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IP Subnetting VLSMs And Troubleshooting IP

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Answer 50 questions in 15 minutes.
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Match each statement with the correct term.
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This is a study tool. The 3 wrong answers for each question are randomly chosen from answers to other questions. So, you might find at times the answers obvious, but you will see it re-enforces your understanding as you take the test each time.

1. Given 192.168.10.0/28 - How many subnets?
2
/25 is 255.255.255.128. The fourth octet is a block size of 128. 0 - 128. The host is in the 0 subnet - broadcast of 127. Valid host 1-126
A Class B network ID with a /22 mask is 255.255.252.0 - with a block size of 4 in the third octet. The network address in the question is in subnet 172.16.16.0 with a broadcast address of 172.16.19.255. Only option E even has the correct subnet mask
Create many networks using subnet masks of different lengths from one network

2. Why subnet?
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
All interfaces within the classful address space have the same subnet mask
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances
A 240 mask is 4 subnet bits and provides 16 subnets - each with 14 hosts. We need more subnets - so let's add subnet bits. One more subnet bit would be a 248 mask. This provides 5 subnet bits (32 subnets) with 3 hosts bits (6 host per subnet).

3. /26
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet
11000000 = 192
/29 is 255.255.255.248. The fourth octet is a block size of 8. 0 - 8 - 16 - 24. The host is in the 16 subnet - broadcast of 23. Valid host 17-22
4

4. How many hosts are available with a Class C /29 mask?
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet
11000000 = 192
Allows you to use the first and last subnet in your network design; turned this command on by default

5. On a VLSM network - which mask should you use on point-to-point WAN links in order to reduce the waste of IP addresses?
RIPv2 - EIGRP - or OSPF
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
A /30 - regardless of the class of address - has a 252 in the fourth octet. This means we have a block size of 4 and our subnets are 0 - 4 - 8 - 12 - 16 - etc. Address 14 is obviously in the 12 subnet.

6. What's the broadcast address of each subnet?
16 -382
The number just before the next subnet; the broadcast of the last subnet is always 255
A /28 is a 255.255.255.240 mask. The first subnet is 16 (remember that the question stated not to use subnet zero) and the next subnet is 32 - so our broadcast address is 31. This makes our host range 17-30. 30 is the last valid host.
11000000 = 192

7. Variable Length Subnet Masks (VLSMs)
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
Each network segment can use a different subnet mask
/27 is 255.255.255.224. The fourth octet is a block size of 32. Count by 32s until you pass the host address of 66. 0 - 32 - 64. The host is in the 32 subnet - broadcast address of 63. Valid host range of 33-62.
A /28 is a 255.255.255.240 mask. We need to count to the eighth subnet - not starting at subnet-zero. 16 - 32 - 48 - 64 - 80 - 96 - 112 - 128. The ninth subnet is 144 (we need this to help us find the 128 subnet broadcast address - which is 143).The

8. Classful Routing Protocols
RIPv1 and IGRP
/29 is 255.255.255.248. The fourth octet is a block size of 8. 0 - 8 - 16 - 24. The host is in the 16 subnet - broadcast of 23. Valid host 17-22
All interfaces within the classful address space have the same subnet mask
/30 is 255.255.255.252. The valid subnet is 192.168.100.24 - broadcast is 192.168.100.27 - and valid hosts are 192.168.100.25 and 26

9. Classless Routing Protocols
.1 - .126 & .129 - .254
RIPv2 - EIGRP - or OSPF
RIPv1 and IGRP
127 & 255

10. Given 172.16.0.0/18 - What are the valid hosts?
/29 is 255.255.255.248. This is a block size of 8 in the fourth octet. 0 - 8 - 16. The host is in the 8 subnet - broadcast is 15
/30 is 255.255.255.252. The valid subnet is 192.168.100.24 - broadcast is 192.168.100.27 - and valid hosts are 192.168.100.25 and 26
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254
/25 is 255.255.255.128. The fourth octet is a block size of 128. 0 - 128. The host is in the 0 subnet - broadcast of 127. Valid host 1-126

