Ip addressing and_subnetting_workbook (1)
- 1. 001 1010100 10011000 10001111100 1011100101011100 101100011101001 1011110100011010 00001010010110010 1001010101100111 1111010101000101 1101001101010011 001010010101010 1010101000110010 010101001011000 110101100011010 11010100001011 001010100110 1001010010 IP Addressing and Subnetting Workbook Version 1.5 11111110 10010101 00011011 10000110 Student Name: 11010011
- 2. IP Address Classes Class A 1 – 127 (Network 127 is reserved for loopback and internal testing) Leading bit pattern 0 00000000.00000000.00000000.00000000 Network . Host . Host . Host Class B 128 – 191 Leading bit pattern 10 10000000.00000000.00000000.00000000 Network . Network . Host . Host Class C 192 – 223 Leading bit pattern 110 11000000.00000000.00000000.00000000 Network . Network . Network . Host Class D 224 – 239 (Reserved for multicast) Class E 240 – 255 (Reserved for experimental, used for research) Private Address Space Class A 10.0.0.0 to 10.255.255.255 Class B 172.16.0.0 to 172.31.255.255 Class C 192.168.0.0 to 192.168.255.255 Default Subnet Masks Class A 255.0.0.0 Class B 255.255.0.0 Class 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, and to everyone who has sent in suggestions to improve the series. Workbooks included in the series: IP Addressing and Subnetting Workbooks ACLs - Access Lists Workbooks VLSM Variable-Length Subnet Mask IWorkbooks Inside Cover
- 3. Binary To Decimal Conversion 128 64 32 16 8 4 2 1 Answers Scratch Area 1 0 0 1 0 0 1 0 146 128 64 16 32 0 1 1 1 0 1 1 1 119 2 16 146 4 1 1 1 1 1 1 1 1 2 1 1 1 0 0 0 1 0 1 119 1 1 1 1 0 1 1 0 0 0 0 1 0 0 1 1 1 0 0 0 0 0 0 1 0 0 1 1 0 0 0 1 0 1 1 1 1 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 1 0 1 1 0 0 0 0 0 1 1 1 00011011 10101010 01101111 11111000 00100000 01010101 00111110 00000011 11101101 11000000 1
- 4. Decimal To Binary Conversion Use all 8 bits for each problem 128 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 _________________________________________ 123 46 0 -32 _________________________________________ 50 14 -8 _________________________________________ 255 6 -4 _________________________________________ 200 2 -2 _________________________________________ 10 0 _________________________________________ 138 _________________________________________ 1 _________________________________________ 13 _________________________________________ 250 _________________________________________ 107 _________________________________________ 224 _________________________________________ 114 _________________________________________ 192 _________________________________________ 172 _________________________________________ 100 _________________________________________ 119 _________________________________________ 57 _________________________________________ 98 _________________________________________ 179 _________________________________________ 2 2
- 5. Address Class Identification Address Class 10.250.1.1 A _____ 150.10.15.0 B _____ 192.14.2.0 _____ 148.17.9.1 _____ 193.42.1.1 _____ 126.8.156.0 _____ 220.200.23.1 _____ 230.230.45.58 _____ 177.100.18.4 _____ 119.18.45.0 _____ 249.240.80.78 _____ 199.155.77.56 _____ 117.89.56.45 _____ 215.45.45.0 _____ 199.200.15.0 _____ 95.0.21.90 _____ 33.0.0.0 _____ 158.98.80.0 _____ 219.21.56.0 _____ 3
- 6. Network & Host Identification Circle the network portion Circle the host portion of of these addresses: these addresses: 177.100.18.4 10.15.123.50 119.18.45.0 171.2.199.31 209.240.80.78 198.125.87.177 199.155.77.56 223.250.200.222 117.89.56.45 17.45.222.45 215.45.45.0 126.201.54.231 192.200.15.0 191.41.35.112 95.0.21.90 155.25.169.227 33.0.0.0 192.15.155.2 158.98.80.0 123.102.45.254 217.21.56.0 148.17.9.155 10.250.1.1 100.25.1.1 150.10.15.0 195.0.21.98 192.14.2.0 25.250.135.46 148.17.9.1 171.102.77.77 193.42.1.1 55.250.5.5 126.8.156.0 218.155.230.14 220.200.23.1 10.250.1.1 4
- 7. Network Addresses Using the IP address and subnet mask shown write out the network address: 188.10.18.2 188 . 10 . 0 . 0 _____________________________ 255.255.0.0 10.10.48.80 10 . 10 . 48 . 0 _____________________________ 255.255.255.0 192.149.24.191 _____________________________ 255.255.255.0 150.203.23.19 _____________________________ 255.255.0.0 10.10.10.10 _____________________________ 255.0.0.0 186.13.23.110 _____________________________ 255.255.255.0 223.69.230.250 _____________________________ 255.255.0.0 200.120.135.15 _____________________________ 255.255.255.0 27.125.200.151 _____________________________ 255.0.0.0 199.20.150.35 _____________________________ 255.255.255.0 191.55.165.135 _____________________________ 255.255.255.0 28.212.250.254 _____________________________ 255.255.0.0 5
- 8. Host Addresses Using the IP address and subnet mask shown write out the host address: 188.10.18.2 0 . 0 . 18 . 2 _____________________________ 255.255.0.0 10.10.48.80 0 . 0 . 0 . 80 _____________________________ 255.255.255.0 222.49.49.11 _____________________________ 255.255.255.0 128.23.230.19 _____________________________ 255.255.0.0 10.10.10.10 _____________________________ 255.0.0.0 200.113.123.11 _____________________________ 255.255.255.0 223.169.23.20 _____________________________ 255.255.0.0 203.20.35.215 _____________________________ 255.255.255.0 117.15.2.51 _____________________________ 255.0.0.0 199.120.15.135 _____________________________ 255.255.255.0 191.55.165.135 _____________________________ 255.255.255.0 48.21.25.54 _____________________________ 255.255.0.0 6
- 9. Default Subnet Masks Write 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 _____________________________ 223.23.223.109 _____________________________ 10.10.250.1 _____________________________ 126.123.23.1 _____________________________ 223.69.230.250 _____________________________ 192.12.35.105 _____________________________ 77.251.200.51 _____________________________ 189.210.50.1 _____________________________ 88.45.65.35 _____________________________ 128.212.250.254 _____________________________ 193.100.77.83 _____________________________ 125.125.250.1 _____________________________ 1.1.10.50 _____________________________ 220.90.130.45 _____________________________ 134.125.34.9 _____________________________ 95.250.91.99 _____________________________ 7
- 10. ANDING With Default subnet masks Every IP address must be accompanied by a subnet mask. By now you should be able to look at 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 255.255.255.0 ANDING Equations: 1 AND 1 = 1 1 AND 0 = 0 0 AND 1 = 0 0 AND 0 = 0 Sample: 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 . 33 In order for you computer to get the same information it must AND the IP address with the subnet mask in binary. Network Host IP Address: 1 1 0 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 . 0 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 . 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. 8
