nine valid subnets. 2N – 2 ≥ 9 N = 4, so you need to borrow 4 H bits and turn them into N bits. Start with 16 H bits HHHHHHHHHHHHHHHH (Remove the decimal point for now) Borrow 4 bits NNNNHHHHHHHHHHHH