Ip addressing and subnetting instructors workbook

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Ip addressing and subnetting instructors workbook

  1. 1. 001 101010010011000 10001111100 1011100101011100 101100011101001 1011110100011010 00001010010110010 1001010101100111 1111010101000101 1101001101010011 001010010101010 1010101000110010 010101001011000 110101100011010 11010100001011 001010100110 1001010010 IP Addressing and Subnetting Workbook Version 1.1 Instructor’s Edition 11111110 10010101 00011011 1000011011010011
  2. 2. IP Address ClassesClass A 1 – 127 (Network 127 is reserved for loopback and internal testing) Leading bit pattern 0 00000000.00000000.00000000.00000000 Network . Host . Host . HostClass B 128 – 191 Leading bit pattern 10 10000000.00000000.00000000.00000000 Network . Network . Host . HostClass C 192 – 223 Leading bit pattern 110 11000000.00000000.00000000.00000000 Network . Network . Network . HostClass D 224 – 239 (Reserved for multicast)Class E 240 – 255 (Reserved for experimental, used for research) Private Address SpaceClass A 10.0.0.0 to 10.255.255.255Class B 172.16.0.0 to 172.31.255.255Class C 192.168.0.0 to 192.168.255.255 Default Subnet MasksClass A 255.0.0.0Class B 255.255.0.0Class C 255.255.255.0 Produced by: Robb Jones jonesr@careertech.net Frederick County Career & Technology Center Cisco Networking Academy Frederick County Public Schools Frederick, Maryland, USA Special Thanks to Melvin Baker and Jim Dorsch for taking the time to check this workbook for errors.Inside Cover
  3. 3. Binary To Decimal Conversion128 64 32 16 8 4 2 1 Answers Scratch Area1 0 0 1 0 0 1 0 146 128 64 16 320 1 1 1 0 1 1 1 119 2 16 146 41 1 1 1 1 1 1 1 255 2 11 1 0 0 0 1 0 1 197 1191 1 1 1 0 1 1 0 2460 0 0 1 0 0 1 1 191 0 0 0 0 0 0 1 1290 0 1 1 0 0 0 1 490 1 1 1 1 0 0 0 1201 1 1 1 0 0 0 0 2400 0 1 1 1 0 1 1 590 0 0 0 0 1 1 1 7 00011011 27 10101010 170 01101111 111 11111000 248 00100000 32 01010101 85 00111110 62 00000011 3 11101101 237 11000000 192 1
  4. 4. Decimal To Binary Conversion Use all 8 bits for each problem128 64 32 16 8 4 2 1 = 255 Scratch Area 1 1 1 0 1 1 1 0_________________________________________ 238 238 34 -128 -32 0 0 1 0 0 0 1 0_________________________________________ 34 110 2 -64 -2 0 1 1 1 1 0 1 1_________________________________________ 123 46 0 -32 0 0 1 1 0 0 1 0_________________________________________ 50 14 -8 1 1 1 1 1 1 1 1_________________________________________ 255 6 -4 1 1 0 0 1 0 0 0_________________________________________ 200 2 -2 0 0 0 0 1 0 1 0_________________________________________ 10 0 1 0 0 0 1 0 1 0_________________________________________ 138 0 0 0 0 0 0 0 1_________________________________________ 1 0 0 0 0 1 1 0 1_________________________________________ 13 1 1 1 1 1 0 1 0_________________________________________ 250 0 1 1 0 1 0 1 1_________________________________________ 107 1 1 1 0 0 0 0 0_________________________________________ 224 0 1 1 1 0 0 1 0_________________________________________ 114 1 1 0 0 0 0 0 0_________________________________________ 192 1 0 1 0 1 1 0 0_________________________________________ 172 0 1 1 0 0 1 0 0_________________________________________ 100 0 1 1 1 0 1 1 1_________________________________________ 119 0 0 1 1 1 0 0 1_________________________________________ 57 0 1 1 0 0 0 1 0_________________________________________ 98 1 0 1 1 0 0 1 1_________________________________________ 179 0 0 0 0 0 0 1 0_________________________________________ 22
  5. 5. Address Class Identification Address Class10.250.1.1 A _____150.10.15.0 B _____192.14.2.0 C _____148.17.9.1 B _____193.42.1.1 C _____126.8.156.0 A _____220.200.23.1 C _____230.230.45.58 D _____177.100.18.4 B _____119.18.45.0 A _____249.240.80.78 E _____199.155.77.56 C _____117.89.56.45 A _____215.45.45.0 C _____199.200.15.0 C _____95.0.21.90 A _____33.0.0.0 A _____158.98.80.0 B _____219.21.56.0 C _____ 3
