IPV4 Subnetting
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Questions


Scenario 1. Find subnets required for this network

128 64 32 16 8 4 2 1

1. An IP address has two parts, the network ID part and the host ID part
2. The network ID for 216.21.5.0 is 216.21.5.0
3. This network can't have a host ID of 0 since the lowest hostnumber of 0 is reserved for network ID
4. This network can't have a host ID of 255 since the highest hostnumber of 255 is reserved for broadcasting
5. The range of host numbers available on this network is 1 to 254
6. There are five networks shown on this diagram
7. There are networks on router1, router2, router3, and between router1 and router2, and between router 2 and router3
8. The five networks on this diagram is represented by the binary number 101
9. The default subnet for this network is 255.255.255.0
10. The binary number for 255.255.255.0 is 11111111. 11111111. 11111111. 00000000
11. The number 255 of the default subnet mask represents the network portion
12. The number 0 of the default subnet mask represents the host portion
13. In binary, each number 1 of the default subnet mask represents the network portion
14. In binary, each number 0 of the default subnet mask represents the host portion
15. The number of networks needed is 5 which is 101 in binary. 101 had three bits
16. The new subnet mask, containing the three bits is 11111111.11111111.11111111.11100000 or 224
17. Each of the segmented networks have 32 IP numbers
18. The segmented subnet is 11100000 which has 5 zeroes. 16+8+4+2+1 is 223 but 0 is also part of the network. Total is 224.
19. The first segment starts at the network ID of 216.21.5.0 and ends with 216.21.5.31 when you add the increment of 32.
20. The next four segments starting ranges are: 216.21.5.32, 216.21.5.64, 216.21.5.96, 216.21.5.128 (increment by 32)
21. The next four segments ending ranges are: 216.21.5.63, 216.21.5.95, 216.21.5.127, 216.21.5.159 (increment by 32)


Scenario 2. Subnet Class C 195.5.20.0 into 50 networks

128 64 32 16 8 4 2 1

22. The number of networks is given at 50. The binary representation of 50 is 00110010 which requires 6 bits.
23. The subnet mask is 255.255.255.252 and the increment is 4.
24. The first two subnet ranges are: 195.5.20.0 to 195.5.20.3 and 195.5.20.4 to 195.5.20.7
25. The number of hosts on each network is two since the increment is 4 and you can't assign the first or the last IP in each range
26. The valid host numbers on the second network segment is 195.5.20.5 and 195.5.20.6
27. A network work with a maximum of two hosts is ideal for assigning to routers which only require two IP numbers.
28. The subnet mask of 252 is the second most popular subnet mask due to their use on routers.
29. A shorthand notation for writing 195.5.20.0 with a subnet of 255.255.255.0 is: 195.5.20.0/24 (24 represents subnet bits)
30. The shorthand notation is known as CIDR Notation. Classless Inter-Domain Routing (CIDR)
31. Class A = /8, Class B = /16, Class C = /24
32. The shorthand notation for this example is /30. 8 for the first three octets and 6 for the last octet = 30.


Scenario 3. Subnet Class B 150.5.0.0 into 100 networks

128 64 32 16 8 4 2 1

33. The number of networks is given as 100. The binary representation of 100 requires 7 bits.
34. The subnet mask is 255.255.254.0 and the increment is 2 since you are only looking at the octet of the subnet (octet 3)
35. The first two subnet ranges are: 150.5.0.0 - 150.5.1.255 and 150.5.2.0 to 150.5.3.255
36. The number of hosts on each network is 512
37. The network ID for the second subnet is 150.5.2.0 and the broadcast ID is 150.5.3.255
38. The actual number of IP addresses you can use in one of these subnets is 512 minus 2 for the network ID and broadcast = 510


