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CCNA 200-301 Portable Command Guide 5th Edition 36

Learn how to subnet IPv4 addresses using binary math, including Class B and C subnetting, binary ANDing, and shortcuts—essential skills for mastering the CCNA 200-301 exam.

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    CHAPTER 2 How to Subnet IPv4 Addresses This chapter provides information concerning the following topics: ■ Subnetting a Class C network using binary ■ Subnetting a Class B network using binary ■ Binary ANDing ■ So Why AND? ■ Shortcuts in binary ANDing In the previous chapter, we looked at how IPv4 addressing works, and the idea that it is possible to break a single large networks into multiple smaller networks for more flexibility in your network design. This chapter shows you how to perform this task. This is known as subnefting. NOTE: Some students (and working IT professionals) are intimidated by subnetting because it deals with math; more specifically, binary math. While some people pick this up quickly, some take more time than others. And this is OK. Just keep practicing. The ability to subnet IPv4 addresses is a key skill that is required to pass the CCNA 200-301 exam. This makes some people nervous during an exam. Just remember that this is math, and therefore there has to be an absolute correct answer. If you follow the steps, you will come up with the correct answer. I always tell my students that subnetting and working with binary should be the easiest questions you have on an exam, because you know that if you follow the steps you will arrive at the correct answer. Keep calm, remember the rules, and you will be fine. After all, it's just math, and math is easy. NOTE: Remember from the previous chapter that there are network bits (N bits) and host bits (H bits) in an IPv4 address and they follow a specific pattern: Octet # I 2 3 4 Class A Address N H H H Class B Address N N H H Class C Address N N N H All Os in host portion = network or subnetwork address All 1s in host portion = broadcast address Combination of 1s and Qs in host portion = valid host address CHAPTER 2 How to Subnet IPv4 Addresses This chapter provides information concerning the following topics: ■ Subnetting a Class C network using binary ■ Subnetting a Class B network using binary ■ Binary ANDing ■ So Why AND? ■ Shortcuts in binary ANDing In the previous chapter, we looked at how IPv4 addressing works, and the idea that it is possible to break a single large networks into multiple smaller networks for more flexibility in your network design. This chapter shows you how to perform this task. This is known as subnefting. NOTE: Some students (and working IT professionals) are intimidated by subnetting because it deals with math; more specifically, binary math. While some people pick this up quickly, some take more time than others. And this is OK. Just keep practicing. The ability to subnet IPv4 addresses is a key skill that is required to pass the CCNA 200-301 exam. This makes some people nervous during an exam. Just remember that this is math, and therefore there has to be an absolute correct answer. If you follow the steps, you will come up with the correct answer. I always tell my students that subnetting and working with binary should be the easiest questions you have on an exam, because you know that if you follow the steps you will arrive at the correct answer. Keep calm, remember the rules, and you will be fine. After all, it's just math, and math is easy. NOTE: Remember from the previous chapter that there are network bits (N bits) and host bits (H bits) in an IPv4 address and they follow a specific pattern: Octet # I 2 3 4 Class A Address N H H H Class B Address N N H H Class C Address N N N H All Os in host portion = network or subnetwork address All 1s in host portion = broadcast address Combination of 1s and Qs in host portion = valid host address