  6. 6. Network & Host IdentificationCircle the network portion Circle the host portion ofof these addresses: these addresses:177.100.18.4 10.15.123.50119.18.45.0 171.2.199.31209.240.80.78 198.125.87.177199.155.77.56 223.250.200.222117.89.56.45 17.45.222.45215.45.45.0 126.201.54.231192.200.15.0 191.41.35.11295.0.21.90 155.25.169.22733.0.0.0 192.15.155.2158.98.80.0 123.102.45.254217.21.56.0 148.17.9.15510.250.1.1 100.25.1.1150.10.15.0 195.0.21.98192.14.2.0 25.250.135.46148.17.9.1 171.102.77.77193.42.1.1 55.250.5.5126.8.156.0 218.155.230.14220.200.23.1 10.250.1.14
  7. 7. Default Subnet MasksWrite the correct default subnet mask for each of the following addresses:177.100.18.4 255 . 255 . 0 . 0 _____________________________119.18.45.0 255 . 0 . 0 . 0 _____________________________191.249.234.191 255 . 255 . 0 . 0 _____________________________223.23.223.109 255 . 255 . 255 . 0 _____________________________10.10.250.1 255 . 0 . 0 . 0 _____________________________126.123.23.1 255 . 255 . 0 . 0 _____________________________223.69.230.250 255 . 255 . 255 . 0 _____________________________192.12.35.105 255 . 255 . 255 . 0 _____________________________77.251.200.51 255 . 0 . 0 . 0 _____________________________189.210.50.1 255 . 255 . 0 . 0 _____________________________88.45.65.35 255 . 0 . 0 . 0 _____________________________128.212.250.254 255 . 255 . 0 . 0 _____________________________193.100.77.83 255 . 255 . 255 . 0 _____________________________125.125.250.1 255 . 0 . 0 . 0 _____________________________1.1.10.50 255 . 0 . 0 . 0 _____________________________220.90.130.45 255 . 255 . 255 . 0 _____________________________134.125.34.9 255 . 255 . 0 . 0 _____________________________95.250.91.99 255 . 0 . 0 . 0 _____________________________ 5
  8. 8. ANDING With Default 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 way.For your computer to determine the network and subnet portion of an IP address it must“AND” the IP address with the subnet mask.Default Subnet Masks: Class A 255.0.0.0 Class B 255.255.0.0 Class C 255.255.255.0ANDING Equations: 1 AND 1 = 1 1 AND 0 = 0 0 AND 1 = 0 0 AND 0 = 0Sample:What you see... IP Address: 192 . 100 . 10 . 33 What you can figure out in your head... Address Class: C Network Portion: 192 . 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. Network Host IP Address: 1 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 1 (192 . 100 . 10 . 33)Default Subnet Mask: 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 0 0 0 0 0 (255 . 255 . 255 . 0) AND: 1 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 (192 . 100 . 10 . 0) ANDING with the default subnet mask allows your computer to figure out the network portion of the address.6
  9. 9. ANDING With Custom subnet masksWhen you take a single network such as 192.100.10.0 and divide it into five smaller networks(192.100.10.16, 192.100.10.32, 192.100.10.48, 192.100.10.64, 192.100.10.80) the outsideworld still sees the network as 192.100.10.0, but the internal computers and routers see fivesmaller subnetworks. 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 to. IP Address: 192 . 100 . 10 . 0 Custom Subnet Mask: 255.255.255.240 Address Ranges: 192.10.10.0 to 192.100.10.15 (Invalid Range) 192.100.10.16 to 192.100.10.31 (1st Usable Range) 192.100.10.32 to 192.100.10.47 (Range in the sample below) 192.100.10.48 to 192.100.10.63 192.100.10.64 to 192.100.10.79 192.100.10.80 to 192.100.10.95 192.100.10.96 to 192.100.10.111 192.100.10.112 to 192.100.10.127 192.100.10.128 to 192.100.10.143 192.100.10.144 to 192.100.10.159 192.100.10.160 to 192.100.10.175 192.100.10.176 to 192.100.10.191 192.100.10.192 to 192.100.10.207 192.100.10.208 to 192.100.10.223 192.100.10.224 to 192.100.10.239 192.100.10.240 to 192.100.10.255 (Invalid Range) Sub Network Network Host IP Address: 1 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 1 (192 . 