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IP Subnetting VLSMs And Troubleshooting IP

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Answer 50 questions in 15 minutes.
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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. 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?
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.
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
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet
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.

2. 192.168.100.37/28
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.
/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
11111100 = 252
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254

3. 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?
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
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination
Local physical network problem between NIC and router
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.

4. Class C Subnet Masks
11111000 = 248
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /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
Tracert - Show ip arp

5. Given 172.16.0.0/18 - How many subnets?
4
Use different size masks on each router interface
To test the local stack on your host - ping the loopback interface of 127.0.0.1
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances

6. 192.168.100.25/30
Traceroute - Arp -a - Ipconfig /all
Problem with the NIC; Replace the NIC
/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

7. What are the valid subnets?
127 & 255
2y - 2 where y is the number of unmasked bits (or 0's)
11000000 = 192
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached

8. If a host on a network has the address 172.16.45.14/30 - what is the subnetwork this host belongs to?
11100000 = 224
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
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 & 128

9. Able to ping but still unable to communicate
11000000 = 192
All interfaces within the classful address space have the same subnet mask
Possible DNS problem
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.

10. 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?
4
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 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
0 & 128

11. Unable to Ping Local Host
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.
Problem with the NIC; Replace the NIC
RIPv2 - EIGRP - or OSPF

12. 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?

13. 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?
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
IP stack failure; Reinstall TCP/IP
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.
11111100 = 252

14. What is the subnetwork number of a host with an IP address of 172.16.66.0/21?
127 & 255
11100000 = 224
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.
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30

15. /25
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.
The number just before the next subnet; the broadcast of the last subnet is always 255
/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
10000000 = 128

16. 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?
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.
11111100 = 252
All nodes in the network use the same subnet mask
The subnet is 80.0 and the broadcast address is 95.255

17. What subnet and broadcast address is the IP address 172.16.46.255 255.255.240.0 (/20) a member of?
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
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
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 /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

18. Given 192.168.10.0/28 - How many subnets?
2
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.
Traceroute - Arp -a - Ipconfig /all
RIPv2 - EIGRP - or OSPF

19. On a VLSM network - which mask should you use on point-to-point WAN links in order to reduce the waste of IP addresses?
Each network segment can use a different subnet mask
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
11000000 = 192
0 & 128

20. /30
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.
Use different size masks on each router interface
/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.
11111100 = 252

21. What is the subnet and broadcast address of the host 172.16.88.255/20?
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
The subnet is 80.0 and the broadcast address is 95.255
Create many networks using subnet masks of different lengths from one network
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached

22. Subnet Mask
127 & 255
11000000 = 192
RIPv1 and IGRP
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

23. 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
16 -382
11000000 = 192
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255

24. Unable to Ping Default Gateway
Each network segment can use a different subnet mask
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
/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
Local physical network problem between NIC and router

25. Classful Routing
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 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
All interfaces within the classful address space have the same subnet mask
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached

26. What is the broadcast address of 192.168.192.10/29?
To test the local stack on your host - ping the loopback interface of 127.0.0.1
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances
/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
/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

27. What's the broadcast address of each subnet?
RIPv1 and IGRP
The number just before the next subnet; the broadcast of the last subnet is always 255
Traceroute - Arp -a - Ipconfig /all
IP stack failure; Reinstall TCP/IP

28. Unable to Ping Loopback
10000000 = 128
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
Tracert - Show ip arp
IP stack failure; Reinstall TCP/IP

29. Unable to Ping Remote Destination
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
IP stack failure; Reinstall TCP/IP
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 is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21

30. Given 172.16.0.0/18 - How many hosts 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.
To test the local stack on your host - ping the loopback interface of 127.0.0.1
/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
16 -382

31. What subnet and broadcast address is the IP address 172.16.10.33 255.255.255.224 (/27) a member of?
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
Problem with the NIC; Replace the NIC
The 10.32 subnet; The broadcast is 10.63

32. Why subnet?
11111000 = 248
To test the local stack on your host - ping the loopback interface of 127.0.0.1
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances
11100000 = 224

33. 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
/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
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
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 /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

34. Additional Windows Troubleshooting
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination
4
11110000 = 240
Traceroute - Arp -a - Ipconfig /all

35. 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?
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 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.
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
11000000 = 192

36. Given 172.16.0.0/18 - What are the valid hosts?
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254
To test the local stack on your host - ping the loopback interface of 127.0.0.1
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 (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.

37. How many hosts are available with a Class C /29 mask?
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
127 & 255
10000000 = 128
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet

38. What are the valid hosts in each subnet?
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
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.
/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

39. Additional Cisco Troubleshooting
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254
Tracert - Show ip arp
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
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

40. /29
11111000 = 248
The number just before the next subnet; the broadcast of the last subnet is always 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).

41. Given 192.168.10.0/28 - What are the valid hosts?
.1 - .126 & .129 - .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
IP stack failure; Reinstall TCP/IP
2y - 2 where y is the number of unmasked bits (or 0's)

42. What subnet and broadcast address is the IP address 172.16.45.14 255.255.255.252 (/30) a member of?
Each network segment can use a different 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.
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
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

43. Classful Routing Protocols
0.0 - 64.0 - 128.0 - and 192.0
11111100 = 252
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
RIPv1 and IGRP

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.
Create many networks using subnet masks of different lengths from one network
/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
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination

45. What subnet and broadcast address is the IP address 172.16.66.10 255.255.192.0 (/18) a member of?
.1 - .126 & .129 - .254
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
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.
IP stack failure; Reinstall TCP/IP

46. 192.168.100.17/29
RIPv2 - EIGRP - or OSPF
/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
Tracert - Show ip arp

47. How many valid hosts per subnet?
All interfaces within the classful address space have the same subnet mask
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)
Local physical network problem between NIC and router

48. 4 Troubleshooting Steps
The number just before the next subnet; the broadcast of the last subnet is always 255
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
11111000 = 248
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination

49. The network address of 172.16.0.0/19 provides how many subnets and hosts?
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet
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.
/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

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