IP Addressing and Subnetting Workbook Version.5 Student Name:
IP Address Classes Class A 27 (Network 27 is reserved for loopback and internal testing) Leading bit pattern... Network. Host. Host. Host Class B 28 9 Leading bit pattern... Network. Network. Host. Host Class C 92 223 Leading bit pattern... Network. Network. Network. Host Class D 224 239 (Reserved for multicast) Class E 24 255 (Reserved for experimental, used for research) Class A... to.255.255.255 Class B 72.6.. to 72.3.255.255 Class C 92.68.. to 92.68.255.255 Private Address Space Class A 255... Class B 255.255.. Class C 255.255.255. Default Subnet Masks 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. Inside Cover Workbooks included in the series: IP Addressing and Subnetting Workbooks ACLs - Access Lists Workbooks VLSM Variable-Length Subnet Mask IWorkbooks
Binary To Decimal Conversion 28 64 32 6 8 4 2 Answers Scratch Area 46 9 28 6 2 46 64 32 6 4 2 9
Decimal To Binary Conversion Use all 8 bits for each problem 28 64 32 6 8 4 2 = 255 Scratch Area 238 238 34-28 -32 34 2-64 -2 23 46-32 5 4-8 255 6-4 2 2-2 38 3 25 7 224 4 92 72 9 57 98 79 2 2
Address Class Identification Address Class.25.. A 5..5. B 92.4.2. 48.7.9. 93.42.. 26.8.56. 22.2.23. 23.23.45.58 77..8.4 9.8.45. 249.24.8.78 99.55.77.56 7.89.56.45 25.45.45. 99.2.5. 95..2.9 33... 58.98.8. 29.2.56. 3
Network & Host Identification Circle the network portion of these addresses: 77..8.4 9.8.45. 29.24.8.78 99.55.77.56 7.89.56.45 25.45.45. 92.2.5. 95..2.9 33... 58.98.8. 27.2.56..25.. 5..5. 92.4.2. 48.7.9. 93.42.. 26.8.56. 22.2.23. Circle the host portion of these addresses:.5.23.5 7.2.99.3 98.25.87.77 223.25.2.222 7.45.222.45 26.2.54.23 9.4.35.2 55.25.69.227 92.5.55.2 23.2.45.254 48.7.9.55.25.. 95..2.98 25.25.35.46 7.2.77.77 55.25.5.5 28.55.23.4.25.. 4
Network Addresses Using the IP address and subnet mask shown write out the network address: 88..8.2 _ 88... 255.255....48.8 _.. 48. 255.255.255. 92.49.24.9 _ 255.255.255. 5.23.23.9 _ 255.255..... _ 255... 86.3.23. _ 255.255.255. 223.69.23.25 _ 255.255.. 2.2.35.5 _ 255.255.255. 27.25.2.5 _ 255... 99.2.5.35 _ 255.255.255. 9.55.65.35 _ 255.255.255. 28.22.25.254 _ 255.255.. 5
Host Addresses Using the IP address and subnet mask shown write out the host address: 88..8.2 _.. 8. 2 255.255....48.8 _... 8 255.255.255. 222.49.49. _ 255.255.255. 28.23.23.9 _ 255.255..... _ 255... 2.3.23. _ 255.255.255. 223.69.23.2 _ 255.255.. 23.2.35.25 _ 255.255.255. 7.5.2.5 _ 255... 99.2.5.35 _ 255.255.255. 9.55.65.35 _ 255.255.255. 48.2.25.54 _ 255.255.. 6
Default Subnet Masks Write the correct default subnet mask for each of the following addresses: 77..8.4 _ 255. 255.. 9.8.45. _ 255... 9.249.234.9 _ 223.23.223.9 _..25. _ 26.23.23. _ 223.69.23.25 _ 92.2.35.5 _ 77.25.2.5 _ 89.2.5. _ 88.45.65.35 _ 28.22.25.254 _ 93..77.83 _ 25.25.25. _...5 _ 22.9.3.45 _ 34.25.34.9 _ 95.25.9.99 _ 7
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... Class B 255.255.. Class C 255.255.255. ANDING Equations: AND = AND = AND = AND = Sample: What you see... IP Address: 92... 33 What you can figure out in your head... Address Class: C Network Portion: 92... 33 Host Portion: 92... 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: Default Subnet Mask: AND:......... (92... 33) (255. 255. 255. ) (92... ) ANDING with the default subnet mask allows your computer to figure out the network portion of the address. 8
ANDING With Custom subnet masks When you take a single network such as 92... and divide it into five smaller networks (92...6, 92...32, 92...48, 92...64, 92...8) the outside world still sees the network as 92..., 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 4 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: 92... Custom Subnet Mask: 255.255.255.24 Address Ranges: 92... to 92...5 92...6 to 92...3 92...32 to 92...47 (Range in the sample below) 92...48 to 92...63 92...64 to 92...79 92...8 to 92...95 92...96 to 92... 92...2 to 92...27 92...28 to 92...43 92...44 to 92...59 92...6 to 92...75 92...76 to 92...9 92...92 to 92...27 92...28 to 92...223 92...224 to 92...239 92...24 to 92...255 IP Address: Custom Subnet Mask: AND: Network Sub Network Host......... (92... 33) (255. 255. 255. 24) (92... 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
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 = 2 s - 2) h Number of hosts per subnet = 2-2 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 3 three bits. This would be 2 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 25or 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. 95. 223. 5. The number of subnets created by borrowing 2 2 bits is 2 or 2 x 2 = 4 subnets. The number of hosts created by leaving 6 bits is 26-2 or 2 x 2 x 2 x 2 x 2 x 2 = 64-2 = 62 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.