11. What subnet and broadcast address is the IP address 172.16.45.14 255.255.255.252 (/30) a member of?
A /28 is a 255.255.255.240 mask. We need to count to the eighth subnet - not starting at subnet-zero. 16 - 32 - 48 - 64 - 80 - 96 - 112 - 128. The ninth subnet is 144 (we need this to help us find the 128 subnet broadcast address - which is 143).The
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
The mask 255.255.254.0 (/23) used with a Class A means that there are 15 subnet bits and 9 hosts bits. The block size in the third octet is 2 (256 - 254). So this makes the subnets in the interesting octet 0 - 2 - 4 - 6 - etc. - all the way to 254. T
11111100 = 252

12. /30
11111100 = 252
A 240 mask is 4 subnet bits and provides 16 subnets - each with 14 hosts. We need more subnets - so let's add subnet bits. One more subnet bit would be a 248 mask. This provides 5 subnet bits (32 subnets) with 3 hosts bits (6 host per subnet).
Each network segment can use a different subnet mask
A /29 is 255.255.255.248 - which is a block size of 8 in the fourth octet. The subnets are 0 - 8 - 16 - 24 - 32 - 40 - etc. 192.168.19.24 is the 24 subnet - and since 32 is the next subnet - the broadcast address for the 24 subnet is 31. 192.168.19.2

13. Given 172.16.0.0/18 - How many hosts per subnet?
16 -382
A 240 mask is 4 subnet bits and provides 16 subnets - each with 14 hosts. We need more subnets - so let's add subnet bits. One more subnet bit would be a 248 mask. This provides 5 subnet bits (32 subnets) with 3 hosts bits (6 host per subnet).
All nodes in the network use the same subnet mask
/23 is 255.255.254.0. The third octet is a block size of 2. 0 - 2 - 4. The subnet is in the 16.2.0 subnet - the broadcast address is 16.3.255

14. You have a Class B network and need 29 subnets. What is your mask?
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
The routers IP address on the E0 interface is 172.16.2.1/23 - which is a 255.255.254.0. This makes the third octet a block size of 2. The routers interface is in the 2.0 subnet - the broadcast address is 3.255 because the next subnet is 4.0. The vali
2
32-bit value that allows the recipient of IP packets to distinguish the network ID portion of the IP address from the host ID portion of the IP address

15. Variable Length Subnet Masks (VLSMs)
11000000 = 192
/23 is 255.255.254.0. The third octet is a block size of 2. 0 - 2 - 4. The subnet is in the 16.2.0 subnet - the broadcast address is 16.3.255
Use different size masks on each router interface
A Class B network ID with a /22 mask is 255.255.252.0 - with a block size of 4 in the third octet. The network address in the question is in subnet 172.16.16.0 with a broadcast address of 172.16.19.255. Only option E even has the correct subnet mask

16. How many subnets?

17. Given 172.16.0.0/18 - What are the valid subnets?
0.0 - 64.0 - 128.0 - and 192.0
IP stack failure; Reinstall TCP/IP
63.255 - 127.255 - 191.255 - 255.255
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances

18. What is the subnet and broadcast address of the host 172.16.88.255/20?
The subnet is 80.0 and the broadcast address is 95.255
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
Local physical network problem between NIC and router
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.

