Test your basic knowledge |

IP Subnetting VLSMs And Troubleshooting IP

Subject :

Instructions:

Answer 50 questions in 15 minutes.
If you are not ready to take this test, you can study here.
Match each statement with the correct term.
Don't refresh. All questions and answers are randomly picked and ordered every time you load a test.

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. How many subnets?

2. If an Ethernet port on a router were assigned an IP address of 172.16.112.1/25 - what would be the valid subnet address of this host?
/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 /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 -
11110000 = 240
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.

3. Variable Length Subnet Masks (VLSMs)
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.
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
127 & 255
Create many networks using subnet masks of different lengths from one network

4. Classful Routing
All nodes in the network use the same subnet mask
2^x where x is the number of masked bits (or 1's)
0 & 128
IP stack failure; Reinstall TCP/IP

5. Given 192.168.10.0/28 - what is the subnet mask?
4
255.255.255.128
2y - 2 where y is the number of unmasked bits (or 0's)
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached

6. Able to ping but still unable to communicate
Possible DNS problem
.1 - .126 & .129 - .254
11000000 = 192
RIPv1 and IGRP

7. 192.168.100.99/26
63.255 - 127.255 - 191.255 - 255.255
/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
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.
/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

8. On a VLSM network - which mask should you use on point-to-point WAN links in order to reduce the waste of IP addresses?
2
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
255.255.255.128
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.

9. /30
.1 - .126 & .129 - .254
11111100 = 252
127 & 255
Tracert - Show ip arp

10. To test the IP stack on your local host - which IP address would you ping?
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
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.
To test the local stack on your host - ping the loopback interface of 127.0.0.1
.1 - .126 & .129 - .254

11. 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?
Possible DNS problem
Allows you to use the first and last subnet in your network design; turned this command on by default
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.
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.

12. What is the broadcast address of 192.168.192.10/29?
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
/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
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
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

13. Unable to Ping Default Gateway
To test the local stack on your host - ping the loopback interface of 127.0.0.1
/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
All interfaces within the classful address space have the same subnet mask
Local physical network problem between NIC and router

14. What are the valid subnets?
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 172.16.64.0. The broadcast must be 172.16.127.255
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
0 & 128

15. What subnet and broadcast address is the IP address 172.16.50.10 255.255.224.0 (/19) a member of?
10000000 = 128
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
16 -382
Local physical network problem between NIC and router

16. What are the valid hosts in each subnet?
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 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.
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
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.

17. /29
11111100 = 252
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances
.1 - .126 & .129 - .254
11111000 = 248

18. Given 172.16.0.0/18 - What are the valid hosts?
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet
11100000 = 224
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254
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.

19. Unable to Ping Loopback
All nodes in the network use the same subnet mask
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
IP stack failure; Reinstall TCP/IP
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

20. 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?

21. You need to subnet a network that has five subnets - each with at least 16 hosts. Which classful subnet mask would you use?
16 -382
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
Possible DNS problem

22. /28
11110000 = 240
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
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 -
Tracert - Show ip arp

23. 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
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 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.
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25

24. Classless Routing Protocols
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
RIPv2 - EIGRP - or OSPF
Use different size masks on each 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

25. Given 192.168.10.0/28 - What's the broadcast address for each subnet?
Discard it; by default will discard any broadcast packets
127 & 255
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.
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.

26. Why subnet?
Use different size masks on each router interface
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances
Possible DNS problem
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination

27. Which two statements describe the IP address 10.16.3.65/23?
16 -382
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 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 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

28. What subnet and broadcast address is the IP address 172.16.45.14 255.255.255.252 (/30) a member of?
/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^x where x is the number of masked bits (or 1's)
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
Allows you to use the first and last subnet in your network design; turned this command on by default

29. The network address of 172.16.0.0/19 provides how many subnets and hosts?
The subnet is 80.0 and the broadcast address is 95.255
/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
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.
126

30. Class C Subnet Masks
Use different size masks on each router interface
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
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.
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

31. 192.168.100.17/29
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
/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
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
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.

32. 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
RIPv1 and IGRP
11110000 = 240
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.

33. 192.168.100.99/25
63.255 - 127.255 - 191.255 - 255.255
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
To test the local stack on your host - ping the loopback interface of 127.0.0.1
/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

34. Given 172.16.0.0/18 - What are the valid subnets?
0.0 - 64.0 - 128.0 - and 192.0
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254
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.

35. Given 192.168.10.0/28 - How many hosts per subnet?
/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
126
The number just before the next subnet; the broadcast of the last subnet is always 255
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.

36. IP Subnet-Zero
Allows you to use the first and last subnet in your network design; turned this command on by default
Create many networks using subnet masks of different lengths from one network
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
2

37. 192.168.100.37/28
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.
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).
/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

38. 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?
11110000 = 240
2^x where x is the number of masked bits (or 1's)
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.
Discard it; by default will discard any broadcast packets

39. What subnet and broadcast address is the IP address 172.16.10.33 255.255.255.224 (/27) a member of?
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.
The 10.32 subnet; The broadcast is 10.63
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15

40. Additional Cisco Troubleshooting
Local physical network problem between NIC and router
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
Tracert - Show ip arp
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21

41. What subnet and broadcast address is the IP address 172.16.66.10 255.255.192.0 (/18) a member of?
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.
Tracert - Show ip arp
To test the local stack on your host - ping the loopback interface of 127.0.0.1

42. Unable to Ping Remote Destination
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.
Possible DNS problem
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination

43. /26
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
11000000 = 192
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25

44. What is the subnet for host ID 10.16.3.65/23?
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
0.0 - 64.0 - 128.0 - and 192.0
Possible DNS problem
/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

45. Variable Length Subnet Masks (VLSMs)
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination
Possible DNS problem
IP stack failure; Reinstall TCP/IP
Use different size masks on each router interface

46. 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?

47. 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
Problem with the NIC; Replace the NIC
/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 /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.
126

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?
Possible DNS problem
2y - 2 where y is the number of unmasked bits (or 0's)
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.
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.

49. If a host on a network has the address 172.16.45.14/30 - what is the subnetwork this host belongs to?
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
2y - 2 where y is the number of unmasked bits (or 0's)
126
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.

50. How many valid hosts per 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
10000000 = 128
2y - 2 where y is the number of unmasked bits (or 0's)
Traceroute - Arp -a - Ipconfig /all