- 11. ANDING With Custom subnet masks When 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 outside world still sees the network as 192.100.10.0, but the internal computers and routers see five smaller subnetworks. Each independent of the other. This can only be accomplished by using a custom subnet mask. A custom subnet mask borrows bits from the host portion of the address to create a subnetwork address between the network and host portions of an IP address. In this example each range has 14 usable addresses in it. The computer must still AND the IP address against the custom subnet mask to see what the network portion is and which subnetwork it belongs to. IP Address: 192 . 100 . 10 . 0 Custom Subnet Mask: 255.255.255.240 Address Ranges: 192.10.10.0 to 192.100.10.15 192.100.10.16 to 192.100.10.31 192.100.10.32 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 Sub Network Network Host IP Address: 1 1 0 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 . 0 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 . 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 the correct subnet mask for a variety of IP addresses. 9
- 12. How to determine the number of subnets and the number of hosts per subnet Two formulas can provide this basic information: Number of subnets = 2 s (Second subnet formula: Number of subnets = 2s - 2) Number of hosts per subnet = 2 h - 2 Both formulas calculate the number of hosts or subnets based on the number of binary bits used. For example if you borrow three bits from the host portion of the address use the number of subnets formula to determine the total number of subnets gained by borrowing the three bits. This would be 23 or 2 x 2 x 2 = 8 subnets To determine the number of hosts per subnet you would take the number of binary bits used in the host portion and apply this to the number of hosts per subnet formula If five bits are in the host portion of the address this would be 2 5 or 2 x 2 x 2 x 2 x 2 = 32 hosts. When dealing with the number of hosts per subnet you have to subtract two addresses from the range. The first address in every range is the subnet number. The last address in every range is the broadcast address. These two addresses cannot be assigned to any device in the network which is why you have to subtract two addresses to find the number of usable addresses in each range. For example if two bits are borrowed for the network portion of the address you can easily determine the number of subnets and hosts per subnets using the two formulas. 195. 223 . 50 . 0 0 0 0 0 0 0 0 The number of subnets The number of hosts created by created by borrowing 2 leaving 6 bits is 26 - 2 or bits is 2 2 or 2 x 2 = 4 2 x 2 x 2 x 2 x 2 x 2 = 64 - 2 = 62 subnets. usable hosts per subnet. What about that second subnet formula: s Number of subnets = 2 - 2 In some instances the first and last subnet range of addresses are reserved. This is similar to the first and last host addresses in each range of addreses. The first range of addresses is the zero subnet. The subnet number for the zero subnet is also the subnet number for the classful subnet address. The last range of addresses is the broadcast subnet. The broadcast address for the last subnet in the broadcast subnet is the same as the classful broadcast address. 10
- 13. Class C Address unsubnetted: 195. 223 . 50 . 0 195.223.50.0 to 195.223.50.255 Notice that the subnet and broadcast addresses match. Class C Address subnetted (2 bits borrowed): 195. 223 . 50 . 0 0 0 0 0 0 0 0 (Invalid range) (0) 195.223.50.0 to 195.223.50.63 (1) 195.223.50.64 to 195.223.50.127 (2) 195.223.50.128 to 195.223.50.191 (Invalid range) (3) 195.223.50.192 to 195.223.50.255 The primary reason the the zero and broadcast subnets were not used had to do pirmarily with the broadcast addresses. If you send a broadcast to 195.223.255 are you sending it to all 255 addresses in the classful C address or just the 62 usable addresses in the broadcast range? The CCNA and CCENT certification exams may have questions which will require you to determine which formula to use, and whehter or not you can use the first and last subnets. Use the chart below to help decide. When to use which formula to determine the number of subnets s s Use the 2 - 2 formula and don’t use the Use the 2 formula and use the zero and zero and broadcast ranges if... broadcast ranges if... Classful routing is used Classless routing or VLSM is used RIP version 1 is used RIP version 2, EIGRP, or OSPF is used The no ip subnet zero command is The ip subnet zero command is configured on your router configured on your router (default setting) No other clues are given Bottom line for the CCNA exams; if a question does not give you any clues as to whether or not to allow these two subnets, assume you can use them. s This workbook has you use the number of subnets = 2 formula. 11
- 14. Custom Subnet Masks Problem 1 Number of needed 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 ___________________ 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 values 192 . 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 32 14 create the custom subnet mask. Subtract 2 for the number of +16 usable hosts. 240 12
- 15. Custom Subnet Masks Problem 2 Number of needed 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 ___________________ 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 8 Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 165 . 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 62 Subtract 2 for the number of create the custom subnet mask. 4 usable hosts. 2 +1 255 13
- 16. Custom Subnet Masks Problem 3 /26 indicates the total number of bits used for the network and subnetwork portion of the Network Address 148.75.0.0 /26 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 ___________________ 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 8 Binary 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 62 Subtract 2 for the number of create the custom subnet mask. 4 usable hosts. 2 1024 +1 Subtract 2 for the total number of -2 255 subnets to get the usable number of 1,022 subnets. 14
- 17. Custom Subnet Masks Problem 4 Number of needed subnets 6 Number of needed usable hosts 30 Network Address 210.100.56.0 Address class _______ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ 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 values 210 . 100 . 56 . 0 0 0 0 0 0 0 0 15
- 18. Custom Subnet Masks Problem 5 Number of needed subnets 6 Number of needed usable hosts 30 Network Address 195.85.8.0 Address class _______ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ 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 16