19. What are the valid hosts in each subnet?
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
Create many networks using subnet masks of different lengths from one network
/25 is 255.255.255.128. The fourth octet is a block size of 128. 0 - 128. The host is in the 0 subnet - broadcast of 127. Valid host 1-126
255.255.255.128

20. Additional Windows Troubleshooting
Allows you to use the first and last subnet in your network design; turned this command on by default
Traceroute - Arp -a - Ipconfig /all
/29 is 255.255.255.248. The fourth octet is a block size of 8. 0 - 8 - 16 - 24. The host is in the 16 subnet - broadcast of 23. Valid host 17-22
Use different size masks on each router interface

21. Given 192.168.10.0/28 - How many hosts per subnet?
126
/26 is 255.255.255.192. The fourth octet has a block size of 64. 0 - 64 - 128. The host is in the 64 subnet - broadcast of 127. Valid host 65-126
.1 - .126 & .129 - .254
Tracert - Show ip arp

22. Given 192.168.10.0/28 - What are the valid subnets?
0 & 128
A /28 is a 255.255.255.240 mask. We need to count to the eighth subnet - not starting at subnet-zero. 16 - 32 - 48 - 64 - 80 - 96 - 112 - 128. The ninth subnet is 144 (we need this to help us find the 128 subnet broadcast address - which is 143).The
11111000 = 248
The 10.32 subnet; The broadcast is 10.63

23. Unable to Ping Local Host
/26 is 255.255.255.192. The fourth octet has a block size of 64. 0 - 64 - 128. The host is in the 64 subnet - broadcast of 127. Valid host 65-126
/28 is 255.255.255.240. The fourth octet is a block size of 16. Just count by 16s until you pass 37. 0 - 16 - 32 - 48. The host is in the 32 subnet - with a broadcast address of 47. Valid hosts 33-46
Problem with the NIC; Replace the NIC
RIPv2 - EIGRP - or OSPF

24. You have a network that needs 29 subnets while maximizing the number of host addresses available on each subnet. How many bits must you borrow from the host field to provide the correct subnet mask?

25. Which configuration command must be in effect to allow the use of 8 subnets if the Class C subnet mask is 255.255.255.224?
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
A Class C subnet mask of 255.255.255.224 is 3 bits on and 5 bits off (11100000) and provides 8 subnets - each with 30 hosts. However - if the command ip subnet-zero is not used - then only 6 subnets would be available for use.
Each network segment can use a different subnet mask
Local physical network problem between NIC and router

26. Using the following illustration - what would be the IP address of E0 if you were using the eighth subnet? The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range. The zero subnet should not be considered vali
A /29 (255.255.255.248) has a block size of 8 in the fourth octet. This means the subnets are 0 - 8 - 16 - 24 - etc. 10 is in the 8 subnet. The next subnet is 16 - so 15 is the broadcast address.
11000000 = 192
A /28 is a 255.255.255.240 mask. We need to count to the eighth subnet - not starting at subnet-zero. 16 - 32 - 48 - 64 - 80 - 96 - 112 - 128. The ninth subnet is 144 (we need this to help us find the 128 subnet broadcast address - which is 143).The
RIPv1 and IGRP

27. Given 192.168.10.0/28 - What are the valid hosts?
Create many networks using subnet masks of different lengths from one network
.1 - .126 & .129 - .254
/29 is 255.255.255.248. The fourth octet is a block size of 8. 0 - 8 - 16 - 24. The host is in the 16 subnet - broadcast of 23. Valid host 17-22
Use different size masks on each router interface

28. What are the valid subnets?
A /30 - regardless of the class of address - has a 252 in the fourth octet. This means we have a block size of 4 and our subnets are 0 - 4 - 8 - 12 - 16 - etc. Address 14 is obviously in the 12 subnet.
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
All interfaces within the classful address space have the same subnet mask
The 10.32 subnet; The broadcast is 10.63

29. The network address of 172.16.0.0/19 provides how many subnets and hosts?
/25 is 255.255.255.128. The fourth octet is a block size of 128. 0 - 128. The host is in the 0 subnet - broadcast of 127. Valid host 1-126
A CIDR address of /19 is 255.255.224.0. This is a Class B address - so that is only 3 subnet bits but provides 13 host bits - or 8 subnets - each with 8 -190 hosts.
0 & 128
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21

30. Variable Length Subnet Masks (VLSMs)
Create many networks using subnet masks of different lengths from one network
This is a pretty simple question. A /28 is 255.255.255.240 - which means that our block size is 16 in the fourth octet. 0 - 16 - 32 - 48 - 64 - 80 - etc. The host is in the 64 subnet.
126
/23 is 255.255.254.0. The third octet is a block size of 2. 0 - 2 - 4. The subnet is in the 16.2.0 subnet - the broadcast address is 16.3.255

31. A network administrator is connecting hosts A and B directly thorough their Ethernet interfaces as shown in the illustration. Ping attempts between the hosts are unsuccessful. What can be done to provide connectivity between the hosts?