Class C Address unsubnetted: 95. 223. 5. 95.223.5. to 95.223.5.255 Class C Address subnetted (2 bits borrowed): Notice that the subnet and broadcast addresses match. 95. 223. 5. (Invalid range) (Invalid range) () () (2) (3) 95.223.5. 95.223.5.64 95.223.5.28 95.223.5.92 to to to to 95.223.5.63 95.223.5.27 95.223.5.9 95.223.5.255 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 95.223.255 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 Use the 2-2 formula and don t use the zero and broadcast ranges if... s Use the 2 formula and use the zero and broadcast ranges if... Classful routing is used RIP version is used The no ip subnet zero command is configured on your router Classless routing or VLSM is used RIP version 2, EIGRP, or OSPF is used The ip subnet zero command is 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.
Problem Number of needed subnets Number of needed usable hosts Network Address Custom Subnet Masks 4 4 92... 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 _ C 255. 255. 255. 255. 255. 255. 24 6 6 4 4 Show your work for Problem in the space below. Number of Number of 256 28 64 32 6 8 4 2 - Hosts Subnets - 2 4 8 6 32 64 28 256 28 64 32 6 8 4 2 - Binary values 92... Add the binary value numbers to the left of the line to create the custom subnet mask. 28 64 32 +6 24 6-2 4 Observe the total number of hosts. Subtract 2 for the number of usable hosts. 2
Problem 2 Number of needed subnets Number of needed usable hosts Network Address Custom Subnet Masks 6 65... 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 _ B 255. 255.. 255. 255. 255. 92,24 64 62 Show your work for Problem 2 in the space below. 32,768 65,536 6,384,24 2,48 4,96 8,92 52 Number of Hosts -. 256 28 64 32 6 8 4 2 65,536 6,384 32,768 4,96 24 248 8,92 Number of Subnets - 2 4 8 6 32 64 28 256. Binary values - 28 64 32 6 8 4 2 65... Add the binary value numbers to the left of the line to create the custom subnet mask. 28 64 32 6 8 4 2. 28 64 32 6 8 4 2 + 255 28 +64 92 52 64-2 62 Observe the total number of hosts. Subtract 2 for the number of usable hosts. 3
Custom Subnet Masks Problem 3 Network Address Address class 48.75.. /26 _ B /26 indicates the total number of bits used for the network and subnetwork portion of the address. All bits remaining belong to the host portion of the address. Default subnet mask Custom subnet mask Total number of subnets Total number of host addresses Number of usable addresses Number of bits borrowed 255. 255.. 255. 255. 255. 92,24 64 62 Show your work for Problem 3 in the space below. 6,384 32,768 65,536,24 2,48 4,96 8,92 52 Number of Hosts -. 256 28 64 32 6 8 4 2 65,536 6,384 32,768 4,96 24 248 8,92 Number of Subnets - 2 4 8 6 32 64 28 256. Binary values - 28 64 32 6 8 4 2 48. 75.. Add the binary value numbers to the left of the line to create the custom subnet mask. 4 28 64 32 6 8 4 2. 28 64 32 6 8 4 2 + 255 28 +64 92 24-2,22 52 64-2 62 Subtract 2 for the total number of subnets to get the usable number of subnets. Observe the total number of hosts. Subtract 2 for the number of usable hosts.