100 . 10 . 33)Custom Subnet Mask: 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 1 0 0 0 0 (255 . 255 . 255 . 240) AND: 1 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 (192 . 100 . 10 . 32) Four bits borrowed from the host portion of the address for the custom subnet mask. The ANDING process of the four borrowed bits shows which range of IP addresses this particular address will fall into.In the next set of problems you will determine the necessary information to determine thecorrect subnet mask for a variety of IP addresses. 7
  10. 10. Custom Subnet MasksProblem 1 Number of needed usable subnets 14 Number of needed usable hosts 14 Network Address 192.10.10.0 C Address class __________ 255 . 255 . 255 . 0 Default subnet mask _______________________________ 255 . 255 . 255 . 240 Custom subnet mask _______________________________ 16 Total number of subnets ___________________ 14 Number of usable subnets ___________________ 16 Total number of host addresses ___________________ 14 Number of usable addresses ___________________ 4 Number of bits borrowed ___________________ Show your work for Problem 1 in the space below. Number of 256 128 64 32 16 8 4 2 - Hosts Number of Subnets - 2 4 8 16 32 64 128 256 128 64 32 16 8 4 2 1 - Binary values192 . 10 . 10 . 0 0 0 0 0 0 0 0 128 16 Observe the total number of Add the binary value 64 hosts. numbers to the left of the line to -2 create the custom subnet mask. 32 Subtract 2 for the number of 14 +16 usable hosts. 240 16 -2 Subtract 2 for the total number of subnets to get the usable number of 14 subnets.8
  11. 11. Custom Subnet Masks Problem 2 Number of needed usable subnets 1000 Number of needed usable hosts 60 Network Address 165.100.0.0 B Address class __________ 255 . 255 . 0 . 0 Default subnet mask _______________________________ 255 . 255 . 255 . 192 Custom subnet mask _______________________________ 1,024 Total number of subnets ___________________ 1,022 Number of usable subnets ___________________ 64 Total number of host addresses ___________________ 62 Number of usable addresses ___________________ 10 Number of bits borrowed ___________________ Show your work for Problem 2 in the space below. 65, 32, 16,3 4,0 2,0 1,02 8,19 512 536 768 Number of 84 96 48 . 256 128 64 32 16 8 4 2 2 4 Hosts - 65, 32, 16,3 4,0 102 204 8,19 536 512 Number of 768 84 96 Subnets - 2 4 8 16 32 64 128 256. 2 4 8Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1165 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 128 128 64 +64 32 192 64 Observe the total number of 16 hosts. Add the binary value numbers to the left of the line to 8 -2 Subtract 2 for the number of create the custom subnet mask. 4 62 usable hosts. 2 1024 +1 Subtract 2 for the total number of -2 subnets to get the usable number of 255 subnets. 1,022 9
  12. 12. Custom Subnet Masks Problem 3 /26 indicates the total number of bits used for the network and Network Address 148.75.0.0 /26 subnetwork portion of the address. All bits remaining belong to the host portion of the address. B Address class __________ 255 . 255 . 0 . 0 Default subnet mask _______________________________ 255 . 255 . 255 . 192 Custom subnet mask _______________________________ 1,024 Total number of subnets ___________________ 1,022 Number of usable subnets ___________________ 64 Total number of host addresses ___________________ 62 Number of usable addresses ___________________ 10 Number of bits borrowed ___________________ Show your work for Problem 3 in the space below. 65, 32, 16,3 4,0 2,0 1,02 8,19 512 536 768 Number of 84 96 48 . 256 128 64 32 16 8 4 2 2 4 Hosts - 65, 32, 16,3 4,0 102 204 8,19 536 512 Number of 768 84 96 Subnets - 2 4 8 16 32 64 128 256. 2 4 8Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 148 . 75 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 128 128 64 +64 32 192 64 Observe the total number of 16 hosts. Add the binary value numbers to the left of the line to 8 -2 Subtract 2 for the number of create the custom subnet mask. 4 62 usable hosts. 2 1024 +1 Subtract 2 for the total number of -2 subnets to get the usable number of 255 subnets. 1,022 10