- 19. Custom Subnet Masks Problem 6 Number of needed subnets 126 Number of needed usable hosts 131,070 Network Address 118.0.0.0 Address class _______ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ 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 6 Binary 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 17
- 20. Custom Subnet Masks Problem 7 Number of needed subnets 2000 Number of needed usable hosts 15 Network Address 178.100.0.0 Address class __________ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ 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 8 Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 178 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 18
- 21. Custom Subnet Masks Problem 8 Number of needed subnets 3 Number of needed usable hosts 45 Network Address 200.175.14.0 Address class _______ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ Number of bits borrowed ___________________ Show your work for Problem 8 in the space below. 19
- 22. Custom Subnet Masks Problem 9 Number of needed subnets 60 Number of needed usable hosts 1,000 Network Address 128.77.0.0 Address class _______ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ Number of bits borrowed ___________________ Show your work for Problem 9 in the space below. 20
- 23. Custom Subnet Masks Problem 10 Number of needed usable hosts 60 Network Address 198.100.10.0 Address class _______ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ Number of bits borrowed ___________________ Show your work for Problem 10 in the space below. 21
- 24. Custom Subnet Masks Problem 11 Number of needed subnets 250 Network Address 101.0.0.0 Address class _______ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ Number of bits borrowed ___________________ Show your work for Problem 11 in the space below. 22
- 25. Custom Subnet Masks Problem 12 Number of needed subnets 5 Network Address 218.35.50.0 Address class _______ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ Number of bits borrowed ___________________ Show your work for Problem 12 in the space below. 23
- 26. Custom Subnet Masks Problem 13 Number of needed usable hosts 25 Network Address 218.35.50.0 Address class _______ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ Number of bits borrowed ___________________ Show your work for Problem 13 in the space below. 24
- 27. Custom Subnet Masks Problem 14 Number of needed subnets 10 Network Address 172.59.0.0 Address class _______ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ Number of bits borrowed ___________________ Show your work for Problem 14 in the space below. 25
- 28. Custom Subnet Masks Problem 15 Number of needed usable hosts 50 Network Address 172.59.0.0 Address class _______ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ Number of bits borrowed ___________________ Show your work for Problem 15 in the space below. 26
- 29. Custom Subnet Masks Problem 16 Number of needed usable hosts 29 Network Address 23.0.0.0 Address class _______ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ Number of bits borrowed ___________________ Show your work for Problem 16 in the space below. 27
- 30. Subnetting Problem 1 Number of needed 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 ___________________ 16 Total number of host addresses ___________________ 14 Number of usable addresses ___________________ 4 Number of bits borrowed ___________________ What is the 4th 192.10.10.48 to 192.10.10.63 subnet range? _______________________________________________ What is the subnet number 192 . 10 . 10 . 112 for the 8th subnet? ________________________ What is the subnet broadcast address for 192 . 10 . 10 . 207 the 13th subnet? ________________________ What are the assignable addresses for the 9th 192.10.10.129 to 192.10.10.142 subnet? ______________________________________ 28
- 31. 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 values 192. 10 . 10 . 0 0 0 0 0 0 0 0 (0) 0 0 0 0 192.10.10.0 to 192.10.10.15 (1) 0 0 0 1 192.10.10.16 to 192.10.10.31 (2) 0 0 1 0 192.10.10.32 to 192.10.10.47 (3) 0 0 1 1 192.10.10.48 to 192.10.10.63 (4) 0 1 0 0 192.10.10.64 to 192.10.10.79 (5) 0 1 0 1 192.10.10.80 to 192.10.10.95 (6) 0 1 1 0 192.10.10.96 to 192.10.10.111 (7) 0 1 1 1 192.10.10.112 to 192.10.10.127 (8) 1 0 0 0 192.10.10.128 to 192.10.10.143 (9) 1 0 0 1 192.10.10.144 to 192.10.10.159 (0 1) 1 0 1 0 192.10.10.160 to 192.10.10.175 (1 1) 1 0 1 1 192.10.10.176 to 192.10.10.191 (2 1) 1 1 0 0 192.10.10.192 to 192.10.10.207 (3 1) 1 1 0 1 192.10.10.208 to 192.10.10.223 (4 1) 1 1 1 0 192.10.10.224 to 192.10.10.239 (5 1) 1 1 1 1 192.10.10.240 to 192.10.10.255 128 64 32 16 16 +16 -2 -2 240 14 14 Custom subnet Usable subnets Usable hosts mask 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 number. The last address in each subnet range is the subnet broadcast address. 29
- 32. Subnetting Problem 2 Number of needed 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 ___________________ 64 Total number of host addresses ___________________ 62 Number of usable addresses ___________________ 10 Number of bits borrowed ___________________ What is the 15th 165.100.3.128 to 165.100.3.191 subnet range? _______________________________________________ What is the subnet number 165 . 100 . 1 . 64 for the 6th subnet? ________________________ What is the subnet broadcast address for 165 . 100 . 1 . 127 the 6th subnet? ________________________ What are the assignable addresses for the 9th 165.100.2.1 to 165.100.0.62 subnet? ______________________________________ 30
- 33. 65, 32, 16,3 4,0 2,0 8,19 1,02 Number of 768 84 96 48 4 512 2 536 Hosts - . 256 128 64 32 16 8 4 2 65, 32, 16,3 4,0 8,19 102 204 Number of 512 96 84 768 4 8 2 536 Subnets - 2 4 8 16 32 64 128 256. Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 165 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 (0) . 0 165.100.0.0 to 165.100.0.63 (1) 1 165.100.0.64 to 165.100.0.127 64 128 1 0 165.100.0.128 to 165.100.0.191 (2) 1 1 165.100.0.192 to 165.100.0.255 Usable -2 64 hosts 62 (3) 32 (4) 1 . 0 0 165.100.1.0 to 165.100.1.63 16 (5) 1 . 0 1 165.100.1.64 to 165.100.1.127 8 1 . 1 0 165.100.1.128 to 165.100.1.191 Custom 128 4 (6) 1 . 1 165.100.1.192 to 165.100.1.255 1 subnet mask +64 (7) 2 192 (8) 1 0 . 0 0 165.100.2.0 to 165.100.0.63 +1 The binary value of the last bit borrowed is (9) 1 0 . 0 1 165.100.2.64 to 165.100.0.127 the range. In this problem the range is 64. 255 (0 1) 1 0 . 1 0 165.100.2.128 to 165.100.0.191 The first address in each subnet range is the 1 0 . 1 1 165.100.2.192 to 165.100.0.255 subnet number. (1 1) The last address in each subnet range is the (2 1) 1 1 . 0 0 165.100.3.0 to 165.100.3.63 subnet broadcast address. (3 1) 1 1 . 0 1 165.100.3.64 to 165.100.3.127 (4 1) 1 1 . 1 0 165.100.3.128 to 165.100.3.191 1) (5 1 1 . 1 1 165.100.3.192 to 165.100.3.255 Show your work for Problem 2 in the space below. Down to (1022) 1 1 1 1 1 1 1 1 . 1 0 165.100.255.128 to 165.100.255.191 (1023) 1 1 1 1 1 1 1 1 . 1 1 165.100.255.192 to 165.100.255.255 31