Problem 4 Number of needed subnets Number of needed usable hosts Network Address Custom Subnet Masks 6 3 2..56. 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 Number of 256 28 64 32 6 8 4 2 - Hosts Subnets - 2 4 8 6 32 64 28 256 28 64 32 6 8 4 2 - Binary values 2.. 56. 5
Problem 5 Number of needed subnets Number of needed usable hosts Network Address Custom Subnet Masks 6 3 95.85.8. 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 Number of 256 28 64 32 6 8 4 2 - Hosts Subnets - 2 4 8 6 32 64 28 256 28 64 32 6 8 4 2 - Binary values 95. 85. 8. 6
Problem 6 Number of needed subnets Number of needed usable hosts Network Address Custom Subnet Masks 26 3,7 8... 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.,48,576 2,97,52 4,94,34 262,44 524,288 3,72 65,536 32,768 6,384,24 2,48 4,96 8,92 Number of Hosts - 52. 256 28 64 32 6 8 4 2 2,97,52,48,576 524,288 262,44 4,94,34 3,72 65,536 6,384 32,768,24 2,48 4,96 Number of Subnets - 2 4 8 6 32 64 28 256. 52 Binary values -28 64 32 6 8 4 2. 28 64 32 6 8 4 2. 28 64 32 6 8 4 2 8... 8,92. 7
Problem 7 Number of needed subnets Number of needed usable hosts Network Address Custom Subnet Masks 2 5 78... 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. 32,768 65,536 6,384,24 2,48 4,96 8,92 52 Number of Hosts -. 256 28 64 32 6 8 4 2 65,536 6,384 32,768 4,96 Number of Subnets - 2 4 8 6 32 64 28 256. Binary values - 28 64 32 6 8 4 2. 28 64 32 6 8 4 2 78... 52 24 248 8,92 8
Problem 8 Number of needed subnets Number of needed usable hosts Network Address Custom Subnet Masks 3 45 2.75.4. 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. 9
Problem 9 Number of needed subnets Number of needed usable hosts Network Address Custom Subnet Masks 6, 28.77.. 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. 2
Problem Number of needed usable hosts Network Address Custom Subnet Masks 6 98... 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 in the space below. 2
Problem Number of needed subnets Network Address Custom Subnet Masks 25... 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 in the space below. 22
Problem 2 Number of needed subnets Network Address Custom Subnet Masks 5 28.35.5. 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 2 in the space below. 23
Problem 3 Number of needed usable hosts Network Address Custom Subnet Masks 25 28.35.5. 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 3 in the space below. 24
Problem 4 Number of needed subnets Network Address Custom Subnet Masks 72.59.. 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. 25
Problem 5 Number of needed usable hosts Network Address Custom Subnet Masks 5 72.59.. 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. 26
Problem 6 Number of needed usable hosts Network Address Custom Subnet Masks 29 23... 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. 27
Problem Number of needed subnets Number of needed usable hosts Network Address Subnetting 4 4 92... 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 _ C 255. 255. 255. 255. 255. 255. 24 6 6 4 4 What is the 4th subnet range? 92...48 to 92...63 What is the subnet number for the 8th subnet? What is the subnet broadcast address for the 3th subnet? 92... 2 92... 27 What are the assignable addresses for the 9th subnet? 92...29 to 92...42 28
Show your work for Problem in the space below. Number of Number of 256 28 64 32 6 8 4 2 - Hosts Subnets - 2 4 8 6 32 64 28 256 28 64 32 6 8 4 2 - Binary values 92... () () (2) (3) (4) (5) (6) (7) (8) (9) () () (2) (3) (4) (5) 92... to 92...5 92...6 to 92...3 92...32 to 92...47 92...48 to 92...63 92...64 to 92...79 92...8 to 92...95 92...96 to 92... 92...2 to 92...27 92...28 to 92...43 92...44 to 92...59 92...6 to 92...75 92...76 to 92...9 92...92 to 92...27 92...28 to 92...223 92...224 to 92...239 92...24 to 92...255 Custom subnet mask 28 64 32 +6 24 Usable subnets 6-2 4 Usable hosts 6-2 4 The binary value of the last bit borrowed is the range. In this problem the range is 6. The first address in each subnet range is the subnet number. The last address in each subnet range is the subnet broadcast address. 29