  13. 13. Custom Subnet MasksProblem 4 Number of needed usable subnets 6 Number of needed usable hosts 30 Network Address 210.100.56.0 C Address class _______ 255 . 255 . 255 . 0 Default subnet mask _______________________________ 255 . 255 . 255 . 224 Custom subnet mask _______________________________ 8 Total number of subnets ___________________ 6 Number of usable subnets ___________________ 32 Total number of host addresses ___________________ 30 Number of usable addresses ___________________ 3 Number of bits borrowed ___________________ Show your work for Problem 4 in the space below. Number of 256 128 64 32 16 8 4 2 - Hosts Number of Subnets - 2 4 8 16 32 64 128 256 128 64 32 16 8 4 2 1 - Binary values210 . 100 . 56 . 0 0 0 0 0 0 0 0 128 64 8 32 +32 -2 -2 224 6 30 11
  14. 14. Custom Subnet MasksProblem 5 Number of needed usable subnets 6 Number of needed usable hosts 30 Network Address 195.85.8.0 C Address class _______ 255 . 255 . 255 . 0 Default subnet mask _______________________________ 255 . 255 . 255 . 252 Custom subnet mask _______________________________ 64 Total number of subnets ___________________ 62 Number of usable subnets ___________________ 4 Total number of host addresses ___________________ 2 Number of usable addresses ___________________ 6 Number of bits borrowed ___________________ Show your work for Problem 5 in the space below. Number of 256 128 64 32 16 8 4 2 - Hosts Number of Subnets - 2 4 8 16 32 64 128 256 128 64 32 16 8 4 2 1 - Binary values 195 . 85 . 8 . 0 0 0 0 0 0 0 0 128 64 32 16 8 64 4 +4 -2 -2 252 60 212
  15. 15. Custom Subnet Masks Problem 6 Number of needed usable subnets 126 Number of needed usable hosts 131,070 Network Address 118.0.0.0 A Address class _______ 255 . 0 . 0 . 0 Default subnet mask _______________________________ 255 . 254. 0 . 0 Custom subnet mask _______________________________ 128 Total number of subnets ___________________ 126 Number of usable subnets ___________________ 131,072 Total number of host addresses ___________________ 131,070 Number of usable addresses ___________________ 7 Number of bits borrowed ___________________ Show your work for Problem 6 in the space below. 4,19 2,09 1,04 52 262 131 65,5 32,7 16,3 4,0 2,0 1,02 4,2 4,30 8,19 8,57 7,15 Number of ,07 ,144 512 . 256 128 64 32 16 8 4 2 36 96 48 84 Hosts 68 - 88 2 2 2 4 4 6 1,04 2,09 4,19 262 52 131 65,5 32,7 16,3 1,02 2,0 4,0 4,2 4,30 8,19 8,57 7,15 Number of ,07 ,144 512 . 36 48 96 84 68 Subnets - 2 4 8 16 32 64 128 256 . 88 2 2 2 4 4 6Binary values -128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 118. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 128 64 32 16 8 4 128 131,072 +2 -2 -2 254 126 131,070 13
  16. 16. Custom Subnet Masks Problem 7 Number of needed usable subnets 2000 Number of needed usable hosts 15 Network Address 178.100.0.0 B Address class __________ 255 . 255 . 0 . 0 Default subnet mask _______________________________ 255 . 255 . 255 . 224 Custom subnet mask _______________________________ 2,048 Total number of subnets ___________________ 2,046 Number of usable subnets ___________________ 32 Total number of host addresses ___________________ 30 Number of usable addresses ___________________ 11 Number of bits borrowed ___________________ Show your work for Problem 7 in the space below. 65, 32, 16,3 4,0 2,0 1,02 8,19 512 536 768 Number of 84 96 48 . 256 128 64 32 16 8 4 2 2 4 Hosts - 65, 32, 16,3 4,0 102 204 8,19 536 512 Number of 768 84 96 Subnets - 2 4 8 16 32 64 128 256. 2 4 8Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1178 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 128 64 32 16 8 4 2,048 32 2 -2 -2 +1 2,046 30 14 255