32. Your router has the following IP address on Ethernet0: 172.16.2.1/23. Which of the following can be valid host ID's on the LAN interface attached to the router?
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
Allows you to use the first and last subnet in your network design; turned this command on by default
The routers IP address on the E0 interface is 172.16.2.1/23 - which is a 255.255.254.0. This makes the third octet a block size of 2. The routers interface is in the 2.0 subnet - the broadcast address is 3.255 because the next subnet is 4.0. The vali
32-bit value that allows the recipient of IP packets to distinguish the network ID portion of the IP address from the host ID portion of the IP address

33. Which two statements describe the IP address 10.16.3.65/23?
126
A 240 mask is 4 subnet bits and provides 16 subnets - each with 14 hosts. We need more subnets - so let's add subnet bits. One more subnet bit would be a 248 mask. This provides 5 subnet bits (32 subnets) with 3 hosts bits (6 host per subnet).
63.255 - 127.255 - 191.255 - 255.255
The mask 255.255.254.0 (/23) used with a Class A means that there are 15 subnet bits and 9 hosts bits. The block size in the third octet is 2 (256 - 254). So this makes the subnets in the interesting octet 0 - 2 - 4 - 6 - etc. - all the way to 254. T

34. You need to subnet a network that has five subnets - each with at least 16 hosts. Which classful subnet mask would you use?
255.255.255.128
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
A Class B network ID with a /22 mask is 255.255.252.0 - with a block size of 4 in the third octet. The network address in the question is in subnet 172.16.16.0 with a broadcast address of 172.16.19.255. Only option E even has the correct subnet mask
4

35. 192.168.100.66/27
/27 is 255.255.255.224. The fourth octet is a block size of 32. Count by 32s until you pass the host address of 66. 0 - 32 - 64. The host is in the 32 subnet - broadcast address of 63. Valid host range of 33-62.
/23 is 255.255.254.0. The third octet is a block size of 2. 0 - 2 - 4. The subnet is in the 16.2.0 subnet - the broadcast address is 16.3.255
Possible DNS problem
All interfaces within the classful address space have the same subnet mask

36. Classful Routing
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination
Possible DNS problem
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet
All interfaces within the classful address space have the same subnet mask

37. /25
Problem with the NIC; Replace the NIC
A /29 (255.255.255.248) - regardless of the class of address - has only three hosts bits. Six hosts is the maximum amount of hosts on this LAN - including the router interface.
10000000 = 128
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25

38. What is the subnetwork address for a host with the IP address 200.10.5.68/28?
0.0 - 64.0 - 128.0 - and 192.0
/29 is 255.255.255.248. The fourth octet is a block size of 8. 0 - 8 - 16 - 24. The host is in the 16 subnet - broadcast of 23. Valid host 17-22
A Class B network ID with a /22 mask is 255.255.252.0 - with a block size of 4 in the third octet. The network address in the question is in subnet 172.16.16.0 with a broadcast address of 172.16.19.255. Only option E even has the correct subnet mask
This is a pretty simple question. A /28 is 255.255.255.240 - which means that our block size is 16 in the fourth octet. 0 - 16 - 32 - 48 - 64 - 80 - etc. The host is in the 64 subnet.

