IP Addressing & Subnetting Handbook - Instructors Version

1 1111111010101001000111110010111001010111 0010110001110100110111101000110100000101 0010110010100101010110011111110101010001 0111010011010100110010100101010101010101 0001100100101010010110001101011000110101 1010100001011001010100110100101001010011 0001101001110000110100101010001101101IP AddressingandSubnettingWorkbookInstructo r s EditionVersion Address ClassesClass A1 127(Network 127 is reserved for loopback and internal testing)Leading bit B128 191 Leading bit C192 223 Leading bit D224 239(Reserved for multicast)Class E240 255(Reserved for experimental, used for research)Private Address SpaceClass to to to Subnet MasksClass . Host . Host . HostNetwork . Network . Host . HostNetwork.

2 Network . Network . HostInside CoverProduced by: Robb County Career & Technology CenterCisco Networking AcademyFrederick County Public SchoolsFrederick, Maryland, USAS pecial Thanks to Melvin Baker and Jim Dorschfor taking the time to check this workbook for To Decimal Conversion1286432168421 Answers Scratch Area10010010 14601110111 11911111111 25511000101 19711110110 24600010011 1910000001 12900110001 4901111000 12011110000 24000111011 5900000111 700011011 2710101010 17001101111 11111111000 24800100000 3201010101 8500111110 6200000011 311101101 23711000000 1921281621466432164211191 Decimal To Binary Conversion1286432168421 = 255 Scratch Area_____ 238_____ 34_____ 123_____ 50_____ 255_____ 200_____ 10_____ 138_____ 1_____ 13_____ 250_____ 107_____ 224_____ 114_____ 192_____ 172_____ 100_____ 119_____ 57_____ 98_____

3 179_____ 2238-128110-6446-3214-86-42-20 1110111034-322-20 00100010 Use all 8 bits for each problem2 01111011 0 0 1 1 0 0 1 0 1 1 1 1 1 1 1 1 1 1 0 0 1 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 1 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 1 1 1 1 1 1 0 1 0 0 1 1 0 1 0 1 1 1 1 1 0 0 0 0 0 0 1 1 1 0 0 1 0 1 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 0 1 1 0 0 1 0 0 0 1 1 1 0 1 1 1 0 0 1 1 1 0 0 1 0 1 1 0 0 0 1 0 1 0 1 1 0 0 1 1 0 0 0 0 0 0 1 0 Address Class Identification & Host IdentificationCircle the network portionof these the host portion ofthese Subnet MasksWrite the correct default subnet mask for each of the following.

4 255 . 0 . 0255 . 0 . 0 . 0255 . 255 . 0 . 0255 . 255 . 255 . 0255 . 0 . 0 . 0255 . 255 . 0 . 0255 . 255 . 255 . 0255 . 255 . 255 . 0255 . 0 . 0 . 0255 . 255 . 0 . 0255 . 0 . 0 . 0255 . 255 . 0 . 0255 . 255 . 255 . 0255 . 0 . 0 . 0255 . 0 . 0 . 0255 . 255 . 255 . 0255 . 255 . 0 . 0255 . 0 . 0 . 05 ANDING WithDefault subnet masksEvery IP address must be accompanied by a subnet mask. By now you should be able to lookat an IP address and tell what class it is. Unfortunately your computer doesn t think that your computer to determine the network and subnet portion of an IP address it must AND the IP address with the subnet Subnet Masks:Class Equations:1 AND 1 = 11 AND 0 = 00 AND 1 = 00 AND 0 = 0 Sample:What you Address:192 . 100 . 10 . 33 What you can figure out in your Class: CNetwork Portion:192.

5 100 . 10 . 33 Host Portion:192 . 100 . 10 . 33In order for you computer to get the same information it must AND the IP address withthe subnet mask in binary. ANDING with the default subnet mask allows your computer to figure out the networkportion of the 1 0 0 0 0 0 0 . 1 1 0 0 1 0 0 . 0 0 0 0 1 0 1 0 . 0 0 1 0 0 0 0 11 1 1 1 1 1 1 1 . 1 1 1 1 1 1 1 . 1 1 1 1 1 1 1 1 . 0 0 0 0 0 0 0 01 1 0 0 0 0 0 0 . 1 1 0 0 1 0 0 . 0 0 0 0 1 0 1 0 . 0 0 0 0 0 0 0 0(255 . 255 . 255 . 0)(192 . 100 . 10 . 33)(192 . 100 . 10 . 0)NetworkHost IP Address:Default Subnet Mask:AND:6 ANDING WithCustom subnet masksWhen you take a single network such as and divide it into five smaller networks( , , , , ) the outsideworld still sees the network as , but the internal computers and routers see fivesmaller subnetworks.