  17. 17. Custom Subnet MasksProblem 8 Number of needed usable subnets 1 Number of needed usable hosts 45 Network Address 200.175.14.0 C Address class _______ 255 . 255 . 255 . 0 Default subnet mask _______________________________ 255 . 255 . 255 . 192 Custom subnet mask _______________________________ 4 Total number of subnets ___________________ 2 Number of usable subnets ___________________ 64 Total number of host addresses ___________________ 62 Number of usable addresses ___________________ 2 Number of bits borrowed ___________________ Show your work for Problem 8 in the space below. Number of 256 128 64 32 16 8 4 2 - Hosts Number of Subnets - 2 4 8 16 32 64 128 256 128 64 32 16 8 4 2 1 - Binary values200 . 175 . 14 . 0 0 0 0 0 0 0 0 128 4 64 +64 -2 -2 240 2 62 15
  18. 18. Custom Subnet Masks Problem 9 Number of needed usable subnets 60 Number of needed usable hosts 1,000 Network Address 128.77.0.0 B Address class _______ 255 . 255 . 0 . 0 Default subnet mask _______________________________ 255 . 255 . 252 . 0 Custom subnet mask _______________________________ 64 Total number of subnets ___________________ 62 Number of usable subnets ___________________ 1,024 Total number of host addresses ___________________ 1,022 Number of usable addresses ___________________ 6 Number of bits borrowed ___________________ Show your work for Problem 9 in the space below. 65, 32, 16,3 4,0 2,0 1,02 8,19 512 536 768 Number of 84 96 48 . 256 128 64 32 16 8 4 2 2 4 Hosts - 65, 32, 16,3 4,0 102 204 8,19 536 512 Number of 768 84 96 Subnets - 2 4 8 16 32 64 128 256. 2 4 8Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 128 . 77 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 128 64 32 16 8 64 1,024 +4 -2 -2 252 62 1,022 16
  19. 19. Custom Subnet MasksProblem 10 Number of needed usable hosts 60 Network Address 198.100.10.0 C Address class _______ 255 . 255 . 255 . 0 Default subnet mask _______________________________ 255 . 255 . 255 . 192 Custom subnet mask _______________________________ 4 Total number of subnets ___________________ 2 Number of usable subnets ___________________ 64 Total number of host addresses ___________________ 62 Number of usable addresses ___________________ 2 Number of bits borrowed ___________________ Show your work for Problem 10 in the space below. Number of 256 128 64 32 16 8 4 2 - Hosts Number of Subnets - 2 4 8 16 32 64 128 256 128 64 32 16 8 4 2 1 - Binary values198 . 100 . 10 . 0 0 0 0 0 0 0 0 128 64 4 +64 -2 -2 192 62 2 17
  20. 20. Custom Subnet Masks Problem 11 Number of needed usable subnets 250 Network Address 101.0.0.0 A Address class _______ 255 . 0 . 0 . 0 Default subnet mask _______________________________ 255 . 255 . 0 . 0 Custom subnet mask _______________________________ 256 Total number of subnets ___________________ 254 Number of usable subnets ___________________ 65,536 Total number of host addresses ___________________ 65,534 Number of usable addresses ___________________ 8 Number of bits borrowed ___________________ Show your work for Problem 11 in the space below. 4,19 2,09 1,04 52 262 131 65,5 32,7 16,3 4,0 2,0 1,02 4,2 4,30 8,19 8,57 7,15 Number of ,07 ,144 512 . 256 128 64 32 16 8 4 2 36 96 48 84 Hosts 68 - 88 2 2 2 4 4 6 1,04 2,09 4,19 262 52 131 65,5 32,7 16,3 1,02 2,0 4,0 4,2 4,30 8,19 8,57 7,15 Number of ,07 ,144 512 . . 36 48 96 84 68 88 2 2 2 4 4 6 Subnets - 2 4 8 16 32 64 128 256Binary values -128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 101. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 128 64 32 16 8 4 2 256 65,536 +1 -2 -2 255 254 65,534 18