Problem 2 Number of needed subnets Number of needed usable hosts Network Address Subnetting 6 65... 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 _ B 255. 255.. 255. 255. 255. 92,24 64 62 What is the 5th subnet range? 65..3.28 to 65..3.9 What is the subnet number for the 6th subnet? What is the subnet broadcast address for the 6th subnet? 65... 64 65... 27 What are the assignable addresses for the 9th subnet? 65..2. to 65...62 3
Show your work for Problem 2 in the space below.. 256 28 64 32 6 8 4 2 52,24 2,48 4,96 8,92 6,384 32,768 65,536 Number of Hosts - 65,536 32,768 6,384 8,92 4,96 248 24 52 Number of Subnets - 2 4 8 6 32 64 28 256. 28 64 32 6 8 4 2 28 64 32 6 8 4 2. Binary values - 65.... 65... to 65...63 65...64 to 65...27 65...28 to 65...9 65...92 to 65...255 64-2 62 Usable hosts 65... to 65...63 65...64 to 65...27 65...28 to 65...9 65...92 to 65...255.... 65..2. to 65...63 65..2.64 to 65...27 65..2.28 to 65...9 65..2.92 to 65...255 28 64 32 6 8 4 2 + 255 28 +64 92 Custom subnet mask.... The binary value of the last bit borrowed is the range. In this problem the range is 64. The first address in each subnet range is the subnet number. 65..3. to 65..3.63 65..3.64 to 65..3.27 65..3.28 to 65..3.9 65..3.92 to 65..3.255.... () () (2) (3) (4) (5) (6) (7) (8) (9) () () (2) (3) (4) (5) The last address in each subnet range is the subnet broadcast address. Down to 65..255.28 to 65..255.9 65..255.92 to 65..255.255.. (22) (23) 3
Problem 3 Number of needed subnets Network Address Subnetting 2 95.223.5. 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 st subnet? What are the assignable addresses for the 3rd subnet? 32
Show your work for Problem 3 in the space below. Number of Number of 256 28 64 32 6 8 4 2 - Hosts Subnets - 2 4 8 6 32 64 28 256 28 64 32 6 8 4 2 - Binary values 95. 223. 5. 33
Problem 4 Number of needed subnets Network Address Subnetting 75 9.35.. 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 3th subnet? What is the subnet broadcast address for the th subnet? What are the assignable addresses for the 6th subnet? 34
Show your work for Problem 4 in the space below. 35
Problem 5 Number of needed usable hosts Network Address Subnetting 6 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 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 th subnet? 36
Show your work for Problem 5 in the space below. 37
Problem 6 Subnetting Number of needed subnets Network Address 92.7.. 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 2th subnet? What are the assignable addresses for the th subnet? 38
Show your work for Problem 6 in the space below. 39
Subnetting Problem 7 Network Address Address class Default subnet mask Custom subnet mask... /6 Total number of subnets Total number of host addresses Number of usable addresses Number of bits borrowed What is the th 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? 4
Show your work for Problem 7 in the space below. 4
Problem 8 Number of needed subnets Network Address Subnetting 5 72.5.. 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
Show your work for Problem 8 in the space below. 43
Problem 9 Number of needed usable hosts Network Address Subnetting 28 72.5.. 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 th subnet? What is the subnet broadcast address for the 4th subnet? What are the assignable addresses for the 6th subnet? 44
Show your work for Problem 9 in the space below. 45
Problem Number of needed subnets Network Address Subnetting 45 22... 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 3th subnet? What are the assignable addresses for the 2th subnet? 46
Show your work for Problem in the space below. 47
Problem Number of needed usable hosts Network Address Subnetting 8, 35.7.. 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
Show your work for Problem in the space below. 49
Problem 2 Number of needed usable hosts Network Address Subnetting 45 98.25.5. 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? 5
Show your work for Problem 2 in the space below. 5
