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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. 192.168.100.25/30
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination
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
Allows you to use the first and last subnet in your network design; turned this command on by default
/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. You have a Class B network and need 29 subnets. What is your mask?
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
/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
Local physical network problem between NIC and router

3. Variable Length Subnet Masks (VLSMs)
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.
RIPv2 - EIGRP - or OSPF
Create many networks using subnet masks of different lengths from one network
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.

4. 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
.1 - .126 & .129 - .254
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

5. Additional Cisco Troubleshooting
63.255 - 127.255 - 191.255 - 255.255
Tracert - Show ip arp
11000000 = 192
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30

6. What subnet and broadcast address is the IP address 172.16.46.255 255.255.240.0 (/20) a member of?
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
Possible DNS problem
Each network segment can use a different subnet mask

7. Subnet Mask
4
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 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
127 & 255

8. The network address of 172.16.0.0/19 provides how many subnets and hosts?
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet
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.
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).

9. /27
11100000 = 224
0.0 - 64.0 - 128.0 - and 192.0
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
RIPv2 - EIGRP - or OSPF

10. /26
11000000 = 192
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances
63.255 - 127.255 - 191.255 - 255.255
/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

11. Given 192.168.10.0/28 - How many subnets?
/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
11000000 = 192
2
/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

12. /28
11110000 = 240
0.0 - 64.0 - 128.0 - and 192.0
.1 - .126 & .129 - .254
/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

13. What subnet and broadcast address is the IP address 172.16.66.10 255.255.192.0 (/18) a member of?
2y - 2 where y is the number of unmasked bits (or 0's)
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
11100000 = 224
/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

14. 192.168.100.66/27
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.
/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.
RIPv1 and IGRP
2y - 2 where y is the number of unmasked bits (or 0's)

15. What is the subnetwork address for a host with the IP address 200.10.5.68/28?
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).
Traceroute - Arp -a - Ipconfig /all
11111100 = 252
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.

16. Why subnet?
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
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254
16 -382

17. What is the subnetwork number of a host with an IP address of 172.16.66.0/21?
Create many networks using subnet masks of different lengths from one network
/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
0 & 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.

18. How many subnets?

19. If a host on a network has the address 172.16.45.14/30 - what is the subnetwork this host belongs to?
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
11000000 = 192
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination

20. 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?

21. You need to subnet a network that has five subnets - each with at least 16 hosts. Which classful subnet mask would you use?
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254
/25 is 255.255.255.128. The fourth octet is a block size of 128. 0 - 128. The host is in the 0 subnet - broadcast of 127. Valid host 1-126

22. What is the subnet for host ID 10.16.3.65/23?
Local physical network problem between NIC and router
/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
Tracert - Show ip arp
11110000 = 240

23. Additional Windows Troubleshooting
127 & 255
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
126
Traceroute - Arp -a - Ipconfig /all

24. 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 /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.
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
/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 /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

25. 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).
/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
IP stack failure; Reinstall TCP/IP
RIPv1 and IGRP

26. Able to ping but still unable to communicate
Possible DNS problem
The number just before the next subnet; the broadcast of the last subnet is always 255
2^x where x is the number of masked bits (or 1's)
RIPv1 and IGRP

27. Variable Length Subnet Masks (VLSMs)
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
Use different size masks on each router interface
The number just before the next subnet; the broadcast of the last subnet is always 255

28. /29
11100000 = 224
11111000 = 248
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.
/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

29. 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?
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
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 /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 -

30. 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?
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
Each network segment can use a different subnet mask
Tracert - Show ip arp
Discard it; by default will discard any broadcast packets

31. /30
/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
The subnet is 80.0 and the broadcast address is 95.255
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.

32. 4 Troubleshooting Steps
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
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
11100000 = 224

33. What subnet and broadcast address is the IP address 172.16.50.10 255.255.224.0 (/19) a member of?
/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
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
4
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25

34. 192.168.100.99/26
The subnet is 80.0 and the broadcast address is 95.255
IP stack failure; Reinstall TCP/IP
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.
/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

35. Classful Routing Protocols
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).
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances
To test the local stack on your host - ping the loopback interface of 127.0.0.1
RIPv1 and IGRP

36. Given 192.168.10.0/28 - What are the valid hosts?
Allows you to use the first and last subnet in your network design; turned this command on by default
.1 - .126 & .129 - .254
127 & 255
IP stack failure; Reinstall TCP/IP

37. Unable to Ping Local Host
11100000 = 224
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).
Problem with the NIC; Replace the NIC
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet

38. 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
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.
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 subnet is 172.16.45.12 - with a broadcast of 172.16.45.15

39. 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?

40. Variable Length Subnet Masks (VLSMs)
/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
11111000 = 248
Each network segment can use a different subnet mask
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.

41. 192.168.100.17/29
11000000 = 192
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances
/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
0 & 128

42. Class C Subnet Masks
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
11111000 = 248
Possible DNS problem
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.

43. 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 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.
2^x where x is the number of masked bits (or 1's)
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
2y - 2 where y is the number of unmasked bits (or 0's)

44. Classful Routing
The 10.32 subnet; The broadcast is 10.63
All nodes in the network use the same subnet mask
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.
/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. 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?
11111000 = 248
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
All nodes in the network use the same subnet mask
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

46. Given 192.168.10.0/28 - How many hosts per subnet?
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.
2y - 2 where y is the number of unmasked bits (or 0's)
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 -

47. Given 172.16.0.0/18 - What are the valid subnets?
Use different size masks on each router interface
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.
0.0 - 64.0 - 128.0 - and 192.0
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s

48. 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.
.1 - .126 & .129 - .254
Discard it; by default will discard any broadcast packets
The number just before the next subnet; the broadcast of the last subnet is always 255

49. 192.168.100.99/25
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
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.
63.255 - 127.255 - 191.255 - 255.255
/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

50. 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
/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 /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 /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.