6 Each independent of the other. This can only be accomplished by usinga custom subnet mask. A custom subnet mask borrows bits from the host portion of theaddress to create a subnetwork address between the network and host portions of an IPaddress. In this example each range has 14 usable addresses in it. The computer must stillAND the IP address against the custom subnet mask to see what the network portion is andwhich subnetwork it belongs Address:192 . 100 . 10 . 0 Custom Subnet to (Invalid Range) to (1st Usable Range) to (Range in the sample below) to to to to to to to to to to to to to (Invalid Range)In the next set of problems you will determine the necessary information to determine thecorrect subnet mask for a variety of IP 1 0 0 0 0 0 0.

7 1 1 0 0 1 0 0 . 0 0 0 0 1 0 1 0 . 0 0 1 0 0 0 0 11 1 1 1 1 1 1 1 . 1 1 1 1 1 1 1 . 1 1 1 1 1 1 1 1 . 1 1 1 1 0 0 0 01 1 0 0 0 0 0 0 . 1 1 0 0 1 0 0 . 0 0 0 0 1 0 1 0 . 0 0 1 0 0 0 0 0(255 . 255 . 255 . 240)(192 . 100 . 10 . 33)(192 . 100 . 10 . 32)Four bits borrowed from the hostportion of the address for thecustom subnet ANDING process of the four borrowed bitsshows which range of IP addresses thisparticular address will fall Address:Custom Subnet Mask:AND:7 Custom Subnet MasksProblem 1 Number of needed usable subnetsNumber of needed usable hostsNetwork AddressAddress classDefault subnet maskCustom subnet maskTotal number of subnetsNumber of usable subnetsTotal number of host addressesNumber of usable addressesNumber of bits . 10.

8 10 . 0 0 0 0 0 0 0 0128 64 32 16 8 4 2 1 - Binary valuesNumber of Subnets - 2 4 8 16 32 64 128 256 Number of256 128 64 32 16 8 4 2 - HostsShow your work for Problem 1 in the space the binary value numbers to the left of the line tocreate the custom subnet . 255 . 255 . 0255 . 255 . 255 . 240161416144 Observe the total number 2 for the number ofusable 2 for the total number ofsubnets to get the usable number +162409 Custom Subnet MasksProblem 2 Number of needed usable subnetsNumber of needed usable hostsNetwork AddressAddress classDefault subnet maskCustom subnet maskTotal number of subnetsNumber of usable subnetsTotal number of host addressesNumber of usable addressesNumber of bits.

9 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0128 64 32 16 8 4 2 1 Number of Subnets - 2 4 8 16 32 64 128 256 128 64 32 16 8 4 2 Show your work for Problem 2 in the space the binary value numbers to the left of the line tocreate the custom subnet ,022B255 . 255 . 0 . 0255 . 255 . 255 . 192 1,024 1,022646210 Observe the total number 2 for the number ofusable 2 for the total number ofsubnets to get the usable number +1255128 64 32 16 8 4 2 1 .512 Binary values -Number ofHosts-128+64192102420484,0968,19216,38 432,76865,5365121,0242,0484,0968,19216,3 8432,76865,536 Custom Subnet MasksProblem 3 Network AddressAddress classDefault subnet maskCustom subnet maskTotal number of subnetsNumber of usable subnetsTotal number of host addressesNumber of usable addressesNumber of bits /26_____Show your work for Problem 3 in the space.

10 255 . 0 . 0255 . 255 . 255 . 192 1,024 1,022646210/26 indicates the total number ofbits used for the network andsubnetwork portion of theaddress. All bits remaining belongto the host portion of the . 75 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0128 64 32 16 8 4 2 1 Number of Subnets - 2 4 8 16 32 64 128 256 128 64 32 16 8 4 2 Add the binary value numbers to the left of the line tocreate the custom subnet ,022 Observe the total number 2 for the number ofusable 2 for the total number ofsubnets to get the usable number +1255128 64 32 16 8 4 2 1 .512 Binary values -Number ofHosts-128+64192102420484,0968,19216,38 432,76865,5365121,0242,0484,0968,19216,3 8432,76865,53610 Custom Subnet MasksProblem 4 Number of needed usable subnetsNumber of needed usable hostsNetwork AddressAddress classDefault subnet maskCustom subnet maskTotal number of subnetsNumber of usable subnetsTotal number of host addressesNumber of usable addressesNumber of bits.