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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 2How to Subnet IPv4 AddressesThis 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 ANDingIn the previous chapter, we looked at how IPv4 addressing works, and the idea that itis possible to break a single large networks into multiple smaller networks for moreflexibility in your network design. This chapter shows you how to perform this task. Thisis known assubnefting.NOTE:Some students (and working IT professionals) are intimidated by subnettingbecause it deals with math; more specifically, binary math. While some people pickthis 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 CCNA200-301 exam. This makes some people nervous during an exam. Just remember thatthis is math, and therefore there has to be an absolute correct answer. If you follow thesteps, you will come up with the correct answer. I always tell my students thatsubnetting and working with binary should be the easiest questions you have on anexam, because you know that if you follow the steps you will arrive at the correctanswer. 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) andhost bits (H bits) in an IPv4 address and they follow a specific pattern:Octet #I234Class A AddressNHHHClass B AddressNNHHClass C AddressNNNHAll Os in host portion = network or subnetwork addressAll 1s in host portion = broadcast addressCombination of 1s and Qs in host portion = valid host addressCHAPTER 2How to Subnet IPv4 AddressesThis 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 ANDingIn the previous chapter, we looked at how IPv4 addressing works, and the idea that itis possible to break a single large networks into multiple smaller networks for moreflexibility in your network design. This chapter shows you how to perform this task. Thisis known assubnefting.NOTE:Some students (and working IT professionals) are intimidated by subnettingbecause it deals with math; more specifically, binary math. While some people pickthis 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 CCNA200-301 exam. This makes some people nervous during an exam. Just remember thatthis is math, and therefore there has to be an absolute correct answer. If you follow thesteps, you will come up with the correct answer. I always tell my students thatsubnetting and working with binary should be the easiest questions you have on anexam, because you know that if you follow the steps you will arrive at the correctanswer. 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) andhost bits (H bits) in an IPv4 address and they follow a specific pattern:Octet #I234Class A AddressNHHHClass B AddressNNHHClass C AddressNNNHAll Os in host portion = network or subnetwork addressAll 1s in host portion = broadcast addressCombination 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:2N(where N is equal to the number of networkbits borrowed)Number of XoXa/ subnets created2n(where El is equal to the number of host bits)Number of tota/ hosts per subnet2" - 2Number ofvalidhosts per subnetSubnetting a Class C Network Using BinaryYou have an address of 192.168.100.0 724. You need nine subnets. What is the IP plan ofnetwork numbers, broadcast numbers, and valid host numbers? What is the subnet maskneeded for this plan?You cannot use N bits, only H bits. Therefore, ignore 192.168.100. These numberscannot change. You only work with host bits. You need to borrow some host bits and lumthem into network bits (or in this ease, subnetwork bits; I use the variable N to refer toboth network and subnetwork bits).Step 1.Determine how many II bits you need to borrow to create nine valid subnets.2* > 9N = 4, so you need to borrow 4 El bits and lum them into N bits.Start with 8 II bitsHHHHHHHHBorrow 4 bitsNNNNHHHHStep 2.Determine the first subnet in binary.OOOOHHHH(XMX)0000All 0s in host portion = subnetwork number00000001First valid host numberoooootioSecond valid host number00000011rITiird valid host number...00001110Ijist valid host number00001111.All Is in host portion = broadcast numberStep3.Convert binary to decimal.(XXXMJOOO = 0Subnetwork number00000001 = 1First valid host number00000010 = 2Second valid host number00000011 = 3I’hird valid host number. ...00001110= 14Ijst valid host number00001111 = 15?MI Is in host portion = broadcast numberTo subnet a network address space, we will use the following formulae:2N(where N is equal to the number of networkbits borrowed)Number of XoXa/ subnets created2n(where El is equal to the number of host bits)Number of tota/ hosts per subnet2" - 2Number ofvalidhosts per subnetSubnetting a Class C Network Using BinaryYou have an address of 192.168.100.0 724. You need nine subnets. What is the IP plan ofnetwork numbers, broadcast numbers, and valid host numbers? What is the subnet maskneeded for this plan?You cannot use N bits, only H bits. Therefore, ignore 192.168.100. These numberscannot change. You only work with host bits. You need to borrow some host bits and lumthem into network bits (or in this ease, subnetwork bits; I use the variable N to refer toboth network and subnetwork bits).Step 1.Determine how many II bits you need to borrow to create nine valid subnets.2* > 9N = 4, so you need to borrow 4 El bits and lum them into N bits.Start with 8 II bitsHHHHHHHHBorrow 4 bitsNNNNHHHHStep 2.Determine the first subnet in binary.OOOOHHHH(XMX)0000All 0s in host portion = subnetwork number00000001First valid host numberoooootioSecond valid host number00000011rITiird valid host number...00001110Ijist valid host number00001111.All Is in host portion = broadcast numberStep3.Convert binary to decimal.(XXXMJOOO = 0Subnetwork number00000001 = 1First valid host number00000010 = 2Second valid host number00000011 = 3I’hird valid host number. ...00001110= 14Ijst valid host number00001111 = 15?MI Is in host portion = broadcast number