39. What is the broadcast address of 192.168.192.10/29?
/29 is 255.255.255.248. This is a block size of 8 in the fourth octet. 0 - 8 - 16. The host is in the 8 subnet - broadcast is 15
A /30 - regardless of the class of address - has a 252 in the fourth octet. This means we have a block size of 4 and our subnets are 0 - 4 - 8 - 12 - 16 - etc. Address 14 is obviously in the 12 subnet.
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254
0 & 128

40. IP Subnet-Zero
Allows you to use the first and last subnet in your network design; turned this command on by default
/23 is 255.255.254.0. The third octet is a block size of 2. 0 - 2 - 4. The subnet is in the 16.2.0 subnet - the broadcast address is 16.3.255
/28 is 255.255.255.240. The fourth octet is a block size of 16. Just count by 16s until you pass 37. 0 - 16 - 32 - 48. The host is in the 32 subnet - with a broadcast address of 47. Valid hosts 33-46
A /30 - regardless of the class of address - has a 252 in the fourth octet. This means we have a block size of 4 and our subnets are 0 - 4 - 8 - 12 - 16 - etc. Address 14 is obviously in the 12 subnet.

41. Classful Routing
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
All nodes in the network use the same subnet mask
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts

42. 192.168.100.99/26
/29 is 255.255.255.248. This is a block size of 8 in the fourth octet. 0 - 8 - 16. The host is in the 8 subnet - broadcast is 15
/26 is 255.255.255.192. The fourth octet has a block size of 64. 0 - 64 - 128. The host is in the 64 subnet - broadcast of 127. Valid host 65-126
2
11000000 = 192

43. Given 172.16.0.0/18 - What's the broadcast address for each subnet?
/30 is 255.255.255.252. The valid subnet is 192.168.100.24 - broadcast is 192.168.100.27 - and valid hosts are 192.168.100.25 and 26
63.255 - 127.255 - 191.255 - 255.255
A /21 is 255.255.248.0 - which means we have a block size of 8 in the third octet - so we just count by 8 until we reach 66. The subnet in this question is 64.0. The next subnet is 72.0 - so the broadcast address of the 64 subnet is 71.255.
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached

44. Additional Cisco Troubleshooting
RIPv2 - EIGRP - or OSPF
0 & 128
Tracert - Show ip arp
A /29 (255.255.255.248) has a block size of 8 in the fourth octet. This means the subnets are 0 - 8 - 16 - 24 - etc. 10 is in the 8 subnet. The next subnet is 16 - so 15 is the broadcast address.

45. 4 Troubleshooting Steps
2
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
/29 is 255.255.255.248. This is a block size of 8 in the fourth octet. 0 - 8 - 16. The host is in the 8 subnet - broadcast is 15

46. Unable to Ping Default Gateway
Local physical network problem between NIC and router
Problem with the NIC; Replace the NIC
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
A /29 (255.255.255.248) - regardless of the class of address - has only three hosts bits. Six hosts is the maximum amount of hosts on this LAN - including the router interface.

47. How many valid hosts per subnet?
16 -382
11100000 = 224
2y - 2 where y is the number of unmasked bits (or 0's)
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25

48. /28
11110000 = 240
/29 is 255.255.255.248. The fourth octet is a block size of 8. 0 - 8 - 16 - 24. The host is in the 16 subnet - broadcast of 23. Valid host 17-22
The 10.32 subnet; The broadcast is 10.63
Create many networks using subnet masks of different lengths from one network

49. Given 192.168.10.0/28 - What's the broadcast address for each subnet?
127 & 255
RIPv2 - EIGRP - or OSPF
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255

50. You have an interface on a router with the IP address of 192.168.192.10/29. Including the router interface - how many hosts can have IP addresses on the LAN attached to router interface?
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
A Class B network ID with a /22 mask is 255.255.252.0 - with a block size of 4 in the third octet. The network address in the question is in subnet 172.16.16.0 with a broadcast address of 172.16.19.255. Only option E even has the correct subnet mask
Create many networks using subnet masks of different lengths from one network
A /29 (255.255.255.248) - regardless of the class of address - has only three hosts bits. Six hosts is the maximum amount of hosts on this LAN - including the router interface.