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IP Subnetting VLSMs And Troubleshooting IP

Subject :

Instructions:

Answer 50 questions in 15 minutes.
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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. Why subnet?
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.
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances
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
0.0 - 64.0 - 128.0 - and 192.0

2. The network address of 172.16.0.0/19 provides how many subnets and hosts?
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.
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
4
16 -382

3. What are the valid subnets?
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
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 -
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s

4. Classless Routing Protocols
/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
RIPv2 - EIGRP - or OSPF
All nodes in the network use the same subnet mask
/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

5. Class C Subnet Masks
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.
63.255 - 127.255 - 191.255 - 255.255
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30

6. Unable to Ping Default Gateway
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
Each network segment can use a different subnet mask
Local physical network problem between NIC and router
0.0 - 64.0 - 128.0 - and 192.0

7. 192.168.100.99/25
The 10.32 subnet; The broadcast is 10.63
/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
RIPv1 and IGRP
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).

8. 192.168.100.66/27
11111000 = 248
/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.
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.

9. Given 172.16.0.0/18 - How many hosts per subnet?
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 -
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
16 -382

10. Given 192.168.10.0/28 - How many hosts per subnet?
/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.
126
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet
0 & 128

11. Variable Length Subnet Masks (VLSMs)
Each network segment can use a different subnet mask
127 & 255
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

12. 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?

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13. 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?
Local physical network problem between NIC and router
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.
Discard it; by default will discard any broadcast packets
Each network segment can use a different subnet mask

14. 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
10000000 = 128
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
IP stack failure; Reinstall TCP/IP

15. 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
2
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 -
/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

16. 192.168.100.25/30
11000000 = 192
0 & 128
/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
2

17. Additional Windows Troubleshooting
/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
Use different size masks on each router interface
Traceroute - Arp -a - Ipconfig /all
11111000 = 248

18. /25
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
All interfaces within the classful address space have the same subnet mask
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
10000000 = 128

19. 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 /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 -
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 /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 subnet is 172.16.45.12 - with a broadcast of 172.16.45.15

20. 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?
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
2y - 2 where y is the number of unmasked bits (or 0's)
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.
/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

21. /26
11000000 = 192
Use different size masks on each router interface
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 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

22. Unable to Ping Local Host
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
Problem with the NIC; Replace the NIC
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 -
126

23. 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 /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 -
63.255 - 127.255 - 191.255 - 255.255
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
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.

24. Unable to Ping Remote Destination
2
/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
11110000 = 240
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination

25. Classful Routing
The number just before the next subnet; the broadcast of the last subnet is always 255
0 & 128
All interfaces within the classful address space have the same subnet mask
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. /30
11111100 = 252
Create many networks using subnet masks of different lengths from one network
16 -382
0 & 128

27. 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
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
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. 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?
2^x where x is the number of masked bits (or 1's)
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.
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. Additional Cisco Troubleshooting
Create many networks using subnet masks of different lengths from one network
Tracert - Show ip arp
/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
11000000 = 192

30. On a VLSM network - which mask should you use on point-to-point WAN links in order to reduce the waste of IP addresses?
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
Local physical network problem between NIC and router
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.

31. Given 172.16.0.0/18 - What's the broadcast address for each subnet?
RIPv1 and IGRP
63.255 - 127.255 - 191.255 - 255.255
Each network segment can use a different subnet mask
127 & 255

32. /29
11111000 = 248
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.
RIPv2 - EIGRP - or OSPF
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.

33. Unable to Ping Loopback
IP stack failure; Reinstall TCP/IP
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 10.32 subnet; The broadcast is 10.63
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

34. What are the valid hosts in each subnet?
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
Use different size masks on each router interface
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
Discard it; by default will discard any broadcast packets

35. 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
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
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.
2

36. What subnet and broadcast address is the IP address 172.16.50.10 255.255.224.0 (/19) a member of?
11100000 = 224
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
/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
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

37. Given 172.16.0.0/18 - What are the valid subnets?
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
0.0 - 64.0 - 128.0 - and 192.0
Create many networks using subnet masks of different lengths from one network
The number just before the next subnet; the broadcast of the last subnet is always 255

38. IP Subnet-Zero
/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
Allows you to use the first and last subnet in your network design; turned this command on by default
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.
16 -382

39. Given 192.168.10.0/28 - What are the valid subnets?
16 -382
/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
All nodes in the network use the same subnet mask
0 & 128

40. 192.168.100.99/26
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
/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
10000000 = 128

41. Given 192.168.10.0/28 - What are the valid hosts?
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
11111000 = 248
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.
.1 - .126 & .129 - .254

42. What is the subnet and broadcast address of the host 172.16.88.255/20?
IP stack failure; Reinstall TCP/IP
The subnet is 80.0 and the broadcast address is 95.255
/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.
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

43. 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?

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44. How many subnets?

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45. You have a Class B network and need 29 subnets. What is your 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).
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
All nodes in the network use the same subnet mask

46. Able to ping but still unable to communicate
Possible DNS problem
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
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

47. Given 192.168.10.0/28 - What's the broadcast address for each subnet?
16 -382
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.
RIPv2 - EIGRP - or OSPF
127 & 255

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?
A /28 is a 255.255.255.240 mask. We need to count to the eighth subnet - not starting at subnet-zero. 16 - 32 - 48 - 64 - 80 - 96 - 112 - 128. The ninth subnet is 144 (we need this to help us find the 128 subnet broadcast address - which is 143).The
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
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

49. What is the subnetwork address for a host with the IP address 200.10.5.68/28?
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.
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
RIPv1 and IGRP

50. Subnet Mask
Discard it; by default will discard any broadcast packets
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 /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-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