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Step 4.Determine the second subnet in binary.0001HHHH00010000All Os in host portion = subnetwork number00010001First valid host number00010010Second valid host number00011110Last valid host number00011111All Is in host portion = broadcast numberStep 5.Convert binary to decimal.00010000= 16Subnetwork number00010001 = 17First valid host number00011 110 = 30Last valid host number00011111 = 31All Is in host portion = broadcast numberStep 6.Create an IP plan table.SubnetNetworkNumberRange ofValid HostsBroadcastNumber101-141521617-303133233-4647Notice a pattern? Counting by 16.Step 7.Verify the pattern in binary. (The third subnet in binary is used here.)OOIOHHHH'ITiird subnet00100000 = 32Subnetwork number00100001 = 33First valid host number00100010 = 34Second valid host number00101110=46Last valid host number00101 II 1 = 47Broadcast numberStep 8.Finish the IP plan tabic.SubnetNetworkAddress (0000)Range of ValidHosts (0001-1110)BroadcastAddress (1111)1 (0000)I92J 68.100.0I92.168.100.1-192.168. J00.14I92J68.I00.152(0001)192.168. J00.16192.168.100.17-192.168.100.30I92J68.I00.31Step 4.Determine the second subnet in binary.0001HHHH00010000All Os in host portion = subnetwork number00010001First valid host number00010010Second valid host number00011110Last valid host number00011111All Is in host portion = broadcast numberStep 5.Convert binary to decimal.00010000= 16Subnetwork number00010001 = 17First valid host number00011 110 = 30Last valid host number00011111 = 31All Is in host portion = broadcast numberStep 6.Create an IP plan table.SubnetNetworkNumberRange ofValid HostsBroadcastNumber101-141521617-303133233-4647Notice a pattern? Counting by 16.Step 7.Verify the pattern in binary. (The third subnet in binary is used here.)OOIOHHHH'ITiird subnet00100000 = 32Subnetwork number00100001 = 33First valid host number00100010 = 34Second valid host number00101110=46Last valid host number00101 II 1 = 47Broadcast numberStep 8.Finish the IP plan tabic.SubnetNetworkAddress (0000)Range of ValidHosts (0001-1110)BroadcastAddress (1111)1 (0000)I92J 68.100.0I92.168.100.1-192.168. J00.14I92J68.I00.152(0001)192.168. J00.16192.168.100.17-192.168.100.30I92J68.I00.31

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SubnetNetworkAddress (0000}Range of ValidHosts (0001-1110)BroadcastAddress (1111}3 {0010}192.168.10032I92J68.10033-I92.I68.J00.46192.168.100.474 (OOH)192.168.100.48I92.l68.l00.49-I92.I68.J00.62192.168.100.635 (0100)192.168.100.64192.168.100:65-192.168.100.78192.168.100.796(0101)192.168.100 80192.168.10081-192.168.100.94192.168.100.957(01 10)192.168.100.96192.168.100.97-192.168.100.110192.168.100.1118(0111)192.168.100.112192.168.100.113-192.168.100.126192.168.100.1279(1000)192.168.100.128192.168.100.129-192.168.100.142192.168.100.14310(1001)192.168.100.144192.168.100.145-192.168.100.158192.168.100.15911 (1010)192.168.100.160192.168.100.161-192.168.100.174192.168.100.17512(101 1)192.168.100.176192.168.100.177-192.168.100.190192.168.100.19113 (1 100)192.168.100.192192.168.100.193-192.168.100206192.168.100.20714(1 101)192.168.100208192.168.100209-192.168.100222192.168.100.22315 (1110)192.168.100224192.168.100225-192.168.100238192.168.100.23916(1 111)192.168.100240192.168.100241-192.168.100254192.168.100.255QuickCheckAlways an evennumberFirst valid host isalways an odd #Last valid host isalways an even #Always an oddnumberUse any nine subncls???the rest are for future growth.Step 9.Calculate the subnet mask. The default subnet mask for a Class C network isas follows:DecimalBinary255.255.255.0H i l lII 1.H H l l l l . i l 1 III 11.000000001 = Network or subnetwork bit0 = Host bitSubnetNetworkAddress (0000}Range of ValidHosts (0001-1110)BroadcastAddress (1111}3 {0010}192.168.10032I92J68.10033-I92.I68.J00.46192.168.100.474 (OOH)192.168.100.48I92.l68.l00.49-I92.I68.J00.62192.168.100.635 (0100)192.168.100.64192.168.100:65-192.168.100.78192.168.100.796(0101)192.168.100 80192.168.10081-192.168.100.94192.168.100.957(01 10)192.168.100.96192.168.100.97-192.168.100.110192.168.100.1118(0111)192.168.100.112192.168.100.113-192.168.100.126192.168.100.1279(1000)192.168.100.128192.168.100.129-192.168.100.142192.168.100.14310(1001)192.168.100.144192.168.100.145-192.168.100.158192.168.100.15911 (1010)192.168.100.160192.168.100.161-192.168.100.174192.168.100.17512(101 1)192.168.100.176192.168.100.177-192.168.100.190192.168.100.19113 (1 100)192.168.100.192192.168.100.193-192.168.100206192.168.100.20714(1 101)192.168.100208192.168.100209-192.168.100222192.168.100.22315 (1110)192.168.100224192.168.100225-192.168.100238192.168.100.23916(1 111)192.168.100240192.168.100241-192.168.100254192.168.100.255QuickCheckAlways an evennumberFirst valid host isalways an odd #Last valid host isalways an even #Always an oddnumberUse any nine subncls???the rest are for future growth.Step 9.Calculate the subnet mask. The default subnet mask for a Class C network isas follows:DecimalBinary255.255.255.0H i l lII 1.H H l l l l . i l 1 III 11.000000001 = Network or subnetwork bit0 = Host bit
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