  21. 21. Custom Subnet MasksProblem 12 Number of needed usable subnets 5 Network Address 218.35.50.0 C Address class _______ 255 . 255 . 255 . 0 Default subnet mask _______________________________ 255 . 255 . 255 . 224 Custom subnet mask _______________________________ 8 Total number of subnets ___________________ 6 Number of usable subnets ___________________ 32 Total number of host addresses ___________________ 30 Number of usable addresses ___________________ 3 Number of bits borrowed ___________________ Show your work for Problem 12 in the space below. Number of 256 128 64 32 16 8 4 2 - Hosts Number of Subnets - 2 4 8 16 32 64 128 256 128 64 32 16 8 4 2 1 - Binary values 218 . 35 . 50 . 0 0 0 0 0 0 0 0 128 64 64 4 +32 -2 -2 224 62 2 19
  22. 22. Custom Subnet MasksProblem 13 Number of needed usable hosts 25 Network Address 218.35.50.0 C Address class _______ 255 . 255 . 255 . 0 Default subnet mask _______________________________ 255 . 255 . 255 . 224 Custom subnet mask _______________________________ 8 Total number of subnets ___________________ 6 Number of usable subnets ___________________ 32 Total number of host addresses ___________________ 30 Number of usable addresses ___________________ 3 Number of bits borrowed ___________________ Show your work for Problem 13 in the space below. Number of 256 128 64 32 16 8 4 2 - Hosts Number of Subnets - 2 4 8 16 32 64 128 256 128 64 32 16 8 4 2 1 - Binary values 218 . 35 . 50 . 0 0 0 0 0 0 0 0 128 64 8 32 +32 -2 -2 224 6 3020
  23. 23. Custom Subnet Masks Problem 14 Number of needed usable subnets 10 Network Address 172.59.0.0 B Address class _______ 255 . 255 . 0 . 0 Default subnet mask _______________________________ 255 . 255 . 240 . 0 Custom subnet mask _______________________________ 16 Total number of subnets ___________________ 14 Number of usable subnets ___________________ 4,096 Total number of host addresses ___________________ 4,094 Number of usable addresses ___________________ 4 Number of bits borrowed ___________________ Show your work for Problem 14 in the space below. 65, 32, 16,3 4,0 2,0 1,02 8,19 512 536 768 Number of 84 96 48 . 256 128 64 32 16 8 4 2 2 4 Hosts - 65, 32, 16,3 4,0 102 204 8,19 536 512 Number of 768 84 96 Subnets - 2 4 8 16 32 64 128 256. 2 4 8Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 172 . 59 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 128 64 32 16 4,096 +16 -2 -2 240 14 4,094 21
  24. 24. Custom Subnet Masks Problem 15 Number of needed usable hosts 50 Network Address 172.59.0.0 B Address class _______ 255 . 255 . 0 . 0 Default subnet mask _______________________________ 255 . 255 . 255 . 192 Custom subnet mask _______________________________ 1,024 Total number of subnets ___________________ 1,022 Number of usable subnets ___________________ 64 Total number of host addresses ___________________ 62 Number of usable addresses ___________________ 10 Number of bits borrowed ___________________ Show your work for Problem 15 in the space below. 65, 32, 16,3 4,0 2,0 1,02 8,19 512 536 768 Number of 84 96 48 . 256 128 64 32 16 8 4 2 2 4 Hosts - 65, 32, 16,3 4,0 102 204 8,19 536 512 Number of 768 84 96 Subnets - 2 4 8 16 32 64 128 256. 2 4 8Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1172 . 59 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 128 64 32 16 8 4 2 128 64 1,024 +1 +64 -2 -2 255 192 62 1,022 22