- 34. Subnetting Problem 3 Number of needed subnets 2 Network Address 195.223.50.0 Address class __________ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ Number of bits borrowed ___________________ What is the 3rd subnet range? _______________________________________________ What is the subnet number for the 2nd subnet? ________________________ What is the subnet broadcast address for the 1st subnet? ________________________ What are the assignable addresses for the 3rd subnet? ______________________________________ 32
- 35. Show your work for Problem 3 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. 223 . 50 . 0 0 0 0 0 0 0 0 33
- 36. Subnetting Problem 4 Number of needed subnets 750 Network Address 190.35.0.0 Address class __________ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ Number of bits borrowed ___________________ What is the 15th subnet range? _______________________________________________ What is the subnet number for the 13th subnet? ________________________ What is the subnet broadcast address for the 10th subnet? ________________________ What are the assignable addresses for the 6th subnet? ______________________________________ 34
- 37. Show your work for Problem 4 in the space below. 35
- 38. Subnetting Problem 5 Number of needed usable hosts 6 Network Address 126.0.0.0 Address class __________ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ________________ Number of usable addresses ___________________ Number of bits borrowed ___________________ What is the 2nd subnet range? _______________________________________________ What is the subnet number for the 5th subnet? ________________________ What is the subnet broadcast address for the 7th subnet? ________________________ What are the assignable addresses for the 10th subnet? ______________________________________ 36
- 39. Show your work for Problem 5 in the space below. 37
- 40. Subnetting Problem 6 Number of needed subnets 10 Network Address 192.70.10.0 Address class __________ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ Number of bits borrowed ___________________ What is the 9th subnet range? _______________________________________________ What is the subnet number for the 4th subnet? ________________________ What is the subnet broadcast address for the 12th subnet? ________________________ What are the assignable addresses for the 10th subnet? ______________________________________ 38
- 41. Show your work for Problem 6 in the space below. 39
- 42. Subnetting Problem 7 Network Address 10.0.0.0 /16 Address class __________ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ Number of bits borrowed ___________________ What is the 11th subnet range? _______________________________________________ What is the subnet number for the 6th subnet? ________________________ What is the subnet broadcast address for the 2nd subnet? ________________________ What are the assignable addresses for the 9th subnet? ______________________________________ 40
- 43. Show your work for Problem 7 in the space below. 41
- 44. Subnetting Problem 8 Number of needed subnets 5 Network Address 172.50.0.0 Address class __________ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ Number of bits borrowed ___________________ What is the 4th subnet range? _______________________________________________ What is the subnet number for the 5th subnet? ________________________ What is the subnet broadcast address for the 6th subnet? ________________________ What are the assignable addresses for the 3rd subnet? ______________________________________ 42
- 45. Show your work for Problem 8 in the space below. 43
- 46. Subnetting Problem 9 Number of needed usable hosts 28 Network Address 172.50.0.0 Address class __________ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ Number of bits borrowed ___________________ What is the 2nd subnet range? _______________________________________________ What is the subnet number for the 10th subnet? ________________________ What is the subnet broadcast address for the 4th subnet? ________________________ What are the assignable addresses for the 6th subnet? ______________________________________ 44
- 47. Show your work for Problem 9 in the space below. 45
- 48. Subnetting Problem 10 Number of needed subnets 45 Network Address 220.100.100.0 Address class __________ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ Number of bits borrowed ___________________ What is the 5th subnet range? _______________________________________________ What is the subnet number for the 4th subnet? ________________________ What is the subnet broadcast address for the 13th subnet? ________________________ What are the assignable addresses for the 12th subnet? ______________________________________ 46
- 49. Show your work for Problem 10 in the space below. 47
- 50. Subnetting Problem 11 Number of needed usable hosts 8,000 Network Address 135.70.0.0 Address class __________ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ Number of bits borrowed ___________________ What is the 6th subnet range? _______________________________________________ What is the subnet number for the 7th subnet? ________________________ What is the subnet broadcast address for the 3rd subnet? ________________________ What are the assignable addresses for the 5th subnet? ______________________________________ 48
- 51. Show your work for Problem 11 in the space below. 49
- 52. Subnetting Problem 12 Number of needed usable hosts 45 Network Address 198.125.50.0 Address class __________ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ Number of bits borrowed ___________________ What is the 2nd subnet range? _______________________________________________ What is the subnet number for the 2nd subnet? ________________________ What is the subnet broadcast address for the 4th subnet? ________________________ What are the assignable addresses for the 3rd subnet? ______________________________________ 50
- 53. Show your work for Problem 12 in the space below. 51
- 54. Subnetting Problem 13 Network Address 165.200.0.0 /26 Address class __________ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ Number of bits borrowed ___________________ What is the 10th subnet range? _______________________________________________ What is the subnet number for the 11th subnet? ________________________ What is the subnet broadcast address for the 1023rd subnet? ________________________ What are the assignable addresses for the 1022nd subnet? ______________________________________ 52
