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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. You have a network with a subnet of 172.16.17.0/22. Which are valid host addresses?
Each network segment can use a different subnet mask
The number just before the next subnet; the broadcast of the last subnet is always 255
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
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached

2. Variable Length Subnet Masks (VLSMs)
126
Create many networks using subnet masks of different lengths from one network
Use different size masks on each router interface
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.

3. Classful Routing
11111100 = 252
126
All nodes in the network use the same subnet mask
63.255 - 127.255 - 191.255 - 255.255

4. Given 172.16.0.0/18 - How many hosts per subnet?
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.
11100000 = 224
16 -382
All interfaces within the classful address space have the same subnet mask

5. Variable Length Subnet Masks (VLSMs)
Create many networks using subnet masks of different lengths from one network
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances
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.
RIPv1 and IGRP

6. How many subnets?

7. Given 172.16.0.0/18 - How many subnets?
11111000 = 248
4
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
/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

8. /25
10000000 = 128
The subnet is 80.0 and the broadcast address is 95.255
63.255 - 127.255 - 191.255 - 255.255
/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.

9. Classful Routing
All interfaces within the classful address space have the same subnet mask
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
.1 - .126 & .129 - .254
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30

10. Given 192.168.10.0/28 - What are the valid hosts?
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
/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
.1 - .126 & .129 - .254
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.

11. Which two statements describe the IP address 10.16.3.65/23?
/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 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
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.
Use different size masks on each router interface

12. On a VLSM network - which mask should you use on point-to-point WAN links in order to reduce the waste of IP addresses?
Tracert - Show ip arp
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
Possible DNS problem
11111000 = 248

13. 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?
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
11111100 = 252
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
Create many networks using subnet masks of different lengths from one network

14. To test the IP stack on your local host - which IP address would you ping?
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
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
To test the local stack on your host - ping the loopback interface of 127.0.0.1
2y - 2 where y is the number of unmasked bits (or 0's)

15. How many hosts are available with a Class C /29 mask?
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
RIPv2 - EIGRP - or OSPF
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
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet

16. 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 /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
First - if you have two hosts directly connected - as shown in the graphic - then you need a crossover cable. A straight-through cable won't work. Second - the hosts have different masks - which puts them in different subnets. The easily solution is
11100000 = 224
2y - 2 where y is the number of unmasked bits (or 0's)

17. What subnet and broadcast address is the IP address 172.16.66.10 255.255.192.0 (/18) a member of?
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
All interfaces within the classful address space have the same subnet mask
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.

18. Additional Windows Troubleshooting
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.
Traceroute - Arp -a - Ipconfig /all
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.
/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

19. Why subnet?
126
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances
A /27 (255.255.255.224) is 3 bits on and 5 bits off. This provides 8 subnets - each with 30 hosts. Does it matter if this mask is used with a Class A - B or C network address? Not at all. The amount of hosts bits would never change.
127 & 255

20. 192.168.100.25/30
/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
11110000 = 240
/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
/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

21. 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?
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.
0.0 - 64.0 - 128.0 - and 192.0
127 & 255
First - if you have two hosts directly connected - as shown in the graphic - then you need a crossover cable. A straight-through cable won't work. Second - the hosts have different masks - which puts them in different subnets. The easily solution is

22. What subnet and broadcast address is the IP address 172.16.45.14 255.255.255.252 (/30) a member of?
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
11111100 = 252
2

23. /30
11111100 = 252
Each network segment can use a different subnet mask
Traceroute - Arp -a - Ipconfig /all
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.

24. 192.168.100.37/28
/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 point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
All nodes in the network use the same subnet mask
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

25. Able to ping but still unable to communicate
11100000 = 224
/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
Possible DNS problem
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

26. If a host on a network has the address 172.16.45.14/30 - what is the subnetwork this host belongs to?
0.0 - 64.0 - 128.0 - and 192.0
2
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.
127 & 255

27. 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?

28. Unable to Ping Local Host
Discard it; by default will discard any broadcast packets
Create many networks using subnet masks of different lengths from one network
Problem with the NIC; Replace the NIC
0 & 128

29. You need to subnet a network that has five subnets - each with at least 16 hosts. Which classful subnet mask would you use?
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.
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
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.
11111100 = 252

30. 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?

31. Subnet Mask
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
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
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.
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

32. 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 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.
Each network segment can use a different subnet mask
255.255.255.128

33. What subnet and broadcast address is the IP address 172.16.50.10 255.255.224.0 (/19) a member of?
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
11111000 = 248

34. Given 192.168.10.0/28 - How many hosts per subnet?
Each network segment can use a different subnet mask
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
126

35. 4 Troubleshooting Steps
16 -382
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.
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination
/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

36. Given 172.16.0.0/18 - What's the broadcast address for each subnet?
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances
63.255 - 127.255 - 191.255 - 255.255
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
Traceroute - Arp -a - Ipconfig /all

37. Given 192.168.10.0/28 - How many subnets?
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
63.255 - 127.255 - 191.255 - 255.255
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.
2

38. You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address. Which of the following should you assign to the server?
2
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.
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
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances

39. What's the broadcast address of each subnet?
The number just before the next subnet; the broadcast of the last subnet is always 255
/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
11111100 = 252
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached

40. What are the valid hosts in each subnet?
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
First - if you have two hosts directly connected - as shown in the graphic - then you need a crossover cable. A straight-through cable won't work. Second - the hosts have different masks - which puts them in different subnets. The easily solution is
The subnet is 80.0 and the broadcast address is 95.255
Traceroute - Arp -a - Ipconfig /all

41. /29
.1 - .126 & .129 - .254
11111000 = 248
11110000 = 240
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.

42. 192.168.100.99/26
0 & 128
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
/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

43. Variable Length Subnet Masks (VLSMs)
A /25 mask is 255.255.255.128. Used with a Class B network - the third and fourth octets are used for subnetting with a total of 9 subnet bits - 8 bits in the third octet and 1 bit in the fourth octet. Since there is only 1 bit in the fourth octet -
Each network segment can use a different subnet mask
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination
A /27 (255.255.255.224) is 3 bits on and 5 bits off. This provides 8 subnets - each with 30 hosts. Does it matter if this mask is used with a Class A - B or C network address? Not at all. The amount of hosts bits would never change.

44. Using the illustration from the previous question - what would be the IP address of S0 if you were using the first subnet? The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range. Again - the zero subnet shoul
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.
63.255 - 127.255 - 191.255 - 255.255
Use different size masks on each router interface
Possible DNS problem

45. What subnet and broadcast address is the IP address 172.16.46.255 255.255.240.0 (/20) a member of?
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
11111100 = 252
All interfaces within the classful address space have the same subnet mask
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s

46. What is the subnetwork address for a host with the IP address 200.10.5.68/28?
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.
/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
The 10.32 subnet; The broadcast is 10.63
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.

47. IP Subnet-Zero
Allows you to use the first and last subnet in your network design; turned this command on by default
/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
Discard it; by default will discard any broadcast packets
All interfaces within the classful address space have the same subnet mask

48. You have an interface on a router with the IP address of 192.168.192.10/29. What is the broadcast address the hosts will use on this LAN?
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
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.
The number just before the next subnet; the broadcast of the last subnet is always 255

49. The network address of 172.16.0.0/19 provides how many subnets and hosts?
To test the local stack on your host - ping the loopback interface of 127.0.0.1
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.
16 -382
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

50. Classless Routing Protocols
2^x where x is the number of masked bits (or 1's)
Allows you to use the first and last subnet in your network design; turned this command on by default
10000000 = 128
RIPv2 - EIGRP - or OSPF