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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.
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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. Unable to Ping Local Host
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.
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
Problem with the NIC; Replace the NIC

2. 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
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.
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
/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. How many subnets?

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4. Classful Routing Protocols
RIPv1 and IGRP
16 -382
Each network segment can use a different subnet mask
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

5. /26
11000000 = 192
IP stack failure; Reinstall TCP/IP
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.
4

6. What is the broadcast address of 192.168.192.10/29?
/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
/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
RIPv2 - EIGRP - or OSPF
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25

7. What is the subnet and broadcast address of the host 172.16.88.255/20?
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
2
The subnet is 80.0 and the broadcast address is 95.255
0 & 128

8. Given 192.168.10.0/28 - How many subnets?
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
127 & 255
2
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.

9. What is the subnetwork address for a host with the IP address 200.10.5.68/28?
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
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.
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.
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.

10. How many hosts are available with a Class C /29 mask?
The number just before the next subnet; the broadcast of the last subnet is always 255
/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
Use different size masks on each router interface
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet

11. 192.168.100.25/30
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
/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
0 & 128
16 -382

12. Unable to Ping Loopback
IP stack failure; Reinstall TCP/IP
127 & 255
63.255 - 127.255 - 191.255 - 255.255
RIPv1 and IGRP

13. 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?
To test the local stack on your host - ping the loopback interface of 127.0.0.1
/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
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 /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.

14. Given 172.16.0.0/18 - What are the valid subnets?
RIPv2 - EIGRP - or OSPF
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.
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
0.0 - 64.0 - 128.0 - and 192.0

15. Additional Windows Troubleshooting
Traceroute - Arp -a - Ipconfig /all
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
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
11100000 = 224

16. Variable Length Subnet Masks (VLSMs)
16 -382
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
Each network segment can use a different subnet mask
127 & 255

17. Given 172.16.0.0/18 - How many hosts per subnet?
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
63.255 - 127.255 - 191.255 - 255.255
16 -382
Local physical network problem between NIC and router

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 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
Use different size masks on each router interface
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 -

19. Given 172.16.0.0/18 - What's the broadcast address for 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
63.255 - 127.255 - 191.255 - 255.255
/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
2y - 2 where y is the number of unmasked bits (or 0's)

20. 192.168.100.17/29
/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 numbers between the subnets and the broadcasts omitting the all 0s and all 1s
11111000 = 248
/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. What is the subnet for host ID 10.16.3.65/23?
To test the local stack on your host - ping the loopback interface of 127.0.0.1
/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
/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
11000000 = 192

22. 192.168.100.66/27
Use different size masks on each router interface
/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.
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination

23. Variable Length Subnet Masks (VLSMs)
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).
/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
Create many networks using subnet masks of different lengths from one network

24. 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?
Discard it; by default will discard any broadcast packets
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
/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
/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

25. 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
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
Possible DNS problem
63.255 - 127.255 - 191.255 - 255.255

26. 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 /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.
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.
All interfaces within the classful address space have the same 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.

27. Unable to Ping Remote Destination
16 -382
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.
Use different size masks on each router interface
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination

28. What subnet and broadcast address is the IP address 172.16.46.255 255.255.240.0 (/20) a member of?
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
11111000 = 248
11110000 = 240
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25

29. What is the subnetwork number of a host with an IP address of 172.16.66.0/21?
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
16 -382
11000000 = 192
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.

30. 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
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
RIPv1 and IGRP
/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

31. Variable Length Subnet Masks (VLSMs)
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
RIPv1 and IGRP
Use different size masks on each router interface
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination

32. What are the valid hosts in each subnet?
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.
RIPv2 - EIGRP - or OSPF
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet

33. Subnet Mask
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
All nodes in the network use the same subnet mask
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 /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.

34. What are the valid subnets?
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
11110000 = 240
2

35. What subnet and broadcast address is the IP address 172.16.66.10 255.255.192.0 (/18) a member of?
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
/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
/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 172.16.32.0 - and the broadcast must be 172.16.63.25

36. IP Subnet-Zero
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.
Allows you to use the first and last subnet in your network design; turned this command on by default
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 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

37. 4 Troubleshooting Steps
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination
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
All nodes in the network use the same subnet mask
/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

38. How many valid hosts per subnet?
2y - 2 where y is the number of unmasked bits (or 0's)
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
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached

39. 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.
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
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 /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

40. 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
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
RIPv1 and IGRP
127 & 255

41. Classless Routing Protocols
RIPv2 - EIGRP - or OSPF
11111100 = 252
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
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances

42. 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 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).
127 & 255
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
Tracert - Show ip arp

43. Classful Routing
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
All nodes in the network use the same subnet mask
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
Local physical network problem between NIC and router

44. /30
11111100 = 252
Possible DNS problem
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 subnet is 172.16.64.0. The broadcast must be 172.16.127.255

45. Additional Cisco Troubleshooting
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
Create many networks using subnet masks of different lengths from one network
Tracert - Show ip arp

46. 192.168.100.99/26
/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
Create many networks using subnet masks of different lengths from one network
255.255.255.128
To test the local stack on your host - ping the loopback interface of 127.0.0.1

47. 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
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
Create many networks using subnet masks of different lengths from one network
/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

48. /25
Problem with the NIC; Replace the NIC
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
Possible DNS problem

49. 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?
0.0 - 64.0 - 128.0 - and 192.0
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.
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
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.

50. Class C Subnet Masks
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.
2
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30