- 55. Show your work for Problem 13 in the space below. 53
- 56. Subnetting Problem 14 Number of needed usable hosts 16 Network Address 200.10.10.0 Address class __________ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ Number of bits borrowed ___________________ What is the 7th subnet range? _______________________________________________ What is the subnet number for the 5th subnet? ________________________ What is the subnet broadcast address for the 4th subnet? ________________________ What are the assignable addresses for the 6th subnet? ______________________________________ 54
- 57. Show your work for Problem 14 in the space below. 55
- 58. Subnetting Problem 15 Network Address 93.0.0.0 19 Address class __________ Default subnet mask _______________________________ Custom subnet mask _______________________________ Total number of subnets ___________________ Total number of host addresses ___________________ Number of usable addresses ___________________ Number of bits borrowed ___________________ What is the 15th subnet range? _______________________________________________ What is the subnet number for the 9th subnet? ________________________ What is the subnet broadcast address for the 7th subnet? ________________________ What are the assignable addresses for the 12th subnet? ______________________________________ 56
- 59. Show your work for Problem 15 in the space below. 57
- 60. Practical Subnetting 1 Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of subnets, and allow enough extra subnets and hosts for 100% growth in both areas. Circle each subnet on the graphic and answer the questions below. IP Address 172.16.0.0 F0/0 Router A S0/0/0 S0/0/1 F0/1 Router B F0/0 Marketing Management 24 Hosts Reasearch 15 Hosts 60 Hosts B Address class _____________________________ 255.255.224.0 Custom subnet mask _____________________________ 4 Minimum number of subnets needed _________ + 4 Extra subnets required for 100% growth _________ (Round up to the next whole number) = 8 Total number of subnets needed _________ Number of host addresses 60 in the largest subnet group _________ Number of addresses needed for + 60 100% growth in the largest subnet _________ (Round up to the next whole number) Total number of address = 120 needed for the largest subnet _________ Start with the first subnet and arrange your sub-networks from the largest group to the smallest. 172.16.0.0 to 172.31.255 IP address range for Research _____________________________ 172.16.32.0 to 172.63.255 IP address range for Marketing _____________________________ 172.16.64.0 to 172.95.255 IP address range for Management _____________________________ IP address range for Router A 172.16.96.0 to 172.127.255 to Router B serial connection _____________________________ 58
- 61. 65, 32, 16,3 4,0 2,0 8,19 1,02 Number of 768 84 96 48 4 512 2 536 Hosts - . 256 128 64 32 16 8 4 2 65, 32, 16,3 4,0 8,19 102 204 Number of 512 96 84 768 4 8 2 536 Subnets - 2 4 8 16 32 64 128 256. Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 172 . 16 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 (0) 0 172.16.0.0 to 172.16.31.255 (1) 1 172.16.32.0 to 172.16.63.255 (2) 1 0 172.16.64.0 to 172.16.95.255 (3) 1 1 172.16.96.0 to 172.16.127.255 (4) 1 0 0 172.16.128.0 to 172.16.159.255 (5) 1 0 1 172.16.160.0 to 172.16.191.255 (6) 1 1 0 172.16.192.0 to 172.16.223.255 4 (7) 1 1 1 172.16.224.0 to 172.16.255.255 x1.0 4 60 x1.0 60 Show your work for Practical Subnetting 1 in the space below. 59
- 62. Practical Subnetting 2 Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of hosts per subnet, and allow enough extra subnets and hosts for 30% growth in all areas. Circle each subnet on the graphic and answer the questions below. IP Address 135.126.0.0 S0/0/0 F0/0 Router A S0/0/1 S0/0/1 F0/0 Router B F0/1 S0/0/0 Tech Ed Lab 20 Hosts Router C F0/1 Science Lab 10 Hosts English Department 15 Hosts B Address class _____________________________ 255.255.255.224 Custom subnet mask _____________________________ 5 Minimum number of subnets needed _________ + 2 Extra subnets required for 30% growth _________ (Round up to the next whole number) = 7 Total number of subnets needed _________ Number of host addresses 20 in the largest subnet group _________ Number of addresses needed for + 6 30% growth in the largest subnet _________ (Round up to the next whole number) Total number of address = 26 needed for the largest subnet _________ Start with the first subnet and arrange your sub-networks from the largest group to the smallest. 135.126.0.0 to 135.126.0.31 IP address range for Tech Ed _____________________________ 135.126.0.32 to 135.126.0.63 IP address range for English _____________________________ 135.126.0.64 to 135.126.0.95 IP address range for Science _____________________________ IP address range for Router A to Router B serial connection 135.126.0.96 to 135.126.0.127 _____________________________ IP address range for Router A to Router B serial connection135.126.0.128 to 135.126.0.159 _____________________________ 60
- 63. 65, 32, 16,3 4,0 2,0 8,19 1,02 Number of 768 84 96 48 4 512 2 536 Hosts - . 256 128 64 32 16 8 4 2 65, 32, 16,3 4,0 8,19 102 204 Number of 512 96 84 768 4 8 2 536 Subnets - 2 4 8 16 32 64 128 256. Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 135. 126 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 (0) . 0 135.126.0.0 to 135.126.0.31 (1) 1 135.126.0.32 to 135.126.0.63 (2) 1 0 135.126.0.64 to 135.126.0.95 5 (3) 1 1 135.126.0.96 to 135.126.0.127 x.3 (4) 1 0 0 135.126.0.128 to 135.126.0.159 1.5 (5) 1 0 1 135.126.0.160 to 135.126.0.191 (Round up to 2) (6) 1 1 0 135.126.0.192 to 135.126.0.223 (7) 1 1 1 135.126.0.224 to 135.126.0.255 (8) 1 . 0 0 0 135.126.1.0 to 135.126.1.31 (9) 1 . 0 0 1 135.126.1.32 to 135.126.1.63 20 . (0 1) 1 0 1 0 135.126.1.64 to 135.126.1.95 x.3 . (1 1) 1 0 1 1 135.126.1.96 to 135.126.1.127 6 (2 1) 1 . 1 0 0 135.126.1.128 to 135.126.1.159 (3 1) 1 . 1 0 1 135.126.1.160 to 135.126.1.191 (4 1) 1 . 1 1 0 135.126.1.192 to 135.126.1.223 Show your work for Problem 2 in the space below. 1) (5 1 . 1 1 1 135.1261.224 to 135.126.1.255 61
- 64. Practical Subnetting 3 Based on the information in the graphic shown, design a classfull network addressing scheme that will supply the minimum number of hosts per subnet, and allow enough extra subnets and hosts for 25% growth in all areas. Circle each subnet on the graphic and answer the questions below. IP Address 172.16.0.0 F0/0 S0/0/1 F0/0 Sales Router A Administrative Router B 185 Hosts 30 Hosts F0/1 S0/0/0 Marketing 50 Hosts Address class _____________________________ Custom subnet mask _____________________________ Minimum number of subnets needed _________ + Extra subnets required for 25% growth _________ (Round up to the next whole number) = Total number of subnets needed _________ Number of host addresses in the largest subnet group _________ Number of addresses needed for + 25% growth in the largest subnet _________ (Round up to the next whole number) Total number of address = needed for the largest subnet _________ Start with the first subnet and arrange your sub-networks from the largest group to the smallest. IP address range for Sales _____________________________ IP address range for Marketing _____________________________ IP address range for Administrative _____________________________ IP address range for Router A to Router B serial connection _____________________________ 62