  25. 25. Custom Subnet Masks Problem 16 Number of needed usable hosts 29 Network Address 23.0.0.0 A Address class _______ 255 . 0 . 0 . 0 Default subnet mask _______________________________ 255 . 255 . 255 . 224 Custom subnet mask _______________________________ 524,288 Total number of subnets ___________________ 524,286 Number of usable subnets ___________________ 32 Total number of host addresses ___________________ 30 Number of usable addresses ___________________ 19 Number of bits borrowed ___________________ Show your work for Problem 16 in the space below. 4,19 2,09 1,04 52 262 131 65,5 32,7 16,3 4,0 2,0 1,02 4,2 4,30 8,19 8,57 7,15 Number of ,07 ,144 512 . 256 128 64 32 16 8 4 2 36 96 48 84 68 Hosts - 88 2 2 2 4 4 6 1,04 2,09 4,19 262 52 131 65,5 32,7 16,3 1,02 2,0 4,0 4,2 4,30 8,19 8,57 7,15 Number of ,07 ,144 512 . . 36 48 96 84 68 88 2 2 2 4 4 6 Subnets - 2 4 8 16 32 64 128 256Binary values -128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 23 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 128 64 32 524,288 +32 -2 -2 224 30 524,286 23
  26. 26. SubnettingProblem 1 Number of needed usable subnets 14 Number of needed usable hosts 14 Network Address 192.10.10.0 C Address class __________ 255 . 255 . 255 . 0 Default subnet mask _______________________________ 255 . 255 . 255 . 240 Custom subnet mask _______________________________ 16 Total number of subnets ___________________ 14 Number of usable subnets ___________________ 16 Total number of host addresses ___________________ 14 Number of usable addresses ___________________ 4 Number of bits borrowed ___________________ What is the 3rd usable 192.10.10.48 to 192.10.10.63 subnet range? _______________________________________________What is the subnet number 192 . 10 . 10 . 112for the 7th usable subnet? ________________________ What is the subnet broadcast address for 192 . 10 . 10 . 207 the 12th usable subnet? ________________________ What are the assignable addresses for the 8th 192.10.10.129 to 192.10.10.142 usable subnet? ______________________________________24
  27. 27. Show your work for Problem 1 in the space below. Number of 256 128 64 32 16 8 4 2 - Hosts Number of Subnets - 2 4 8 16 32 64 128 256 128 64 32 16 8 4 2 1 - Binary values192. 10 . 10 . 0 0 0 0 0 0 0 0 (Invalid range) 0 192.10.10.0 to 192.10.10.15 1 192.10.10.16 to 192.10.10.31 1 0 192.10.10.32 to 192.10.10.47 1 1 192.10.10.48 to 192.10.10.63 1 0 0 192.10.10.64 to 192.10.10.79 1 0 1 192.10.10.80 to 192.10.10.95 1 1 0 192.10.10.96 to 192.10.10.111 1 1 1 192.10.10.112 to 192.10.10.127 1 0 0 0 192.10.10.128 to 192.10.10.143 1 0 0 1 192.10.10.144 to 192.10.10.159 1 0 1 0 192.10.10.160 to 192.10.10.175 1 0 1 1 192.10.10.176 to 192.10.10.191 1 1 0 0 192.10.10.192 to 192.10.10.207 1 1 0 1 192.10.10.208 to 192.10.10.223 1 1 1 0 192.10.10.224 to 192.10.10.239 (Invalid range) 1 1 1 1 192.10.10.240 to 192.10.10.255 128 64 32 16 16 Custom subnet +16 Usable subnets -2 Usable hosts -2 mask 240 14 14 The binary value of the last bit borrowed is the range. In this problem the range is 16. The first address in each subnet range is the subnet The first and last range of addresses are not usable. number. The first usable range of addresses is: 192.10.10.16 to The last address in each subnet range is the subnet 192.10.10.31. broadcast address. 25
  28. 28. SubnettingProblem 2 Number of needed usable subnets 1000 Number of needed usable hosts 60 Network Address 165.100.0.0 B Address class __________ 255 . 255 . 0 . 0 Default subnet mask _______________________________ 255 . 255 . 255 . 192 Custom subnet mask _______________________________ 1,024 Total number of subnets ___________________ 1,022 Number of usable subnets ___________________ 64 Total number of host addresses ___________________ 62 Number of usable addresses ___________________ 10 Number of bits borrowed ___________________ What is the 14th usable 165.100.3.128 to 165.100.3.191 subnet range? _______________________________________________What is the subnet number 165 . 100 . 1 . 64for the 5th usable subnet? ________________________ What is the subnet broadcast address for 165 . 100 . 1 . 127 the 5th usable subnet? ________________________ What are the assignable addresses for the 8th 165.100.2.1 to 165.100.0.62 usable subnet? ______________________________________26

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