Scenario 4. Subnet Class A 10.0.0.0 into 500 networks

128 64 32 16 8 4 2 1

39. The number of networks is given as 500. The binary representation of 500 requires 9 bits
40. The subnet mask is 255.255.128.0 and the increment is 128.
41. The first two subnet ranges are 10.0.0.0 to 10.0.127.255 and 10.0.128.0 to 10.0.255.255
42. The range for the third subnet is 10.1.0.0 to 10.1.127.255
43. The beginning ranges for the first six subnets are 10.0.0.0, 10.0.128.0, 10.1.0.0, 10.1.128, 10.2.0.0 and 10.2.128.0


Scenario 5. Subnet Class C 216.21.5.0 into segments of 30 hosts

128 64 32 16 8 4 2 1

44. The number of hosts is given as 30. The binary representation of 30 requires 5 bits.
45. The subnet equals 8 minus the host bits of 5 = 3 bits. 11100000 is the subnet of 255.255.255.224. The increment is 32
46. The first two subnet ranges are 216.21.5.0 to 216.21.5.31 and 216.21.5.32 to 216.21.5.63


Scenario 6. Subnet Class C 195.5.20.0 into segments of 50 hosts

128 64 32 16 8 4 2 1

47. The number of hosts is given as 50. The binary representation of 50 requires 6 bits.
48. The subnet equals 8 minus the host bits of 6 = 2 bits. 11000000 is the subnet of 255.255.255.192 The increment is 64
49. The first two subnet ranges are 195.5.20.0 to 195.5.20.63 and 195.5.20.64. to 195.5.20.127


Scenario 7. Subnet Class B 150.5.0.0 into segments of 500 hosts

128 64 32 16 8 4 2 1

50. The number of hosts is given as 500. The binary representation of 500 requires 9 bits.
51. The subnet equals 255.255.255.254. The increment is 2
52. The first two subnet ranges are 150.5.0.0 to 150.5.1.255 to 150.5.2.0 to 150.5.3.255


Scenario 8. Subnet Class B 10.0.0.0 into segments of 100 hosts

128 64 32 16 8 4 2 1

53. The number of hosts is given as 100. The binary representation of 100 requires 8 bits.
54. The subnet equals 255.255.255.128 The increment is 128
55. The first six beginning ranges are 10.0.0.0, 10.0.0.128, 10.0.1.0, 10.0.1.128, 10.0.2.0, 10.0.2.128


Scenario 9. Reverse Engineering 1

Given: IP 192.168.1.127 and Mask: 255.255.255.224

128 64 32 16 8 4 2 1

56. The binary representation of subnet mask 255.255.255.224 is 11111111 . 11111111 . 11111111 . 11100000
57. The increment number is 32
58. The first five beginning subnet ranges are 192.168.1.0, 192.168.1.32, 192.168.1.64, 192.168.1.96, 192.168.1.128
59. The IP number 192.168.1.127 is the broadcast number for the fifth segment and can't be assigned to a computer, router or device.


Scenario 10. Reverse Engineering 2 - PC1 can't reach the Gateway

Given PC1: IP 172.16.68.65, Mask: 255.255.255.240, Gateway 172.16.68.62

Given Gateway: IP 172.16.68.62, Mask 255.255.255.240

128 64 32 16 8 4 2 1

60. The binary representation of subnet mask 255.255.255.240 is 11111111 . 11111111 . 11111111 . 11110000
61. The increment number is 16
62. The first six beginning subnet ranges are 172.16.68.0, 172.16.68.16, 172.16.68.32, 172.16.68.48, 172.16.68.64, 172.16.68.80
63. The Gateway IP number of 172.16.68.62 is on another network from PC1 IP number of 172.16.68.65


Common Subnetting Problems

64. The Class C mask if you want to subnet into four segments is 255.255.255.192. You can identify four segments with only two bits. 00, 01, 10, 11
65. The Class C mask if you want as many segments of two IP numbers is 255.255.255.192. You have to remember that you need to subtract the broadcast number and the network number when calculating how many IP numbers are available for assignment.