- 65. Show your work for Problem 3 in the space below. 63
- 66. Practical Subnetting 4 Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of subnets, and allow enough extra subnets and hosts for 70% growth in all areas. Circle each subnet on the graphic and answer the questions below. IP Address 135.126.0.0 F0/0 S0/0/0 Router A S0/0/1 Router B S0/0/1 F0/0 S0/0/0 Router C F0/0 F0/1 Dallas 150 Hosts New York Washington D.C. 325 Hosts 220 Hosts Address class _____________________________ Custom subnet mask _____________________________ Minimum number of subnets needed _________ + Extra subnets required for 70% growth _________ (Round up to the next whole number) = Total number of subnets needed _________ Number of host addresses in the largest subnet group _________ Number of addresses needed for + 70% growth in the largest subnet _________ (Round up to the next whole number) Total number of address = needed for the largest subnet _________ Start with the first subnet and arrange your sub-networks from the largest group to the smallest. IP address range for New York _____________________________ IP address range for Washington D. C. _____________________________ IP address range for Dallas _____________________________ IP address range for Router A to Router B serial connection _____________________________ IP address range for Router A to Router C serial connection _____________________________ 64
- 67. Show your work for Problem 4 in the space below. 65
- 68. Practical Subnetting 5 Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of hosts per subnet, and allow enough extra subnets and hosts for 100% growth in all areas. Circle each subnet on the graphic and answer the questions below. IP Address 210.15.10.0 F0/1 F0/0 Science Room Tech Ed Lab 10 Hosts 18 Hosts English classroom 15 Hosts Art Classroom 12 Hosts Address class _____________________________ Custom subnet mask _____________________________ Minimum number of subnets needed _________ + Extra subnets required for 100% growth _________ (Round up to the next whole number) = Total number of subnets needed _________ Number of host addresses in the largest subnet group _________ Number of addresses needed for + 100% growth in the largest subnet _________ (Round up to the next whole number) Total number of address = needed for the largest subnet _________ Start with the first subnet and arrange your sub-networks from the largest group to the smallest. IP address range for Router F0/0 Port _____________________________ IP address range for Router F0/1 Port _____________________________ 66
- 69. Show your work for Problem 5 in the space below. 67
- 70. Practical Subnetting 6 Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of subnets, and allow enough extra subnets and hosts for 20% growth in all areas. Circle each subnet on the graphic and answer the questions below. IP Address 10.0.0.0 S0/0/0 Router A S0/0/1 Technology S0/0/1 S0/0/0 Router B Building F0/0 F0/1 320 Hosts S0/0/0 S0/0/1 Art & Drama Router C Administration 75 Hosts 35 Hosts F0/0 F0/1 Science Building 225 Hosts Address class _____________________________ Custom subnet mask _____________________________ Minimum number of subnets needed _________ + Extra subnets required for 20% growth _________ (Round up to the next whole number) = Total number of subnets needed _________ Start with the first subnet and arrange your sub-networks from the largest group to the smallest. IP address range for Technology _____________________________ IP address range for Science _____________________________ IP address range for Arts & Drama _____________________________ IP Address range Administration _____________________________ IP address range for Router A to Router B serial connection _____________________________ IP address range for Router A to Router C serial connection _____________________________ IP address range for Router B to Router C serial connection _____________________________ 68
- 71. Show your work for Problem 6 in the space below. 69
- 72. Practical Subnetting 7 Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of hosts per subnet, and allow enough extra subnets and hosts for 125% growth in all areas. Circle each subnet on the graphic and answer the questions below. IP Address 177.135.0.0 S0/0/0 Router A S0/0/0 F0/0 Router B F0/0 F0/1 Administration Research Deployment Marketing 33 Hosts Sales 135 Hosts 63 Hosts 75 Hosts 255 Hosts Address class _____________________________ Custom subnet mask _____________________________ Minimum number of subnets needed _________ + Extra subnets required for 125% growth _________ (Round up to the next whole number) = Total number of subnets needed _________ Number of host addresses in the largest subnet group _________ Number of addresses needed for + 125% growth in the largest subnet _________ (Round up to the next whole number) Total number of address = needed for the largest subnet _________ Start with the first subnet and arrange your sub-networks from the largest group to the smallest. IP address range for Router A Port F0/0 _____________________________ IP address range for Research _____________________________ IP address range for Deployment _____________________________ IP address range for Router A to Router B serial connection _____________________________ 70
- 73. Show your work for Problem 7 in the space below. 71
- 74. Practical Subnetting 8 Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number subnets, and allow enough extra subnets and hosts for 85% growth in all areas. Circle each subnet on the graphic and answer the questions below. IP Address 192.168.1.0 F0/0 S0/0/0 Router A S0/0/1 F0/1 Router B F0/0 New York 8 Hosts Boston 5 Hosts Research & Development 8 Hosts Address class _____________________________ Custom subnet mask _____________________________ Minimum number of subnets needed _________ + Extra subnets required for 85% growth _________ (Round up to the next whole number) = Total number of subnets needed _________ Number of host addresses in the largest subnet group _________ Number of addresses needed for + 85% growth in the largest subnet _________ (Round up to the next whole number) Total number of address = needed for the largest subnet _________ Start with the first subnet and arrange your sub-networks from the largest group to the smallest. IP address range for Router A F0/0 _____________________________ IP address range for New York _____________________________ IP address range for Router A to Router B serial connection _____________________________ 72