Subnetting Problem 3 Network Address Address class Default subnet mask Custom subnet mask 65.2.. /26 Total number of subnets Total number of host addresses Number of usable addresses Number of bits borrowed What is the th subnet range? What is the subnet number for the th subnet? What is the subnet broadcast address for the 23rd subnet? What are the assignable addresses for the 22nd subnet? 52
Show your work for Problem 3 in the space below. 53
Problem 4 Number of needed usable hosts Network Address Subnetting 6 2... 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
Show your work for Problem 4 in the space below. 55
Subnetting Problem 5 Network Address Address class Default subnet mask Custom subnet mask 93... \9 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 9th subnet? What is the subnet broadcast address for the 7th subnet? What are the assignable addresses for the 2th subnet? 56
Show your work for Problem 5 in the space below. 57
F/ Practical Subnetting 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 % growth in both areas. Circle each subnet on the graphic and answer the questions below. Router A IP Address 72.6.. S// S// F/ Router B F/ Marketing 24 Hosts Reasearch 6 Hosts Management 5 Hosts Address class Custom subnet mask Minimum number of subnets needed Extra subnets required for % 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 % growth in the largest subnet (Round up to the next whole number) _ B _ 255.255.224. 4 + 4 = 8 6 + 6 Total number of address needed for the largest subnet 2 Start with the first subnet and arrange your sub-networks from the largest group to the smallest. = IP address range for Research IP address range for Marketing IP address range for Management IP address range for Router A to Router B serial connection _ 72.6.. to 72.3.255 _ 72.6.32. to 72.63.255 _ 72.6.64. to 72.95.255 _ 72.6.96. to 72.27.255 58
Show your work for Practical Subnetting in the space below. Number of Subnets - 2 4 8 6 32 64 28 256.. 256 28 64 32 6 8 4 2 52,24 2,48 4,96 8,92 6,384 32,768 65,536 Number of Hosts - 4,96 248 24 52 65,536 32,768 6,384 8,92 Binary values - 28 64 32 6 8 4 2. 28 64 32 6 8 4 2 72. 6.. 4 x. 4 () () (2) (3) (4) (5) (6) (7) 72.6.. 72.6.32. 72.6.64. 72.6.96. 72.6.28. 72.6.6. 72.6.92. 72.6.224. to to to to to to to to 72.6.3.255 72.6.63.255 72.6.95.255 72.6.27.255 72.6.59.255 72.6.9.255 72.6.223.255 72.6.255.255 6 x. 6 59
F/ 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 3% growth in all areas. Circle each subnet on the graphic and answer the questions below. Router A S// IP Address 35.26.. S// S// F/ Router B F/ Science Lab Hosts Address class Custom subnet mask Minimum number of subnets needed Extra subnets required for 3% 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 3% growth in the largest subnet (Round up to the next whole number) Router C Tech Ed Lab 2 Hosts English Department 5 Hosts _ B _ 255.255.255.224 5 + 2 = S// F/ 7 2 + 6 Total number of address needed for the largest subnet 26 Start with the first subnet and arrange your sub-networks from the largest group to the smallest. = 6 IP address range for Tech Ed IP address range for English IP address range for Science IP address range for Router A to Router B serial connection _ 35.26.. to 35.26..3 _ 35.26..32 to 35.26..63 _ 35.26..64 to 35.26..95 35.26..96 _ to 35.26..27 IP address range for Router A to Router B serial connection35.26..28 _ to 35.26..59
Show your work for Problem 2 in the space below.. 256 28 64 32 6 8 4 2 52,24 2,48 4,96 8,92 6,384 32,768 65,536 Number of Hosts - 65,536 32,768 6,384 8,92 4,96 248 24 52 Number of Subnets - 2 4 8 6 32 64 28 256. 28 64 32 6 8 4 2 28 64 32 6 8 4 2. 35.26..3 35.26..63 35.26..95 35.26..27 35.26..59 35.26..9 35.26..223 35.26..255 35.26..3 35.26..63 35.26..95 35.26..27 35.26..59 35.26..9 35.26..223 35.26..255 Binary values - 35. 26.. to to to to to to to to to to to to to to to to 35.26.. 35.26..32 35.26..64 35.26..96 35.26..28 35.26..6 35.26..92 35.26..224 35.26.. 35.26..32 35.26..64 35.26..96 35.26..28 35.26..6 35.26..92 35.26.224......... () () (2) (3) (4) (5) (6) (7) (8) (9) () () (2) (3) (4) (5) 5 x.3.5 (Round up to 2) 2 x.3 6 6
