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IP Subnetting VLSMs And Troubleshooting IP

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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. 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?
11000000 = 192
RIPv1 and IGRP
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.
.1 - .126 & .129 - .254

2. Unable to Ping Local Host
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
Tracert - Show ip arp
Problem with the NIC; Replace the NIC

3. 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?
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.
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254
Allows you to use the first and last subnet in your network design; turned this command on by default
11111000 = 248

4. What subnet and broadcast address is the IP address 172.16.45.14 255.255.255.252 (/30) a member of?
/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 subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
Possible DNS problem

5. What subnet and broadcast address is the IP address 172.16.46.255 255.255.240.0 (/20) a member of?
IP stack failure; Reinstall TCP/IP
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
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 10.32 subnet; The broadcast is 10.63

6. What subnet and broadcast address is the IP address 172.16.10.33 255.255.255.224 (/27) a member of?
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 10.32 subnet; The broadcast is 10.63
11000000 = 192

7. /29
11111000 = 248
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
63.255 - 127.255 - 191.255 - 255.255
Traceroute - Arp -a - Ipconfig /all

8. 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
Problem with the NIC; Replace the NIC
/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
Possible DNS problem

9. If a host on a network has the address 172.16.45.14/30 - what is the subnetwork this host belongs to?
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.
Create many networks using subnet masks of different lengths from one network
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.
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances

10. /26
RIPv1 and IGRP
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.
11000000 = 192
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255

11. What is the subnetwork address for a host with the IP address 200.10.5.68/28?
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
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 172.16.64.0. The broadcast must be 172.16.127.255
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination

12. /27
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
11100000 = 224
Traceroute - Arp -a - Ipconfig /all
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.

13. What's the broadcast address of 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
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
The number just before the next subnet; the broadcast of the last subnet is always 255

14. What is the subnet and broadcast address of the host 172.16.88.255/20?
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
RIPv1 and IGRP
To test the local stack on your host - ping the loopback interface of 127.0.0.1
The subnet is 80.0 and the broadcast address is 95.255

15. Unable to Ping Default Gateway
Local physical network problem between NIC and router
0.0 - 64.0 - 128.0 - and 192.0
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.
RIPv2 - EIGRP - or OSPF

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

17. Variable Length Subnet Masks (VLSMs)
0.0 - 64.0 - 128.0 - and 192.0
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.
Use different size masks on each router interface
11111000 = 248

18. Variable Length Subnet Masks (VLSMs)
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.
Each network segment can use a different subnet mask
/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

19. 192.168.100.17/29
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 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 /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.
Discard it; by default will discard any broadcast packets

20. 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
126
/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

21. 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
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
/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 /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

22. Subnet Mask
2
11111100 = 252
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

23. Given 192.168.10.0/28 - what is the subnet mask?
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.
4
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.

24. What is the broadcast address of 192.168.192.10/29?
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
/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
.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

25. How many hosts are available with a Class C /29 mask?
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.
11100000 = 224
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254

26. Classful Routing
The subnet is 80.0 and the broadcast address is 95.255
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.
All interfaces within the classful address space have the same subnet mask
11100000 = 224

27. /28
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet
10000000 = 128
Traceroute - Arp -a - Ipconfig /all
11110000 = 240

28. Classful Routing Protocols
/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.
11111100 = 252
RIPv1 and IGRP
126

29. Given 192.168.10.0/28 - How many hosts per subnet?
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
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.
Each network segment can use a different subnet mask

30. Given 172.16.0.0/18 - How many subnets?
IP stack failure; Reinstall TCP/IP
0 & 128
Allows you to use the first and last subnet in your network design; turned this command on by default
4

31. Unable to Ping Loopback
IP stack failure; Reinstall TCP/IP
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
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15

32. Class C Subnet Masks
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
The subnet is 80.0 and the broadcast address is 95.255
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
/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.

33. 192.168.100.37/28
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
63.255 - 127.255 - 191.255 - 255.255
/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
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances

34. 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?
2y - 2 where y is the number of unmasked bits (or 0's)
Discard it; by default will discard any broadcast packets
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
Use different size masks on each router interface

35. Unable to Ping Remote Destination
11110000 = 240
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
The subnet is 80.0 and the broadcast address is 95.255
Problem with the NIC; Replace the NIC

36. 192.168.100.99/25
4
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
Allows you to use the first and last subnet in your network design; turned this command on by default
/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

37. /25
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.
10000000 = 128
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
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.

38. What subnet and broadcast address is the IP address 172.16.66.10 255.255.192.0 (/18) a member of?
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.
Possible DNS problem
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
Create many networks using subnet masks of different lengths from one network

39. How many subnets?

40. Given 192.168.10.0/28 - What's the broadcast address for each subnet?
The 10.32 subnet; The broadcast is 10.63
16 -382
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
127 & 255

41. Given 192.168.10.0/28 - What are the valid subnets?
16 -382
0 & 128
/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) 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.

42. Classful Routing
All nodes in the network use the same subnet mask
2y - 2 where y is the number of unmasked bits (or 0's)
Traceroute - Arp -a - Ipconfig /all
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts

43. Why 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.
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
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances
63.255 - 127.255 - 191.255 - 255.255

44. Given 172.16.0.0/18 - What are the valid subnets?
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
/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
0.0 - 64.0 - 128.0 - and 192.0
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s

45. What is the subnet for host ID 10.16.3.65/23?
Local physical network problem between NIC and router
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.
/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
63.255 - 127.255 - 191.255 - 255.255

46. /30
Possible DNS problem
/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
11111100 = 252
/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

47. You have a network with a subnet of 172.16.17.0/22. Which are valid host addresses?
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
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
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination

48. Which two statements describe the IP address 10.16.3.65/23?
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
11111000 = 248
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

49. 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?
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
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.
Local physical network problem between NIC and router

50. 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.
RIPv2 - EIGRP - or OSPF
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 172.16.64.0. The broadcast must be 172.16.127.255