- 75. Show your work for Problem 8 in the space below. 73
- 76. Practical Subnetting 9 Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of hosts per subnet, and allow enough extra subnets and hosts for 15% growth in all areas. Circle each subnet on the graphic and answer the questions below. IP Address 148.55.0.0 S0/0/0 Router A S0/0/1 F0/1 Router B S0/0/1 F0/0 S0/0/0 Dallas 1500 Hosts Router C F0/0 S0/0/1 Router D S0/0/0 Ft. Worth 2300 Hosts Address class _____________________________ Custom subnet mask _____________________________ Minimum number of subnets needed _________ + Extra subnets required for 15% growth _________ (Round up to the next whole number) = Total number of subnets needed _________ Number of host addresses in the largest subnet group _________ Number of addresses needed for + 15% growth in the largest subnet _________ (Round up to the next whole number) Total number of address = needed for the largest subnet _________ Start with the first subnet and arrange your sub-networks from the largest group to the smallest. IP address range for Ft. Worth _____________________________ IP address range for Dallas _____________________________ IP address range for Router A _____________________________ to Router B serial connection IP address range for Router A _____________________________ to Router C serial connection IP address range for Router C _____________________________ 74 to Router D serial connection
- 77. Show your work for Problem 9 in the space below. 75
- 78. Practical Subnetting 10 Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of subnets, and allow enough extra subnets and hosts for 110% growth in all areas. Circle each subnet on the graphic and answer the questions below. IP Address 172.16.0.0 Marketing Sales 56 Hosts 115 Hosts S0/0/0 F0/0 F0/0 Router A S0/0/1 Router B F0/1 Management Research 25 Hosts 35 Hosts Address class _____________________________ Custom subnet mask _____________________________ Minimum number of subnets needed _________ + Extra subnets required for 110% growth _________ (Round up to the next whole number) = Total number of subnets needed _________ Number of host addresses in the largest subnet group _________ Number of addresses needed for + 110% growth in the largest subnet _________ (Round up to the next whole number) Total number of address = needed for the largest subnet _________ Start with the first subnet and arrange your sub-networks from the largest group to the smallest. IP address range for Sales/Managemnt _____________________________ IP address range for Marketing _____________________________ IP address range for Research _____________________________ IP address range for Router A to Router B serial connection _____________________________ 76
- 79. Show your work for Problem 10 in the space below. 77
- 80. Valid and Non-Valid IP Addresses Using the material in this workbook identify which of the addresses below are correct and usable. If they are not usable addresses explain why. IP Address: 0.230.190.192 The network ID cannot be 0. ________________________________ Subnet Mask: 255.0.0.0 ________________________________ Reference Page Inside Front Cover IP Address: 192.10.10.1 OK ________________________________ Subnet Mask: 255.255.255.0 ________________________________ Reference Pages 28-29 IP Address: 245.150.190.10 ________________________________ Subnet Mask: 255.255.255.0 ________________________________ Reference Page Inside Front Cover IP Address: 135.70.191.255 ________________________________ Subnet Mask: 255.255.254.0 ________________________________ Reference Pages 48-49 IP Address: 127.100.100.10 ________________________________ Subnet Mask: 255.0.0.0 ________________________________ Reference Pages Inside Front Cover IP Address: 93.0.128.1 ________________________________ Subnet Mask: 255.255.224.0 ________________________________ Reference Pages 56-57 IP Address: 200.10.10.128 ________________________________ Subnet Mask: 255.255.255.224 ________________________________ Reference Pages 54-55 IP Address: 165.100.255.189 ________________________________ Subnet Mask: 255.255.255.192 ________________________________ Reference Pages 30-31 IP Address: 190.35.0.10 ________________________________ Subnet Mask: 255.255.255.192 ________________________________ Reference Pages 34-35 IP Address: 218.35.50.195 ________________________________ Subnet Mask: 255.255.0.0 ________________________________ Reference Page Inside Front Cover IP Address: 200.10.10.175 /22 ________________________________ Reference Pages 54-55 and/or Inside Front Cover ________________________________ IP Address: 135.70.255.255 ________________________________ Subnet Mask: 255.255.224.0 ________________________________ Reference Pages 48-49 78
- 81. IP Address Breakdown /24 /25 /26 /27 /28 /29 /30 8+8+8 8+8+8+1 8+8+8+2 8+8+8+3 8+8+8+4 8+8+8+5 8+8+8+6 255.255.255.0 255.255.255.128 255.255.255.192 255.255.255.224 255.255.255.240 255.255.255.248 255.255.255.252 256 Hosts 128 Hosts 64 Hosts 32 Hosts 16 Hosts 8 Hosts 4 Hosts 0-3 0-7 4-7 0-15 8-11 8-15 12-15 16-19 16-23 20-23 16-31 24-27 24-31 28-31 0-63 32-35 32-39 36-39 32-47 40-43 40-47 44-47 48-51 48-55 52-55 48-63 56-59 56-63 60-63 0-127 64-67 64-71 68-71 64-79 72-75 72-79 76-79 80-83 80-87 84-87 80-95 88-91 88-95 92-95 64-127 96-99 96-103 100-103 96-111 104-107 104-111 108-111 112-115 112-119 116-119 112-127 120-123 120-127 124-127 0-255 128-131 128-135 132-135 128-143 136-139 136-143 140-143 144-147 144-151 148-151 144-159 152-155 152-159 156-159 128-191 160-163 16-167 164-167 160-175 168-171 168-175 172-175 176-179 176-183 180-183 176-191 184-187 184-191 188-191 128-255 192-195 192-199 196-199 192-207 200-203 200-207 204-207 208-211 208-215 212-215 208-223 216-219 216-223 220-223 192-255 224-227 224-231 228-231 224-239 232-235 232-239 236-239 240-243 240-247 244-247 240-255 248-251 248-255 252-255 79
- 82. Visualizing Subnets Using The Box Method The box method is the simplest way to visualize the breakdown of subnets and addresses into smaller sizes. Start with a square. The whole square is a single subnet comprised of 256 addresses. /24 255.255.255.0 256 Hosts 1 Subnet Split the box in half and you get two subnets with 128 addresses, /25 255.255.255.128 128 Hosts 2 Subnets Divide the box into quarters and you get four subnets with 64 addresses, /26 255.255.255.192 64 Hosts 4 Subnets 80