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 72.6.. Administrative 3 Hosts F/ S// Router A F/ S// Router B F/ Sales 85 Hosts Marketing 5 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
Show your work for Problem 3 in the space below. 63
F/ S// Router A 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 7% growth in all areas. Circle each subnet on the graphic and answer the questions below. S// S// F/ F/ Dallas 5 Hosts New York Washington D.C. 325 Hosts 22 Hosts Address class _ Custom subnet mask Minimum number of subnets needed Extra subnets required for 7% 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 7% growth in the largest subnet (Round up to the next whole number) Total number of address needed for the largest subnet IP Address 35.26.. S// _ Router B Router C F/ Start with the first subnet and arrange your sub-networks from the largest group to the smallest. + = + = 64 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 _
Show your work for Problem 4 in the space below. 65
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 % growth in all areas. Circle each subnet on the graphic and answer the questions below. IP Address 2.5.. F/ F/ Science Room Hosts Tech Ed Lab 8 Hosts English classroom 5 Hosts Art Classroom 2 Hosts Address class Custom subnet mask Minimum number of subnets needed Extra subnets required for % 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 % 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 F/ Port IP address range for Router F/ Port 66
Show your work for Problem 5 in the space below. 67
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 2% growth in all areas. Circle each subnet on the graphic and answer the questions below. Router A F/ S// S// S// Art & Drama 75 Hosts F/ IP Address... Router C S// F/ S// S// Administration 35 Hosts Router B F/ Technology Building 32 Hosts Science Building 225 Hosts Address class Custom subnet mask Minimum number of subnets needed Extra subnets required for 2% 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
Show your work for Problem 6 in the space below. 69
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 25% growth in all areas. Circle each subnet on the graphic and answer the questions below. IP Address 77.35.. S// Router A S// F/ Router B F/ F/ Marketing 75 Hosts Administration 33 Hosts Sales 255 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. + = + = Research 35 Hosts Deployment 63 Hosts IP address range for Router A Port F/ IP address range for Research IP address range for Deployment IP address range for Router A to Router B serial connection 7
Show your work for Problem 7 in the space below. 7
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 92.68.. F/ S// Router A S// F/ Router B F/ 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 F/ IP address range for New York IP address range for Router A to Router B serial connection _ 72
Show your work for Problem 8 in the space below. 73
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 5% growth in all areas. Circle each subnet on the graphic and answer the questions below. Router A S// IP Address 48.55.. S// S// F/ Router B F/ F/ Router D S// S// Router C S// Dallas 5 Hosts Ft. Worth 23 Hosts Address class Custom subnet mask Minimum number of subnets needed Extra subnets required for 5% 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 5% 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. = 74 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 to Router D serial connection _
Show your work for Problem 9 in the space below. 75
Practical Subnetting 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 % growth in all areas. Circle each subnet on the graphic and answer the questions below. Sales 5 Hosts F/ Router A IP Address 72.6.. S// S// Marketing 56 Hosts F/ Router B F/ Management 25 Hosts Research 35 Hosts Address class Custom subnet mask Minimum number of subnets needed Extra subnets required for % 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 % 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