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    To subnet a network address space, we will use the following formulae: 2 N (where N is equal to the number of network bits borrowed) Number of XoXa/ subnets created 2 n (where El is equal to the number of host bits) Number of tota/ hosts per subnet 2" - 2 Number of valid hosts per subnet Subnetting a Class C Network Using Binary You have an address of 192.168.100.0 724. 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. You only work with host bits. You need to borrow some host bits and lum them into network bits (or in this ease, subnetwork bits; I use the variable N to refer to both network and subnetwork bits). Step 1. Determine how many II bits you need to borrow to create nine valid subnets. 2* > 9 N = 4, so you need to borrow 4 El bits and lum them into N bits. Start with 8 II bits HHHHHHHH Borrow 4 bits NNNNHHHH Step 2. Determine the first subnet in binary. OOOOHHHH (XMX)0000 All 0s in host portion = subnetwork number 00000001 First valid host number oooootio Second valid host number 00000011 r ITiird valid host number ... 00001110 Ijist valid host number 00001111 .All Is in host portion = broadcast number Step 3. Convert binary to decimal. (XXXMJOOO = 0 Subnetwork number 00000001 = 1 First valid host number 00000010 = 2 Second valid host number 00000011 = 3 I’hird valid host number . ... 00001110= 14 Ijst valid host number 00001111 = 15 ?MI Is in host portion = broadcast number To subnet a network address space, we will use the following formulae: 2 N (where N is equal to the number of network bits borrowed) Number of XoXa/ subnets created 2 n (where El is equal to the number of host bits) Number of tota/ hosts per subnet 2" - 2 Number of valid hosts per subnet Subnetting a Class C Network Using Binary You have an address of 192.168.100.0 724. 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. You only work with host bits. You need to borrow some host bits and lum them into network bits (or in this ease, subnetwork bits; I use the variable N to refer to both network and subnetwork bits). Step 1. Determine how many II bits you need to borrow to create nine valid subnets. 2* > 9 N = 4, so you need to borrow 4 El bits and lum them into N bits. Start with 8 II bits HHHHHHHH Borrow 4 bits NNNNHHHH Step 2. Determine the first subnet in binary. OOOOHHHH (XMX)0000 All 0s in host portion = subnetwork number 00000001 First valid host number oooootio Second valid host number 00000011 r ITiird valid host number ... 00001110 Ijist valid host number 00001111 .All Is in host portion = broadcast number Step 3. Convert binary to decimal. (XXXMJOOO = 0 Subnetwork number 00000001 = 1 First valid host number 00000010 = 2 Second valid host number 00000011 = 3 I’hird valid host number . ... 00001110= 14 Ijst valid host number 00001111 = 15 ?MI Is in host portion = broadcast number

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    Step 4. Determine the second subnet in binary. 0001 HHHH 00010000 All Os in host portion = subnetwork number 00010001 First valid host number 00010010 Second valid host number 00011110 Last valid host number 00011111 All Is in host portion = broadcast number Step 5. Convert binary to decimal. 00010000= 16 Subnetwork number 00010001 = 17 First valid host number 00011 110 = 30 Last valid host number 00011111 = 31 All Is in host portion = broadcast number Step 6. Create an IP plan table. Subnet Network Number Range of Valid Hosts Broadcast Number 1 0 1-14 15 2 16 17-30 31 3 32 33-46 47 Notice a pattern? Counting by 16. Step 7. Verify the pattern in binary. (The third subnet in binary is used here.) OOIOHHHH 'ITiird subnet 00100000 = 32 Subnetwork number 00100001 = 33 First valid host number 00100010 = 34 Second valid host number 00101110=46 Last valid host number 00101 II 1 = 47 Broadcast number Step 8. Finish the IP plan tabic. Subnet Network Address (0000) Range of Valid Hosts (0001-1110) Broadcast Address (1111) 1 (0000) I92J 68. 100.0 I92.168.100.1- 192. 168. J 00.14 I92J68.I00.15 2(0001) 192. 168. J 00.16 192. 168. 100.17- 192.168.100.30 I92J68.I00.31 Step 4. Determine the second subnet in binary. 0001 HHHH 00010000 All Os in host portion = subnetwork number 00010001 First valid host number 00010010 Second valid host number 00011110 Last valid host number 00011111 All Is in host portion = broadcast number Step 5. Convert binary to decimal. 00010000= 16 Subnetwork number 00010001 = 17 First valid host number 00011 110 = 30 Last valid host number 00011111 = 31 All Is in host portion = broadcast number Step 6. Create an IP plan table. Subnet Network Number Range of Valid Hosts Broadcast Number 1 0 1-14 15 2 16 17-30 31 3 32 33-46 47 Notice a pattern? Counting by 16. Step 7. Verify the pattern in binary. (The third subnet in binary is used here.) OOIOHHHH 'ITiird subnet 00100000 = 32 Subnetwork number 00100001 = 33 First valid host number 00100010 = 34 Second valid host number 00101110=46 Last valid host number 00101 II 1 = 47 Broadcast number Step 8. Finish the IP plan tabic. Subnet Network Address (0000) Range of Valid Hosts (0001-1110) Broadcast Address (1111) 1 (0000) I92J 68. 100.0 I92.168.100.1- 192. 168. J 00.14 I92J68.I00.15 2(0001) 192. 168. J 00.16 192. 168. 100.17- 192.168.100.30 I92J68.I00.31