- 83. Split each individual square and you get eight subnets with 32 addresses, /27 255.255.255.224 32 Hosts 8 Subnets Split the boxes in half again and you get sixteen subnets with sixteen addresses, /28 255.255.255.240 16 Hosts 16 Subnets The next split gives you thirty two subnets with eight addresses, /29 255.255.255.248 8 Hosts 32 Subnets The last split gives sixty four subnets with four addresses each, /30 255.255.255.252 4 Hosts 64 Subnets 81
- 84. Class A Addressing Guide # of Bits Subnet Total # of Total # of Usable # of CIDR Borrowed Mask Subnets Hosts Hosts ______________________________________________________________________________________________ /8 0 255.0.0.0 1 16,777,216 16,777,214 _____________________________________________________________________________________________ /9 1 255.128.0.0 2 8,388,608 8,388,606 _____________________________________________________________________________________________ /10 2 255.192.0.0 4 4,194,304 4,194,302 __________________________________________________________________________________________________ /11 3 255.224.0.0 8 2,097,152 2,097,150 ______________________________________________________________________________________________ /12 4 255.240.0.0 16 1,048,576 1,048,574 _____________________________________________________________________________________________ /13 5 255.248.0.0 32 524,288 524,286 ________________________________________________________________________________________________ /14 6 255.252.0.0 64 262,144 262,142 ______________________________________________________________________________________________ /15 7 255.254.0.0 128 131,072 131,070 __________________________________________________________________________________________________ /16 8 255.255.0.0 256 65,536 65,534 ___________________________________________________________________________________________________ /17 9 255.255.128.0 512 32,768 32,766 _______________________________________________________________________________________________ /18 10 255.255.192.0 1,024 16,384 16,382 _____________________________________________________________________________________________________ /19 11 255.255.224.0 2,048 8,192 8,190 ______________________________________________________________________________________________ /20 12 255.255.240.0 4,096 4,096 4,094 __________________________________________________________________________________________________ /21 13 255.255.248.0 8,192 2,048 2,046 _________________________________________________________________________________________________ /22 14 255.255.252.0 16,384 1,024 1,022 ________________________________________________________________________________________________ /23 15 255.255.254.0 32,768 512 510 ____________________________________________________________________________________________________ /24 16 255.255.255.0 65,536 256 254 _____________________________________________________________________________________________________ /25 17 255.255.255.128 131,072 128 126 ____________________________________________________________________________________________________ /26 18 255.255.255.192 262,144 64 62 ___________________________________________________________________________________________________ /27 19 255.255.255.224 524,288 32 30 ____________________________________________________________________________________________________ /28 20 255.255.255.240 1,048,576 16 14 ____________________________________________________________________________________________________ /29 21 255.255.255.248 2,097,152 8 6 ________________________________________________________________________________________________ /30 22 255.255.255.252 4,194,304 4 2 Class B Addressing Guide # of Bits Subnet Total # of Total # of Usable # of CIDR Borrowed Mask Subnets Hosts Hosts ______________________________________________________________________________________________ /16 0 255.255.0.0 1 65,536 65,534 _____________________________________________________________________________________________ /17 1 255.255.128.0 2 32,768 32,766 _____________________________________________________________________________________________ /18 2 255.255.192.0 4 16,384 16,382 __________________________________________________________________________________________________ /19 3 255.255.224.0 8 8,192 8,190 ______________________________________________________________________________________________ /20 4 255.255.240.0 16 4,096 4,094 _____________________________________________________________________________________________ /21 5 255.255.248.0 32 2,048 2,046 ________________________________________________________________________________________________ /22 6 255.255.252.0 64 1,024 1,022 ______________________________________________________________________________________________ /23 7 255.255.254.0 128 512 510 __________________________________________________________________________________________________ /24 8 255.255.255.0 256 256 254 ___________________________________________________________________________________________________ /25 9 255.255.255.128 512 128 126 _______________________________________________________________________________________________ /26 10 255.255.255.192 1,024 64 62 _____________________________________________________________________________________________________ /27 11 255.255.255.224 2,048 32 30 ______________________________________________________________________________________________ /28 12 255.255.255.240 4,096 16 14 ______________________________________________________________________________________________ /29 13 255.255.255.248 8,192 8 6 ________________________________________________________________________________________________ /30 14 255.255.255.252 16,384 4 2 Class C Addressing Guide # of Bits Subnet Total # of Total # of Usable # of CIDR Borrowed Mask Subnets Hosts Hosts ______________________________________________________________________________________________ /24 0 255.255.255.0 1 256 254 _____________________________________________________________________________________________ /25 1 255.255.255.128 2 128 126 _____________________________________________________________________________________________ /26 2 255.255.255.192 4 64 62 __________________________________________________________________________________________________ /27 3 255.255.255.224 8 32 30 ______________________________________________________________________________________________ /28 4 255.255.255.240 16 16 14 _____________________________________________________________________________________________ /29 5 255.255.255.248 32 8 6 ________________________________________________________________________________________________ /30 6 255.255.255.252 64 4 2 82
- 85. Inside Cover