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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 172.16.0.0/18 - How many hosts per subnet?
Traceroute - Arp -a - Ipconfig /all
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
16 -382
/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

2. Variable Length Subnet Masks (VLSMs)
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
Use different size masks on each router interface
Create many networks using subnet masks of different lengths from one network
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet

3. 192.168.100.99/25
Discard it; by default will discard any broadcast packets
/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
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet
16 -382

4. 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 /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.
RIPv1 and IGRP
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

5. 192.168.100.66/27
Local physical network problem between NIC and router
/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.
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
/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

6. Given 192.168.10.0/28 - what is the subnet mask?
Use different size masks on each router interface
Allows you to use the first and last subnet in your network design; turned this command on by default
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances
255.255.255.128

7. 192.168.100.37/28
0.0 - 64.0 - 128.0 - and 192.0
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
/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
/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

8. How many subnets?

9. Given 172.16.0.0/18 - What's the broadcast address for each subnet?
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
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
63.255 - 127.255 - 191.255 - 255.255
Create many networks using subnet masks of different lengths from one network

10. If a host on a network has the address 172.16.45.14/30 - what is the subnetwork this host belongs to?
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.
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
Possible DNS problem

11. 192.168.100.25/30
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
/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
2y - 2 where y is the number of unmasked bits (or 0's)

12. 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) - 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.
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).
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30

13. 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?

14. Unable to Ping Loopback
2y - 2 where y is the number of unmasked bits (or 0's)
IP stack failure; Reinstall TCP/IP
127 & 255
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15

15. What's the broadcast address of each subnet?
The number just before the next subnet; the broadcast of the last subnet is always 255
Tracert - Show ip arp
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.
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances

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 /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
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
16 -382
11100000 = 224

17. To test the IP stack on your local host - which IP address would you ping?
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.
0 & 128
To test the local stack on your host - ping the loopback interface of 127.0.0.1
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).

18. What is the subnet and broadcast address of the host 172.16.88.255/20?
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
The number just before the next subnet; the broadcast of the last subnet is always 255
The subnet is 80.0 and the broadcast address is 95.255
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances

19. Given 172.16.0.0/18 - What are the valid hosts?
/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
11100000 = 224
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254
IP stack failure; Reinstall TCP/IP

20. 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
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
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.
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.

21. /29
11110000 = 240
The subnet is 80.0 and the broadcast address is 95.255
11111000 = 248
2^x where x is the number of masked bits (or 1's)

22. How many hosts are available with a Class C /29 mask?
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet
255.255.255.128
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
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.

23. What subnet and broadcast address is the IP address 172.16.50.10 255.255.224.0 (/19) a member of?
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).
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
4
The subnet is 80.0 and the broadcast address is 95.255

24. 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?
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.
/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 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.
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).

25. /28
11110000 = 240
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
Allows you to use the first and last subnet in your network design; turned this command on by default

26. What is the maximum number of IP addresses that can be assigned to hosts on a local subnet that uses the 255.255.255.224 subnet mask?
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
11100000 = 224
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.
All interfaces within the classful address space have the same subnet mask

27. What subnet and broadcast address is the IP address 172.16.66.10 255.255.192.0 (/18) a member of?
/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
Use different size masks on each router interface
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
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. /30
63.255 - 127.255 - 191.255 - 255.255
16 -382
Tracert - Show ip arp
11111100 = 252

29. 4 Troubleshooting Steps
Create many networks using subnet masks of different lengths from one network
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination
0.0 - 64.0 - 128.0 - and 192.0

30. You have a Class B network and need 29 subnets. What is your mask?
2
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21

31. Classful Routing
/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
11110000 = 240
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

32. How many valid hosts per subnet?
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
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
2y - 2 where y is the number of unmasked bits (or 0's)
2^x where x is the number of masked bits (or 1's)

33. 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?
/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 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).
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.
Create many networks using subnet masks of different lengths from one network

34. Given 192.168.10.0/28 - How many hosts per subnet?
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
10000000 = 128
126

35. /27
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254
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 point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
11100000 = 224

36. Class C Subnet Masks
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
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 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
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30

37. /25
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.
10000000 = 128
126

38. On a VLSM network - which mask should you use on point-to-point WAN links in order to reduce the waste of IP addresses?
126
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
The 10.32 subnet; The broadcast is 10.63
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.

39. 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?
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
Traceroute - Arp -a - Ipconfig /all
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21

40. 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
RIPv2 - EIGRP - or OSPF
11111000 = 248
127 & 255

41. Given 192.168.10.0/28 - How many subnets?
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
2
RIPv1 and IGRP
2^x where x is the number of masked bits (or 1's)

42. A router receives a packet on an interface with a destination address of 172.16.46.191/26. What will the router do with this packet?
Tracert - Show ip arp
126
10000000 = 128
Discard it; by default will discard any broadcast packets

43. Given 192.168.10.0/28 - What are the valid hosts?
.1 - .126 & .129 - .254
The subnet is 80.0 and the broadcast address is 95.255
/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
255.255.255.128

44. What subnet and broadcast address is the IP address 172.16.45.14 255.255.255.252 (/30) a member of?
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.
11111000 = 248
10000000 = 128
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15

45. 192.168.100.99/26
/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
All interfaces within the classful address space have the same subnet mask
11111000 = 248

46. Given 192.168.10.0/28 - What are the valid subnets?
Local physical network problem between NIC and router
0 & 128
11110000 = 240
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21

47. You have a network with a subnet of 172.16.17.0/22. Which are valid host addresses?
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
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
11111000 = 248

48. Classful Routing
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.
All nodes in the network use 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.
/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

49. Variable Length Subnet Masks (VLSMs)
Use different size masks on each router interface
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.
Each network segment can use a different subnet mask
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).

50. Unable to Ping Default Gateway
/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
Local physical network problem between NIC and router
Each network segment can use a different subnet mask
Use different size masks on each router interface