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    Subnet Network Address (0000} Range of Valid Hosts (0001-1110) Broadcast Address (1111} 3 {0010} 192. 168.10032 I92J68.10033- I92.I68.J00.46 192.168. 100.47 4 (OOH) 192.168.100.48 I92.l68.l00.49- I92.I68.J00.62 192.168.100.63 5 (0100) 192.168.100.64 192.168.100:65- 192.168.100.78 192.168.100.79 6(0101) 192. 168.100 80 192.168.10081- 192.168.100.94 192.168.100.95 7(01 10) 192.1 68.100.96 192.168.100.97- 192.168. 100.110 192.168.100.111 8(0111) 192.168.100.112 192.168. 100.113- 192.168.100.126 192.168.100.127 9(1000) 192.168.100.128 192.168.100.129- 192.168.100.142 192.168.100.143 10(1001) 192.168.100.144 192.168.100.145- 192.168.100.158 192.168.100.159 11 (1010) 192.168.100.160 192.168.100.161- 192.168.100.174 192.168.100.175 12(101 1) 192.168.100.176 192.168. 100.177- 192.168.100.190 192.168.100.191 13 (1 100) 192.168.100.192 192.168.100.193- 192.168.100206 192.168. 100.207 14(1 101) 192.1 68.100208 192.168.100209- 192. 168. 100222 192.168.100.223 15 (1110) 192. 168.100224 192.168.100225- 192.168. 100238 192.168.100.239 16(1 111) 192.168.100240 192.168.100241- 192.168.100254 192.168.100.255 Quick Check Always an even number First valid host is always an odd # Last valid host is always an even # Always an odd number Use any nine subncls???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 H i l l II 1. H H l l l l . i l 1 III 11.00000000 1 = Network or subnetwork bit 0 = Host bit Subnet Network Address (0000} Range of Valid Hosts (0001-1110) Broadcast Address (1111} 3 {0010} 192. 168.10032 I92J68.10033- I92.I68.J00.46 192.168. 100.47 4 (OOH) 192.168.100.48 I92.l68.l00.49- I92.I68.J00.62 192.168.100.63 5 (0100) 192.168.100.64 192.168.100:65- 192.168.100.78 192.168.100.79 6(0101) 192. 168.100 80 192.168.10081- 192.168.100.94 192.168.100.95 7(01 10) 192.1 68.100.96 192.168.100.97- 192.168. 100.110 192.168.100.111 8(0111) 192.168.100.112 192.168. 100.113- 192.168.100.126 192.168.100.127 9(1000) 192.168.100.128 192.168.100.129- 192.168.100.142 192.168.100.143 10(1001) 192.168.100.144 192.168.100.145- 192.168.100.158 192.168.100.159 11 (1010) 192.168.100.160 192.168.100.161- 192.168.100.174 192.168.100.175 12(101 1) 192.168.100.176 192.168. 100.177- 192.168.100.190 192.168.100.191 13 (1 100) 192.168.100.192 192.168.100.193- 192.168.100206 192.168. 100.207 14(1 101) 192.1 68.100208 192.168.100209- 192. 168. 100222 192.168.100.223 15 (1110) 192. 168.100224 192.168.100225- 192.168. 100238 192.168.100.239 16(1 111) 192.168.100240 192.168.100241- 192.168.100254 192.168.100.255 Quick Check Always an even number First valid host is always an odd # Last valid host is always an even # Always an odd number Use any nine subncls???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 H i l l II 1. H H l l l l . i l 1 III 11.00000000 1 = Network or subnetwork bit 0 = Host bit

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