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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. Given 172.16.0.0/18 - What are the valid subnets?
0.0 - 64.0 - 128.0 - and 192.0
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
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.

2. Class C Subnet Masks
Problem with the NIC; Replace the NIC
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
4
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25

3. 4 Troubleshooting Steps
/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
Discard it; by default will discard any broadcast packets
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination
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.

4. What's the broadcast address of each subnet?
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
The number just before the next subnet; the broadcast of the last subnet is always 255
All interfaces within the classful address space have the same subnet mask
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.

5. 192.168.100.17/29
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.
/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
To test the local stack on your host - ping the loopback interface of 127.0.0.1
/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

6. What subnet and broadcast address is the IP address 172.16.10.33 255.255.255.224 (/27) a member of?
The 10.32 subnet; The broadcast is 10.63
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet
Use different size masks on each router interface

7. 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 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 /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 numbers between the subnets and the broadcasts omitting the all 0s and all 1s
10000000 = 128

8. 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?

9. Unable to Ping Local Host
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.
/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.
Problem with the NIC; Replace the NIC
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255

10. Classless Routing Protocols
RIPv2 - EIGRP - or OSPF
11110000 = 240
Possible DNS problem
Discard it; by default will discard any broadcast packets

11. What subnet and broadcast address is the IP address 172.16.50.10 255.255.224.0 (/19) a member of?
The number just before the next subnet; the broadcast of the last subnet is always 255
11111100 = 252
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

12. Given 172.16.0.0/18 - What's the broadcast address for each subnet?
The 10.32 subnet; The broadcast is 10.63
/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
2^x where x is the number of masked bits (or 1's)
63.255 - 127.255 - 191.255 - 255.255

13. Additional Windows Troubleshooting
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
/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
Traceroute - Arp -a - Ipconfig /all

14. /27
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet
11100000 = 224
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.
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination

15. What is the subnetwork address for a host with the IP address 200.10.5.68/28?
Each network segment can use a different subnet mask
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
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.
Use different size masks on each router interface

16. What is the subnet for host ID 10.16.3.65/23?
/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
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
Each network segment can use a different subnet mask
10000000 = 128

17. If a host on a network has the address 172.16.45.14/30 - what is the subnetwork this host belongs to?
Local physical network problem between NIC and 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.
10000000 = 128
/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

18. Subnet Mask
11110000 = 240
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
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15

19. What is the subnet and broadcast address of the host 172.16.88.255/20?
The subnet is 80.0 and the broadcast address is 95.255
/25 is 255.255.255.128. The fourth octet is a block size of 128. 0 - 128. The host is in the 0 subnet - broadcast of 127. Valid host 1-126
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances
11111100 = 252

20. Able to ping but still unable to communicate
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
Possible DNS problem
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
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

21. 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?
All interfaces within the classful address space have the same subnet mask
10000000 = 128
To test the local stack on your host - ping the loopback interface of 127.0.0.1
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 -

22. /26
IP stack failure; Reinstall TCP/IP
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
11000000 = 192

23. Unable to Ping Default Gateway
Local physical network problem between NIC and router
/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
RIPv2 - EIGRP - or OSPF
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30

24. Unable to Ping Loopback
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 /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 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
IP stack failure; Reinstall TCP/IP

25. Given 192.168.10.0/28 - what is the subnet mask?
The 10.32 subnet; The broadcast is 10.63
255.255.255.128
11111100 = 252
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. How many valid 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.
All interfaces within the classful address space have the same subnet mask
The number just before the next subnet; the broadcast of the last subnet is always 255
2y - 2 where y is the number of unmasked bits (or 0's)

27. Given 192.168.10.0/28 - How many hosts per subnet?
127 & 255
Each network segment can use a different subnet mask
126
/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.

28. 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?
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
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
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.
/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

29. On a VLSM network - which mask should you use on point-to-point WAN links in order to reduce the waste of IP addresses?
All nodes in the network use the same 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
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet

30. How many hosts are available with a Class C /29 mask?
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet
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
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

31. What is the subnetwork number of a host with an IP address of 172.16.66.0/21?
All interfaces within the classful address space have the same subnet mask
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
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.

32. What subnet and broadcast address is the IP address 172.16.45.14 255.255.255.252 (/30) a member of?
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.
RIPv1 and IGRP
10000000 = 128
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15

33. Given 172.16.0.0/18 - How many subnets?
/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
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 -
To test the local stack on your host - ping the loopback interface of 127.0.0.1
4

34. /29
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
11111000 = 248
2^x where x is the number of masked bits (or 1's)
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.

35. Unable to Ping Remote Destination
/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 /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.
Tracert - Show ip arp
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination

36. 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
2
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

37. /25
10000000 = 128
126
Local physical network problem between NIC and router
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.

38. 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
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).
Possible DNS problem
Tracert - Show ip arp

39. Given 192.168.10.0/28 - What are the valid hosts?
16 -382
127 & 255
Each network segment can use a different subnet mask
.1 - .126 & .129 - .254

40. 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
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
0.0 - 64.0 - 128.0 - and 192.0
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.
11100000 = 224

41. To test the IP stack on your local host - which IP address would you ping?
To test the local stack on your host - ping the loopback interface of 127.0.0.1
A 240 mask is 4 subnet bits and provides 16 subnets - each with 14 hosts. We need more subnets - so let's add subnet bits. One more subnet bit would be a 248 mask. This provides 5 subnet bits (32 subnets) with 3 hosts bits (6 host per subnet).
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination

42. Additional Cisco Troubleshooting
Use different size masks on each router interface
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.
Tracert - Show ip arp
11111000 = 248

43. IP Subnet-Zero
126
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
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.
Allows you to use the first and last subnet in your network design; turned this command on by default

44. 192.168.100.99/25
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
/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
2

45. Classful Routing Protocols
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
/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
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances

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. Classful Routing
IP stack failure; Reinstall TCP/IP
All interfaces within the classful address space have the same subnet mask
11100000 = 224
2^x where x is the number of masked bits (or 1's)

48. Given 172.16.0.0/18 - How many hosts per subnet?
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
/28 is 255.255.255.240. The fourth octet is a block size of 16. Just count by 16s until you pass 37. 0 - 16 - 32 - 48. The host is in the 32 subnet - with a broadcast address of 47. Valid hosts 33-46
All interfaces within the classful address space have the same subnet mask
16 -382

49. Variable Length Subnet Masks (VLSMs)
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
Use different size masks on each router interface
127 & 255
11000000 = 192

50. 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.
/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
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