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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. Variable Length Subnet Masks (VLSMs)
IP stack failure; Reinstall TCP/IP
/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
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254
Each network segment can use a different subnet mask

2. 192.168.100.66/27
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
/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
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
/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.

3. Able to ping but still unable to communicate
2^x where x is the number of masked bits (or 1's)
2y - 2 where y is the number of unmasked bits (or 0's)
The subnet is 80.0 and the broadcast address is 95.255
Possible DNS problem

4. 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?
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
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
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254

5. You need to subnet a network that has five subnets - each with at least 16 hosts. Which classful subnet mask would you 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
0 & 128
/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.
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts

6. You have a Class B network and need 29 subnets. What is your mask?
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
11110000 = 240
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
IP stack failure; Reinstall TCP/IP

7. /25
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
11000000 = 192
RIPv1 and IGRP
10000000 = 128

8. What subnet and broadcast address is the IP address 172.16.66.10 255.255.192.0 (/18) a member of?
11110000 = 240
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 -
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
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255

9. Variable Length Subnet Masks (VLSMs)
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254
10000000 = 128
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
Create many networks using subnet masks of different lengths from one network

10. 192.168.100.25/30
Each network segment can use a different subnet mask
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
/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

11. What is the subnetwork number of a host with an IP address of 172.16.66.0/21?
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 /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 -
Discard it; by default will discard any broadcast packets
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.

12. /26
11000000 = 192
/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
Discard it; by default will discard any broadcast packets
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.

13. Additional Windows Troubleshooting
A /30 - regardless of the class of address - has a 252 in the fourth octet. This means we have a block size of 4 and our subnets are 0 - 4 - 8 - 12 - 16 - etc. Address 14 is obviously in the 12 subnet.
11111100 = 252
Traceroute - Arp -a - Ipconfig /all
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30

14. Given 172.16.0.0/18 - How many subnets?
4
The 10.32 subnet; The broadcast is 10.63
11100000 = 224
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255

15. 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?
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 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.
All nodes in the network use the same subnet mask
11100000 = 224

16. What subnet and broadcast address is the IP address 172.16.46.255 255.255.240.0 (/20) a member of?
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
/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.
To test the local stack on your host - ping the loopback interface of 127.0.0.1

17. 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 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 10.32 subnet; The broadcast is 10.63
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
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.

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

19. Given 172.16.0.0/18 - What are the valid 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.
RIPv2 - EIGRP - or OSPF
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254
11110000 = 240

20. Given 192.168.10.0/28 - What are the valid subnets?
The number just before the next subnet; the broadcast of the last subnet is always 255
0 & 128
Traceroute - Arp -a - Ipconfig /all
126

21. What's the broadcast address of each subnet?
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
The number just before the next subnet; the broadcast of the last subnet is always 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.
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.

22. 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?
Possible DNS problem
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
Local physical network problem between NIC and router
Discard it; by default will discard any broadcast packets

23. The network address of 172.16.0.0/19 provides how many subnets and hosts?
A /30 - regardless of the class of address - has a 252 in the fourth octet. This means we have a block size of 4 and our subnets are 0 - 4 - 8 - 12 - 16 - etc. Address 14 is obviously in the 12 subnet.
.1 - .126 & .129 - .254
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 172.16.45.12 - with a broadcast of 172.16.45.15

24. Given 172.16.0.0/18 - How many hosts per subnet?
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
16 -382
To test the local stack on your host - ping the loopback interface of 127.0.0.1
11100000 = 224

25. If a host on a network has the address 172.16.45.14/30 - what is the subnetwork this host belongs to?
.1 - .126 & .129 - .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.
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
Each network segment can use a different subnet mask

26. Given 192.168.10.0/28 - what is the subnet mask?
Traceroute - Arp -a - Ipconfig /all
255.255.255.128
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
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

27. /28
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.
Local physical network problem between NIC and router
10000000 = 128
11110000 = 240

28. Unable to Ping Local Host
11100000 = 224
Problem with the NIC; Replace the NIC
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
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

29. What subnet and broadcast address is the IP address 172.16.10.33 255.255.255.224 (/27) a member of?
Problem with the NIC; Replace the NIC
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
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255

30. 4 Troubleshooting Steps
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination
/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
2^x where x is the number of masked bits (or 1's)
/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

31. 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?
11111000 = 248
RIPv2 - EIGRP - or OSPF
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.
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

32. To test the IP stack on your local host - which IP address would you ping?
Create many networks using subnet masks of different lengths from one network
The subnet is 80.0 and the broadcast address is 95.255
To test the local stack on your host - ping the loopback interface of 127.0.0.1
RIPv2 - EIGRP - or OSPF

33. Which two statements describe the IP address 10.16.3.65/23?
63.255 - 127.255 - 191.255 - 255.255
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
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

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

35. Class C Subnet Masks
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
11000000 = 192
Allows you to use the first and last subnet in your network design; turned this command on by default
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30

36. Given 192.168.10.0/28 - How many hosts per subnet?
All interfaces within the classful address space have the same subnet mask
0.0 - 64.0 - 128.0 - and 192.0
126
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.

37. IP Subnet-Zero
All nodes in the network use the same subnet mask
Problem with the NIC; Replace the NIC
Allows you to use the first and last subnet in your network design; turned this command on by default
63.255 - 127.255 - 191.255 - 255.255

38. 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
Use different size masks on each router interface
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
Allows you to use the first and last subnet in your network design; turned this command on by default

39. Classless Routing Protocols
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
/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.
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
RIPv2 - EIGRP - or OSPF

40. Classful Routing
11100000 = 224
/29 is 255.255.255.248. This is a block size of 8 in the fourth octet. 0 - 8 - 16. The host is in the 8 subnet - broadcast is 15
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
All nodes in the network use the same subnet mask

41. 192.168.100.37/28
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
RIPv1 and IGRP
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination

42. 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 /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.
.1 - .126 & .129 - .254
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
RIPv1 and IGRP

43. What subnet and broadcast address is the IP address 172.16.50.10 255.255.224.0 (/19) a member of?
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
4
All nodes in the network use the same subnet mask
RIPv2 - EIGRP - or OSPF

44. /30
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
11111100 = 252
Problem with the NIC; Replace the NIC

45. Given 192.168.10.0/28 - What are the valid hosts?
127 & 255
255.255.255.128
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
.1 - .126 & .129 - .254

46. What are the valid subnets?
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
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).
0 & 128
Tracert - Show ip arp

47. /27
16 -382
Tracert - Show ip arp
11100000 = 224
Each network segment can use a different subnet mask

48. Given 172.16.0.0/18 - What's the broadcast address for each 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
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.
63.255 - 127.255 - 191.255 - 255.255
Each network segment can use a different subnet mask

49. What are the valid hosts in each subnet?
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
The subnet is 80.0 and the broadcast address is 95.255
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s

50. Additional Cisco Troubleshooting
Tracert - Show ip arp
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.
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
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 -