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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 hosts?
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254
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.
/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

2. Given 172.16.0.0/18 - How many hosts per subnet?
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.
Problem with the NIC; Replace the NIC
16 -382
Local physical network problem between NIC and router

3. How many valid hosts per subnet?
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
2y - 2 where y is the number of unmasked bits (or 0's)
0.0 - 64.0 - 128.0 - and 192.0
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.

4. What subnet and broadcast address is the IP address 172.16.66.10 255.255.192.0 (/18) a member of?
/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 subnet is 172.16.64.0. The broadcast must be 172.16.127.255
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
11111100 = 252

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

6. Variable Length Subnet Masks (VLSMs)
Allows you to use the first and last subnet in your network design; turned this command on by default
Each network segment can use a different subnet mask
2
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination

7. What subnet and broadcast address is the IP address 172.16.45.14 255.255.255.252 (/30) a member of?
Problem with the NIC; Replace the NIC
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
2^x where x is the number of masked bits (or 1's)
Traceroute - Arp -a - Ipconfig /all

8. 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
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
RIPv2 - EIGRP - or OSPF

9. 192.168.100.66/27
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 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).
/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.
/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

10. /27
11110000 = 240
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).
/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
11100000 = 224

11. Why subnet?
Traceroute - Arp -a - Ipconfig /all
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances
16 -382
2

12. What's the broadcast address of each 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
The number just before the next subnet; the broadcast of the last subnet is always 255
0 & 128
2^x where x is the number of masked bits (or 1's)

13. What subnet and broadcast address is the IP address 172.16.46.255 255.255.240.0 (/20) a member of?
/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
0 & 128
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts

14. Variable Length Subnet Masks (VLSMs)
Discard it; by default will discard any broadcast packets
127 & 255
Use different size masks on each router interface
255.255.255.128

15. /30
0.0 - 64.0 - 128.0 - and 192.0
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances
11111100 = 252

16. /29
Discard it; by default will discard any broadcast packets
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
The subnet is 80.0 and the broadcast address is 95.255
11111000 = 248

17. 192.168.100.37/28
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 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
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.
/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

18. 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 -
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
11100000 = 224
All interfaces within the classful address space have the same subnet mask

19. 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
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
126
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30

20. What is the subnetwork number of a host with an IP address of 172.16.66.0/21?
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
/26 is 255.255.255.192. The fourth octet has a block size of 64. 0 - 64 - 128. The host is in the 64 subnet - broadcast of 127. Valid host 65-126
A /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.

21. 192.168.100.99/26
Allows you to use the first and last subnet in your network design; turned this command on by default
/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
255.255.255.128
The subnet is 80.0 and the broadcast address is 95.255

22. /28
RIPv1 and IGRP
11110000 = 240
11111000 = 248
All interfaces within the classful address space have the same subnet mask

23. What subnet and broadcast address is the IP address 172.16.10.33 255.255.255.224 (/27) a member of?
/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
11111000 = 248
The 10.32 subnet; The broadcast is 10.63
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21

24. Subnet Mask
/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
11100000 = 224
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
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. 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
4
/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
63.255 - 127.255 - 191.255 - 255.255
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.

26. Given 192.168.10.0/28 - what is the subnet mask?
255.255.255.128
11100000 = 224
Create many networks using subnet masks of different lengths from one network
Each network segment can use a different subnet mask

27. 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.
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
11110000 = 240
IP stack failure; Reinstall TCP/IP

28. Classful Routing
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 /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.
All interfaces within the classful address space have 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

29. Unable to Ping Default Gateway
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
/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
Local physical network problem between NIC and router

30. You need to subnet a network that has five subnets - each with at least 16 hosts. Which classful subnet mask would you use?
2^x where x is the number of masked bits (or 1's)
RIPv1 and IGRP
Each network segment can use a different subnet mask
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts

31. What are the valid subnets?
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
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination
The number just before the next subnet; the broadcast of the last subnet is always 255

32. IP Subnet-Zero
All interfaces within the classful address space have the same subnet mask
Allows you to use the first and last subnet in your network design; turned this command on by default
Problem with the NIC; Replace the NIC
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.

33. Classless Routing Protocols
126
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
4
RIPv2 - EIGRP - or OSPF

34. 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?
Local physical network problem between NIC and router
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
Problem with the NIC; Replace the NIC

35. Unable to Ping Local Host
All interfaces within the classful address space have the same subnet mask
Problem with the NIC; Replace the NIC
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
16 -382

36. /25
255.255.255.128
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.
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

37. How many subnets?

38. /26
The 10.32 subnet; The broadcast is 10.63
11000000 = 192
4
IP stack failure; Reinstall TCP/IP

39. Given 192.168.10.0/28 - What are the valid hosts?
0.0 - 64.0 - 128.0 - and 192.0
.1 - .126 & .129 - .254
4
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254

40. 192.168.100.99/25
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances
11110000 = 240
/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

41. 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
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
Possible DNS problem

42. Given 192.168.10.0/28 - How many subnets?
/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
2
/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

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?
/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
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) 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 point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.

44. What is the broadcast address of 192.168.192.10/29?
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
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
/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
10000000 = 128

45. Unable to Ping Remote Destination
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
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

46. Given 192.168.10.0/28 - What are the valid subnets?
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).
16 -382
0 & 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

47. Additional Windows Troubleshooting
0 & 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
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
Traceroute - Arp -a - Ipconfig /all

48. Classful Routing
Traceroute - Arp -a - Ipconfig /all
126
All nodes in the network use the same subnet mask
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet

49. Which two statements describe the IP address 10.16.3.65/23?
The mask 255.255.254.0 (/23) used with a Class A means that there are 15 subnet bits and 9 hosts bits. The block size in the third octet is 2 (256 - 254). So this makes the subnets in the interesting octet 0 - 2 - 4 - 6 - etc. - all the way to 254. T
A /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
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached

50. Given 192.168.10.0/28 - What's the broadcast address for each subnet?
127 & 255
11000000 = 192
RIPv1 and IGRP
63.255 - 127.255 - 191.255 - 255.255