Show your work for Problem in the space below. 77
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:.23.9.92 Subnet Mask: 255... Reference Page Inside Front Cover IP Address: 92... Subnet Mask: 255.255.255. Reference Pages 28-29 IP Address: 245.5.9. Subnet Mask: 255.255.255. Reference Page Inside Front Cover IP Address: 35.7.9.255 Subnet Mask: 255.255.254. Reference Pages 48-49 IP Address: 27... Subnet Mask: 255... Reference Pages Inside Front Cover IP Address: 93..28. Subnet Mask: 255.255.224. Reference Pages 56-57 IP Address: 2...28 Subnet Mask: 255.255.255.224 Reference Pages 54-55 IP Address: 65..255.89 Subnet Mask: 255.255.255.92 Reference Pages 3-3 IP Address: 9.35.. Subnet Mask: 255.255.255.92 Reference Pages 34-35 IP Address: 28.35.5.95 Subnet Mask: 255.255.. Reference Page Inside Front Cover IP Address: 2...75 /22 Reference Pages 54-55 and/or Inside Front Cover IP Address: 35.7.255.255 Subnet Mask: 255.255.224. Reference Pages 48-49 78 The network ID cannot be. OK
-27 28-255 -3 4-7 8-2-5 6-9 2-23 24-27 28-3 32-35 36-39 4-43 44-47 48-5 52-55 56-59 6-63 64-67 68-7 72-75 76-79 8-83 84-87 88-9 92-95 96-99 -3 4-7 8-2-5 6-9 2-23 24-27 28-3 32-35 36-39 4-43 44-47 48-5 52-55 56-59 6-63 64-67 68-7 72-75 76-79 8-83 84-87 88-9 92-95 96-99 2-23 24-27 28-2 22-25 26-29 22-223 224-227 228-23 232-235 236-239 24-243 244-247 248-25 252-255 /3 8+8+8+6 255.255.255.252 4 Hosts /29 8+8+8+5 255.255.255.248 8 Hosts /28 8+8+8+4 255.255.255.24 6 Hosts /27 8+8+8+3 255.255.255.224 32 Hosts /26 8+8+8+2 255.255.255.92 64 Hosts /25 8+8+8+ 255.255.255.28 28 Hosts /24 8+8+8 255.255.255. 256 Hosts -7 8-5 6-23 24-3 32-39 4-47 48-55 56-63 64-7 72-79 8-87 88-95 96-3 4-2-9 2-27 28-35 36-43 44-5 52-59 6-67 68-75 76-83 84-9 92-99 2-27 28-25 26-223 224-23 232-239 24-247 248-255 -5 6-3 32-47 48-63 64-79 8-95 96-2-27 28-43 44-59 6-75 76-9 92-27 28-223 224-239 24-255 -63 64-27 28-9 92-255 -255 IP Address Breakdown 79
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. 256 Hosts Subnet Split the box in half and you get two subnets with 28 addresses, /25 255.255.255.28 28 Hosts 2 Subnets Divide the box into quarters and you get four subnets with 64 addresses, 8 /26 255.255.255.92 64 Hosts 4 Subnets
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, /28 255.255.255.24 6 Hosts 6 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, /3 255.255.255.252 4 Hosts 64 Subnets 8
82 # of Bits Borrowed 2 3 4 5 6 7 8 9 2 3 4 5 6 7 8 9 2 2 22 Subnet Mask 255... 255.28.. 255.92.. 255.224.. 255.24.. 255.248.. 255.252.. 255.254.. 255.255.. 255.255.28. 255.255.92. 255.255.224. 255.255.24. 255.255.248. 255.255.252. 255.255.254. 255.255.255. 255.255.255.28 255.255.255.92 255.255.255.224 255.255.255.24 255.255.255.248 255.255.255.252 Total # of Subnets 2 4 8 6 32 64 28 256 52,24 2,48 4,96 8,92 6,384 32,768 65,536 3,72 262,44 524,288,48,576 2,97,52 4,94,34 Total # of Hosts 6,777,26 8,388,68 4,94,34 2,97,52,48,576 524,288 262,44 3,72 65,536 32,768 6,384 8,92 4,96 2,48,24 52 256 28 64 32 6 8 4 Usable # of Hosts 6,777,24 8,388,66 4,94,32 2,97,5,48,574 524,286 262,42 3,7 65,534 32,766 6,382 8,9 4,94 2,46,22 5 254 26 62 3 4 6 2 # of Bits Borrowed 2 3 4 5 6 7 8 9 2 3 4 Subnet Mask 255.255.. 255.255.28. 255.255.92. 255.255.224. 255.255.24. 255.255.248. 255.255.252. 255.255.254. 255.255.255. 255.255.255.28 255.255.255.92 255.255.255.224 255.255.255.24 255.255.255.248 255.255.255.252 Total # of Subnets 2 4 8 6 32 64 28 256 52,24 2,48 4,96 8,92 6,384 Total # of Hosts 65,536 32,768 6,384 8,92 4,96 2,48,24 52 256 28 64 32 6 8 4 Usable # of Hosts 65,534 32,766 6,382 8,9 4,94 2,46,22 5 254 26 62 3 4 6 2 Class C Addressing Guide # of Bits Borrowed 2 3 4 5 6 Subnet Mask 255.255.255. 255.255.255.28 255.255.255.92 255.255.255.224 255.255.255.24 255.255.255.248 255.255.255.252 Total # of Subnets 2 4 8 6 32 64 Total # of Hosts 256 28 64 32 6 8 4 Usable # of Hosts 254 26 62 3 4 6 2 CIDR /8 /9 / / /2 /3 /4 /5 /6 /7 /8 /9 /2 /2 /22 /23 /24 /25 /26 /27 /28 /29 /3 CIDR /6 /7 /8 /9 /2 /2 /22 /23 /24 /25 /26 /27 /28 /29 /3 CIDR /24 /25 /26 /27 /28 /29 /3 Class B Addressing Guide Class A Addressing Guide
Inside Cover